Java Structures

Data Structures in Java for the Principled Programmer

√

The

7 Edition

(Software release 33)

Duane A. Bailey

Williams College
September 2007

√

This

7 text copyrighted 2005-2007 by

All rights are reserved by The Author.
No part of this draft publiciation may be reproduced or distributed in any form
without prior, written consent of the author.

Contents

Preface to First Edition

Preface to the Second Edition

Preface to the “Root 7” Edition

0 Introduction

0.1 Read Me . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0.2 He Can’t Say That, Can He? . . . . . . . . . . . . . . . . . . . . .

xi

xiii

xv

1
1
2

1 The Object-Oriented Method

5
6
1.1 Data Abstraction and Encapsulation . . . . . . . . . . . . . . . . .
7
1.2 The Object Model . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Object-Oriented Terminology . . . . . . . . . . . . . . . . . . . .
8
1.4 A Special-Purpose Class: A Bank Account . . . . . . . . . . . . . . 11
1.5 A General-Purpose Class: An Association . . . . . . . . . . . . . . 14
1.6 Sketching an Example: A Word List . . . . . . . . . . . . . . . . . 18
1.7 Sketching an Example: A Rectangle Class
. . . . . . . . . . . . . 20
1.8 Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1.9 Who Is the User? . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.10 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.11 Laboratory: The Day of the Week Calculator . . . . . . . . . . . . 29

2 Comments, Conditions, and Assertions

33
2.1 Pre- and Postconditions
. . . . . . . . . . . . . . . . . . . . . . . 34
2.2 Assertions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.3 Craftsmanship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.5 Laboratory: Using Javadoc Commenting . . . . . . . . . . . . . . 39

3 Vectors

43
3.1 The Interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.2 Example: The Word List Revisited . . . . . . . . . . . . . . . . . . 47
3.3 Example: Word Frequency . . . . . . . . . . . . . . . . . . . . . . 48
3.4 The Implementation . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.5 Extensibility: A Feature . . . . . . . . . . . . . . . . . . . . . . . . 53
. . . . . . . . . . . . . . . . . . . . . . . . . 56
3.6 Example: L-Systems
3.7 Example: Vector-Based Sets . . . . . . . . . . . . . . . . . . . . . 57
3.8 Example: The Matrix Class . . . . . . . . . . . . . . . . . . . . . . 60
3.9 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

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3.10 Laboratory: The Silver Dollar Game . . . . . . . . . . . . . . . . . 67

4 Generics

69
4.1 Motivation (in case we need some) . . . . . . . . . . . . . . . . . 70
4.1.1 Possible Solution: Specialization . . . . . . . . . . . . . . 71
4.2 Implementing Generic Container Classes . . . . . . . . . . . . . . 72
4.2.1 Generic Associations . . . . . . . . . . . . . . . . . . . . 72
4.2.2 Parameterizing the Vector Class
. . . . . . . . . . . . . . 74
4.2.3 Restricting Parameters . . . . . . . . . . . . . . . . . . . . 79
4.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

5 Design Fundamentals

81
5.1 Asymptotic Analysis Tools . . . . . . . . . . . . . . . . . . . . . . 81
5.1.1 Time and Space Complexity . . . . . . . . . . . . . . . . . 82
5.1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
5.1.3 The Trading of Time and Space . . . . . . . . . . . . . . . 91
5.1.4 Back-of-the-Envelope Estimations . . . . . . . . . . . . . . 92
5.2 Self-Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
5.2.1 Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
5.2.2 Mathematical Induction . . . . . . . . . . . . . . . . . . . 101
5.3 Properties of Design . . . . . . . . . . . . . . . . . . . . . . . . . 108
5.3.1 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
5.3.2 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
5.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
5.5 Laboratory: How Fast Is Java? . . . . . . . . . . . . . . . . . . . . 115

6 Sorting

119
6.1 Approaching the Problem . . . . . . . . . . . . . . . . . . . . . . 119
6.2 Selection Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
6.3 Insertion Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
6.4 Mergesort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
6.5 Quicksort
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
6.6 Radix Sort . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
6.7 Sorting Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
6.8 Ordering Objects Using Comparators . . . . . . . . . . . . . . . . 140
6.9 Vector-Based Sorting . . . . . . . . . . . . . . . . . . . . . . . . . 143
6.10 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
6.11 Laboratory: Sorting with Comparators . . . . . . . . . . . . . . . 147

7 A Design Method

149
7.1 The Interface-Based Approach . . . . . . . . . . . . . . . . . . . . 149
7.1.1 Design of the Interface . . . . . . . . . . . . . . . . . . . . 150
7.1.2 Development of an Abstract Implementation . . . . . . . . 151
7.1.3 Implementation . . . . . . . . . . . . . . . . . . . . . . . . 152
7.2 Example: Development of Generators . . . . . . . . . . . . . . . . 152
. . . . . . . . . . . . . . . . . . . . . . . 155
7.3 Example: Playing Cards

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7.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

8 Iterators

161
8.1 Java’s Enumeration Interface . . . . . . . . . . . . . . . . . . . . 161
8.2 The Iterator Interface . . . . . . . . . . . . . . . . . . . . . . . . . 163
. . . . . . . . . . . . . . . . . . . . . . 165
8.3 Example: Vector Iterators
8.4 Example: Rethinking Generators
. . . . . . . . . . . . . . . . . . 167
8.5 Example: Filtering Iterators . . . . . . . . . . . . . . . . . . . . . 170
8.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
8.7 Laboratory: The Two-Towers Problem . . . . . . . . . . . . . . . 175

9 Lists

179
9.1 Example: A Unique Program . . . . . . . . . . . . . . . . . . . . . 182
9.2 Example: Free Lists . . . . . . . . . . . . . . . . . . . . . . . . . . 183
. . . . . . . . . . . . . . . 186
9.3 Partial Implementation: Abstract Lists
9.4 Implementation: Singly Linked Lists
. . . . . . . . . . . . . . . . 188
9.5 Implementation: Doubly Linked Lists . . . . . . . . . . . . . . . . 201
. . . . . . . . . . . . . . 206
9.6 Implementation: Circularly Linked Lists
9.7 Implementation: Vectors . . . . . . . . . . . . . . . . . . . . . . . 209
9.8 List Iterators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
9.9 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
9.10 Laboratory: Lists with Dummy Nodes . . . . . . . . . . . . . . . . 215

10 Linear Structures

10.2 Queues

219
10.1 Stacks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
10.1.1 Example: Simulating Recursion . . . . . . . . . . . . . . . 222
10.1.2 Vector-Based Stacks
. . . . . . . . . . . . . . . . . . . . . 225
10.1.3 List-Based Stacks . . . . . . . . . . . . . . . . . . . . . . . 227
. . . . . . . . . . . . . . . . . . . . . . . . . 228
10.1.4 Comparisons
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
10.2.1 Example: Solving a Coin Puzzle . . . . . . . . . . . . . . . 231
10.2.2 List-Based Queues
. . . . . . . . . . . . . . . . . . . . . . 234
10.2.3 Vector-Based Queues . . . . . . . . . . . . . . . . . . . . . 235
10.2.4 Array-Based Queues . . . . . . . . . . . . . . . . . . . . . 238
10.3 Example: Solving Mazes . . . . . . . . . . . . . . . . . . . . . . . 242
10.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
10.5 Laboratory: A Stack-Based Language . . . . . . . . . . . . . . . . 247
10.6 Laboratory: The Web Crawler . . . . . . . . . . . . . . . . . . . . 251

11 Ordered Structures

253
11.1 Comparable Objects Revisited . . . . . . . . . . . . . . . . . . . . 253
11.1.1 Example: Comparable Ratios
. . . . . . . . . . . . . . . . 254
11.1.2 Example: Comparable Associations . . . . . . . . . . . . . 256
11.2 Keeping Structures Ordered . . . . . . . . . . . . . . . . . . . . . 258
11.2.1 The OrderedStructure Interface . . . . . . . . . . . . . . . 258
11.2.2 The Ordered Vector and Binary Search . . . . . . . . . . . 259

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11.2.3 Example: Sorting Revisited . . . . . . . . . . . . . . . . . 264
11.2.4 A Comparator-based Approach . . . . . . . . . . . . . . . 265
11.2.5 The Ordered List
. . . . . . . . . . . . . . . . . . . . . . . 267
11.2.6 Example: The Modiﬁed Parking Lot . . . . . . . . . . . . . 270
11.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
11.4 Laboratory: Computing the “Best Of” . . . . . . . . . . . . . . . . 275

12 Binary Trees

277
12.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
12.2 Example: Pedigree Charts . . . . . . . . . . . . . . . . . . . . . . 280
12.3 Example: Expression Trees . . . . . . . . . . . . . . . . . . . . . . 281
12.4 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282
12.4.1 The BinaryTree Implementation . . . . . . . . . . . . . . . 284
12.5 Example: An Expert System . . . . . . . . . . . . . . . . . . . . . 287
. . . . . . . . . . . . . . . . . . . . . . 290
12.6 Traversals of Binary Trees
. . . . . . . . . . . . . . . . . . . . . . 291
12.6.1 Preorder Traversal
12.6.2 In-order Traversal
. . . . . . . . . . . . . . . . . . . . . . 293
12.6.3 Postorder Traversal . . . . . . . . . . . . . . . . . . . . . . 295
12.6.4 Level-order Traversal . . . . . . . . . . . . . . . . . . . . . 296
. . . . . . . . . . . . . . . . . . . . 297
12.6.5 Recursion in Iterators
12.7 Property-Based Methods . . . . . . . . . . . . . . . . . . . . . . . 299
12.8 Example: Huffman Compression . . . . . . . . . . . . . . . . . . 303
12.9 Example Implementation: Ahnentafel . . . . . . . . . . . . . . . . 307
12.10Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
12.11Laboratory: Playing Gardner’s Hex-a-Pawn . . . . . . . . . . . . . 313

13 Priority Queues

315
13.1 The Interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
13.2 Example: Improving the Huffman Code
. . . . . . . . . . . . . . 317
13.3 A Vector-Based Implementation . . . . . . . . . . . . . . . . . . . 318
13.4 A Heap Implementation . . . . . . . . . . . . . . . . . . . . . . . 319
13.4.1 Vector-Based Heaps
. . . . . . . . . . . . . . . . . . . . . 320
13.4.2 Example: Heapsort . . . . . . . . . . . . . . . . . . . . . . 326
13.4.3 Skew Heaps . . . . . . . . . . . . . . . . . . . . . . . . . . 329
13.5 Example: Circuit Simulation . . . . . . . . . . . . . . . . . . . . . 333
13.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
. . . . . . . . . . . . . . . . . . 341
13.7 Laboratory: Simulating Business

14 Search Trees

343
14.1 Binary Search Trees . . . . . . . . . . . . . . . . . . . . . . . . . . 343
14.2 Example: Tree Sort . . . . . . . . . . . . . . . . . . . . . . . . . . 345
14.3 Example: Associative Structures . . . . . . . . . . . . . . . . . . . 345
14.4 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348
14.5 Splay Trees
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354
14.6 Splay Tree Implementation . . . . . . . . . . . . . . . . . . . . . 357
14.7 An Alternative: Red-Black Trees . . . . . . . . . . . . . . . . . . . 361

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14.8 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
14.9 Laboratory: Improving the BinarySearchTree . . . . . . . . . . . . 367

15 Maps

369
15.1 Example Revisited: The Symbol Table . . . . . . . . . . . . . . . . 369
15.2 The Interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
15.3 Simple Implementation: MapList
. . . . . . . . . . . . . . . . . . 372
15.4 Constant Time Maps: Hash Tables . . . . . . . . . . . . . . . . . . 374
15.4.1 Open Addressing . . . . . . . . . . . . . . . . . . . . . . . 375
15.4.2 External Chaining . . . . . . . . . . . . . . . . . . . . . . 383
15.4.3 Generation of Hash Codes . . . . . . . . . . . . . . . . . . 385
15.4.4 Hash Codes for Collection Classes . . . . . . . . . . . . . . 391
15.4.5 Performance Analysis . . . . . . . . . . . . . . . . . . . . . 392
. . . . . . . . . . . . . . . . . . . . . . 392
15.5 Ordered Maps and Tables
15.6 Example: Document Indexing . . . . . . . . . . . . . . . . . . . . 395
15.7 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
15.8 Laboratory: The Soundex Name Lookup System . . . . . . . . . . 401

16 Graphs

403
16.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403
16.2 The Graph Interface . . . . . . . . . . . . . . . . . . . . . . . . . 404
16.3 Implementations . . . . . . . . . . . . . . . . . . . . . . . . . . . 408
16.3.1 Abstract Classes Reemphasized . . . . . . . . . . . . . . . 408
16.3.2 Adjacency Matrices . . . . . . . . . . . . . . . . . . . . . . 410
16.3.3 Adjacency Lists . . . . . . . . . . . . . . . . . . . . . . . . 416
16.4 Examples: Common Graph Algorithms . . . . . . . . . . . . . . . 422
16.4.1 Reachability . . . . . . . . . . . . . . . . . . . . . . . . . . 422
16.4.2 Topological Sorting . . . . . . . . . . . . . . . . . . . . . . 424
16.4.3 Transitive Closure . . . . . . . . . . . . . . . . . . . . . . 427
16.4.4 All Pairs Minimum Distance . . . . . . . . . . . . . . . . . 428
16.4.5 Greedy Algorithms . . . . . . . . . . . . . . . . . . . . . . 429
16.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434
16.6 Laboratory: Converting Between Units . . . . . . . . . . . . . . . 439

A Answers

441
A.1 Solutions to Self Check Problems . . . . . . . . . . . . . . . . . . 441
. . . . . . . . . . . . . . . 451
A.2 Solutions to Odd-Numbered Problems

B Beginning with Java

489
B.1 A First Program . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
B.2 Declarations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491
B.2.1 Primitive Types . . . . . . . . . . . . . . . . . . . . . . . . 491
. . . . . . . . . . . . . . . . . . . . . . . 493
B.2.2 Reference Types
B.3 Important Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . 494
B.3.1 The structure.ReadStream Class . . . . . . . . . . . . . . . 494
. . . . . . . . . . . . . . . . . 495
B.3.2 The java.util.Scanner Class

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B.4.1 Conditional Statements
B.4.2 Loops

B.3.3 The PrintStream Class . . . . . . . . . . . . . . . . . . . . 496
B.3.4 Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497
B.4 Control Constructs . . . . . . . . . . . . . . . . . . . . . . . . . . 498
. . . . . . . . . . . . . . . . . . . 498
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499
B.5 Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
B.6 Inheritance and Subtyping . . . . . . . . . . . . . . . . . . . . . . 502
B.6.1 Inheritance . . . . . . . . . . . . . . . . . . . . . . . . . . 502
B.6.2 Subtyping . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
. . . . . . . . . . . . . . . 504
B.6.3 Interfaces and Abstract Classes
B.7 Use of the Assert Command . . . . . . . . . . . . . . . . . . . . . 506
B.8 Use of the Keyword Protected . . . . . . . . . . . . . . . . . . . 507

C Collections

511
C.1 Collection Class Features . . . . . . . . . . . . . . . . . . . . . . . 511
C.2 Parallel Features
. . . . . . . . . . . . . . . . . . . . . . . . . . . 511
C.3 Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512

D Documentation

513
D.1 Structure Package Hierarchy . . . . . . . . . . . . . . . . . . . . . 513
D.2 Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515

Index

517

for Mary,
my wife and best friend

without
the model of my mentors,
the comments of my colleagues,
the support of my students,
the friendship of my family
this book would never be

thank you!

Preface to the First Edition

“IT’S A WONDERFUL TIME TO BE ALIVE.” At least that’s what I’ve found myself
saying over the past couple of decades. When I ﬁrst started working with com-
puters, they were resources used by a privileged (or in my case, persistent) few.
They were physically large, and logically small. They were cast from iron. The
challenge was to make these behemoths solve complex problems quickly.

Today, computers are everywhere. They are in the ofﬁce and at home. They
speak to us on telephones; they zap our food in the microwave. They make
starting cars in New England a possibility. Everyone’s using them. What has
aided their introduction into society is their diminished size and cost, and in-
creased capability. The challenge is to make these behemoths solve complex
problems quickly.

Thus, while the computer and its applications have changed over time, the
challenge remains the same: How can we get the best performance out of the
current technology? The design and analysis of data structures lay the funda-
mental groundwork for a scientiﬁc understanding of what computers can do
efﬁciently. The motivations for data structure design work accomplished three
decades ago in assembly language at the keypunch are just as familiar to us to-
day as we practice our craft in modern languages on computers on our laps. The
focus of this material is the identiﬁcation and development of relatively abstract
principles for structuring data in ways that make programs efﬁcient in terms of
their consumption of resources, as well as efﬁcient in terms of “programmability.”
In the past, my students have encountered this material in Pascal, Modula-2,
and, most recently, C++. None of these languages has been ideal, but each has
been met with increasing expectation. This text uses The Java Programming
Language1—“Java”—to structure data. Java is a new and exciting language
that has received considerable public attention. At the time of this writing, for
example, Java is one of the few tools that can effectively use the Internet as a
computing resource. That particular aspect of Java is not touched on greatly
in this text. Still, Internet-driven applications in Java will need supporting data
structures. This book attempts to provide a fresh and focused approach to the
design and implementation of classic structures in a manner that meshes well
with existing Java packages.
It is hoped that learning this material in Java
will improve the way working programmers craft programs, and the way future
designers craft languages.

Pedagogical Implications. This text was developed speciﬁcally for use with
CS2 in a standard Computer Science curriculum. It is succinct in its approach,
and requires, perhaps, a little more effort to read. I hope, though, that this text

1 Java is a trademark of Sun Microsystems, Incorporated.

xii

Preface to the First Edition

becomes not a brief encounter with object-oriented data structure design, but a
touchstone for one’s programming future.

The material presented in this text follows the syllabus I have used for sev-
eral years at Williams. As students come to this course with experience using
Java, the outline of the text may be followed directly. Where students are new
to Java, a couple of weeks early in the semester will be necessary with a good
companion text to introduce the student to new concepts, and an introductory
Java language text or reference manual is recommended. For students that need
a quick introduction to Java we provide a tutorial in Appendix B. While the text
was designed as a whole, some may wish to eliminate less important topics
and expand upon others. Students may wish to drop (or consider!) the sec-
tion on induction (Section 5.2.2). The more nontraditional topics—including,
for example, iteration and the notions of symmetry and friction—have been in-
cluded because I believe they arm programmers with important mechanisms for
implementing and analyzing problems. In many departments the subtleties of
more advanced structures—maps (Chapter 15) and graphs (Chapter 16)—may
be considered in an algorithms course. Chapter 6, a discussion of sorting, pro-
vides very important motivating examples and also begins an early investigation
of algorithms. The chapter may be dropped when better examples are at hand,
but students may ﬁnd the reﬁnements on implementing sorting interesting.

Associated with this text is a Java package of data structures that is freely
available over the Internet for noncommercial purposes. I encourage students,
educators, and budding software engineers to download it, tear it down, build it
up, and generally enjoy it. In particular, students of this material are encouraged
to follow along with the code online as they read. Also included is extensive
documentation gleaned from the code by javadoc. All documentation—within
the book and on the Web—includes pre- and postconditions. The motivation for
this style of commenting is provided in Chapter 2. While it’s hard to be militant
about commenting, this style of documentation provides an obvious, structured
approach to minimally documenting one’s methods that students can appreciate
and users will welcome. These resources, as well as many others, are available
from McGraw-Hill at http://www.mhhe.com/javastructures.

Three icons appear throughout the text, as they do in the margin. The
top “compass” icon highlights the statement of a principle—a statement that
encourages abstract discussion. The middle icon marks the ﬁrst appearance of
a particular class from the structure package. Students will ﬁnd these ﬁles at
McGraw-Hill, or locally, if they’ve been downloaded. The bottom icon similarly
marks the appearance of example code.

Finally, I’d like to note an unfortunate movement away from studying the
implementation of data structures, in favor of studying applications.
In the
extreme this is a disappointing and, perhaps, dangerous precedent. The design
of a data structure is like the solution to a riddle: the process of developing the
answer is as important as the answer itself. The text may, however, be used as a
reference for using the structure package in other applications by selectively
avoiding the discussions of implementation.

List

nim

NNWSWSENEWSEPreface to the Second Edition

Since the ﬁrst edition of Java Structures support for writing programs in Java2
has grown considerably. At that time the Java Development Toolkit consisted
of 504 classes in 23 packages3 In Java 1.2 (also called Java 2) Sun rolled out
1520 classes in 59 packages. This book is ready for Java 1.4, where the number
of classes and packages continues to grow.

Most computer scientists are convinced of the utility of Java for program-
ming in a well structured and platform independent manner. While there are
still signiﬁcant arguments about important aspects of the language (for exam-
ple, support for generic types), the academic community is embracing Java, for
example, as the subject of the Computer Science Advanced Placement Exami-
nation.

It might seem somewhat perplexing to think that many aspects of the origi-
nal Java environment have been retracted (or deprecated) or reconsidered. The
developers at Sun have one purpose in mind: to make Java the indispensable
language of the current generation. As a result, documenting their progress on
the development of data structures gives us valuable insight into the process of
designing useful data structures for general purpose programming. Those stu-
dents and faculty considering a move to this second edition of Java Structures
will see ﬁrst-hand some of the decisions that have been made in the interven-
ing years. During that time, for example, the Collection-based classes were
introduced, and are generally considered an improvement. Another force—
one similar to calciﬁcation—has left a trail of backwards compatible features
that are sometimes difﬁcult to understand. For example, the Iterator class
was introduced, but the Enumeration class was not deprecated. One subject of
the ﬁrst edition—the notion of Comparable classes—has been introduced into
a number of important classes including String and Integer. This is a step
forward and a reconsideration of what we have learned about that material has
lead to important improvements in the text.

Since the main purpose of the text is to demonstrate the design and behavior
of traditional data structures, we have not generally tracked the progress of
Java where it blurs the view. For example, Java 2 introduces a List interface
(we applaud) but the Vector class has been extended to include methods that
are, essentially, motivated by linked lists (we wonder). As this text points out
frequently, the purpose of an interface is often to provide reduced functionality.
If the data structure does not naturally provide the functionality required by the
application, it is probably not an effective tool for solving the problem: search
elsewhere for an effective structure.

2 The Java Programming Language is a trademark of Sun Microsystems, Incorporated.
3 David Flanagan, et al., Java in a Nutshell, O’Reilly & Associates.

xiv

Preface to the Second Edition

As of this writing, more than 100, 000 individuals have searched for and
downloaded the structure package. To facilitate using the comprehensive set
of classes with the Java 2 environment, we have provided a number of features
that support the use of the structure package in more concrete applications.
Please see Appendix C.

Also new to this edition are more than 200 new problems, several dozen

exercises, and over a dozen labs we regularly use at Williams.

Acknowledgments. Several students, instructors, and classes have helped to
shape this edition of Java Structures. Parth Doshi and Alex Glenday—diligent
Williams students—pointed out a large number of typos and stretches of logic.
Kim Bruce, Andrea Danyluk, Jay Sachs, and Jim Teresco have taught this course
at Williams over the past few years, and have provided useful feedback. I tip
my hat to Bill Lenhart, a good friend and advisor, who has helped improve this
text in subtle ways. To Sean Sandys I am indebted for showing me new ways to
teach new minds.

The various reviewers have made, collectively, hundreds of pages of com-
ments that have been incorporated (as much as possible) into this edition:
Eleanor Hare and David Jacobs (Clemson University), Ram Athavale (North
Carolina State University), Yannick Daoudi (McGill University), Walter Daugh-
erty (Texas A&M University), Subodh Kumar (Johns Hopkins University), Toshimi
Minoura (Oregon State University), Carolyn Schauble (Colorado State Univer-
sity), Val Tannen (University of Pennsylvania), Frank Tompa (University of Wa-
terloo), Richard Wiener (University of Colorado at Colorado Springs), Cynthia
Brown Zickos (University of Mississippi), and my good friend Robbie Moll (Uni-
versity of Massachusetts). Deborah Trytten (University of Oklahoma) has re-
viewed both editions! Still, until expert authoring systems are engineered, au-
thors will remain human. Any mistakes left behind or introduced are purely
those of the author.

The editors and staff at McGraw-Hill–Kelly Lowery, Melinda Dougharty, John
Wannemacher, and Joyce Berendes–have attempted the impossible: to keep me
within a deadline. David Hash, Phil Meek, and Jodi Banowetz are responsible
for the look and feel of things. I am especially indebted to Lucy Mullins, Judy
Gantenbein, and Patti Evers whose red pens have often shown me a better way.
Betsy Jones, publisher and advocate, has seen it all and yet kept the faith:

thanks.

Be aware, though: long after these pages are found to be useless folly, my
best work will be recognized in my children, Kate, Megan, and Ryan. None
of these projects, of course, would be possible without the support of my best
friend, my north star, and my partner, Mary.

Enjoy!

Duane A. Bailey
Williamstown, May 2002

Preface to the

√

7 Edition

In your hand is a special edition of Java Structures designed for use with two
semesters of Williams’ course on data structures, Computer Science 136. This
version is only marginally different than the preceding edition, but is positioned
to make use of Java 5 (the trademarked name for version 1.5 of the JDK).
Because Java 5 may not be available (yet) on the platform you use, most of the
code available in this book will run on older JDK’s. The one feature that would
not be available is Java’s new Scanner class from the java.util package; an
alternative is my ReadStream class, which is lightly documented in Section B.3.1
on page 494. It is a feature of the structure package soon to be removed.

In making this book available in this paperbound format, my hope is that
you ﬁnd it a more inviting place to write notes: additions, subtractions, and
updates that you’re likely to have discussed in class. Sometimes you’ll identify
improvements, and I hope you’ll pass those along to me. In any case, you can
download the software (as hundreds of thousands have done in the past) and
modify it as you desire.

On occasion, I will release new sections you can incorporate into your text,
including a discussion of how the structure package can make use of generic
types.

I have spent a considerable amount of time designing the structure pack-
age. The ﬁrst structures were available 8 years ago when Java was still in its
infancy. Many of the structures have since been incorporated (directly or indi-
rectly) into Sun’s own JDK. (Yes, we’ve sold a few books in California.) Still, I
feel the merit of my approach is a slimness that, in the end, you will not ﬁnd
surprising.

Meanwhile, for those of you keeping track, the following table (adapted
from the 121 cubic inch, 3 pound 6 ounce, Fifth edition of David Flanagan’s
essential Java in a Nutshell) demonstrates the growth of Java’s support:

JDK
1.0
1.1
1.2 (Java 2)
1.3
1.4
1.5 (Java 5)

Packages Classes

8
23
59
76
135
166

212
504
1520
1842
2991
3562

Features
First public version
Inner classes
Collection classes
A “maintenance” release.
Improvments, including assert
Generics, autoboxing, and “varargs.”

Seeing this reminds me of the comment made by Niklaus Wirth, designer of
Pascal and the ﬁrst two releases of Modula. After the design team briefed him
on the slew of new features to be incorporated into Modula 3, he parried: “But,
what features have you removed?” A timeless question.

xvi

Preface to the

√

7 Edition

Acknowledgments. This book was primarily written for students of Williams
College. The process of publishing and reviewing a text tends to move the focus
off campus and toward the bottom line. The Route 7 edition4—somewhere
between editions 2 and 3—is an initial attempt to bring that focus back to those
students who made it all possible.

For nearly a decade, students at many institutions have played an important
role in shaping these resources. In this edition, I’m especially indebted to Katie
Creel ’10 (Williams) and Brian Bargh ’07 (Messiah): thanks!

Many colleagues, including Steve Freund ’95 (Stanford, now at Williams),
Jim Teresco ’92 (Union, now at Mount Holyoke), and especially Gene Chase ’65
(M.I.T., now at Messiah) continue to nudge this text in a better direction. Brent
Heeringa ’99 (Morris, now at Williams) showers all around him with youthful
enthusiasm.

And a most special thanks to Bill Mueller for the shot heard around the
world—the game-winning run that showed all things were possible. Called by
Joe Castiglione ’68 (Colgate, now at Fenway):

“Three-and-one to Mueller. One out, nineth inning. 10-9 Yankees,
runner at ﬁrst. Here’s the pitch...swing and a High Drive Deep to
Right...Back Goes Shefﬁeld to the Bullpen...AND IT IS GONE!...AND
THE RED SOX HAVE WON IT!...ON A WALKOFF TWO RUN HOMER
BY BILL MUELLER OFF MARIANO RIVERA! CAN YOU BELIEVE IT?!”

Have I been a Red Sox fan all my life? Not yet.

Finally, nothing would be possible without my running mate, my Sox buddy,

and my best friend, Mary.

Cheers!

Duane A. Bailey ’82 (Amherst, now at Williams)
Williamstown, September 2007

4 Route 7 is a scenic byway through the Berkshires and Green Mountains that eddies a bit as it
passes through Williamstown and Middlebury.

Chapter 0

Introduction

Concepts:
. Approaches to this material
.

Principles

This is an important notice.
Please have it translated.
—The Phone Company

YOUR MOTHER probably provided you with constructive toys, like blocks or
Tinkertoys1 or Lego bricks. These toys are educational: they teach us to think
spatially and to build increasingly complex structures. You develop modules
that can be stuck together and rules that guide the building process.

If you are reading this book, you probably enjoyed playing with construc-
tive toys. You consider writing programs an artistic process. You have grown
from playing with blocks to writing programs. The same guidelines for building
structures apply to writing programs, save one thing: there is, seemingly, no
limit to the complexity of the programs you can write.

Well, almost. When writing large programs, the data structures that main-
tain the data in your program govern the space and time consumed by your
running program. In addition, large programs take time to write. Using differ-
ent structures can actually have an impact on how long it takes to write your
program. Choosing the wrong structures can cause your program to run poorly
or be difﬁcult or impossible to implement effectively.

Thus, part of the program-writing process is choosing between different
structures.
Ideally you arrive at solutions by analyzing and comparing their
various merits. This book focuses on the creation and analysis of traditional
data structures in a modern programming environment, The Java Programming
Language, or Java for short.

I lie.

0.1 Read Me

As might be expected, each chapter is dedicated to a speciﬁc topic. Many of the
topics are concerned with speciﬁc data structures. The structures we will inves-
tigate are abstracted from working implementations in Java that are available
to you if you have access to the Internet.2 Other topics concern the “tools of the

1 All trademarks are recognized.
2 For more information, see http://www.cs.williams.edu/JavaStructures.

2

Introduction

Unicycles: the
ultimate riding
structure.

Structure

Example

trade.” Some are mathematical and others are philosophical, but all consider
the process of programming well.

The topics we cover are not all-inclusive. Some useful structures have been
left out. Instead, we will opt to learn the principles of programming data struc-
tures, so that, down the road, you can design newer and better structures your-
self.

Perhaps the most important aspect of this book is the set of problems at the
end of each section. All are important for you to consider. For some problems
I have attempted to place a reasonable hint or answer in the back of the book.
Why should you do problems? Practice makes perfect. I could show you how to
ride a unicycle, but if you never practiced, you would never learn. If you study
and understand these problems, you will ﬁnd your design and analytical skills
are improved. As for your mother, she’ll be proud of you.

Sometimes we will introduce problems in the middle of the running text—
these problems do not have answers (sometimes they are repeated as formal
problems in the back of the chapter, where they do have answers)—they should
be thought about carefully as you are reading along. You may ﬁnd it useful to
have a pencil and paper handy to help you “think” about these problems on the
ﬂy.

Exercise 0.1 Call3 your Mom and tell her you’re completing your ﬁrst exercise. If
you don’t have a phone handy, drop her a postcard. Ask her to verify that she’s
proud of you.

This text is brief and to the point. Most of us are interested in experimenting.
We will save as much time as possible for solving problems, perusing code, and
practicing writing programs. As you read through each of the chapters, you
might ﬁnd it useful to read through the source code online. As we ﬁrst consider
the text of ﬁles online, the ﬁle name will appear in the margin, as you see here.
The top icon refers to ﬁles in the structure package, while the bottom icon
refers to ﬁles supporting examples.

One more point—this book, like most projects, is an ongoing effort, and
the latest thoughts are unlikely to have made it to the printed page.
If you
are in doubt, turn to the website for the latest comments. You will also ﬁnd
online documentation for each of the structures, generated from the code using
javadoc.
It is best to read the online version of the documentation for the
most up-to-date details, as well as the documentation of several structures not
formally presented within this text.

0.2 He Can’t Say That, Can He?

Sure! Throughout this book are little political comments. These remarks may
seem trivial at ﬁrst blush. Skip them! If, however, you are interested in ways

3 Don’t e-mail her. Call her. Computers aren’t everything, and they’re a poor medium for a mother’s
pride.

0.2 He Can’t Say That, Can He?

3

to improve your skills as a programmer and a computer scientist, I invite you
to read on. Sometimes these comments are so important that they appear as
principles:

Principle 1 The principled programmer understands a principle well enough to
form an opinion about it.

Self Check Problems

Solutions to these problems begin on page 441.
0.1
0.2
0.3
0.4

Where are the answers for “self check” problems found?
What are features of large programs?
Should you read the entire text?
Are principles statements of truth?

Problems

All odd problems have answers. Where do you ﬁnd answers to prob-

Surf to the website associated with this text and review the resources

You are an experienced programmer. What ﬁve serious pieces of advice

Which of the following structures are described in this text (see Append-

Surf to http://www.javasoft.com and review the Java resources avail-

Solutions to the odd-numbered problems begin on page 451.
0.1
lems? (Hint: See page 451.)
0.2
would you give a new programmer?
0.3
available to you.
0.4
ix D): BinarySearchTree, BinaryTree, BitSet, Map, Hashtable, List?
0.5
able from Sun, the developers of Java.
Review documentation for Sun’s java.util package. (See the Core
0.6
API Documentation at http://www.javasoft.com.) Which of the following
data structures are available in this package: BinarySearchTree, BinaryTree,
BitSet, Dictionary, Hashtable, List?
Check your local library or bookstore for Java reference texts.
0.7
If you haven’t done so already, learn how to use your local Java pro-
0.8
gramming environment by writing a Java application to write a line of text.
(Hint: Read Appendix B.)
Find the local documentation for the structure package. If none is to
0.9
be found, remember that the same documentation is available over the Internet
from http://www.cs.williams.edu/JavaStructures.
0.10
age. Many of these examples are discussed later in this text.

Find the examples electronically distributed with the structure pack-

NNWSWSENEWSEChapter 1

The Object-Oriented Method

Concepts:
. Data structures
. Abstract data types
. Objects
. Classes
.

Interfaces

I will pick up the hook.
You will see something new.
Two things. And I call them
Thing One and Thing Two.
These Things will not bite you.
They want to have fun.
—Theodor Seuss Geisel

COMPUTER SCIENCE DOES NOT SUFFER the great history of many other disci-
plines. While other subjects have well-founded paradigms and methods, com-
puter science still struggles with one important question: What is the best method
to write programs? To date, we have no best answer. The focus of language de-
signers is to develop programming languages that are simple to use but provide
the power to accurately and efﬁciently describe the details of large programs
and applications. The development of Java is one such effort.

Throughout this text we focus on developing data structures using object-

oriented programming. Using this paradigm the programmer spends time devel- OOP:
oping templates for structures called classes. The templates are then used to
construct instances or objects. A majority of the statements in object-oriented
programs involve sending messages to objects to have them report or change
their state. Running a program involves, then, the construction and coordina-
tion of objects. In this way languages like Java are object-oriented.

Object-oriented
programming.

In all but the smallest programming projects, abstraction is a useful tool
In programming languages including Pascal,
for writing working programs.
Scheme, and C, the details of a program’s implementation are hidden away in
its procedures or functions. This approach involves procedural abstraction. In
object-oriented programming the details of the implementation of data struc-
tures are hidden away within its objects. This approach involves data abstrac-
tion. Many modern programming languages use object orientation to support
basic abstractions of data. We review the details of data abstraction and the
design of formal interfaces for objects in this chapter.

6

The Object-Oriented Method

Macintosh and
UNIX store
strings
differently.

1.1 Data Abstraction and Encapsulation

If you purchase a donut from Morningside Bakery in Pittsﬁeld, Massachusetts,
you can identify it as a donut without knowing its ingredients. Donuts are
circular, breadlike, and sweet. The particular ingredients in a donut are of little
concern to you. Of course, Morningside is free to switch from one sweetener to
another, as long as the taste is preserved.1 The donut’s ingredients list and its
construction are details that probably do not interest you.

Likewise, it is often unimportant to know how data structures are imple-
mented in order to appreciate their use. For example, most of us are familiar
with the workings or semantics of strings or arrays, but, if pressed, we might
ﬁnd it difﬁcult to describe their mechanics: Do all consecutive locations in the
array appear close together in memory in your computer, or are they far apart?
The answer is: it is unimportant. As long as the array behaves like an array or
the string behaves like a string we are happy. The less one knows about how
arrays or strings are implemented, the less one becomes dependent on a partic-
ular implementation. Another way to think about this abstractly is that the data
structure lives up to an implicit “contract”: a string is an ordered list of charac-
ters, or elements of an array may be accessed in any order. The implementor of
the data structure is free to construct it in any reasonable way, as long as all the
terms of the contract are met. Since different implementors are in the habit of
making very different implementation decisions, anything that helps to hide the
implementation details—any means of using abstraction—serves to make the
world a better place to program.

When used correctly, object-oriented programming allows the programmer
to separate the details that are important to the user from the details that are
only important to the implementation. Later in this book we shall consider very
general behavior of data structures; for example, in Section 10.1 we will study
structures that allow the user only to remove the most recently added item.
Such behavior is inherent to our most abstract understanding of how the data
structure works. We can appreciate the unique behavior of this structure even
though we haven’t yet discussed how these structures might be implemented.
Those abstract details that are important to the user of the structure—including
abstract semantics of the methods—make up its contract or interface. The in-
terface describes the abstract behavior of the structure. Most of us would agree
that while strings and arrays are very similar structures, they behave differently:
you can shrink or expand a string, while you cannot directly do the same with
an array; you can print a string directly, while printing an array involves explic-
itly printing each of its elements. These distinctions suggest they have distinct
abstract behaviors; there are distinctions in the design of their interfaces.

The unimportant details hidden from the user are part of what makes up
the implementation. We might decide (see Figure 1.1) that a string is to be

1 Apple cider is often used to ﬂavor donuts in New England, but that decision decidedly changes
the ﬂavor of the donut for the better. Some of the best apple cider donuts can be found at Atkin’s
apple farm in Amherst, Massachusetts.

1.2 The Object Model

7

Figure 1.1
its length. The terminated string’s length is determined by an end-of-string mark.

Two methods of implementing a string. A counted string explicitly records

constructed from a large array of characters with an attendant character count.
Alternatively, we might specify the length implicitly by terminating the string
with a special end-of-string mark that is not used for any other purpose. Both
of these approaches are perfectly satisfactory, but there are trade-offs. The ﬁrst
implementation (called a counted string) has its length stored explicitly, while
the length of the second implementation (called a terminated string) is implied.
It takes longer to determine the length of a terminated string because we have to
search for the end-of-string mark. On the other hand, the size of a terminated
string is limited only by the amount of available memory, while the longest
counted string is determined by the range of integers that can be stored in its
length ﬁeld (often this is only several hundred characters). If implementors can
hide these details, users do not have to be distracted from their own important
design work. As applications mature, a ﬁxed interface to underlying objects
allows alternative implementations of the object to be considered.

Data abstraction in languages like Java allows a structure to take responsibil-
ity for its own state. The structure knows how to maintain its own state without
bothering the programmer. For example, if two strings have to be concatenated
into a single string structure, a request might have to be made for a new allot-
ment of memory. Thankfully, because strings know how to perform operations
on themselves, the user doesn’t have to worry about managing memory.

1.2 The Object Model

To facilitate the construction of well-designed objects, it is useful to have a de-
sign method in mind. As alluded to earlier, we will often visualize the data for
our program as being managed by its objects. Each object manages its own data
that determine its state. A point on a screen, for example, has two coordinates.

Data1234567891011121314n01234567891011121314nLICKETYSPLIT!E  O    SLICKETYSPLIT!13Counted stringTerminated stringDataCount08

The Object-Oriented Method

A medical record maintains a name, a list of dependents, a medical history, and
a reference to an insurance company. A strand of genetic material has a se-
quence of base pairs. To maintain a consistent state we imagine the program
manipulates the data within its objects only through messages or method calls
to the objects. A string might receive a message “tell me your length,” while
a medical record might receive a “change insurance” message. The string mes-
sage simply accesses information, while the medical record method may involve
changing several pieces of information in this and other objects in a consistent
manner. If we directly modify the reference to the insurance company, we may
forget to modify similar references in each of the dependents. For large applica-
tions with complex data structures, it can be extremely difﬁcult to remember to
coordinate all the operations that are necessary to move a single complex object
from one consistent state to another. We opt, instead, to have the designer of
the data structure provide us a method for carefully moving between states; this
method is activated in response to a high-level message sent to the object.

This text, then, focuses on two important topics: (1) how we implement and
evaluate objects with methods that are logically complex and (2) how we might
use the objects we create. These objects typically represent data structures, our
primary interest. Occasionally we will develop control structures—structures
whose purpose is to control the manipulation of other objects. Control struc-
tures are an important concept and are described in detail in Chapter 8.

1.3 Object-Oriented Terminology

In Java, data abstraction is accomplished through encapsulation of data in an
object—an instance of a class. Like a record in other languages, an object has
ﬁelds. Unlike records, objects also contain methods. Fields and methods of an
object may be declared public, which means that they are visible to entities
outside the class, or protected, in which case they may only be accessed by
code within methods of the class.2 A typical class declaration is demonstrated
by the following simple class that keeps track of the ratio of two integer values:

public class Ratio
{

Ratio

protected int numerator;
protected int denominator; // denominator of ratio

// numerator of ratio

public Ratio(int top, int bottom)
// pre: bottom != 0
// post: constructs a ratio equivalent to top::bottom
{

numerator = top;
denominator = bottom;
reduce();

2 This is not quite the truth. For a discussion of the facts, see Appendix B.8.

1.3 Object-Oriented Terminology

9

}

public int getNumerator()
// post: return the numerator of the fraction
{

return numerator;

}

public int getDenominator()
// post: return the denominator of the fraction
{

return denominator;

}

public double getValue()
// post: return the double equivalent of the ratio
{

return (double)numerator/(double)denominator;

}

public Ratio add(Ratio other)
// pre: other is nonnull
// post: return new fraction--the sum of this and other
{

return new Ratio(this.numerator*other.denominator+
this.denominator*other.numerator,
this.denominator*other.denominator);

}

protected void reduce()
// post: numerator and denominator are set so that
// the greatest common divisor of the numerator and denominator is 1
{

int divisor = gcd(numerator,denominator);
if (denominator < 0) divisor = -divisor;
numerator /= divisor;
denominator /= divisor;

}

protected static int gcd(int a, int b)
// post: computes the greatest integer value that divides a and b
{

if (a < 0) return gcd(-a,b);
if (a == 0) {

if (b == 0) return 1;
else return b;

}
if (b < a) return gcd(b,a);
return gcd(b%a,a);

}

10

The Object-Oriented Method

public String toString()
// post: returns a string that represents this fraction.
{

return getNumerator()+"/"+getDenominator();

}

}

First, a Ratio object maintains the numerator and denominator as protected
ints that are not directly modiﬁable by the user. The Ratio method is a con-
structor: a method whose name is the same as that of the class. (The formal
comments at the top of methods are pre- and postconditions; we discuss these
in detail in Chapter 2.) The constructor is called whenever a new Ratio object is
formed. Constructors initialize all the ﬁelds of the associated object, placing the
object into a predictable and consistent initial state. We declare the construc-
tors for a class public. To construct a new Ratio object, users will have to call
these methods. The value method returns a double that represents the ratio,
while the getNumerator and getDenominator methods fetch the current values
of the numerator and denominator of the fraction. The add method is useful for
adding one Ratio to another; the result is a newly constructed Ratio object.
Finally, the toString method generates the preferred printable representation
of the object; we have chosen to represent it in fractional form.

Two methods, reduce and gcd, are utility methods. The gcd method com-
putes the greatest common divisor of two values using Euclid’s method, one of
the oldest numerical algorithms used today. It is used by the reduce method to
reduce the numerator and denominator to lowest terms by removing any com-
mon factors. Both are declared protected because computing the reduction is
not a necessary (or obvious) operation to be performed on ratios of integers;
it’s part of the implementation. The gcd method is declared static because
the algorithm can be used at any time—its utility is independent of the number
of Ratio objects that exist in our program. The reduce method, on the other
hand, works only with a Ratio object.

Exercise 1.1 Nearly everything can be improved. Are there improvements that
might be made to the gcd method? Can you write the method iteratively? Is
iteration an improvement?

As with the Ratio class, data ﬁelds are usually declared protected. To ma-
nipulate protected ﬁelds the user must invoke public methods. The following
example demonstrates the manipulation of the Ratio class:

public static void main(String[] args)
{

Ratio r = new Ratio(1,1);
r = new Ratio(1,2);
r.add(new Ratio(1,3));
r = r.add(new Ratio(2,8));
System.out.println(r.getValue()); // 0.75 printed

// r == 1.0
// r == 0.5
// sum computed, but r still 0.5
// r == 0.75

1.4 A Special-Purpose Class: A Bank Account

11

System.out.println(r.toString());
System.out.println(r);

// calls toString()

// calls toString()

}

To understand the merit of this technique of class design, we might draw an
analogy between a well-designed object and a lightbulb for your back porch.
The protected ﬁelds and methods of an object are analogous to the internal de-
sign of the bulb. The observable features, including the voltage and the size of
the socket, are provided without giving any details about the implementation
If light socket manufacturers depended on a particular imple-
of the object.
mentation of lightbulbs—for example the socket only supported bright xenon
bulbs—it might ultimately restrict the variety of suppliers of lightbulbs in the
future. Likewise, manufacturers of lightbulbs should be able to have a cer-
tain freedom in their implementation: as long as they only draw current in an
agreed-upon way and as long as their bulb ﬁts the socket, they should be free
to use whatever design they want. Today, most lamps take either incandescent
or ﬂuorescent bulbs.

In the same way that ﬁelds are encapsulated by a class, classes may be encap-
sulated by a package. A package is a collection of related classes that implement
some set of structures with a common theme. The classes of this text, for ex-
ample, are members of the structure package. In the same way that there are
users of classes, there are users of packages, and much of the analogy holds. In
particular, classes may be declared public, in which case they may be used by
anyone who imports the package into their program. If a class is not public, it
is automatically considered protected. These protected classes may only be
constructed and used by other classes within the same package.

1.4 A Special-Purpose Class: A Bank Account

We now look at the detailed construction of a simplistic class: a BankAccount.
Many times, it is necessary to provide a tag associated with an instance of a
data structure. You might imagine that your bank balance is kept in a database
at your bank. When you get money for a trip through the Berkshires, you swipe
your card through an automated teller bringing up your account. Your account Automated
number, presumably, is unique to your account. Nothing about you or your
banking history is actually stored in your account number. Instead, that num-
ber is used to ﬁnd the record linked to your account: the bank searches for a
structure associated with the number you provide. Thus a BankAccount is a sim-
ple, but important, data structure. It has an account (an identiﬁer that never
changes) and a balance (that potentially does change). The public methods of
such a structure are as follows:

teller: a robotic
palm reader.

public class BankAccount
{

public BankAccount(String acc, double bal)
// pre: account is a string identifying the bank account

BankAccount

12

The Object-Oriented Method

// balance is the starting balance
// post: constructs a bank account with desired balance

public boolean equals(Object other)
// pre: other is a valid bank account
// post: returns true if this bank account is the same as other

public String getAccount()
// post: returns the bank account number of this account

public double getBalance()
// post: returns the balance of this bank account

public void deposit(double amount)
// post: deposit money in the bank account

public void withdraw(double amount)
// pre: there are sufficient funds in the account
// post: withdraw money from the bank account

}

The substance of these methods has purposefully been removed because, again,
it is unimportant for us to know exactly how a BankAccount is implemented.
We have ways to construct and compare BankAccounts, as well as ways to read
the account number or balance, or update the balance.

Let’s look at the implementation of these methods, individually. To build a
new bank account, you must use the new operator to call the constructor with
two parameters. The account number provided never changes over the life of
the BankAccount—if it were necessary to change the value of the account num-
ber, a new BankAccount would have to be made, and the balance would have to
be transferred from one to the other. The constructor plays the important role
of performing the one-time initialization of the account number ﬁeld. Here is
the code for a BankAccount constructor:

protected String account;
protected double balance; // the balance associated with account

// the account number

public BankAccount(String acc, double bal)
// pre: account is a string identifying the bank account
// balance is the starting balance
// post: constructs a bank account with desired balance
{

account = acc;
balance = bal;

}

Two ﬁelds—account and balance—of the BankAccount object are responsible
for maintaining the object’s state. The account keeps track of the account num-
ber, while the balance ﬁeld maintains the balance.

1.4 A Special-Purpose Class: A Bank Account

13

Since account numbers are unique to BankAccounts, to check to see if two
accounts are “the same,” we need only compare the account ﬁelds. Here’s the
code:

public boolean equals(Object other)
// pre: other is a valid bank account
// post: returns true if this bank account is the same as other
{

BankAccount that = (BankAccount)other;
// two accounts are the same if account numbers are the same
return this.account.equals(that.account);

}

Notice that the BankAccount equals method calls the equals method of the key,
a String. Both BankAccount and String are nonprimitive types, or examples
of Objects. Every object in Java has an equals method. If you don’t explicitly
provide one, the system will write one for you. Generally speaking, one should
assume that the automatically written or default equals method is of little use.
This notion of “equality” of objects is often based on the complexities of our
abstraction; its design must be considered carefully.

One can ask the BankAccount about various aspects of its state by calling its

getAccount or getBalance methods:

public String getAccount()
// post: returns the bank account number of this account
{

return account;

}

public double getBalance()
// post: returns the balance of this bank account
{

return balance;

}

These methods do little more than pass along the information found in the
account and balance ﬁelds, respectively. We call such methods accessors. In a
different implementation of the BankAccount, the balance would not have to be
explicitly stored—the value might be, for example, the difference between two
ﬁelds, deposits and drafts. Given the interface, it is not much of a concern to
the user which implementation is used.

We provide two more methods, deposit and withdraw, that explicitly mod-

ify the current balance. These are mutator methods:

public void deposit(double amount)
// post: deposit money in the bank account
{

balance = balance + amount;

}

14

The Object-Oriented Method

public void withdraw(double amount)
// pre: there are sufficient funds in the account
// post: withdraw money from the bank account
{

balance = balance - amount;

}

Because we would like to change the balance of the account, it is important to
have a method that allows us to modify it. On the other hand, we purposefully
don’t have a setAccount method because we do not want the account number
to be changed without a considerable amount of work (work that, by the way,
models reality).

Here is a simple application that determines whether it is better to deposit
$100 in an account that bears 5 percent interest for 10 years, or to deposit $100
in an account that bears 2 1
2 percent interest for 20 years. It makes use of the
BankAccount object just outlined:

public static void main(String[] args)
{

// Question: is it better to invest $100 over 10 years at 5%
//
BankAccount jd = new BankAccount("Jain Dough",100.00);
BankAccount js = new BankAccount("Jon Smythe",100.00);

or to invest $100 over 20 years at 2.5% interest?

for (int years = 0; years < 10; years++)
{

jd.deposit(jd.getBalance() * 0.05);

}
for (int years = 0; years < 20; years++)
{

js.deposit(js.getBalance() * 0.025);

}
System.out.println("Jain invests $100 over 10 years at 5%.");
System.out.println("After 10 years " + jd.getAccount() +

" has $" + jd.getBalance());

System.out.println("Jon invests $100 over 20 years at 2.5%.");
System.out.println("After 20 years " + js.getAccount() +

" has $" + js.getBalance());

}

Exercise 1.2 Which method of investment would you pick?

1.5 A General-Purpose Class: An Association

At least Dr.
Seuss started
with 50 words!

The following small application implements a Pig Latin translator based on a
dictionary of nine words. The code makes use of an array of Associations,
each of which establishes a relation between an English word and its Pig Latin

1.5 A General-Purpose Class: An Association

15

translation. For each string passed as the argument to the main method, the
dictionary is searched to determine the appropriate translation.

public class atinLay {

// a pig latin translator for nine words
public static void main(String args[])
{

// build and fill out an array of nine translations
Association dict[] = new Association[9];
dict[0] = new Association("a","aay");
dict[1] = new Association("bad","adbay");
dict[2] = new Association("had","adhay");
dict[3] = new Association("dad","adday");
dict[4] = new Association("day","ayday");
dict[5] = new Association("hop","ophay");
dict[6] = new Association("on","onay");
dict[7] = new Association("pop","oppay");
dict[8] = new Association("sad","adsay");

atinlay

for (int argn = 0; argn < args.length; argn++)
{

// for each argument
for (int dictn = 0; dictn < dict.length; dictn++)
{

// check each dictionary entry
if (dict[dictn].getKey().equals(args[argn]))

System.out.println(dict[dictn].getValue());

}

}

}

}

When this application is run with the arguments hop on pop, the results are

ophay
onay
oppay

While this application may seem rather trivial, it is easy to imagine a large-scale
application with similar needs.3

We now consider the design of the Association. Notice that while the type
of data maintained is different, the purpose of the Association is very similar
to that of the BankAccount class we discussed in Section 1.4. An Association
is a key-value pair such that the key cannot be modiﬁed. Here is the interface
for the Association class:

import java.util.Map;

3 Pig Latin has played an important role in undermining court-ordered restrictions placed on music
piracy. When Napster—the rebel music trading ﬁrm—put in checks to recognize copyrighted music
by title, traders used Pig Latin translators to foil the recognition software!

Association

16

The Object-Oriented Method

public class Association implements Map.Entry
{

public Association(Object key, Object value)
// pre: key is non-null
// post: constructs a key-value pair

public Association(Object key)
// pre: key is non-null
// post: constructs a key-value pair; value is null

public boolean equals(Object other)
// pre: other is non-null Association
// post: returns true iff the keys are equal

public Object getValue()
// post: returns value from association

public Object getKey()
// post: returns key from association

public Object setValue(Object value)
// post: sets association’s value to value

}

For the moment, we will ignore the references to Map and Map.entry; these will
be explained later, in Chapter 15. What distinguishes an Association from a
more specialized class, like BankAccount, is that the ﬁelds of an Association
are type Object. The use of the word Object in the deﬁnition of an Association
makes the deﬁnition very general: any value that is of type Object—any non-
primitive data type in Java—can be used for the key and value ﬁelds.

Unlike the BankAccount class, this class has two different constructors:

protected Object theKey; // the key of the key-value pair
protected Object theValue; // the value of the key-value pair

public Association(Object key, Object value)
// pre: key is non-null
// post: constructs a key-value pair
{

Assert.pre(key != null, "Key must not be null.");
theKey = key;
theValue = value;

}

public Association(Object key)
// pre: key is non-null
// post: constructs a key-value pair; value is null
{

this(key,null);

}

1.5 A General-Purpose Class: An Association

17

The ﬁrst constructor—the constructor distinguished by having two parame-
ters—allows the user to construct a new Association by initializing both ﬁelds.
On occasion, however, we may wish to have an Association whose key ﬁeld is
set, but whose value ﬁeld is left referencing nothing. (An example might be a
medical record: initially the medical history is incomplete, perhaps waiting to
be forwarded from a previous physician.) For this purpose, we provide a sin-
gle parameter constructor that sets the value ﬁeld to null. Note that we use
this(key,null) as the body. The one-parameter constructor calls this object’s
two-parameter constructor with null as the second parameter. We write the
constructors in this dependent manner so that if the underlying implementation
of the Association had to be changed, only the two-parameter method would
have to be updated. It also reduces the complexity of the code and saves your
ﬁngerprints!

Now, given a particular Association, it is useful to be able to retrieve the
key or value. Since the implementation is hidden, no one outside the class is
able to see it. Users must depend on the accessor methods to observe the data.

public Object getValue()
// post: returns value from association
{

return theValue;

}

public Object getKey()
// post: returns key from association
{

return theKey;

}

When necessary, the method setValue can be used to change the value associ-
ated with the key. Thus, the setValue method simply takes its parameter and
assigns it to the value ﬁeld:

public Object setValue(Object value)
// post: sets association’s value to value
{

Object oldValue = theValue;
theValue = value;
return oldValue;

}

There are other methods that are made available to users of the Association
class, but we will not discuss the details of that code until later. Some of the
methods are required, some are useful, and some are just nice to have around.
While the code may look complicated, we take the time to implement it cor-
rectly, so that we will not have to reimplement it in the future.

Principle 2 Free the future: reuse code.

NNWSWSENEWSE18

The Object-Oriented Method

It is difﬁcult to ﬁght the temptation to design data structures from scratch. We
shall see, however, that many of the more complex structures would be very
difﬁcult to construct if we could not base our implementations on the results of
previous work.

1.6 Sketching an Example: A Word List

Suppose we’re interested in building a game of Hangman. The computer selects
random words and we try to guess them. Over several games, the computer
should pick a variety of words and, as each word is used, it should be removed
from the word list. Using an object-oriented approach, we’ll determine the
essential features of a WordList, the Java object that maintains our list of words.
Our approach to designing the data structures has the following ﬁve informal

steps:

1. Identify the types of operations you expect to perform on your object.
What operations access your object only by reading its data? What opera-
tions might modify or mutate your objects?

2. Identify, given your operations, those data that support the state of your
Information about an object’s state is carried within the object
object.
between operations that modify the state. Since there may be many ways
to encode the state of your object, your description of the state may be
very general.

3. Identify any rules of consistency. In the Ratio class, for example, it would
not be good to have a zero denominator. Also, the numerator and denom-
inator should be in lowest terms.

4. Determine the number and form of the constructors. Constructors are
synthetic: their sole responsibility is to get a new object into a good initial
and consistent state. Don’t forget to consider the best state for an object
constructed using the parameterless default constructor.

5. Identify the types and kinds of information that, though declared pro-
tected, can efﬁciently provide the information needed by the public
methods.
Important choices about the internals of a data structure are
usually made at this time. Sometimes, competing approaches are devel-
oped until a comparative evaluation can be made. That is the subject of
much of this book.

The operations necessary to support a list of words can be sketched out
easily, even if we don’t know the intimate details of constructing the Hangman
game itself. Once we see how the data structure is used, we have a handle on
the design of the interface. Thinking about the overall design of Hangman, we
can identify the following general use of the WordList object:

1.6 Sketching an Example: A Word List

19

WordList list;
String targetWord;

// declaration

list = new WordList(10);
list.add("disambiguate");
list.add("inputted");
list.add("subbookkeeper"); // now that’s coollooking!
while (!list.isEmpty())
{

// game loop

// construction

// is this a word? how about ambiguate?
// really? what verbification!

WordList

targetWord = list.selectAny();
// ...play the game using target word...
list.remove(targetWord);

// update

// selection

}

Let’s consider these lines. One of the ﬁrst lines (labeled declaration) de-
clares a reference to a WordList. For a reference to refer to an object, the object
must be constructed. We require, therefore, a constructor for a WordList. The
construction line allocates an initially empty list of words ultimately contain-
ing as many as 10 words. We provide an upper limit on the number of words
that are potentially stored in the list. (We’ll see later that providing such infor-
mation can be useful in designing efﬁcient data structures.) On the next three
lines, three (dubious) words are added to the list.

The while loop accomplishes the task of playing Hangman with the user.
This is possible as long as the list of words is not empty. We use the isEmpty
method to test this fact. At the beginning of each round of Hangman, a random
word is selected (selectAny), setting the targetWord reference. To make things
interesting, we presume that the selectAny method selects a random word each
time. Once the round is ﬁnished, we use the remove method to remove the word
from the word list, eliminating it as a choice in future rounds.

There are insights here. First, we have said very little about the Hangman
game other than its interaction with our rather abstract list of words. The details
of the screen’s appearance, for example, do not play much of a role in under-
standing how the WordList structure works. We knew that a list was necessary
for our program, and we considered the program from the point of view of the
object. Second, we don’t really know how the WordList is implemented. The
words may be stored in an array, or in a ﬁle on disk, or they may use some tech-
nology that we don’t currently understand. It is only important that we have
faith that the structure can be implemented. We have sketched out the method
headers, or signatures, of the WordList interface, and we have faith that an im-
plementation supporting the interface can be built. Finally we note that what
we have written is not a complete program. Still, from the viewpoint of the
WordList structure, there are few details of the interface that are in question.
A reasoned individual should be able to look at this design and say “this will
work—provided it is implemented correctly.” If a reviewer of the code were to
ask a question about how the structure works, it would lead to a reﬁnement of
our understanding of the interface.

We have, then, the following required interface for the WordList class:

20

The Object-Oriented Method

public class WordList
{

public WordList(int size)
// pre: size >= 0
// post: construct a word list capable of holding "size" words

public boolean isEmpty()
// post: return true iff the word list contains no words

public void add(String s)
// post: add a word to the word list, if it is not already there

public String selectAny()
// pre: the word list is not empty
// post: return a random word from the list

public void remove(String word)
// pre: word is not null
// post: remove the word from the word list

}

We will leave the implementation details of this example until later. You might
consider various ways that the WordList might be implemented. As long as
the methods of the interface can be supported by your data structure, your
implementation is valid.

Exercise 1.3 Finish the sketch of the WordList class to include details about the
state variables.

1.7 Sketching an Example: A Rectangle Class

Suppose we are developing a graphics system that allows the programmer to
draw on a DrawingWindow. This window has, associated with it, a Cartesian
coordinate system that allows us to uniquely address each of the points within
the window. Suppose, also, that we have methods for drawing line segments,
say, using the Line object. How might we implement a rectangle—called a
Rect—to be drawn in the drawing window?

One obvious goal would be to draw a Rect on the DrawingWindow. This
might be accomplished by drawing four line segments. It would be useful to
be able to draw a ﬁlled rectangle, or to erase a rectangle (think: draw a ﬁlled
rectangle in the background color). We’re not sure how to do this efﬁciently, but
these latter methods seem plausible and consistent with the notion of drawing.
(We should check to see if it is possible to draw in the background color.) This
leads to the following methods:

Rect

public void fillOn(DrawingTarget d)
// pre: d is a valid drawing window
// post: the rectangle is filled on the drawing window d

1.7 Sketching an Example: A Rectangle Class

21

public void clearOn(DrawingTarget d)
// pre: d is a valid drawing window
// post: the rectangle is erased from the drawing window

public void drawOn(DrawingTarget d)
// pre: d is a valid drawing window
// post: the rectangle is drawn on the drawing window

It might be useful to provide some methods to allow us to perform basic calcu-
lations—for example, we might want to ﬁnd out if the mouse arrow is located
within the Rect. These require accessors for all the obvious data. In the hope
that we might use a Rect multiple times in multiple locations, we also provide
methods for moving and reshaping the Rect.

public boolean contains(Pt p)
// pre: p is a valid point
// post: true iff p is within the rectangle

public int left()
// post: returns left coordinate of the rectangle

public void left(int x)
// post: sets left to x; dimensions remain unchanged

public int width()
// post: returns the width of the rectangle

public void width(int w)
// post: sets width of rectangle, center and height unchanged

public void center(Pt p)
// post: sets center of rect to p; dimensions remain unchanged

public void move(int dx, int dy)
// post: moves rectangle to left by dx and down by dy

public void moveTo(int left, int top)
// post: moves left top of rectangle to (left,top);
//

dimensions are unchanged

public void extend(int dx, int dy)
// post: moves sides of rectangle outward by dx and dy

Again, other approaches might be equally valid. No matter how we might rep-
resent a Rect, however, it seems that all rectangular regions with horizontal
and vertical sides can be speciﬁed with four integers. We can, then, construct a
Rect by specifying, say, the left and top coordinates and the width and height.
For consistency’s sake, it seems appropriate to allow rectangles to be drawn
anywhere (even off the screen), but the width and height should be non-negative

22

The Object-Oriented Method

values. We should make sure that these constraints appear in the documenta-
tion associated with the appropriate constructors and methods. (See Section 2.2
for more details on how to write these comments.)

Given our thinking, we have some obvious Rect constructors:

public Rect()
// post: constructs a trivial rectangle at origin

public Rect(Pt p1, Pt p2)
// post: constructs a rectangle between p1 and p2

public Rect(int x, int y, int w, int h)
// pre: w >= 0, h >= 0
// post: constructs a rectangle with upper left (x,y),
//

width w, height h

We should feel pleased with the progress we have made. We have developed
the signatures for the rectangle interface, even though we have no immediate
application. We also have some emerging answers on approaches to implement-
ing the Rect internally. If we declare our Rect data protected, we can insulate
ourselves from changes suggested by inefﬁciencies we may yet discover.

Exercise 1.4 Given this sketch of the Rect interface, how would you declare the
private data associated with the Rect object? Given your approach, describe how
you might implement the center(int x, int y) method.

1.8 Interfaces

Sometimes it is useful to describe the interface for a number of different classes,
without committing to an implementation. For example, in later sections of this
text we will implement a number of data structures that are able to be modiﬁed
by adding or removing values. We can, for all of these classes, specify a few of
their fundamental methods by using the Java interface declaration:

public interface Structure
{

Structure

public int size();
// post: computes number of elements contained in structure

public boolean isEmpty();
// post: return true iff the structure is empty

public void clear();
// post: the structure is empty

public boolean contains(Object value);
// pre: value is non-null
// post: returns true iff value.equals some value in structure

1.8 Interfaces

23

public void add(Object value);
// pre: value is non-null
// post: value has been added to the structure
//

replacement policy is not specified

public Object remove(Object value);
// pre: value is non-null
// post: an object equal to value is removed and returned, if found

public java.util.Enumeration elements();
// post: returns an enumeration for traversing structure;
//
//

all structure package implementations return
an AbstractIterator

public Iterator iterator();
// post: returns an iterator for traversing structure;
//
//

all structure package implementations return
an AbstractIterator

public Collection values();
// post: returns a Collection that may be used with
//

Java’s Collection Framework

}

Notice that the body of each method has been replaced by a semicolon.
It
is, in fact, illegal to specify any code in a Java interface. Specifying just the
method signatures in an interface is like writing boilerplate for a contract with-
out committing to any implementation. When we decide that we are interested
in constructing a new class, we can choose to have it implement the Structure
interface. For example, our WordList structure of Section 1.6 might have made
use of our Structure interface by beginning its declaration as follows:

public class WordList implements Structure

When the WordList class is compiled by the Java compiler, it checks to see that
each of the methods mentioned in the Structure interface—add, remove, size,
and the others—is actually implemented. In this case, only isEmpty is part of
the WordList speciﬁcation, so we must either (1) not have WordList implement
the Structure interface or (2) add the methods demanded by Structure.

Interfaces may be extended. Here, we have a possible deﬁnition of what it

WordList

means to be a Set:

public interface Set extends Structure
{

public void addAll(Structure other);
// pre: other is non-null
// post: values from other are added into this set

Set

24

The Object-Oriented Method

public boolean containsAll(Structure other);
// pre: other is non-null
// post: returns true if every value in set is in other

public void removeAll(Structure other);
// pre: other is non-null
// post: values of this set contained in other are removed

public void retainAll(Structure other);
// pre: other is non-null
// post: values not appearing in the other structure are removed

}

A Set requires several set-manipulation methods—addAll (i.e., set union) retain-
All (set intersection), and removeAll (set difference)—as well as the meth-
ods demanded by being a Structure. If we implement these methods for the
WordList class and indicate that WordList implements Set, the WordList class
could be used wherever either a Structure or Set is required. Currently, our
WordList is close to, but not quite, a Structure. Applications that demand
the functionality of a Structure will not be satisﬁed with a WordList. Having
the class implement an interface increases the ﬂexibility of its use. Still, it may
require considerable work for us to upgrade the WordList class to the level of
a Structure. It may even work against the design of the WordList to provide
the missing methods. The choices we make are part of an ongoing design pro-
cess that attempts to provide the best implementations of structures to meet the
demands of the user.

1.9 Who Is the User?

When implementing data structures using classes and interfaces, it is sometimes
hard to understand why we might be interested in hiding the implementation.
After all, perhaps we know that ultimately we will be the only programmers
making use of these structures. That might be a good point, except that if
you are really a successful programmer, you will implement the data structure
ﬂawlessly this week, use it next week, and not return to look at the code for
a long time. When you do return, your view is effectively that of a user of the
code, with little or no memory of the implementation.

One side effect of this relationship is that we have all been reminded of the
need to write comments. If you do not write comments, you will not be able to
read the code. If, however, you design, document, and implement your interface
carefully, you might not ever have to look at the implementation! That’s good
news because, for most of us, in a couple of months our code is as foreign to
us as if someone else had implemented it. The end result: consider yourself a
user and design and abide by your interface wherever possible. If you know of
some public ﬁeld that gives a hint of the implementation, do not make use of it.
Instead, access the data through appropriate methods. You will be happy you

1.10 Conclusions

25

did later, when you optimize your implementation.

Principle 3 Design and abide by interfaces as though you were the user.

A quick corollary to this statement is the following:

Principle 4 Declare data ﬁelds protected.

If the data are protected, you cannot access them from outside the class, and
you are forced to abide by the restricted access of the interface.

1.10 Conclusions

The construction of substantial applications involves the development of com-
plex and interacting structures. In object-oriented languages, we think of these
structures as objects that communicate through the passing of messages or,
more formally, the invocation of methods.

We use object orientation in Java to write the structures found in this book.
It is possible, of course, to design data structures without object orientation, but
any effective data structuring model ultimately depends on the use of some form
of abstraction that allows the programmer to avoid considering the complexities
of particular implementations.

In many languages, including Java, data abstraction is supported by sepa-
rating the interface from the implementation of the data structure. To ensure
that users cannot get past the interface to manipulate the structure in an uncon-
trolled fashion, the system controls access to ﬁelds, methods, and classes. The
implementor plays an important role in making sure that the structure is usable,
given the interface. This role is so important that we think of implementation
as supporting the interface—sometimes usefully considered a contract between
the implementor and the user. This analogy is useful because, as in the real
world, if contracts are violated, someone gets upset!

Initial design of the interfaces for data structures arises from considering
how they are used in simple applications. Those method calls that are required
by the application determine the interface for the new structure and constrain,
in various ways, the choices we make in implementing the object.

In our implementation of an Association, we can use the Object class—
that class inherited by all other Java classes—to write very general data struc-
tures. The actual type of value that is stored in the Association is determined
by the values passed to the constructors and mutators of the class. This abil-
ity to pass a subtype to any object that requires a super type is a strength of
object-oriented languages—and helps to reduce the complexity of code.

NNWSWSENEWSENNWSWSENEWSE26

The Object-Oriented Method

Self Check Problems

Solutions to these problems begin on page 441.
What is meant by abstraction?
1.1
What is procedural abstraction?
1.2
What is data abstraction?
1.3
How does Java support the concept of a message?
1.4
What is the difference between an object and a class?
1.5
What makes up a method’s signature?
1.6
What is the difference between an interface and an implementation?
1.7
What is the difference between an accessor and a mutator?
1.8
A general purpose class, such as an Association, often makes use of
1.9
parameters of type Object. Why?
1.10 What is the difference between a reference and an object?
1.11 Who uses a class?

Problems

Which of the following are primitive Java types: int, Integer, double,

Solutions to the odd-numbered problems begin on page 451.
1.1
Double, String, char, Association, BankAccount, boolean, Boolean?
1.2
calls?

Which of the following variables are associated with valid constructor

BankAccount a,b,c,d,e,f;
Association g,h;
a = new BankAccount("Bob",300.0);
b = new BankAccount(300.0,"Bob");
c = new BankAccount(033414,300.0);
d = new BankAccount("Bob",300);
e = new BankAccount("Bob",new Double(300));
f = new BankAccount("Bob",(double)300);
g = new Association("Alice",300.0);
h = new Association("Alice",new Double(300));

1.3

For each pair of classes, indicate which class extends the other:

a. java.lang.Number, java.lang.Double

b. java.lang.Number, java.lang.Integer

c. java.lang.Number, java.lang.Object

d. java.util.Stack, java.util.Vector

1.10 Conclusions

27

e. java.util.Hashtable, java.util.Dictionary

Rewrite the compound interest program (discussed when considering

1.4
BankAccounts in Section 1.4) so that it uses Associations.

Write a program that attempts to modify one of the private ﬁelds of
1.5
an Association. When does your environment detect the violation? What
happens?

Finish the design of a Ratio class that implements a ratio between
1.6
two integers. The class should support standard math operations: addition,
subtraction, multiplication, and division. You should also be able to construct
Ratios from either a numerator-denominator pair, or a single integer, or with
no parameter at all (what is a reasonable default value?).

1.7
a, b, the probability that a
write a program to compute an approximation to π.

Amazing fact: If you construct a Ratio from two random integers, 0 <
b is already in reduced terms is 6
π2 . Use this fact to

Design a class to represent a U.S. telephone number.

It should sup-
1.8
port three types of constructors—one that accepts three numbers, represent-
ing area code, exchange, and extension; another that accepts two integers,
representing a number within your local area code; and a third constructor
that accepts a string of letters and numbers that represent the number (e.g.,
"900-410-TIME"). Provide a method that determines if the number is provided
toll-free (such numbers have area codes of 800, 866, 877, 880, 881, 882, or
888).

Sometimes it is useful to measure the length of time it takes for a piece
1.9
of code to run. (For example, it may help determine where optimizations of
your code would be most effective.) Design a Stopwatch class to support tim-
ing of events. You should consider use of the nanosecond clock in the Java
environment, System.nanoTime(). Like many stopwatches, it should support
starting, temporary stopping, and a reset. The design of the protected section
of the stopwatch should hide the implementation details.

Design a data structure in Java that represents a musical tone. A tone
1.10
can be completely speciﬁed as a number of cycles per second (labeled Hz for
hertz), or the number of half steps above a commonly agreed upon tone, such
as A (in modern times, in the United States, considered to be 440 Hz). Higher
tones have higher frequencies. Two tones are an octave (12 semitones) apart
if one has a frequency twice the other. A half step or semitone increase in tone
√
is 12
2 ≈ 1.06 times higher. Your tone constructors should accept a frequency
(a double) or a number of half steps (an int) above A. Imperfect frequencies
should be tuned to the nearest half step. Once constructed, a tone should be
able to provide its frequency in either cycles per second or half-steps above A.

Extend Problem 1.10 to allow a second parameter to each constructor
1.11
to specify the deﬁnition of A upon which the tone’s deﬁnition is based. What
modern tone most closely resembles that of modern middle C (9 semitones
below A) if A is deﬁned to be 415 Hz?

28

The Object-Oriented Method

Design a data structure to represent a combination lock. When the
1.12
lock is constructed, it is provided with an arbitrary length array of integers
between 0 and 25 specifying a combination (if no combination is provided,
9 − 0 − 21 − 0 is the default).
Initially, it is locked. Two methods—press
and reset—provide a means of entering a combination: press enters the next
integer to be used toward matching the combination, while reset re-readies
the lock for accepting the ﬁrst integer of the combination. Only when press is
used to match the last integer of the combination does the lock silently unlock.
Mismatched integers require a call to the reset method before the combination
can again be entered. The isLocked method returns true if and only if the lock
is locked. The lock method locks and resets the lock. In the unlocked state only
the isLocked and lock methods have effect. (Aside: Because of the physical
construction of many combination locks, it is often the case that combinations
have patterns. For example, a certain popular lock is constructed with a three-
number combination. The ﬁrst and last numbers result in the same remainder x
when divided by 4. The middle number has remainder (x + 2)%4 when divided
by 4!)

Design a data structure to simulate the workings of a car radio. The
1.13
state of the radio is on or off, and it may be used to listen to an AM or FM
station. A dozen modiﬁable push buttons (identiﬁed by integers 1 through 12)
allow the listener to store and recall AM or FM frequencies. AM frequencies can
be represented by multiples of 10 in the range 530 to 1610. FM frequencies are
found at multiples of 0.2 in the range 87.9 to 107.9.

Design a data structure to maintain the position of m coins of radius 1
1.14
through m on a board with n ≥ m squares numbered 0 through n − 1. You may
provide whatever interface you ﬁnd useful to allow your structure to represent
any placement of coins, including stacks of coins in a single cell. A conﬁguration
is valid only if large coins are not stacked on small coins. Your structure should
have an isValid method that returns true if the coins are in a valid position.
(A problem related to this is discussed in Section 10.2.1.)

Top viewSide view1.11 Laboratory: The Day of the Week

Calculator

Objective. To (re)establish ties with Java: to write a program that reminds us
of the particulars of numeric calculations and array manipulation in Java.

Discussion. In this lab we learn to compute the day of the week for any date
between January 1, 1900, and December 31, 2099.4 During this period of time,
the only calendar adjustment is a leap-year correction every 4 years. (Years
divisible by 100 are normally not leap years, but years divisible by 400 always
are.) Knowing this, the method essentially computes the number of days since
the beginning of the twentieth century in modulo 7 arithmetic. The computed
remainder tells us the day of the week, where 0 is Saturday.

An essential feature of this algorithm involves remembering a short table of
monthly adjustments. Each entry in the table corresponds to a month, where
January is month 1 and December is month 12.

Month
Adjustment

1
1

2
4

3
4

4
0

5
2

6
5

7
0

8
3

9
6

10
1

11
4

12
6

If the year is divisible by 4 (it’s a leap year) and the date is January or February,
you must subtract 1 from the adjustment.

Remembering this table is equivalent to remembering how many days are in
each month. Notice that 144 is 122, 025 is 52, 036 is 62, and 146 is a bit more
than 122. Given this, the algorithm is fairly simple:

1. Write down the date numerically. The date consists of a month between
1 and 12, a day of the month between 1 and 31, and the number of
years since 1900. Grace Hopper, computer language pioneer, was born
December 9, 1906. That would be represented as year 6. Jana the Giraffe,
of the National Zoo, was born on January 18, 2001. That year would be
represented as year 101.

2. Compute the sum of the following quantities:

• the month adjustment from the given table (e.g., 6 for Admiral Hop-

per)

• the day of the month

• the year

4 This particular technique is due to John Conway, of Princeton University. Professor Conway
answers 10 day of the week problems before gaining access to his computer. His record is at the
time of this writing well under 15 seconds for 10 correctly answered questions. See “Scientist
at Work: John H. Conway; At Home in the Elusive World of Mathematics,” The New York Times,
October 12, 1993.

30

The Object-Oriented Method

• the whole number of times 4 divides the year (e.g., 25 for Jana the

Giraffe)

3. Compute the remainder of the sum of step 2, when divided by 7. The
remainder gives the day of the week, where Saturday is 0, Sunday is 1, etc.
Notice that we can compute the remainders before we compute the sum.
You may also have to compute the remainder after the sum as well, but if
you’re doing this in your head, this considerably simpliﬁes the arithmetic.

What day of the week was Tiger Woods born?

1. Tiger’s birth date is 12-30-75.

2. Remembering that 18 × 4 = 72, we write the sum as follows:

6 + 30 + 75 + 18

which is equivalent to the following sum, modulo 7:

6 + 2 + 5 + 4 = 17 ≡ 3 mod 7

3. He was born on day 3, a Tuesday.

Now you practice: Which of Grace and Jana was born on a Thursday? (The
other was born on a Sunday.)
Procedure. Write a Java program that performs Conway’s day of the week chal-
lenge:

1. Develop an object that can hold a date.

2. Write a method to compute a random date between 1900 and 2099. How
will you limit the range of days potentially generated for any particular
month?

3. Write a method of your date class to compute the day of the week associ-
ated with a date. Be careful: the table given in the discussion has January
as month 1, but Java would prefer it to be month 0! Don’t forget to handle
the birthday of Jimmy Dorsey (famous jazzman), February 29, 1904.

4. Your main method should repeatedly (1) print a random date, (2) read a
predicted day of the week (as an integer/remainder), and (3) check the
correctness of the guess. The program should stop when 10 dates have
been guessed correctly and print the elapsed time. (You may wish to set
this threshold lower while you’re testing the program.)

Helpful Hints. You may ﬁnd the following Java useful:

1. Random integers may be selected using the java.util.Random class:

Random r = new Random();
int month = (Math.abs(r.nextInt()) % 12) + 1;

Jimmy was a
Monday’s child.

1.11 Laboratory: The Day of the Week Calculator

31

You will need to import java.util.Random; at the top of your program
to make use of this class. Be aware that you need to only construct one
random number generator per program run. Also, the random number
generator potentially returns negative numbers. If Math.abs is not used,
these values generate negative remainders.

2. You can ﬁnd out how many thousandths of seconds have elapsed since
the 1960s, by calling the Java method, System.currentTimeMillis(). It
returns a value of type long. We can use this to measure the duration of
an experiment, with code similar to the following:

In 2001,
1 trillion millis
since the ’60s.
Dig that!

long start = System.currentTimeMillis();
//
// place experiment to be timed here
//
long duration = System.currentTimeMillis()-start;
System.out.println("time: "+(duration/1000.0)+" seconds.");

The granularity of this timer isn’t any better than a thousandth of a second.
Still, we’re probably not in Conway’s league yet.

After you ﬁnish your program, you will ﬁnd you can quickly learn to answer

10 of these day of the week challenges in less than a minute.
Thought Questions. Consider the following questions as you complete the lab:

1. True or not: In Java is it true that (a % 7) == (a - a/7*7) for a >= 0?

2. It’s rough to start a week on Saturday. What adjustments would be nec-
essary to have a remainder of 0 associated with Sunday? (This might
allow a mnemonic of Nun-day, One-day, Twos-day, Wednesday, Fours-day,
Fives-day, Saturday.)

3. Why do you subtract 1 in a leap year if the date falls before March?

4. It might be useful to compute the portion of any calculation associated
with this year, modulo 7. Remembering that value will allow you to opti-
mize your most frequent date calculations. What is the remainder associ-
ated with this year?

For years
divisible by 28:
think zero!

Notes:

Chapter 2

Comments, Conditions,
and Assertions

Preconditions
Postconditions

Concepts:
.
.
. Assertions
. Copyrighting code

Wizards are invited to improve it.

/* This is bogus code.
*/
—Anonymous

CONSIDER THIS: WE CALL OUR PROGRAMS “CODE”! The features of computer
languages, including Java, are designed to help express algorithms in a manner
that a machine can understand. Making a program run more efﬁciently often
makes it less understandable.
If language design was driven by the need to
make the program readable by programmers, it would be hard to argue against
programming in English.

A comment is a carefully crafted piece of text that describes the state of the
machine, the use of a variable, or the purpose of a control construct. Many
of us, though, write comments for the same reason that we exercise: we feel
guilty. You feel that, if you do not write comments in your code, you “just know”
something bad is going to happen. Well, you are right. A comment you write
today will help you out of a hole you dig tomorrow.

All too often comments are hastily written after the fact, to help under-
stand the code. The time spent thinking seriously about the code has long since
passed, and the comment might not be right. If you write comments before-
hand, while you are designing your code, it is more likely your comments will
describe what you want to do as you carefully think it out. Then, when some-
thing goes wrong, the comment is there to help you ﬁgure out the code.
In
fairness, the code and the comment have a symbiotic relationship. Writing one
or the other does not really feel complete, but writing both provides you with
the redundancy of concept: one lucid and one as clear as Java.

The one disadvantage of comments is that, unlike code, they cannot be
checked. Occasionally, programmers come across comments such as “If you
think you understand this, you don’t!” or “Are you reading this?” One could, of
course, annotate programs with mathematical formulas. As the program is com-
piled, the mathematical comments are distilled into very concise descriptions of

Okay, perhaps
French!

Ruth Krauss: “A
hole
is to dig.”

34

Comments, Conditions, and Assertions

Semiformal
convention: a
meeting of tie
haters.

what should be going on. When the output from the program’s code does not
match the result of the formula, something is clearly wrong with your logic. But
which logic? The writing of mathematical comments is a level of detail most
programmers would prefer to avoid.

A compromise is a semiformal convention for comments that provide a rea-
sonable documentation of when and what a program does. In the code associ-
ated with this book, we see one or two comments for each method or function
that describe its purpose. These important comments are the precondition and
postcondition.

2.1 Pre- and Postconditions

The precondition describes, as succinctly as possible in your native tongue, the
conditions under which a method may be called and expected to produce correct
results. Ideally the precondition expresses the state of the program. This state
is usually cast in terms of the parameters passed to the routine. For example,
the precondition on a square root function might be

// pre: x is nonnegative

sqrt

The authors of this square root function expect that the parameter is not a
negative number. It is, of course, legal in Java to call a function or method if
the precondition is not met, but it might not produce the desired result. When
there is no precondition on a procedure, it may be called without failure.

The postcondition describes the state of the program once the routine has
been completed, provided the precondition was met. Every routine should have
some postcondition. If there were not a postcondition, then the routine would
not change the state of the program, and the routine would have no effect!
Always provide postconditions.

Pre- and postconditions do not force you to write code correctly. Nor do they
help you ﬁnd the problems that do occur. They can, however, provide you with
a uniform method for documenting the programs you write, and they require
more thought than the average comment. More thought put into programs
lowers your average blood pressure and ultimately saves you time you might
spend more usefully playing outside, visiting museums, or otherwise bettering
your mind.

2.2 Assertions

And the
batteries never
needed
replacing.

In days gone by, homeowners would sew ﬁrecrackers in their curtains. If the
house were to catch ﬁre, the curtains would burn, setting off the ﬁrecrackers. It
was an elementary but effective ﬁre alarm.

An assertion is an assumption you make about the state of your program. In
Java, we will encode the assertion as a call to a function that veriﬁes the state
of the program. That function does nothing if the assertion is true, but it halts

2.2 Assertions

35

your program with an error message if it is false. It is a ﬁrecracker to sew in
your program.
If you sew enough assertions into your code, you will get an
early warning if you are about to be burned by your logic.

Principle 5 Test assertions in your code.

The Assert class provides several functions to help you test the state of your

program as it runs:

public class Assert
{

static public void pre(boolean test, String message)
// pre: result of precondition test
// post: does nothing if test true, otherwise abort w/message

Assert

static public void post(boolean test, String message)
// pre: result of postcondition test
// post: does nothing if test true, otherwise abort w/message

static public void condition(boolean test, String message)
// pre: result of general condition test
// post: does nothing if test true, otherwise abort w/message

static public void invariant(boolean test, String message)
// pre: result of an invariant test
// post: does nothing if test true, otherwise abort w/message

static public void fail(String message)
// post: throws error with message

}

Each of pre, post, invariant, and condition methods test to see if its ﬁrst
argument—the assertion—is true. The message is used to indicate the condition
tested by the assertion. Here’s an example of a check to make sure that the
precondition for the sqrt function was met:

public static double sqrt(double x)
// pre: x is nonnegative
// post: returns the square root of x
{

Assert.pre(x >= 0,"the value is nonnegative.");
double guess = 1.0;
double guessSquared = guess * guess;

while (Math.abs(x-guessSquared) >= 0.00000001) {

// guess is off a bit, adjust
guess += (x-guessSquared)/2.0/guess;
guessSquared = guess*guess;

}
return guess;

}

NNWSWSENEWSE36

Comments, Conditions, and Assertions

Should we call sqrt with a negative value, the assertion fails, the message
is printed out, and the program comes to a halt. Here’s what appears at the
display:

structure.FailedPrecondition:
Assertion that failed: A precondition: the value is nonnegative.

at Assert.pre(Assert.java:17)
at sqrt(examples.java:24)
at main(examples.java:15)

The ﬁrst two lines of this message indicate that a precondition (that x was non-
negative) failed. This message was printed within Assert.pre on line 17 of the
source, found in Assert.java. The next line of this stack trace indicates that
the call to Assert.pre was made on line 24 of examples.java at the start of
the sqrt function. This is the ﬁrst line of the sqrt method. The problem is
(probably) on line 15 of the main procedure of examples.java. Debugging our
code should probably start in the main routine.

Beginning with Java 1.4, assertion testing is part of the formal Java language
speciﬁcation. The assert keyword can be used to perform many of the types of
checks we have described. If, however, you are using an earlier version of Java,
or you expect your users may wish to use a version of Java before version 1.4,
you may ﬁnd the Assert class to be a more portable approach to the testing of
the conditions of one’s code. A feature of language-based assertion testing is
that the tests can be automatically removed at compile time when one feels se-
cure about the way the code works. This may signiﬁcantly improve performance
of classes that heavily test conditions.

2.3 Craftsmanship

If you really desire to program well, a ﬁrst step is to take pride in your work—
pride enough to sign your name on everything you do. Through the centuries,
ﬁne furniture makers signed their work, painters ﬁnished their efforts by dab-
bing on their names, and authors inscribed their books. Programmers should
stand behind their creations.

Computer software has the luxury of immediate copyright protection—it is
a protection against piracy, and a modern statement that you stand behind the
belief that what you do is worth ﬁghting for. If you have crafted something as
best you can, add a comment at the top of your code:

// Image compression barrel for downlink to robotic cow tipper.
// (c) 2001, 2002 duane r. bailey

If, of course, you have stolen work from another, avoid the comment and

consider, heavily, the appropriate attribution.

2.4 Conclusions

2.4 Conclusions

37

Effective programmers consider their work a craft. Their constructions are well
considered and documented. Comments are not necessary, but documentation
makes working with a program much easier. One of the most important com-
ments you can provide is your name—it suggests you are taking credit and re-
sponsibility for things you create. It makes our programming world less anony-
mous and more humane.

Special comments, including conditions and assertions, help the user and
implementor of a method determine whether the method is used correctly.
While it is difﬁcult for compilers to determine the “spirit of the routine,” the
implementor is usually able to provide succinct checks of the sanity of the func-
tion. Five minutes of appropriate condition description and checking provided I’ve done my
by the implementor can prevent hours of debugging by the user.

time!

Self Check Problems

Solutions to these problems begin on page 442.

Why is it necessary to provide pre- and postconditions?
What can be assumed if a method has no precondition?
Why is it not possible to have a method with no postcondition?
Object orientation allows us to hide unimportant details from the user.

2.1
2.2
2.3
2.4
Why, then, must we put pre- and postconditions on hidden code?

Problems

Solutions to the odd-numbered problems begin on page 457.

What are the pre- and postconditions for the length method of the

What are the pre- and postconditions for String’s charAt method?
What are the pre- and postconditions for String’s concat method?
What are the pre- and postconditions for the floor function in the

2.1
java.lang.String class?
2.2
2.3
2.4
java.lang.Math class?
Improve the comments on an old program.
2.5
Each of the methods of Assert (pre, post, and condition) takes the
2.6
same parameters. In what way do the methods function differently? (Write a
test program to ﬁnd out!)

2.7

What are the pre- and postconditions for java.lang.Math.asin class?

2.5 Laboratory: Using Javadoc Commenting

Objective. To learn how to generate formal documentation for your programs.
Discussion. The Javadoc program1 allows the programmer to write comments
in a manner that allows the generation web-based documentation. Program-
mers generating classes to be used by others are particularly encouraged to
consider using Javadoc-based documentation. Such comments are portable,
web-accessible, and they are directly extracted from the code.

In this lab, we will write documentation for an extended version of the Ratio

class we ﬁrst met in Chapter 1.

Comments used by Javadoc are delimited by a /** */ pair. Note that there
are two asterisks in the start of the comment. Within the comment are a number
of keywords, identiﬁed by a leading “at-sign” (@). These keywords identify
the purpose of different comments you right. For example, the text following
an @author comment identiﬁes the programmer who originally authored the
code. These comments, called Javadoc comments, appear before the objects
they document. For example, the ﬁrst few lines of the Assert class are:

package structure;
/**

Each of the

An assertion is a condition that is expected to

* A library of assertion testing and debugging procedures.
* <p>
* This class of static methods provides basic assertion testing
* facilities.
* be true at a certain point in the code.
* assertion-based routines in this class perform a verification
* of the condition and do nothing (aside from testing side-effects)
* if the condition holds. If the condition fails, however, the
* assertion throws an exception and prints the associated message,
* that describes the condition that failed.
* provided for testing general conditions, and pre- and
* postconditions. There is also a facility for throwing a
* failed condition for code that should not be executed.
* <p>
* Features similar to assertion testing are incorporated
* in the Java 2 language beginning in SDK 1.4.
* @author duane a. bailey
*/

Basic support is

public class Assert
{

. . .

}

For each class you should provide any class-wide documentation, including
@author and @version-tagged comments.

1 Javadoc is a feature of command-line driven Java environments. Graphical environments likely
provide Javadoc-like functionality, but pre- and postcondition support may not be available.

40

Comments, Conditions, and Assertions

Within the class deﬁnition, there should be a Javadoc comment for each in-
stance variable and method. Typically, Javadoc comments for instance variables
are short comments that describe the role of the variable in supporting the class
state:

/**

* Size of the structure.
*/

int size;

Comments for methods should include a description of the method’s purpose.
A comment should describe the purpose of each parameter (@param), as well as
the form of the value returned (@return) for function-like methods. Program-
mers should also provide pre- and postconditions using the @pre and @post
keywords.2 Here is the documentation for a square root method.

/**
*
* This method computes the square root of a double value.
* @param x The value whose root is to be computed.
* @return The square root of x.
* @pre x >= 0
* @post computes the square root of x
*/

To complete this lab, you are to

1. Download a copy of the Ratio.java source from the Java Structures web-

site. This version of the Ratio class does not include full comments.

2. Review the code and make sure that you understand the purpose of each

of the methods.

3. At the top of the Ratio.java ﬁle, place a Javadoc comment that describes
the class. The comment should describe the features of the class and an
example of how the class might be used. Make sure that you include an
@author comment (use your name).

4. Run the documentation generation facility for your particular environ-
ment. For Sun’s Java environment, the Javadoc command takes a parame-
ter that describes the location of the source code that is to be documented:

javadoc prog.java

2 In this book, where there are constraints on space, the pre- and postconditions are provided in
non-Javadoc comments. Code available on the web, however, is uniformly commented using the
Javadoc comments. Javadoc can be upgraded to recognize pre- and postconditions; details are
available from the Java Structures website.

2.5 Laboratory: Using Javadoc Commenting

41

The result is an index.html ﬁle in the current directory that contains links
to all of the necessary documentation. View the documentation to make
sure your description is formatted correctly.

5. Before each instance variable write a short Javadoc comment. The com-
ment should begin with /** and end with */. Generate and view the
documentation and note the change in the documentation.

6. Directly before each method write a Javadoc comment that includes, at
a minimum, one or two sentences that describe the method, a @param
comment for each parameter in the method, a @return comment describ-
ing the value returned, and a @pre and @post comment describing the
conditions.

Generate and view the documentation and note the change in the doc-
umentation. If the documentation facility does not appear to recognize
the @pre and @post keywords, the appropriate Javadoc doclet software
has not been installed correctly. More information on installation of the
Javadoc software can be found at the Java Structures website.

Notes:

Chapter 3

Vectors

Concepts:
. Vectors
.
. Matrices

Extending arrays

Climb high, climb far,
your goal the sky, your aim the star.
—Inscription on a college staircase

THE BEHAVIOR OF A PROGRAM usually depends on its input. Suppose, for ex-
ample, that we wish to write a program that reads in n String values. One
approach would keep track of the n values with n String variables:

public static void main(String args[])
{

// read in n = 4 strings
Scanner s = new Scanner(System.in);
String v1, v2, v3, v4;
v1 = s.next();
v2 = s.next();
v3 = s.next();
v4 = s.next();

// read a space-delimited word

}

This approach is problematic for the programmer of a scalable application—an
application that works with large sets of data as well as small. As soon as n
changes from its current value of 4, it has to be rewritten. Scalable applications
are not uncommon, and so we contemplate how they might be supported.

One approach is to use arrays. An array of n values acts, essentially, as a
collection of similarly typed variables whose names can be computed at run
time. A program reading n values is shown here:

StringReader

public static void main(String args[])
{

// read in n = 4 strings
Scanner s = new Scanner(System.in);
String data[];
int n = 4;
// allocate array of n String references:
data = new String[n];
for (int i = 0; i < n; i++)

44

Vectors

{

}

}

data[i] = s.next();

Here, n is a constant whose value is determined at compile time. As the program
starts up, a new array of n integers is constructed and referenced through the
variable named data.

All is ﬁne, unless you want to read a different number of values. Then n
has to be changed, and the program must be recompiled and rerun. Another
solution is to pick an upper bound on the length of the array and only use the
portion of the array that is necessary. Here’s a modiﬁed procedure that uses up
to one million array elements:

public static void main(String args[])
{

// read in up to 1 million Strings
Scanner s = new Scanner(System.in);
String data[];
int n = 0;
data = new String[1000000];
// read in strings until we hit end of file
while (s.hasNext())
{

data[n] = s.next();
n++;

}

}

Unfortunately, if you are running your program on a small machine and have
small amounts of data, you are in trouble (see Problem 3.9). Because the array
is so large, it will not ﬁt on your machine—even if you want to read small
amounts of data. You have to recompile the program with a smaller upper
bound and try again. All this seems rather silly, considering how simple the
problem appears to be.

We might, of course, require the user to specify the maximum size of the
array before the data are read, at run time. Once the size is speciﬁed, an appro-
priately sized array can be allocated. While this may appear easier to program,
the burden has shifted to the user of the program: the user has to commit to a
speciﬁc upper bound—beforehand:

public static void main(String args[])
{

// read in as many Strings as demanded by input
Scanner s = new Scanner(System.in);
String data[];
int n;
// read in the number of strings to be read
n = s.nextInt();

3.1 The Interface

45

// allocate references for n strings
data = new String[n];
// read in the n strings
for (int i = 0; i < n; i++)
{

data[i] = s.next();

}

}

A nice solution is to build a vector—an array whose size may easily be
changed. Here is our String reading program retooled one last time, using
Vectors:

public static void main(String args[])
{

// read in an arbitrary number of strings
Scanner s = new Scanner(System.in);
Vector data;
// allocate vector for storage
data = new Vector();
// read strings, adding them to end of vector, until eof
while (s.hasNext())
{

String st = s.next();
data.add(st);

}

}

The Vector starts empty and expands (using the add method) with every String
read from the input. Notice that the program doesn’t explicitly keep track of the
number of values stored in data, but that the number may be determined by a
call to the size method.

3.1 The Interface

The semantics of a Vector are similar to the semantics of an array. Both can
store multiple values that may be accessed in any order. We call this property
random access. Unlike the array, however, the Vector starts empty and is ex-
tended to hold object references. In addition, values may be removed from the
Vector causing it to shrink. To accomplish these same size-changing operations
in an array, the array would have to be reallocated.

With these characteristics in mind, let us consider a portion of the “inter-

face”1 for this structure:

1 Stricktly speaking, constructors cannot be speciﬁed in formal Javainterfaces. Nonetheless,
adopt a convention of identifying constructors as part of the public view of structures (often called
the Application Program Interface or API).

Vector

46

Vectors

public class Vector extends AbstractList implements Cloneable
{

public Vector()
// post: constructs a vector with capacity for 10 elements

public Vector(int initialCapacity)
// pre: initialCapacity >= 0
// post: constructs an empty vector with initialCapacity capacity

public void add(Object obj)
// post: adds new element to end of possibly extended vector

public Object remove(Object element)
// post: element equal to parameter is removed and returned

public Object get(int index)
// pre: 0 <= index && index < size()
// post: returns the element stored in location index

public void add(int index, Object obj)
// pre: 0 <= index <= size()
// post: inserts new value in vector with desired index,
moving elements from index to size()-1 to right
//

public boolean isEmpty()
// post: returns true iff there are no elements in the vector

public Object remove(int where)
// pre: 0 <= where && where < size()
// post: indicated element is removed, size decreases by 1

public Object set(int index, Object obj)
// pre: 0 <= index && index < size()
// post: element value is changed to obj; old value is returned

public int size()
// post: returns the size of the vector

}

First, the constructors allow construction of a Vector with an optional initial
capacity. The capacity is the initial number of Vector locations that are reserved
for expansion. The Vector starts empty and may be freely expanded to its
capacity. At that point the Vector’s memory is reallocated to handle further
expansion. While the particulars of memory allocation and reallocation are
hidden from the user, there is obvious beneﬁt to specifying an appropriate initial
capacity.

The one-parameter add method adds a value to the end of the Vector, ex-
panding it. To insert a new value in the middle of the Vector, we use the
two-parameter add method, which includes a location for insertion. To access

3.2 Example: The Word List Revisited

47

an existing element, one calls get. If remove is called with an Object, it re-
moves at most one element, selected by value. Another remove method shrinks
the logical size of the Vector by removing an element from an indicated loca-
tion. The set method is used to change a value in the Vector. Finally, two
methods provide feedback about the current logical size of the Vector: size
and isEmpty. The size method returns the number of values stored within
the Vector. As elements are added to the Vector, the size increases from zero
up to the capacity of the Vector. When the size is zero, then isEmpty returns
true. The result is a data structure that provides constant-time access to data
within the structure, without concern for determining explicit bounds on the
structure’s size.

There are several ways that a Vector is different than its array counterpart.
First, while both the array and Vector maintain a number of references to ob-
jects, the Vector typically grows with use and stores a non-null reference in
each entry. An array is a static structure whose entries may be initialized and
used in any order and are often null. Second, the Vector has an end where ele-
ments can be appended, while the array does not directly support the concept of
appending values. There are times, of course, when the append operation might
not be a feature desired in the structure; either the Vector or array would be a
suitable choice.

The interface for Vectors in the structure package was driven, almost ex-
clusively, by the interface for Java’s proprietary java.util.Vector class. Thus,
while we do not have access to the code for that class, any program written to
use Java’s Vector class can be made to use the Vector class described here;
their interfaces are consistent.

3.2 Example: The Word List Revisited

We now reconsider an implementation of the word list part of our Hangman
program of Section 1.6 implemented directly using Vectors:

Vector list;
String targetWord;
java.util.Random generator = new java.util.Random();
list = new Vector(10);
list.add("clarify");
list.add("entered");
list.add("clerk");
while (list.size() != 0)
{

{

// select a word from the list
int index = Math.abs(generator.nextInt())%list.size();
targetWord = (String)list.get(index);

}
// ... play the game using targetWord ...
list.remove(targetWord);

}

WordList

48

Vectors

Here, the operations of the Vector are seen to be very similar to the opera-
tions of the WordList program fragment shown on page 19. The Vector class,
however, does not have a selectAny method. Instead, the bracketed code ac-
complishes that task. Since only Strings are placed within the Vector, the
assignment of targetWord involves a cast from Object (the type of value re-
turned from the get method of Vector) to String. This cast is necessary for
Java to be reassured that you’re expecting an element of type String to be
returned.
If the cast were not provided, Java would complain that the types
involved in the assignment were incompatible.

Now that we have an implementation of the Hangman code in terms of both
the WordList and Vector structures, we can deduce an implementation of the
WordList structure in terms of the Vector class. In this implementation, the
WordList contains a Vector that is used to hold the various words, as well
as the random number generator (the variable generator in the code shown
above). To demonstrate the implementation, we look at the implementation of
the WordList’s constructor and selectAny method:

protected Vector theList;
protected java.util.Random generator;

public WordList(int n)
{

theList = new Vector(n);
generator = new java.util.Random();

}

public String selectAny()
{

int i = Math.abs(generator.nextInt())%theList.size();
return (String)theList.get(i);

}

Clearly, the use of a Vector within the WordList is an improvement over the
direct use of an array, just as the use of WordList is an improvement over the
complications of directly using a Vector in the Hangman program.

3.3 Example: Word Frequency

Suppose one day you read a book, and within the ﬁrst few pages you read
“behemoth” twice. A mighty unusual writing style! Word frequencies within
documents can yield interesting information.2 Here is a little application for
computing the frequency of words appearing on the input:

public static void main(String args[])

WordFreq

2 Recently, using informal “literary forensics,” Don Foster has identiﬁed the author of the anony-
mously penned book Primary Colors and is responsible for new attributions of poetry to Shake-
speare. Foster also identiﬁed Major Henry Livingston Jr. as the true author of “The Night Before
Christmas.”

3.3 Example: Word Frequency

49

{

Vector vocab = new Vector(1000);
Scanner s = new Scanner(System.in);
int i;

// for each word on input
while (s.hasNext())
{

Association wordInfo; // word-frequency association
String vocabWord;

// word in the list

// read in and tally instance of a word
String word = s.next();
for (i = 0; i < vocab.size(); i++)
{

// get the association
wordInfo = (Association)vocab.get(i);
// get the word from the association
vocabWord = (String)wordInfo.getKey();
if (vocabWord.equals(word))
{

// match: increment integer in association
Integer f = (Integer)wordInfo.getValue();
wordInfo.setValue(new Integer(f.intValue() + 1));
break;

}

}
// mismatch: add new word, frequency 1.
if (i == vocab.size())
{

vocab.add(new Association(word,new Integer(1)));

}

}
// print out the accumulated word frequencies
for (i = 0; i < vocab.size(); i++)
{

Association wordInfo = (Association)vocab.get(i);
System.out.println(

wordInfo.getKey()+" occurs "+
wordInfo.getValue()+" times.");

}

}

First, for each word found on the input, we maintain an Association between
the word (a String) and its frequency (an Integer). Each element of the
Vector is such an Association. Now, the outer loop at the top reads in each
word. The inner loop scans through the Vector searching for matching words
that might have been read in. Matching words have their values updated. New
words cause the construction of a new Association. The second loop scans
through the Vector, printing out each of the Associations.

50

Vectors

Each of these applications demonstrates the most common use of Vectors—
keeping track of data when the number of entries is not known far in advance.
When considering the List data structure we will consider the efﬁciency of
these algorithms and, if necessary, seek improvements.

3.4 The Implementation

Clearly, the Vector must be able to store a large number of similar items. We
choose, then, to have the implementation of the Vector maintain an array of
Objects, along with an integer that describes its current size or extent. When
the size is about to exceed the capacity (the length of the underlying array), the
Vector’s capacity is increased to hold the growing number of elements.

The constructor is responsible for allocation of the space and initializing the
local variables. The number of elements initially allocated for expansion can be
speciﬁed by the user:

protected Object elementData[];
protected int elementCount;

// the data
// number of elements in vector

Vector

public Vector()
// post: constructs a vector with capacity for 10 elements
{

this(10); // call one-parameter constructor

}

public Vector(int initialCapacity)
// pre: initialCapacity >= 0
// post: constructs an empty vector with initialCapacity capacity
{

Assert.pre(initialCapacity >= 0, "Initial capacity should not be negative.");
elementData = new Object[initialCapacity];
elementCount = 0;

}

Unlike other languages, all arrays within Java must be explicitly allocated. At
the time the array is allocated, the number of elements is speciﬁed. Thus, in the
constructor, the new operator allocates the number of elements desired by the
user. Since the size of an array can be gleaned from the array itself (by asking
for elementData.length), the value does not need to be explicitly stored within
the Vector object.3

To access and modify elements within a Vector, we use the following oper-

ations:

public Object get(int index)

3 It could, of course, but explicitly storing it within the structure would mean that the implementor
would have to ensure that the stored value was always consistent with the value accessible through
the array’s length variable.

3.4 The Implementation

51

// pre: 0 <= index && index < size()
// post: returns the element stored in location index
{

return elementData[index];

}

public Object set(int index, Object obj)
// pre: 0 <= index && index < size()
// post: element value is changed to obj; old value is returned
{

Object previous = elementData[index];
elementData[index] = obj;
return previous;

}

The arguments to both methods identify the location of the desired element. Be-
cause the index should be within the range of available values, the precondition
states this fact.

For the accessor (get), the desired element is returned as the result. The set
method allows the Object reference to be changed to a new value and returns
the old value. These operations, effectively, translate operations on Vectors
into operations on arrays.

Now consider the addition of an element to the Vector. One way this can
be accomplished is through the use of the one-parameter add method. The task
requires extending the size of the Vector and then storing the element at the
location indexed by the current number of elements (this is the ﬁrst free location
within the Vector). Here is the Java method:

public void add(Object obj)
// post: adds new element to end of possibly extended vector
{

ensureCapacity(elementCount+1);
elementData[elementCount] = obj;
elementCount++;

}

(We will discuss the method ensureCapacity later. Its purpose is simply to en-
sure that the data array actually has enough room to hold the indicated number
of values.) Notice that, as with many modern languages, arrays are indexed
starting at zero. There are many good reasons for doing this. There are prob-
ably just as many good reasons for not doing this, but the best defense is that
this is what programmers are currently used to.

Principle 6 Maintaining a consistent interface makes a structure useful.

If one is interested in inserting an element in the middle of the Vector, it is
necessary to use the two-parameter add method. The operation ﬁrst creates an
unused location at the desired point by shifting elements out of the way. Once
the opening is created, the new element is inserted.

NNWSWSENEWSE52

Vectors

Figure 3.1
room for an inserted value.

The incorrect (a) and correct (b) way of moving values in an array to make

public void add(int index, Object obj)
// pre: 0 <= index <= size()
// post: inserts new value in vector with desired index,
//
moving elements from index to size()-1 to right
{

int i;
ensureCapacity(elementCount+1);
// must copy from right to left to avoid destroying data
for (i = elementCount; i > index; i--) {

elementData[i] = elementData[i-1];

}
// assertion: i == index and element[index] is available
elementData[index] = obj;
elementCount++;

}

Note that the loop that moves the elements higher in the array runs backward.
To see why, it is only necessary to see what happens if the loop runs forward (see
Figure 3.1a): the lowest element gets copied into higher and higher elements,
ultimately copying over the entire Vector to the right of the insertion point.
Figure 3.1b demonstrates the correct technique.

Removing an element from a speciﬁc location in the Vector is very similar,
reversing the effect of add. Here, using an argument similar to the previous one,
the loop moves in the forward direction:

public Object remove(int where)
// pre: 0 <= where && where < size()
// post: indicated element is removed, size decreases by 1
{

2345678900123456789090000000003456789000(a)1234567890123456780123456701088876543217(b)3.5 Extensibility: A Feature

53

Object result = get(where);
elementCount--;
while (where < elementCount) {

elementData[where] = elementData[where+1];
where++;

}
elementData[elementCount] = null; // free reference
return result;

}

We also allow the removal of a speciﬁc value from the Vector, by passing an
example Object to remove (not shown). Within this code, the equals method
of the value passed to the routine is used to compare it to values within the
Vector. When (and if) a match is found, it is removed using the technique just
described.

The methods having to do with size are relatively straightforward:

public boolean isEmpty()
// post: returns true iff there are no elements in the vector
{

return size() == 0;

}

public int size()
// post: returns the size of the vector
{

return elementCount;

}

The logical size of the Vector is the number of elements stored within the
Vector, and it is empty when this size is zero.

3.5 Extensibility: A Feature

Sometimes, our initial estimate of the maximum number of values is too small.
In this case, it is necessary to extend the capacity of the Vector, carefully main-
taining the values already stored within the Vector. Fortunately, because we
have packaged the implementation within an interface, it is only necessary to
extend the functionality of the existing operations and provide some additional
methods to describe the features.

A ﬁrst approach might be to extend the Vector to include just as many
elements as needed. Every time an element is added to the Vector, the number
of elements is compared to the capacity of the array. If the capacity is used up,
an array that is one element longer is allocated. This reallocation also requires
copying of the existing data from one array to the other. Of course, for really
long arrays, these copying operations would take a proportionally long time.
Over time, as the array grows to n elements, the array data get copied many
times. At the beginning, the array holds a single element, but it is expanded to

54

Vectors

hold two. The original element must be copied to the new space to complete
the operation. When a third is added, the ﬁrst two must be copied. The result
is that

1 + 2 + 3 + · · · + (n − 1) =

n(n − 1)
2

elements are copied as the array grows to size n. (Proving this last formula is
the core of Problem 3.8.) This is expensive since, if in the beginning we had
just allocated the Vector with a capacity of n elements, none of the data items
would have to be copied during extension!

It turns out there is a happy medium: every time you extend the array, just
double its capacity. Now, if we reconsider the number of times that an item
gets copied during the extension process, the result is dramatically different.
Suppose, for neatness only, that n is a power of 2, and that the Vector started
with a capacity of 1. What do we know? When the Vector was extended from
capacity 1 to capacity 2, one element was copied. When the array was extended
from capacity 2 to capacity 4, two elements were copied. When the array was
extended from capacity 4 to capacity 8, four elements were copied. This contin-
ues until the last extension, when the Vector had its capacity extended from n
2
to n: then n
2 elements had to be preserved. The total number of times elements
were copied is

1 + 2 + 4 + · · · +

= n − 1

n
2

Thus, extension by doubling allows unlimited growth of the Vector with an
overhead that is proportional to the ultimate length of the array. Another way
to think about it is that there is a constant overhead in supporting each element
of a Vector extended in this way.

The Java language speciﬁes a Vector interface that allows the user to specify
how the Vector is to be extended if its capacity is not sufﬁcient for the current
operation. When the Vector is constructed, a capacityIncrement is speciﬁed.
This is simply the number of elements to be added to the underlying array
when extension is required. A nonzero value for this increment leads to the n2
behavior we saw before, but it may be useful if, for example, one does not have
the luxury of being able to double the size of a large array. If the increment is
zero, the doubling strategy is used.

Our design, then, demands another protected value to hold the increment;
we call this capacityIncrement. This value is speciﬁed in a special constructor
and is not changed during the life of the Vector:

protected int capacityIncrement;

// the rate of growth for vector

public Vector(int initialCapacity, int capacityIncr)
// pre: initialCapacity >= 0, capacityIncr >= 0
// post: constructs an empty vector with initialCapacity capacity
that extends capacity by capacityIncr, or doubles if 0
//
{

Assert.pre(initialCapacity >= 0 && capacityIncr >= 0,

"Neither capacity nor increment should be negative.");

3.5 Extensibility: A Feature

55

elementData = new Object[initialCapacity];
elementCount = 0;
capacityIncrement = capacityIncr;

}

We are now prepared to investigate ensureCapacity, a method that, if nec-

essary, resizes Vector to have a capacity of at least minCapacity:

public void ensureCapacity(int minCapacity)
// post: the capacity of this vector is at least minCapacity
{

if (elementData.length < minCapacity) {

int newLength = elementData.length; // initial guess
if (capacityIncrement == 0) {

// increment of 0 suggests doubling (default)
if (newLength == 0) newLength = 1;
while (newLength < minCapacity) {

newLength *= 2;

}
} else {

// increment != 0 suggests incremental increase
while (newLength < minCapacity)
{

newLength += capacityIncrement;

}

}
// assertion: newLength > elementData.length.
Object newElementData[] = new Object[newLength];
int i;
// copy old data to array
for (i = 0; i < elementCount; i++) {

newElementData[i] = elementData[i];

}
elementData = newElementData;
// garbage collector will (eventually) pick up old elementData

}
// assertion: capacity is at least minCapacity

}

This code deserves careful investigation. If the current length of the underlying
array is already sufﬁcient to provide minCapacity elements, then the method
does nothing. On the other hand, if the Vector is too short, it must be ex-
tended. We use a loop here that determines the new capacity by doubling (if
capacityIncrement is zero) or by directly incrementing if capacityIncrement
is nonzero. In either case, by the time the loop is ﬁnished, the desired capacity
is determined. At that point, an array of the appropriate size is allocated, the
old values are copied over, and the old array is dereferenced in favor of the new.

56

Vectors

3.6 Example: L-Systems

In the late 1960s biologists began to develop computational models for growth.
One of the most successful models, L-systems, was developed by Aristid Lin-
denmayer. An L-system consists of a seed or start string of symbols derived
from an alphabet, along with a number of rules for changing or rewriting the
symbols, called productions. To simulate an interval of growth, strings are com-
pletely rewritten using the productions. When the rewriting begins with the
start string, it is possible to iteratively simulate the growth of a simple organ-
ism. To demonstrate the complexity of this approach, we can use an alphabet
of two characters—S (for stem) and L (for leaf). If the two productions

Before After

S
L

L
SL

are used, we can generate the following strings over 6 time steps:

Time
0
1
2
3
4
5
6

String
S
L
SL
LSL
SLLSL
LSLSLLSL
SLLSLLSLSLLSL

Although there are some observations that might be made (there are never two
consecutive Ss), any notion of a pattern in this string quickly breaks down. Still,
many organisms display patterns that are motivated by the seemingly simple
production system.

We can use Vectors to help us perform this rewriting process. By construct-
ing two Character objects, L and S, we can store patterns in a Vector of refer-
ences. The rewriting process involves constructing a new result Vector. Here is
a program that would verify the growth pattern suggested in the table:

public class LSystem
{

// constants that define the alphabet
final static Character L = new Character(’L’);
final static Character S = new Character(’S’);

LSystem

public static Vector rewrite(Vector s)
// pre: s is a string of L and S values
// post: returns a string rewritten by productions
{

Vector result = new Vector();
for (int pos = 0; pos < s.size(); pos++)
{

3.7 Example: Vector-Based Sets

57

// rewrite according to two different rules
if (S == s.get(pos)) {
result.add(L);

} else if (L == s.get(pos)) {

result.add(S); result.add(L);

}

}
return result;

}

public static void main(String[] args)
{

Vector string = new Vector();
string.add(S);

// determine the number of strings
Scanner s = new Scanner(System.in);
int count = s.nextInt();

// write out the start string
System.out.println(string);
for (int i = 1; i <= count; i++)
{

string = rewrite(string);
System.out.println(string); // print it out

// rewrite the string

}

}

}

L-systems are an interesting example of a grammar system. The power of a
grammar to generate complex structures—including languages and, biologi-
cally, plants—is of great interest to theoretical computer scientists.

3.7 Example: Vector-Based Sets

In Section 1.8 we discussed Java’s interface for a Set. Mathematically, it is an
unordered collection of unique values. The set abstraction is an important fea-
ture of many algorithms that appear in computer science, and so it is important
that we actually consider a simple implementation before we go much further.
As we recall, the Set is an extension of the Structure interface. It demands
that the programmer implement not only the basic Structure methods (add,
contains, remove, etc.), but also the following methods of a Set. Here is the
interface associated with a Vector-based implementation of a Set:

public class SetVector extends AbstractSet
{

public SetVector()
// post: constructs a new, empty set

SetVector

58

Vectors

public SetVector(Structure other)
// post: constructs a new set with elements from other

public void clear()
// post: elements of set are removed

public boolean isEmpty()
// post: returns true iff set is empty

public void add(Object e)
// pre: e is non-null object
// post: adds element e to set

public Object remove(Object e)
// pre: e is non-null object
// post: e is removed from set, value returned

public boolean contains(Object e)
// pre: e is non-null
// post: returns true iff e is in set

public boolean containsAll(Structure other)
// pre: other is non-null reference to set
// post: returns true iff this set is subset of other

public Object clone()
// post: returns a copy of set

public void addAll(Structure other)
// pre: other is a non-null structure
// post: add all elements of other to set, if needed

public void retainAll(Structure other)
// pre: other is non-null reference to set
// post: returns set containing intersection of this and other

public void removeAll(Structure other)
// pre: other is non-null reference to set
// post: returns set containing difference of this and other

public Iterator iterator()
// post: returns traversal to traverse the elements of set

public int size()
// post: returns number of elements in set

}

A SetVector might take the approach begun by the WordList implementation

3.7 Example: Vector-Based Sets

59

we have seen in Section 3.2: each element of the Set would be stored in a
location in the Vector. Whenever a new value is to be added to the Set, it
is only added if the Set does not already contain the value. When values are
removed from the Set, the structure contracts. At all times, we are free to keep
the order of the data in the Vector hidden from the user since the ordering of
the values is not part of the abstraction.

We construct a SetVector using the following constructors, which initialize

a protected Vector:

protected Vector data; // the underlying vector

public SetVector()
// post: constructs a new, empty set
{

data = new Vector();

}

public SetVector(Structure other)
// post: constructs a new set with elements from other
{

this();
addAll(other);

}

The second constructor is a copy constructor that makes use of the union op-
erator, addAll. Since the initial set is empty (the call to this() calls the ﬁrst
constructor), the SetVector essentially picks up all the values found in the other
structure.

Most methods of the Set are adopted from the underlying Vector class. For

example, the remove method simply calls the remove method of the Vector:

public Object remove(Object e)
// pre: e is non-null object
// post: e is removed from set, value returned
{

return data.remove(e);

}

The add method, though, is responsible for ensuring that duplicate values are
not added to the Set. It must ﬁrst check to see if the value is already a member:

public void add(Object e)
// pre: e is non-null object
// post: adds element e to set
{

if (!data.contains(e)) data.add(e);

}

60

Vectors

To perform the more complex Set-speciﬁc operations (addAll and others), we
must perform the speciﬁed operation for all the values of the other set. To ac-
complish this, we make use of an Iterator, a mechanism we will not study
until Chapter 8, but which is nonetheless simple to understand. Here, for ex-
ample, is the implementation of addAll, which attempts to add all the values
found in the other structure:

public void addAll(Structure other)
// pre: other is a non-null structure
// post: add all elements of other to set, if needed
{

Iterator yourElements = other.iterator();
while (yourElements.hasNext())
{

add(yourElements.next());

}

}

Other methods are deﬁned in a straightforward manner.

3.8 Example: The Matrix Class

One application of the Vector class is to support a two-dimensional Vector-like
object: the matrix. Matrices are used in applications where two dimensions of
data are needed. Our Matrix class has the following methods:

public class Matrix
{

Matrix

public Matrix(int h, int w)
// pre: h >= 0, w >= 0;
// post: constructs an h row by w column matrix

public Object get(int row, int col)
// pre: 0 <= row < height(), 0 <= col < width()
// post: returns object at (row, col)

public void set(int row, int col, Object value)
// pre: 0 <= row < height(), 0 <= col < width()
// post: changes location (row, col) to value

public void addRow(int r)
// pre: 0 <= r < height()
// post: inserts row of null values to be row r

public void addCol(int c)
// pre: 0 <= c < width()
// post: inserts column of null values to be column c

3.8 Example: The Matrix Class

61

Figure 3.2
Vector of references to Objects. Elements are labeled with their indices.

The Matrix class is represented as a Vector of rows, each of which is a

public Vector removeRow(int r)
// pre: 0 <= r < height()
// post: removes row r and returns it as a Vector

public Vector removeCol(int c)
// pre: 0 <= c < width()
// post: removes column c and returns it as a vector

public int width()
// post: returns number of columns in matrix

public int height()
// post: returns number of rows in matrix

}

The two-parameter constructor speciﬁes the width and height of the Matrix. El-
ements of the Matrix are initially null, but may be reset with the set method.
This method, along with the get method, accepts two parameters that identify
the row and the column of the value. To expand and shrink the Matrix, it is
possible to insert and remove both rows and columns at any location. When a
row or column is removed, a Vector of removed values is returned. The meth-
ods height and width return the number of rows and columns found within
the Matrix, respectively.

To support this interface, we imagine that a Matrix is a Vector of rows,
which are, themselves, Vectors of values (see Figure 3.2). While it is not strictly
necessary, we explicitly keep track of the height and width of the Matrix (if we
determine at some later date that keeping this information is unnecessary, the
interface would hide the removal of these ﬁelds). Here, then, is the constructor
for the Matrix class:

protected int height, width; // size of matrix

Rows(0,0)(0,1)(0,2)(0,3)(1,0)(1,1)(1,2)(1,3)(2,0)(2,1)(2,2)(2,3)(3,0)(3,1)(3,2)(3,3)(4,0)(4,1)(4,2)(4,3)01230123462

Vectors

protected Vector rows;

// vector of row vectors

public Matrix(int h, int w)
// pre: h >= 0, w >= 0;
// post: constructs an h row by w column matrix
{

// initialize height and width

height = h;
width = w;
// allocate a vector of rows
rows = new Vector(height);
for (int r = 0; r < height; r++)
{

// each row is allocated and filled with nulls
Vector theRow = new Vector(width);
rows.add(theRow);
for (int c = 0; c < width; c++)
{

theRow.add(null);

}

}

}

We allocate a Vector for holding the desired number of rows, and then, for each
row, we construct a new Vector of the appropriate width. All the elements are
initialized to null. It’s not strictly necessary to do this initialization, but it’s a
good habit to get into.

The process of manipulating individual elements of the matrix is demon-

strated by the get and set methods:

public Object get(int row, int col)
// pre: 0 <= row < height(), 0 <= col < width()
// post: returns object at (row, col)
{

Assert.pre(0 <= row && row < height, "Row in bounds.");
Assert.pre(0 <= col && col < width, "Col in bounds.");
Vector theRow = (Vector)rows.get(row);
return theRow.get(col);

}

public void set(int row, int col, Object value)
// pre: 0 <= row < height(), 0 <= col < width()
// post: changes location (row, col) to value
{

Assert.pre(0 <= row && row < height, "Row in bounds.");
Assert.pre(0 <= col && col < width, "Col in bounds.");
Vector theRow = (Vector)rows.get(row);
theRow.set(col,value);

}

The process of manipulating an element requires looking up a row within the
rows table and ﬁnding the element within the row. It is also important to notice

3.8 Example: The Matrix Class

63

Figure 3.3
associated with matrix before addRow. Compare with Figure 3.2.

The insertion of a new row (gray) into an existing matrix. Indices are those

that in the set method, the row is found using the get method, while the
element within the row is changed using the set method. Although the element
within the row changes, the row itself is represented by the same vector.

Many of the same memory management issues discussed in reference to
Vectors hold as well for the Matrix class. When a row or column needs to be
expanded to make room for new elements (see Figure 3.3), it is vital that the
management of the arrays within the Vector class be hidden. Still, with the
addition of a row into the Matrix, it is necessary to allocate the new row object
and to initialize each of the elements of the row to null:

public void addRow(int r)
// pre: 0 <= r < height()
// post: inserts row of null values to be row r
{

Assert.pre(0 <= r && r < width, "Row in bounds.");
height++;
Vector theRow = new Vector(width);
for (int c = 0; c < width; c++)
{

theRow.add(null);

}
rows.add(r,theRow);

}

We leave it to the reader to investigate the implementation of other Matrix
methods. In addition, a number of problems consider common extensions to
the Matrix class.

(3,1)(3,2)(3,3)(3,0)(0,0)(0,1)(0,2)(0,3)(1,0)(1,1)(1,2)(1,3)(2,0)(2,1)(2,2)(2,3)0123012345(4,0)(4,1)(4,2)(4,3)Rows64

Vectors

3.9 Conclusions

Most applications that accept data are made more versatile by not imposing
constraints on the number of values processed by the application. Because the
size of an array is ﬁxed at the time it is allocated, programmers ﬁnd it difﬁcult
to create size-independent code without the use of extensible data structures.
The Vector and Matrix classes are examples of extensible structures.

Initially Vectors are empty, but can be easily expanded when necessary.
When a programmer knows the upper bound on the Vector size, this infor-
mation can be used to minimize the amount of copying necessary during the
entire expansion process. When a bound is not known, we saw that doubling
the allocated storage at expansion time can reduce the overall copying cost.

The implementation of Vector and Matrix classes is not trivial. Data ab-
straction hides many important housekeeping details. Fortunately, while these
details are complex for the implementor, they can considerably reduce the com-
plexity of applications that make use of the Vector and Matrix structures.

Self Check Problems

Solutions to these problems begin on page 442.

3.1

How are arrays and Vectors the same? How do they differ?

What is the difference between the add(v) and add(i,v) methods of

3.2
Vector?

What is the difference between the add(i,v) method and the set(i,v)

3.3
method?

What is the difference between the remove(v) method (v is an Object

3.4
value), and the remove(i) (i is an int)?

3.5

What is the distinction between the capacity and size of a Vector?

Why is the use of a Vector an improvement over the use of an array in

3.6
the implementation of Hangman in Section 3.2?

3.7
ments to the right starting at the high end of the Vector? (See Figure 3.1.)

When inserting a value into a Vector why is it necessary to shift ele-

By default, when the size ﬁrst exceeds the capacity, the capacity of the

3.8
Vector is doubled. Why?
3.9

What is the purpose of the following code?

elementData = new Object[initialCapacity];

What can be said about the values found in elementData after this code is
executed?

3.10 When there is more than one constructor for a class, when and how do
we indicate the appropriate method to use? Compare, for example,

3.9 Conclusions

65

Vector v = new Vector();
Vector w = new Vector(1000);

Is the row index of the Matrix bounded by the matrix height or width?

3.11
When indexing a Matrix which is provided ﬁrst, the row or the column?

Problems

Solutions to the odd-numbered problems begin on page 457.

Explain the difference between the size and capacity of a vector. Which

3.1
is more important to the user?

The default capacity of a Vector in a structure package implementa-
3.2
tion is 10. It could have been one million. How would you determine a suitable
value?

The implementation of java.util.Vector provides a method trimTo-
3.3
Size. This method ensures that the capacity of the Vector is the same as its
size. Why is this useful? Is it possible to trim the capacity of a Vector without
using this method?

The implementation of java.util.Vector provides a method setSize.
3.4
This method explicitly sets the size of the Vector. Why is this useful? Is it
possible to set the size of the Vector without using this method?

Write a Vector method, indexOf, that returns the index of an object in
3.5
the Vector. What should the method return if no object that is equals to this
object can be found? What does java.lang.Vector do in this case? How long
does this operation take to perform, on average?

Write a class called BitVector that has an interface similar to Vector,
3.6
but the values stored within the BitVector are all known to be boolean (the
primitive type). What is the primary advantage of having a special-purpose
vector, like BitVector?

Suppose we desire to implement a method reverse for the Vector class.
3.7
One approach would be to remove location 0 and to use add near the end or tail
of the Vector. Defend or reject this suggested implementation. In either case,
write the best method you can.

Suppose that a precisely sized array is used to hold data, and that each
3.8
time the array size is to be increased, it is increased by exactly one and the data
are copied over. Prove that, in the process of growing an array incrementally
from size 0 to size n, approximately n2 values must be copied.
What is the maximum length array of Strings you can allocate on your
3.9
machine? (You needn’t initialize the array.) What is the maximum length array
of boolean you can allocate on your machine? What is to be learned from the
ratio of these two values?

3.10

Implement the Object-based remove method for the Vector class.

66

Vectors

In our discussion of L-systems, the resulting strings are always linear.
3.11
Plants, however, often branch. Modify the LSystem program so that it includes
the following ﬁve productions:

Before After

Before After

S
T

T
U

U
V

V
W

Before After
[S]U

W

where [S] is represented by a new Vector that contains a single S. (To test to
see if an Object, x, is a Vector, use the test x instanceof Vector.)
Finish the two-dimensional Vector-like structure Matrix. Each element
3.12
of the Matrix is indexed by two integers that identify the row and column
containing the value. Your class should support a constructor, methods addRow
and addCol that append a row or column, the get and set methods, and width
and height methods. In addition, you should be able to use the removeRow and
removeCol methods.
3.13 Write Matrix methods for add and multiply. These methods should
implement the standard matrix operations from linear algebra. What are the
preconditions that are necessary for these methods?
A Matrix is useful for nonmathematical applications. Suppose, for ex-
3.14
ample, that the owners of cars parked in a rectangular parking lot are stored
in a Matrix. How would you design a new Matrix method to return the lo-
cation of a particular value in the Matrix? (Such an extension implements an
associative memory. We will discuss associative structures when we consider
Dictionarys.)
An m × n Matrix could be implemented using a single Vector with
3.15
mn locations. Assuming that this choice was made, implement the get method.
What are the advantages and disadvantages of this implementation over the
Vector of Vectors approach?
A triangular matrix is a two-dimensional structure with n rows. Row i
3.16
has i + 1 columns (numbered 0 through i) in row i. Design a class that supports
all the Matrix operations, except addRow, removeRow, addCol, and removeCol.
You should also note that when a row and column must be speciﬁed, the row
must be greater than or equal to the column.
A symmetric matrix is a two-dimensional Matrix-like structure such that
3.17
the element at [i][j] is the same element found at [j][i]. How would you imple-
ment each of the Matrix operations? The triangular matrix of Problem 3.16
may be useful here. Symmetric matrices are useful in implementing undirected
graph structures.
Sometimes it is useful to keep an unordered list of characters (with
3.18
ASCII codes 0 through 127), with no duplicates. Java, for example, has a
Implement a class, CharSet, using
CharSet class in the java.util package.
a Vector. Your class should support (1) the creation of an empty set, (2) the
addition of a single character to the set, (3) the check for a character in the set,
(4) the union of two sets, and (5) a test for set equality.

3.10 Laboratory: The Silver Dollar Game

Objective. To implement a simple game using Vectors or arrays.
Discussion. The Silver Dollar Game is played between two players. An arbitrar-
ily long strip of paper is marked off into squares:

The game begins by placing silver dollars in a few of the squares. Each square
holds at most one coin. Interesting games begin with some pairs of coins sepa-
rated by one or more empty squares.

The goal is to move all the n coins to the leftmost n squares of the paper.
This is accomplished by players alternately moving a single coin, constrained by
the following rules:

1. Coins move only to the left.

2. No coin may pass another.

3. No square may hold more than one coin.

The last person to move is the winner.
Procedure. Write a program to facilitate playing the Silver Dollar Game. When
the game starts, the computer has set up a random strip with 3 or more coins.
Two players are then alternately presented with the current game state and are
allowed to enter moves. If the coins are labeled 0 through n−1 from left to right,
a move could be speciﬁed by a coin number and the number of squares to move
the coin to the left. If the move is illegal, the player is repeatedly prompted to
enter a revised move. Between turns the computer checks the board state to
determine if the game has been won.

Here is one way to approach the problem:

1. Decide on an internal representation of the strip of coins. Does your rep-
resentation store all the information necessary to play the game? Does
your representation store more information than is necessary? Is it easy to
test for a legal move? Is it easy to test for a win?

2. Develop a new class, CoinStrip, that keeps track of the state of the play-
ing strip. There should be a constructor, which generates a random board.
Another method, toString, returns a string representation of the coin
strip. What other operations seem to be necessary? How are moves per-
formed? How are rules enforced? How is a win detected?

68

Vectors

Hint: When
ﬂipped, the
Belgian Euro is
heads
149 times
out of 250.

3. Implement an application whose main method controls the play of a single

game.

Thought Questions. Consider the following questions as you complete the lab:

1. How might one pick game sizes so that, say, one has a 50 percent chance
of a game with three coins, a 25 percent chance of a game with four coins,
a 12 1
2 percent chance of a game with ﬁve coins, and so on? Would your
technique bias your choice of underlying data structure?

2. How might one generate games that are not immediate wins? Suppose
you wanted to be guaranteed a game with the possibility of n moves?

3. Suppose the computer could occasionally provide good hints. What op-

portunities appear easy to recognize?

4. How might you write a method, computerPlay, where the computer plays

to win?

5. A similar game, called Welter’s Game (after C. P. Welter, who analyzed the
game), allows the coins to pass each other. Would this modiﬁcation of the
rules change your implementation signiﬁcantly?

Notes:

Chapter 4

Generics

Concepts:
. Motivation for Parameterized Types
.
.

Simple Parameterized Types
Parameterization in Vector

Thank God that
the people running this world
are not smart enough
to keep running it forever.
—Arlo Guthrie

THE MAIN PURPOSE OF A LANGUAGE is to convey information from one party
to another. Programming languages are no exception. Ideally, a language like
Java lets the programmer express the logic of an algorithm in a natural way,
and the language would identify errors of grammar (i.e., errors of syntax) and
meaning (semantics). When these errors are ﬂagged as the program is com-
piled, we call them compile-time errors. Certain logical errors, of course, do not
appear until the program is run. When these errors are detected (for example,
a division-by-zero, or an object reference through a null pointer) Java generates
runtime errors.1 It is generally believed that the sooner one can detect an error,
the better. So, for example, it is useful to point out a syntax error (“You’ve for-
gotten to declare the variable, cheatCode, you use, here, on line 22.”) where
it occurs rather than a few lines later (“After compiling Lottery.java, we no-
ticed that you mentioned 4 undeclared variables: random, range, trapdoor, and
winner.”). Because a runtime error may occur far away and long after the pro-
gram was written, compile-time errors (that can be ﬁxed by the programming
staff) are considered to be preferable to runtime errors (that must be ﬁxed by
the public relations staff).

Java 5 provides some important language features that shift the detection of
certain important errors from runtime to compile-time. The notion of a generic
class is one of them. This chapter is about building generic classes.

1 Even after considering compile-time and runtime errors, there are many errors that even a run-
ning system cannot detect. For example, undesirable inﬁnite loops (some are desirable) are often
the result of errors in logic that the computer cannot detect. The word “cannot” is quite strong here:
it is impossible to build an environment that correctly identiﬁes any program that loops inﬁnitely
(or even more broadly–fails to halt). This result is due to Alan Turing.

70

Generics

4.1 Motivation (in case we need some)

In Chapter 3 we constructed the Vector class, a general purpose array. Its pri-
mary advantage over arrays is that it is extensible: the Vector can be efﬁciently
It suffers one important disadvantage:
lengthened and shortened as needed.
it is impossible to declare a Vector that contains exactly one type of object.
Ponder this: we almost always declare variables and arrays to contain values of
one particular type, but Vectors hold Objects. As we have seen, an Object can
hold any reference type, so it is hardly a restriction. How can we construct a
Vector restricted to holding just String references?

Of course, you could depend on your naturally good behavior. As long as
the programs you write only add and remove Strings from the Vector, life will
be good. As an example, consider the following (silly) program:

LongWords

public static void main(String[] args)
{

Vector longWords = new Vector();
int i;
for (i = 0; i < args.length; i++) {
if (args[i].length() > 4) {
longWords.add(args[i]);

// line 12

}

}

...

for (i = 0; i < longWords.size(); i++) {

String word = (String)longWords.get(i); // line 31
System.out.println(word+", length "+word.length());

}

}

This main method builds a Vector of long arguments—Strings longer than 4
characters—and, sometime later, prints out the list of words along with their
respective lengths. If we type

java LongWords Fuzzy Wozzie was a bear

we expect to get

Fuzzy, length 5
Wozzie, length 6

But programming is rarely so successful. Suppose we had instead written

for (i = 0; i < args.length; i++) {
if (args[i].length() > 4) {
longWords.add(args);

// line 12

}

}

This mistake is
silly, but most
mistakes are.

On line 12 we are missing the index from the args array in the call to add.
Instead of adding the ith argument, we have added the entire argument array,
every time. Because we have no way to restrict the Vector class, the program
compiles correctly. We only notice the mistake when we type

4.1 Motivation (in case we need some)

71

java LongWords Fuzzy Wozzie had no hair

and get

Exception in thread "main" java.lang.ClassCastException: [Ljava.lang.String;

at LongWords.main(LongWords.java:31)

The problem is eventually recognized (on line 31) when what we thought was
a String (but is actually an array) is removed as an Object (O.K.) and cast to a
String (not O.K.). Notice if our broken program is run with only short words:

java LongWords aye but what a bear that Fuzz was

no runtime error occurs because no words are added to the Vector.

4.1.1 Possible Solution: Specialization

Another solution to our problem would be to create a specialized class, a String-
Vector that was similar to the Vector class in every way, except that Objects
are replaced with Strings. The add method, for example, would be declared:

public class StringVector implements Cloneable
{

protected String elementData[];
protected int elementCount;
...
public void add(String obj)
{

// the data
// number of elements in vector

StringVector

ensureCapacity(elementCount+1);
elementData[elementCount] = obj;
elementCount++;

}

}

Compare this with the code that appears in Chapter 3 on page 51.

There are several things to note about this approach. First, it works. For
fun, we create a similar (erroneous) program, LongWords2, that makes use of
the StringVector:

public static void main(String[] args)
{

StringVector longWords = new StringVector();
int i;
for (i = 0; i < args.length; i++) {
if (args[i].length() > 4) {
longWords.add(args);

// line 12

}

}

...

for (i = 0; i < longWords.size(); i++) {

String word = longWords.get(i); // line 31
System.out.println(word+", length "+word.length());

}

}

LongWords2

72

Generics

Instead of using Vector we use StringVector. The compiler knows that the
add method must take a String, so passing an array of Strings to the add
method generates the error:

LongWords2.java:12: cannot find symbol
symbol
location: class StringVector

: method add(java.lang.String[])

longWords.add(args);
^

The success of this technique leads us to the following principle:

Principle 7 Write code to identify errors as soon as possible.

Here, (1) the actual source of the logical error was identiﬁed, and (2) it was
identiﬁed at compile-time.

To accomplish this, however, we had to write a new class and modify it
appropriately. This is, itself, an error-prone task. Also, we must rewrite the
class for every every new contained type. Soon we would be overwhelmed by
special-purpose classes. These classes are not related in any way and, it should
be noted, our original class, Vector, would never be used. Imagine the difﬁculty
of adding a new method to the Vector class that you wished to make available
to each of your specialized versions! A better solution is to create a generic class
or a class parameterized by the type of data it holds.

4.2 Implementing Generic Container Classes

Ideally, we would like to be able to construct container types, like Associations
and Vectors, that hold objects of one or more speciﬁc types. At the same time,
we would like to write each of these classes once, probably without prior knowl-
edge of their client applications. If this were possible, we could provide much of
the polymorphic utility of these classes as they were implemented in Chapters 1
and 3, and still identify errors in data types at compile-time. In Java 5, generic
or parameterized data types provide the programmer the necessarily ﬂexibility.

4.2.1 Generic Associations

For simplicity, we ﬁrst consider the signatures of the protected data and several
of the methods of the Association class, declared using parameterized types:2

package structure5;
public class Association<K,V>
{

Association

2 The remainder of this text describes classes of the structure5 package. The structure package
provides Object-based implementation, for backward comparability with pre-Java5 compilers. Both
the structure and structure5 packages are available in a single jar ﬁle, bailey.jar, that may
be inserted in your CLASSPATH.

NNWSWSENEWSE4.2 Implementing Generic Container Classes

73

protected K theKey; // the key of the key-value pair
protected V theValue; // the value of the key-value pair

/*

for example:
Association<String,Integer> personAttribute =

new Assocation<String,Integer>("Age",34);

*/

public Association(K key, V value)
// pre: key is non-null
// post: constructs a key-value pair

public V getValue()
// post: returns value from association

public K getKey()
// post: returns key from association

public V setValue(V value)
// post: sets association’s value to value

}

At the time of their declaration, parameterized class name are followed by a list
of comma-separated type parameters in angle brackets. This book follows the
common convention that type parameters are indicated by single capital Latin
letters.3 In this example, K is a place holder for the type of the association’s
key, while V will be used to represent the actual type of the association’s value.
Within the class deﬁnition, these type variables may be used wherever class
names are used: declaring data variables and declaring method parameters and
return values. The one constraint is that

Generic types may not appear in array allocations.

The reason is technical and obviously not of concern in our declaration of
Associations.

The code associated with the generic implementation of the Association
class is, at this point, straightforward. For example, the setValue method would
be written:

public V setValue(V value)
{

V oldValue = theValue;
theValue = value;
return oldValue;

}

3 There is some dispute about this convention. See your instructor. Professional driver, closed
course. Your mileage may vary. Danger, Will Robinson. Some identiﬁers too small for three year
olds.

74

Generics

Notice that there are no casts in this code: since the value is declared to be of
type V, and since the return value for setValue is the same, no cast is necessary.
The removal of casts is a sign that type checking will happen in the compiler
and not while the program is running.

To make use of the generic version of a class, we must specify the actual
parameter types associated with the formal types demanded by Association.
For example, a new Association between a String and an Integer would be
declared:

Association<String,Integer> personAttribute =

new Assocation<String,Integer>("Age",34);

(Hotdoggers: the 34 here is, yes, autoboxed4 into an Integer.) Of course, if
the Association is declared within a class that is itself parameterized, formal
type parameters may be speciﬁed. We will see this on many occasions in future
chapters.

4.2.2 Parameterizing the Vector Class

We now consider the parameterization of a signiﬁcant container class, Vector.
Because of the way that Vectors are implemented, there will be special techni-
cal details (associated with arrays) that pertain, for the most part, only to the
Vector class. Most other structures will avoid these difﬁculties because they
will make use of parameterized Vectors, and therefore build on the work we
do here. Still, for the purposes of efﬁciency, it may be useful to declare struc-
tures that make direct use of the array-based techniques described here.

(The reader may ﬁnd it useful to review the implementation of the non-
generic Vector as declared in Chapter 3 before following along here, where the
technical details of the Vector implementation will not be reviewed.)

One thing seems fairly obvious: the parameterized Vector class should be
deﬁned in terms of a single type—the type of each element within the Vector.
The declaration of the major method signatures of the Vector class is as follows:

public class Vector<E> extends AbstractList<E> implements Cloneable
{

private Object elementData[];
protected int elementCount;
protected int capacityIncrement;
protected E initialValue;
protected final static int defaultCapacity = 10; // def’t capacity, must be>0

// the data
// number of elements in vector
// the rate of growth for vector

// new elements have this value

public Vector()
// post: constructs a vector with capacity for 10 elements

4 Two syntactic features of Java 5 include autoboxing and its inverse, -unboxing. With autobox-
ing, primitive types are automatically “boxed” into their object equivalents if doing so would ﬁx
a type incompatibility. In addition, the corresponding object types will be converted to primitive
equivalents, if necessary.

Yeah, and if we
eat our spinach
now, we won’t
get E. coli later.

Vector

4.2 Implementing Generic Container Classes

75

public Vector(int initialCapacity)
// pre: initialCapacity >= 0
// post: constructs an empty vector with initialCapacity capacity

public Vector(int initialCapacity, int capacityIncr)
// pre: initialCapacity >= 0, capacityIncr >= 0
// post: constructs an empty vector with initialCapacity capacity
that extends capacity by capacityIncr, or doubles if 0
//

public Vector(int initialCapacity, int capacityIncr, E initValue)
// pre: initialCapacity, capacityIncr >= 0
// post: constructs empty vector with capacity that begins at
//
//

initialCapacity and extends by capacityIncr or doubles
if 0.

New entries in vector are initialized to initValue.

public void add(E obj)
// post: adds new element to end of possibly extended vector

public E remove(E element)
// post: element equal to parameter is removed and returned

public boolean contains(E elem)
// post: returns true iff Vector contains the value
//

(could be faster, if orderedVector is used)

public E get(int index)
// pre: 0 <= index && index < size()
// post: returns the element stored in location index

public void insertElementAt(E obj, int index)
// pre: 0 <= index <= size()
// post: inserts new value in vector with desired index,
moving elements from index to size()-1 to right
//

public E remove(int where)
// pre: 0 <= where && where < size()
// post: indicated element is removed, size decreases by 1

public E set(int index, E obj)
// pre: 0 <= index && index < size()
// post: element value is changed to obj; old value is returned

}

First, it is important to note that the Vector holding objects of type E, the
Vector<E> class, implements AbstractList<E>. Anywhere where an Abstract-
List<E> is required, Vector<E> may be provided.5 The E in the Vector<E>
identiﬁes the formal type parameter, while the E of AbstractList<E> is used

5 This reads like a law journal. A technical point important to note here is that a Vector<E>
does not have any relation to Vector<S> if S is a subclass of E. For example, you cannot assign a

76

Generics

to provide an actual type parameter to the AbstractList class. This, of course,
presumes that the AbstractList class has been implemented using a single
generic type parameter.

At the top of the declaration, we see the instance variables and constants
that are found in a Vector. The initialValue variable holds, for each object,
the value that is used to ﬁll out the Vector as it is extended. Typically (and, in
most constructors, by default) this is null. Clearly, if it is non-null, it should
be a value of type E. The values elementCount, capacityIncrement, and the
constant defaultCapacity are all ints, unconstrained by the choice of the pa-
rameterized type.

The variable elementData is an extensible array (the array the supports the
Vector’s store): it should contain only objects of type E. Again, because of a
technical constraint in Java6, it is impossible to construct new arrays of any
type that involves a type parameter. We must know, at the time the class is
compiled, the exact type of the array constructed.

But this goes against our desire to construct a parameterized array of type E.
Our “workaround” is to construct an array of type Object, limit the type of the
elements to E with appropriate casts within mutator methods, and to declare the
array private eliminating the possibility that subclasses would be able to store
non-E values within the array.7 When values are retrieved from the array, we
can be conﬁdent that they are of the right type, and using casts (that we know
are trivial) we return values of type E.

Let’s look at the details. Here is the implementation of the most general

Vector constructor:

public Vector(int initialCapacity, int capacityIncr, E initValue)
{

Assert.pre(initialCapacity >= 0, "Nonnegative capacity.");
capacityIncrement = capacityIncr;
elementData = new Object[initialCapacity];
elementCount = 0;
initialValue = initValue;

}

The Vector constructor does not mention the <E> type parameter in its name.
Why? The type parameter is obvious because the constructor declaration ap-
pears within the class Vector<E>. When the value is constructed, of course,

Vector<Integer> to a Vector<Number>, even though Integer values may be assigned to Numbers.
You might think about why this is the case: Why is this the case?
6 This constraint is related to the fact that every assignment to the array at runtime forces a runtime
check of the data type. This check is necessitated by the fact that, over time, alternative arrays—
arrays of different base types—can be assigned to a single array reference. Unfortunately Java 5
does not transmit parameterized type information to the running program. This means that param-
eterized types cannot be exactly checked at runtime, but they should be. The easiest way to enforce
correct checking is to make it impossible to construct arrays of parameterized types. We expect
future implementations of Java will address this issue, and that David Ortiz will be the American
League Most Valuable Player, but good things take time.
7 See more about privacy rights in Appendix B.8.

4.2 Implementing Generic Container Classes

77

the actual parameter appears within the new statement (see the example on
page 78). Within the constructor, the supporting array is allocated as a logically
empty array of Object, and the initial value (type E) is cached away within the
type. Is there any way that a non-E value can be stored in elementData, using
this method? No. The method ensures the safety of the types of Vector’s values.
If we ask this question for each mutator method, and the answer is uniformly
No, we can be sure the Vector contains only the correct types.
Now, consider a method that adds a value to the Vector:

public void add(E obj)
// post: adds new element to end of possibly extended vector
{

ensureCapacity(elementCount+1);
elementData[elementCount] = obj;
elementCount++;

}

The Vector is expanded and the new value is appended to the end of the array.
Is there a possibility of a type error, here? In a more extensive review, we might
check to make sure ensureCapacity allows only type E values in elementData.
(It does.) The second and third statements append an object of type E to the ar-
ray, so there is no opportunity for a non-E value to be assigned. Incidentally, one
might wonder what happens if the add method were called with a non-E type. It
is precisely what happens in the StringVector example at the beginning of the
chapter: an exception is raised at the site of the call, during compilation. We’ll
review this in a moment.

Next, we consider the get method. This accessor method retrieves a value

from within the array.

public E get(int index)
{

return (E)elementData[index];

}

By assumption (and eventually through proof) we believe the safety of the array
is intact to this point of access. Every element is of type E, even though it appears
in a more general array of Objects. The cast of the value, while required, will
always hold; runtime type errors will never occur at this point. The method,
then, returns a value of type E.

The set method is a combination of the prior two methods. It is useful to
convince yourself that (1) the method is type-safe, and (2) the cast is necessary
but always successful.

public E set(int index, E obj)
{

Assert.pre(0 <= index && index < elementCount,"index is within bounds");
E previous = (E)elementData[index];
elementData[index] = obj;
return previous;

}

78

Generics

It is, of course, possible to make type mistakes when using the Vector class.
They fall, however, into two distinct categories. Vector implementation mis-
takes may allow elements of types other than E into the array. These mistakes
are always possible, but they are the responsiblity of a single programmer. Good
design and testing will detect and elminate these compile-time and runtime er-
rors quickly. The second class of errors are abuses of the Vector<E> class by
the client application. These errors are unfortunate, but they will be reported
at compile-time at the appropriate location, one that focuses the programmer’s
attention on the mistake.

We can now test our new implementation on working code and then attempt
to break it. In LongWords3, we consider, yet again, the implementation of our
program that checks for long argument words. The implementation makes use
of a Vector<String> as follows:

LongWords3

public static void main(String[] args)
{

Vector<String> longWords = new Vector<String>();
int i;
for (i = 0; i < args.length; i++) {
if (args[i].length() > 4) {
longWords.add(args[i]);

// line 12

}

}

...

for (i = 0; i < longWords.size(); i++) {

String word = longWords.get(i); // line 31
System.out.println(word+", length "+word.length());

}

}

Because the Vector<String> only contains String elements, there is no need
for a cast to a String on line 31. This makes the running program safer and
faster.

If the add statement were incorrectly written as:

longWords.add(args);

// line 12: woops!

the compiler would be confused—we’re passing an array of Strings to a class
that has only methods to add Strings. The resulting error is, again, a missing
symbol (i.e. a missing method) error:

LongWords3.java:12: cannot find symbol
symbol
location: class java.util.Vector<java.lang.String>

: method add(java.lang.String[])

longWords.add(args);
^

The compiler cannot ﬁnd an appropriate add method.

By the way, you can get the behavior of an unparameterized Vector class by

simply using Vector<Object>.

4.2 Implementing Generic Container Classes

79

4.2.3 Restricting Parameters

There are times when it is useful to place restrictions on type parameters to
generic classes. For example, we may wish to implement a class called a Num-
ericalAssociation, whose keys are always some speciﬁc subtype of java.lang.Number.
This would, for example, allow us to make an assumption that we could com-
pute the integer value by calling intValue on the key. Examples of declarations
of NumericalAssociations include:

“numerical”
means
“numeric”

NumericalAssociation<Integer,String> i;
NumericalAssociation<Double,Integer> d;
NumericalAssociation<Number,Object> n;
NumericalAssociation<BigInteger,Association<String,Object>> I;

We declare the NumericalAssociation class as follows:

public class NumericalAssociation<K extends Number,V>

extends Association<K,V>

{

public NumericalAssociation(K key, V value)
// pre: key is a non-null value of type K
// post: constructs a key-value pair
{

super(key,value);

}

public int keyValue()
// post: returns the integer that best approximates the key
{

return getKey().intValue();

}

}

The extends keyword in the type parameter for the class indicates that type K is
any subclass of Number (including itself). Such a constraint makes it impossible
to construct an class instance where K is a String because String is not a
subclass of Number. Since the Number class implements the intValue method,
we can call intValue on the key of our new class. This type restriction is called
a type bound.

Realize, of course, that the type bound bounds the type of class constructed.
Once the class is constructed, the particular types used may be more restric-
tive. The class NumericalAssociation<Double,Object> supports keys of type
Double, but not Integer. The NumericalAssociation<Number,Object> is the
most general form of the class.

By the way, the same extends type bound can be used to indicate that a type
variable implements a speciﬁc interface. An Association whose key is a Number
and whose value implements a Set of keys is declared as

public class NSAssociation<K extends Number, V extends Set<K>>

extends Association<K,V>

We have only touched on the complexities of parametric type speciﬁcation.

80

Generics

4.3 Conclusions

When we write data structures, we often write them for others. When we write
for others, we rarely know the precise details of the application. It is important
that we write our code as carefully as possible so that semantic errors associated
with our data structures occur as early as possible in development and as close
as possible to the mistake.

We have seen, here, several examples of generic data types. The Association
class has two type parameters, and is typical in terms of the simplicity of writ-
ing generalized code that can be tailored for use in speciﬁc applications. Formal
type parameters hold the place for actual type parameters to be speciﬁed later.
Code is written just as before, but with the knowledge that our structures will
not sacriﬁce the type-safety of their client applications.

There is one “ﬂy in the ointment”: we cannot construct arrays of any type
that involves a formal type parameter. These situations are relatively rare and,
for the most part, can be limited to the some carefully crafted code in the Vector
class. When array-like storage is required in the future, use of the extensible
Vector class has one more advantage over arrays: it hides the difﬁculties of
manipulating generic arrays.

Type bounds provide a mechanism to constrain the type parameters that ap-
pear in generic class speciﬁcations. These bounds can be disturbingly complex.
The remainder of this text demonstrates the power of the generic approach
to writing data structures. While the structure of these data types becomes
increasingly technical, the nature of specifying these structures in a generic and
ﬂexible manner remains quite simple, if not natural.

Chapter 5

Design Fundamentals

Concepts:
. Asymptotic analysis and big-O notation
.
. Back-of-the-envelope estimations
. Recursion and Induction

Time-space trade-off

We shape clay into a pot,
but it is the emptiness inside
that holds whatever we want.
—Lao Tzu

PROGRAMMERS ARE CRAFTSMEN. Their medium—their programming language—
often favors no particular design and pushes for an individual and artistic de-
cision. Given the task of implementing a simple program, any two individuals
are likely to make different decisions about their work. Because modern lan-
guages allow programmers a great deal of expression, implementations of data
structures reﬂect considerable personal choice.

Some aspects of writing programs are, of course, taught and learned. For
example, everyone agrees that commenting code is good. Programmers should
write small and easily understood procedures. Other aspects of the design of
programs, however, are only appreciated after considerable design experience.
In fact, computer science as a whole has only recently developed tools for un-
derstanding what it means to say that an algorithm is “implemented nicely,” or
that a data structure “works efﬁciently.” Since many data structures are quite
subtle, it is important to develop a rich set of tools for developing and analyzing
their performance.

In this chapter, we consider several important conceptual tools. Big-O com-
plexity analysis provides a means of classifying the growth of functions and,
therefore, the performance of the structures they describe. The concepts of
recursion and self-reference make it possible to concisely code solutions to com-
plex problems, and mathematical induction helps us demonstrate the important
properties—including trends in performance—of traditional data structures. Fi-
nally, notions of symmetry and friction help us understand how to design data
structures so that they have a reasonable look and feel.

5.1 Asymptotic Analysis Tools

We might be satisﬁed with evaluating the performance or complexity of data
structures by precisely counting the number of statements executed or objects

82

Design Fundamentals

referenced. Yet, modern architectures may have execution speeds that vary as
much as a factor of 10 or more. The accurate counting of any speciﬁc kind of
operation alone does not give us much information about the actual running
time of a speciﬁc implementation. Thus, while detailed accounting can be use-
ful for understanding the ﬁne distinctions between similar implementations, it
is not generally necessary to make such detailed analyses of behavior. Distinc-
tions between structures and algorithms, however, can often be identiﬁed by
observing patterns of performance over a wide variety of problems.

5.1.1 Time and Space Complexity

What concerns the designer most are trends suggested by the various perfor-
mance metrics as the problem size increases. Clearly, an algorithm that takes
time proportional to the problem size degrades more slowly than algorithms
that decay quadratically. Likewise, it is convenient to agree that any algorithm
that takes time bounded by a polynomial in the problem size, is better than
one that takes exponential time. Each of these rough characterizations—linear,
quadratic, and so on—identiﬁes a class of functions with similar growth behav-
ior. To give us a better grasp on these classiﬁcations, we use asymptotic or big-O
analysis to help us to describe and evaluate a function’s growth.

Deﬁnition 5.1 A function f (n) is O(g(n)) (read “order g” or “big-O of g”), if and
only if there exist two positive constants, c and n0, such that

|f (n)| ≤ c · g(n)

for all n ≥ n0.

In this text, f will usually be a function of problem size that describes the utiliza-
tion of some precious resource (e.g., time or space). This is a subtle deﬁnition
(and one that is often stated incorrectly), so we carefully consider why each of
the parts of the deﬁnition is necessary.

Most importantly, we would like to think of g(n) as being proportional to an
upper bound for f (n) (see Figure 5.1). After some point, f (n) does not exceed
an “appropriately scaled” g(n). The selection of an appropriate c allows us to
enlarge g(n) to the extent necessary to develop an upper bound. So, while g(n)
may not directly exceed f (n), it might if it is multiplied by a constant larger
than 1. If so, we would be happy to say that f (n) has a trend that is no worse
than that of g(n). You will note that if f (n) is O(g(n)), it is also O(10 · g(n)) and
O(5 + g(n)). Note, also, that c is positive. Generally we will attempt to bound
f (n) by positive functions.

Second, we are looking for long-term behavior. Since the most dramatic
growth in functions is most evident for large values, we are happy to ignore
“glitches” and anomalous behavior up to a certain point. That point is n0. We
do not care how big n0 must be, as long as it can be nailed down to some ﬁxed
value when relating speciﬁc functions f and g.

Nails = proofs.

5.1 Asymptotic Analysis Tools

83

Figure 5.1

Examples of functions, f (n), that are O(g(n)).

Third, we are not usually interested in whether the function f (n) is negative
or positive; we are just interested in the magnitude of its growth. In reality, most
of the resources we consider (e.g., time and space) are measured as positive
values and larger quantities of the resource are consumed as the problem grows
in size; growth is usually positive.

Most functions we encounter fall into one of a few categories. A function
that is bounded above by a constant is classiﬁed as O(1).1 The constant factor
can be completely accounted for in the value of c in Deﬁnition 5.1. These func-
tions measure size-independent characteristics of data structures. For example,
the time it takes to assign a value to an arbitrary element of an array of size n
is constant.

What’s your
best guess for
the time to
assign a value?
1
1000 second?
1000000 sec.?
1
1000000000 s.?

1

When a function grows proportionately to problem size, or linearly, we ob-
serve it is O(n). Depending on what’s being measured, this can be classiﬁed as
“nice behavior.” Summing the values in an n-element array, for example, can be
accomplished in linear time. If we double the size of the array, we expect the
time of the summation process to grow proportionately. Similarly, the Vector
takes linear space. Most methods associated with the Vector class, if not con-
stant, are linear in time and space.
If we develop methods that manipulate
the n elements of a Vector of numbers in superlinear time—faster than linear
growth—we’re not pleased, as we know it can be accomplished more efﬁciently.
Other functions grow polynomially and are O(nc), where c is some constant
greater than 1. The function n2 + n is O(n2) (let c = 2 and n0 = 1) and
therefore grows as a quadratic. Many simple methods for sorting n elements of
an array are quadratic. The space required to store a square matrix of size n
takes quadratic space. Usually, we consider functions with polynomial growth
to be fairly efﬁcient, though we would like to see c remain small in practice. Grass could be
Because a function nc−1 is O(n · nc−1) (i.e., O(nc)), we only need consider the
growth of the most signiﬁcant term of a polynomial function. (It is, after all,
most signiﬁcant!) The less signiﬁcant terms are ultimately outstripped by the
leading term.

greener.

1 It is also O(13), but we try to avoid such distractions.

nnn g(n)f(n)f(n)g(n)g(n)f(n)84

Design Fundamentals

Figure 5.2 Near-origin details of common curves. Compare with Figure 5.3.

Figure 5.3

Long-range trends of common curves. Compare with Figure 5.2.

22log(  )nnn!3sqrt(  )1001234545nnnnlog(  )2n22n!log(  )nnsqrt(  )nlog(  )nn604020801000020406080100nn5.1 Asymptotic Analysis Tools

85

Some functions experience exponential growth (see Figures 5.2 and 5.3).
The functions are O(cn), where c is a constant greater than 1. Enumerating
all strings of length n or checking topological equivalence of circuits with n
devices are classic examples of exponential algorithms. Constructing a list of
the n-digit palindromes requires exponential time and space. The demands of
an exponential process grow too quickly to make effective use of resources.
As a result, we often think of functions with exponential behavior as being
intractable.
In the next section we will see that some recursive solutions to
problems are exponential. While these solutions are not directly useful, simple
insights can sometimes make these algorithms efﬁcient.

“2002” is a
palindrome.

5.1.2 Examples

A “Difference Table”

Suppose we’re interested in printing a 10 by 10 table of differences between
two integers (row-col) values. Each value in the table corresponds to the result
of subtracting the row number from the column number:

0 -1 -2 -3 -4 -5 -6 -7 -8 -9
1 0 -1 -2 -3 -4 -5 -6 -7 -8
2 1 0 -1 -2 -3 -4 -5 -6 -7
3 2 1 0 -1 -2 -3 -4 -5 -6
4 3 2 1 0 -1 -2 -3 -4 -5
5 4 3 2 1 0 -1 -2 -3 -4
6 5 4 3 2 1 0 -1 -2 -3
7 6 5 4 3 2 1 0 -1 -2
8 7 6 5 4 3 2 1 0 -1
9 8 7 6 5 4 3 2 1 0

Analysis

As with most programs that generate two-dimensional output, we consider the
use of a nested pair of loops:

public static void diffTable(int n)
// pre: n >= 0
// post: print difference table of width n
{

for (int row = 1; row <= n; row++)
{

for (int col = 1; col <= n; col++)
{

// 1

// 2

System.out.print(row-col+" ");

// 3

}
System.out.println();

// 4

}

}

Each of the loops executes n times. Since printing a value (line 3) takes constant
time c1, the inner loop at line 2 takes c1n time. If line 4 takes constant time c2,

86

Design Fundamentals

then the outer loop at line 1 takes n(c1n+c2) = c1n2+c2n time. This polynomial
is clearly O(n2) (take c = c1 + c2 and n0 = 1). Doubling the problem size
approximately quadruples the running time.

As a rule of thumb, each loop that performs n iterations multiplies the com-
plexity of each iteration by a factor of n. Nested loops multiply the complexity
of the most deeply nested code by another power of n. As we have seen, loops
doubly nested around a simple statement often consume quadratic time.

Since there are only three variables in our difference method, it takes con-

stant space—an amount of space that is independent of problem size.

A Multiplication Table

Unlike the difference operator, the multiplication operator is commutative, and
the multiplication table is symmetric. Therefore, when printing a multiplication
table of size n, only the “lower triangular” region is necessary:

1
2 4
3 6 9
4 8 12 16
5 10 15 20 25
6 12 18 24 30 36
7 14 21 28 35 42 49
8 16 24 32 40 48 56 64
9 18 27 36 45 54 63 72 81
10 20 30 40 50 60 70 80 90 100

Here is a Java method to print the above table:

public static void multTable(int n)
// pre: n >= 0
// post: print multiplication table
{

for (int row = 1; row <= n; row++)
{

// 1

for (int col = 1; col <= row; col++) // 2
{

System.out.print(row*col+" ");

// 3

}
System.out.println();

// 4

}

}

Clearly, this table can be printed at least as fast as our difference table—it has
about half the entries—so it is, similarly, O(n2). Can this limit be improved? If
lines 3 and 4 take constant times c1 and c2, respectively, then the overall time is
approximately

(1c1 + c2) + (2c1 + c2) + · · · + (nc1 + c2) =

c1n(n + 1)
2

+ nc2 =

c1
2

n2 + (c2 +

c1
2

)n

5.1 Asymptotic Analysis Tools

87

Clearly, no linear function will bound this function above, so the bound of O(n2)
is a good estimate. Notice that we have, essentially, used the fastest growing
term of the polynomial—n2.

Notice that both of these programs print an “area” of values, each of which
can be computed in constant time. Thus, the growth rate of the function is the
growth rate of the area of the output—which is O(n2).

Building a Vector of Values

Often, it is useful to build a Vector containing speciﬁc values. For the purposes
of this problem, we will assume the values are integers between 0 and n −
1, inclusive. Our ﬁrst (and best) attempt expands the Vector in the natural
manner:

public static Vector<Integer> buildVector1(int n)
// pre: n >= 0
// post: construct a vector of size n of 1..n
{

Vector<Integer> v = new Vector<Integer>(n); // 1
for (int i = 0; i < n; i++)
{

// 2

v.add(i);

}
return v;

}

// 3

// 4

We will assume (correctly) that lines 1 and 4 take constant time. The loop at
line 2, however, takes n times the length of time it takes to add a single element.
Review of that code will demonstrate that the addition of a new element to a
Vector takes constant time, provided expansion is not necessary. Thus, the total
running time is linear, O(n). Notice that the process of building this Vector
requires space that is linear as well. Clearly, if the method’s purpose is to spend
time initializing the elements of a Vector, it would be difﬁcult for it to consume
space at a faster rate than time.

A slightly different approach is demonstrated by the following code:

public static Vector<Integer> buildVector2(int n)
// pre: n >= 0
// post: construct a vector of size n of 1..n
{

Vector<Integer> v = new Vector<Integer>(n);
for (int i = 0; i < n; i++)
{

// 2

v.add(0,i);

// 3

}
return v;

}

// 4

// 1

88

Design Fundamentals

All the assumptions of buildVector1 hold here, except that the cost of inserting
a value at the beginning of a Vector is proportional to the Vector’s current
length. On the ﬁrst insertion, it takes about 1 unit of time, on the second, 2
units, and so on. The analysis of this method, then, is similar to that of the
triangular multiplication table. Its running time is O(n2). Its space utilization,
however, remains linear.

Printing a Table of Factors

Suppose we are interested in storing a table of factors of numbers between 1
and n. The beginning of such a table—the factors of values between 1 and
10—includes the following values:

1
1 2
1 3
1 2 4
1 5
1 2 3 6
1 7
1 2 4 8
1 3 9
1 2 5 10

How much space must be reserved for storing this table? This problem looks a
little daunting because the number of factors associated with each line varies,
but without any obvious pattern. Here’s a program that generates the desired
table:

public static Vector<Vector<Integer>> factTable(int n)
// pre: n > 0
// post: returns a table of factors of values 1 through n
{

Vector<Vector<Integer>> table = new Vector<Vector<Integer>>();
for (int i = 1; i <= n; i++)
{

Vector<Integer> factors = new Vector<Integer>();
for (int f = 1; f <= i; f++)
{

if ((i % f) == 0) {
factors.add(f);

}

}
table.add(factors);

}
return table;

}

To measure the table size we consider those lines that mention f as a factor.
Clearly, f appears on every f th line. Thus, over n lines of the table there are

5.1 Asymptotic Analysis Tools

89

Figure 5.4
than 1 + R 8

1

Estimating the sum of reciprocal values. Here, P8
1
x dx = 1 + ln 8 ≈ 3.08.

x=1

1
x ≈ 2.72 is no more

no more than n
table size:

f lines that include f . Thus, we have as an upper bound on the

n
1

+

n
2

+

n
3

+ · · · +

n
n − 1

+

n
n

Factoring out n we have:

(cid:18) 1
1

n

+

1
2

+

1
3

+ · · · +

1
n − 1

+

(cid:19)

1
n

We note that these fractions fall on the curve 1

x (see Figure 5.4). We may

compute the area of the curve—an upper bound on the sum—as:

n

n
X

x=1

1
x

(cid:18)

≤ n

1 +

Z n

1

(cid:19)

dx

1
x

≤ n(1 + ln n − ln 1)
≤ O(n ln n)

The size of the table grows only a little faster than linearly. The time necessary
to create the table, of course, is O(n2) since we check n factors for number n.

Exercise 5.1 Slightly modify the method to construct the same table, but in O(n
time.

√

n)

Exercise 5.2 Rewrite the method to construct the same table, but in O(n ln n)
time.

012012345678910111415161718131290

Design Fundamentals

Finding a Space in a String

Some problems appear to have behavior that is more variable than the examples
we have seen so far. Consider, for example, the code to locate the ﬁrst space in
a string:

static int findSpace(String s)
// pre: s is a string, possibly containing a space
// post: returns index of first space, or -1 if none found
{

int i;
for (i = 0; i < s.length(); i++)
{

if (’ ’ == s.charAt(i)) return i;

}
return -1;

}

This simple method checks each of the characters within a string. When one
is found to be a space, the loop is terminated and the index is returned. If, of
course, there is no space within the string, this must be veriﬁed by checking
each character. Clearly, the time associated with this method is determined by
the number of loops executed by the method. As a result, the time taken is
linear in the length of the string.

We can, however, be more precise about its behavior using best-, worst-, and

average-case analyses:

Best case. The best-case behavior is an upper bound on the shortest time that
any problem of size n might take. Usually, best cases are associated with
particularly nice arrangements of values—here, perhaps, a string with a
space in the ﬁrst position. In this case, our method takes at most constant
time! It is important to note that the best case must be a problem of size n.

Worst case. The worst-case behavior is the longest time that any problem of
size n might take. In our string-based procedure, our method will take
the longest when there is no space in the string. In that case, the method
consumes at most linear time. Unless we specify otherwise, we will use
the worst-case consumption of resources to determine the complexity.

Average case. The average-case behavior is the complexity of solving an “av-
erage” problem of size n. Analysis involves computing a weighted sum of
the cost (in time or space) of problems of size n. The weight of each prob-
lem is the probability that the problem would occur. If, in our example,
we knew (somehow) that there was exactly one space in the string, and
that it appears in any of the n positions with equal probability, we would
deduce that, on average,

n
X

i=1

1
n

i =

1
n

· 1 +

1
n

· 2 + · · · +

1
n

· n =

1
n

·

n(n + 1)
2

=

n + 1
2

5.1 Asymptotic Analysis Tools

91

iterations would be necessary to locate the space. Our method has linear
average-time complexity. If, however, we knew that the string was English
prose of length n, the average complexity would be related to the average
length of the ﬁrst word, a value easily bounded above by a constant (say,
10). The weights of the ﬁrst few terms would be large, while the weights
associated with a large number of iterations or more would be zero. The German prose
average complexity would be constant. (In this case, the worst case would
be constant as well.) Obviously determining the average-case complexity
requires some understanding of the desired distributions of data.

may require
larger
constants.

Best-, worst-, and average-case analyses will be important in helping us eval-
uate the theoretical complexities of the structures we develop. Some care, how-
ever, must be used when determining the growth rates of real Java. It is tempt-
ing, for example, to measure the space or time used by a data structure and ﬁt a
curve to it in hopes of getting a handle on its long-term growth. This approach
should be avoided, if possible, as such statements can rarely be made with much
security. Still, such techniques can be fruitfully used to verify that there is no
unexpected behavior.

5.1.3 The Trading of Time and Space

Two resources coveted by programmers are time and space. When programs
are run, the algorithms they incorporate and the data structures they utilize
work together to consume time. This time is directly due to executing machine
instructions. The fewer instructions executed, the faster the program goes.

Most of us have had an opportunity to return to old code and realize that
useless instructions can be removed. For example, when we compute the ta-
ble factors, we realized that we could speed up the process by checking fewer
values for divisibility. Arguably, most programs are susceptible to some of this
“instruction weeding,” or optimization. On the other hand, it is clear that there
must be a limit to the extent that an individual program can be improved. For
some equivalent program, the removal of any statement causes the program to
run incorrectly. This limit, in some sense, is an information theoretic limit: given
the approach of the algorithm and the design of a data structure, no improve-
ments can be made to the program to make it run faster. To be convinced that
there is a ﬁrm limit, we would require a formal proof that no operation could be
avoided. Such proofs can be difﬁcult, especially without intimate knowledge of
the language, its compiler, and the architecture that supports the running code.
Nonetheless, the optimization of code is an important feature of making pro-
grams run quickly. Engineers put considerable effort into designing compilers
to make automated optimization decisions. Most compilers, for example, will
not generate instructions for dead code—statements that will never be executed.
In the following Java code, for example, it is clear that the “then” portion of this
code may be removed without fear:

92

Design Fundamentals

if (false)
{

System.out.println("Man in the moon.");

} else {

System.out.println("Pie in the sky.");

}

After compiler optimizations have been employed, though, there is a limit that
can be placed on how fast the code can be made to run. We will assume—
whenever we consider a time–space trade-off—that all reasonable efforts have
been made to optimize the time and space utilization of a particular approach.
Notice, however, that most optimizations performed by a compiler do not sig-
niﬁcantly affect the asymptotic running time of an algorithm. At most, they tend
to speed up an algorithm by a constant factor, an amount that is easily absorbed
in any theoretical analysis using big-O methods.

Appropriately implemented data structures can, however, yield signiﬁcant
performance improvements. Decisions about data structure design involve weigh-
ing—often using results of big-O analysis—the time and space requirements of
a structure used to solve a problem. For example, in the Vector class, we opted
to maintain a ﬁeld, elementCount, that kept track of how many elements within
the underlying array are actually being used. This variable became necessary
when we realized that as the Vector expanded, the constant reallocation of the
underlying memory could lead to quadratic time complexity over the life of the
Vector. By storing a little more information (here, elementCount) we reduce
the total complexity of expanding the Vector—our implementation, recall, re-
quires O(1) data-copying operations as the Vector expands. Since Vectors are
very likely to expand in this way, we ﬁnd it worthwhile to use this extra space.
In other situations we will see that the trade-offs are less obvious and sometimes
lead to the development of several implementations of a single data structure
designed for various uses by the application designer.

The choice between implementations is sometimes difﬁcult and may require
analysis of the application: if Vector’s add method is to be called relatively in-
frequently, the time spent resizing the structure is relatively insigniﬁcant. On the
other hand, if elements are to be added frequently, maintaining elementCount
saves time. In any case, the careful analysis of trade-off between time and space
is an important part of good data structure design.

5.1.4 Back-of-the-Envelope Estimations

A skill that is useful to the designer is the ability to develop good estimates
of the time and space necessary to run an algorithm or program.
It is one
thing to develop a theoretical analysis of an algorithm, but it is quite another to
develop a sense of the actual performance of a system. One useful technique is
to apply any of a number of back-of-the-envelope approximations to estimating
the performance of an algorithm.

The numbers that programmers work with on a day-to-day basis often vary
in magnitude so much that it is difﬁcult to develop much of a common sense

5.1 Asymptotic Analysis Tools

93

for estimating things. It is useful, then, to keep a store of some simple ﬁgures
that may help you to determine the performance—either in time or space—of a
project. Here are some useful rules of thumb:

• Light travels one foot in a nanosecond (one billionth of a second).

• Approximately π (≈ 3.15) hundredths of a second is a nanoyear (one

billionth of a year).

• It takes between 1 and 10 nanoseconds (ns) to store a value in Java. Basic

math operations take a similar length of time.

• An array assignment is approximately twice as slow as a regular assign-

ment.

• A Vector assignment is approximately 50 times slower than a regular

assignment.

• Modern computers execute 1 billion instructions per second.

• A character is represented by 8 bits (approximately 10).

• An Ethernet network can transmit at 100 million bits per second (expected

throughput is nearer 10 million bits).

• Fewer than 100 words made up 50 percent of Shakespeare’s writing; they
have an average length of π. A core of 3000 words makes up 90 percent
of his vocabulary; they have an average of 5.5 letters.

As an informal example of the process, we might attempt to answer the
question: How many books can we store on a 10 gigabyte hard drive? First
we will assume that 1 byte is used to store a character. Next, assuming that an
average word has about 5 characters, and that a typewritten page has about 500
words per typewritten page, we have about 2500 characters per page. Another
approximation might suggest 40 lines per page with 60 characters per line,
or 2400 characters per page. For computational simplicity, we keep the 2500
character estimate. Next, we assume the average book has, say, 300 pages, so
that the result is 0.75 million bytes required to store a text. Call it 1 million. A
10 gigabyte drive contains approximately 10 billion characters; this allows us
to store approximately 10 thousand books.

A dictionary is a collection of approximately 250,000 words. How long
might it take to compute the average length of words appearing in the dic-
tionary? Assume that the dictionary is stored in memory and that the length
of a word can be determined in constant time—perhaps 10 microseconds (µs).
The length must be accumulated in a sum, taking an additional microsecond
per word—let’s ignore that time. The entire summation process takes, then, 2.5
seconds of time. (On the author’s machine, it took 3.2 seconds.)

94

Design Fundamentals

Exercise 5.3 How many dots can be printed on a single sheet of paper? Assume,
for example, your printer prints at 500 dots per inch. If a dot were used to represent
a bit of information, how much text could be encoded on one page?

As you gain experience designing data structures you will also develop a
sense of the commitments necessary to support a structure that takes O(n2)
space, or an algorithm that uses O(n log n) time.

5.2 Self-Reference

One of the most elegant techniques for constructing algorithms, data structures,
and proofs is to utilize self-reference in the design. In this section we discuss ap-
plications of self-reference in programming—called recursion—and in proofs—
called proof by induction. In both cases the difﬁculties of solving the problem
outright are circumvented by developing a language that is rich enough to sup-
port the self-reference. The result is a compact technique for solving complex
problems.

5.2.1 Recursion

When faced with a difﬁcult problem of computation or structure, often the best
solution can be speciﬁed in a self-referential or recursive manner. Usually, the dif-
ﬁculty of the problem is one of management of the resources that are to be used
by the program. Recursion helps us tackle the problem by focusing on reducing
the problem to one that is more manageable in size and then building up the
answer. Through multiple, nested, progressive applications of the algorithm, a
solution is constructed from the solutions of smaller problems.

Summing Integers

We ﬁrst consider a simple, but classic, problem: suppose we are interested in
computing the sum of the numbers from 0 through n.

n
X

i=0

i = 0 + 1 + 2 + 3 + · · · + n

One approach to the problem is to write a simple loop that over n iterations
accumulates the result.

Recursion

public static int sum1(int n)
// pre: n >= 0
// post: compute the sum of 0..n
{

int result = 0;
for (int i = 1; i <= n; i++)
{

5.2 Self-Reference

95

result = result + i;

}
return result;

}

The method starts by setting a partial sum to 0. If n is a value that is less than
1, then the loop will never execute. The result (0) is what we expect if n = 0. If
n is greater than 0, then the loop executes and the initial portion of the partial
sum is computed. After n − 1 loops, the sum of the ﬁrst n − 1 terms is computed.
The nth iteration simply adds in n. When the loop is ﬁnished, result holds the
sum of values 1 through n. We see, then, that this method works as advertised
in the postcondition.

Suppose, now, that a second programmer is to solve the same problem. If
the programmer is particularly lazy and has access to the sum1 solution the
following code also solves the problem:

public static int sum2(int n)
// pre: n >= 0
// post: compute the sum of 0..n
{

if (n < 1) return 0;
else return sum1(n-1) + n;

}

For the most trivial problem (any number less than 1), we return 0. For all other
values of n, the programmer turns to sum1 to solve the next simplest problem
(the sum of integers 0 through n − 1) and then adds n. Of course, this algorithm
works as advertised in the postcondition, because it depends on sum1 for all but
the last step, and it then adds in the correct ﬁnal addend, n.

Actually, if sum2 calls any method that is able to compute the sum of numbers
0 through n−1, sum2 works correctly. But, wait! The sum of integers is precisely
what sum2 is supposed to be computing! We use this observation to derive, then,
the following self-referential method:

public static int sum3(int n)
// pre: n >= 0
// post: compute the sum of 0..n
{

if (n < 1) return 0;
else return sum3(n-1) + n;

// base case
// reduction, progress, solution

}

This code requires careful inspection (Figure 5.5). First, in the simplest or base
cases (for n < 1), sum3 returns 0. The second line is only executed when n ≥ 1.
It reduces the problem to a simpler problem—the sum of integers between 0 and
n − 1. As with all recursive programs, this requires a little work (a subtraction)
to reduce the problem to one that is closer to the base case. Considering the
problem n + 1 would have been fatal because it doesn’t make suitable progress

96

Design Fundamentals

Figure 5.5
arrows break the problem down; leftward arrows build up the solution.

The “unrolling” of a procedure to recursively sum integers. Rightward

toward the base case. The subproblem is passed off to another invocation of
sum3. Once that procedure computes its result (either immediately or, if nec-
essary, through further recursion), a little more work is necessary to convert
the solution of the problem of size n − 1 into a solution for a problem of size
n. Here, we have simply added in n. Notice the operation involved in build-
ing the answer (addition) opposes the operation used to reduce the problem
(subtraction). This is common in recursive procedures.

Principle 8 Recursive structures must make “progress” toward a “base case.”

We cast this principle in terms of “structures” because much of what we say
about self-referential execution of code can be applied to self-referential struc-
turing of data. Most difﬁculties with recursive structures (including recursive
methods) stem from either incorrectly stating the base case or failing to make
proper progress.

Inserting a Value into a Vector

Recursion is a natural method for accomplishing many complicated tasks on
Vectors and arrays. For example, the add(index,object) method of the Vector
class discussed on page 51 can be written as a recursive procedure. The essen-
tial concept is to insert the value into the Vector only after having moved the
previous value out of the way. That value is inserted at the next larger location.
This leads us to the following alternative to the standard Vector method:

Vector

Work3. Return 1+..+98    +99+1003. Return 1+...+98+992. Add in 983. Return 1+...+981. Compute sum3(97)Compute sum3(98):Compute sum3(0):Trivially return 0Compute sum3(100):1. Compute sum3(99)2. Add in 100Compute sum3(99):1. Compute sum3(98)ProgressBase case2. Add in 99NNWSWSENEWSE5.2 Self-Reference

97

public void add(int index, E value)
// pre: 0 <= index <= size()
// post: inserts new value in vector with desired index
//
{

moving elements from index to size()-1 to right

if (index >= size()) {

add(value); // base case: add at end

} else {

E previous = get(index); // work
add(index+1,previous);
set(index,value);

// work

// progress through recursion

}

}

Note that the base case is identiﬁed through the need to apply a trivial operation
rather than, say, the size of the index. Indeed, progress is determined by how
close the index gets to the size of the Vector. Again, this is a linear or O(n)
process.

Printing a Vector of Values

In the previous example, the recursive routine was suitable for direct use by
the user. Often, though, recursion demands additional parameters that encode,
in some way, progress made toward the solution. These parameters can be
confusing to users who, after all, are probably unaware of the details of the
recursion. To avoid this confusion, we “wrap” the call to a protected recursive
method in a public method. This hides the details of the initial recursive method
call. Here, we investigate a printing extension to the Vector class:

public void print()
// post: print the elements of the vector
{

printFrom(0);

}

protected void printFrom(int index)
// pre: index <= size()
// post: print elements indexed between index and size()
{

if (index < size()) {

System.out.println(get(index));
printFrom(index+1);

}

}

The print method wraps or hides the call to the recursive printFrom method.
The recursive method accepts a single parameter that indicates the index of
the ﬁrst element that should be printed out. As progress is made, the initial

98

Design Fundamentals

index increases, leading to linear performance. To print the entire Vector, the
recursive method is called with a value of zero.

It would appear that the base case is missing. In fact, it is indicated by the
failure of the if statement. Even though the base case is to “do nothing,” the if
statement is absolutely necessary. Every terminating recursive method should
have some conditional statement.

PrintFrom is an example of a tail recursive method. Any recursion happens
just before exiting from the method. Tail recursive methods are particularly nice
because good compilers can translate them into loops. Each iteration of the loop
simulates the computation and return of another of the nested recursive proce-
dure calls. Since there is one call for each of the n values, and the procedure
performs a constant amount of work, the entire process takes O(n) time.

Exercise 5.4 Write a recursive method to print out the characters of a string with
spaces between characters. Make sure your method does not print a leading or
tailing space, unless it is a leading or trailing character of the original string.

Computing Change in Postage Stamps

Suppose, when receiving change at the post ofﬁce, you wished to be paid your
change in various (useful) stamps. For example, at current rates, you might
be interested in receiving either 39 cent stamps, 24 cent postcards, or penny
stamps (just in case of a postage increase). For a particular amount, what is the
smallest number of stamps necessary to make the change?

This problem is fairly complex because, after all, the minimum number of
stamps needed to make 50 cents involves 4 stamps—two postcard stamps and
two penny stamps—and not 12—a 39 cent stamp and 11 penny stamps. (The
latter solution might be suggested by postal clerks used to dealing with U.S.
coinage, which is fairly easily minimized.) We will initially approach this prob-
lem using recursion. Our solution will only report the minimum number of
stamps returned. We leave it as an exercise to report the number of each type
of stamp (consider Problem 5.22); that solution does not greatly change the
approach of the problem.

If no change is required—a base case—the solution is simple: hand the cus-
tomer zero stamps. If the change is anything more, we’ll have to do some work.
Consider the 70 cent problem. We know that some stamps will have to be given
to the customer, but not the variety. We do know that the last stamp handed to
the customer will either be a penny stamp, a 26 cent step, or a 41 cent stamp.
If we could only solve three smaller minimization problems—the 69 cent prob-
lem, the 34 cent problem, and the 29 cent problem—then our answer would
be one stamp more than the minimum of the answers to those three problems.
(The answers to the three problems are 4, 9, and 4, respectively, so our answer
should be 5.) Of course, we should ignore meaningless reduced problems: the
−6 cent problem results from considering handing a 26 cent stamp over to solve
the 20 cent problem.

Here is the stampCount method that computes the solution:

Recursive-
Postage

5.2 Self-Reference

99

public final static int LETTER=41;
public final static int CARD=26;
public final static int PENNY=1;

public static int stampCount(int amount)
// pre: amount >= 0
// post: return *number* of stamps needed to make change
//
{

(only use letter, card, and penny stamps)

int minStamps;
Assert.pre(amount >= 0,"Reasonable amount of change.");
if (amount == 0) return 0;
// consider use of a penny stamp
minStamps = 1+stampCount(amount-PENNY);
// consider use of a post card stamp
if (amount >= CARD) {

int possible = 1+stampCount(amount-CARD);
if (minStamps > possible) minStamps = possible;

}
// consider use of a letter stamp
if (amount >= LETTER) {

int possible = 1+stampCount(amount-LETTER);
if (minStamps > possible) minStamps = possible;

}
return minStamps;

}

For the nontrivial cases, the variable minStamps keeps track of the minimum
number of stamps returned by any of these three subproblems. Since each
method call potentially results in several recursive calls, the method is not tail
recursive. While it is possible to solve this problem using iteration, recursion
presents a very natural solution.

An Efﬁcient Solution to the Postage Stamp Problem

If the same procedure were used to compute the minimum number of stamps Making
to make 70 cents change, the stampCount procedure would be called 2941
times. This number increases exponentially as the size of the problem in-
creases (it is O(3n)). Because 2941 is greater than 70—the number of distinct
subproblems—some subproblems are recomputed many times. For example,
the 2 cent problem must be re-solved by every larger problem.

currency is
illegal.
Making change
is not!

To reduce the number of calls, we can incorporate an array into the method.
Each location n of the array stores either 0 or the answer to the problem of size
n. If, when looking for an answer, the entry is 0, we invest time in computing the
answer and cache it in the array for future use. This technique is called dynamic
programming and yields an efﬁcient linear algorithm. Here is our modiﬁed
solution:

FullPostage

100

Design Fundamentals

public static final int LETTER = 41; // letter rate
public static final int CARD = 26;
public static final int PENNY = 1;

// post card rate
// penny stamp

public static int stampCount(int amount)
// pre: amount >= 0
// post: return *number* of stamps needed to make change
//
{

(only use letter, post card, and penny stamps)

return stampCount(amount, new int[amount+1]);

}

protected static int stampCount(int amount, int answer[])
// pre: amount >= 0; answer array has length >= amount
// post: return *number* of stamps needed to make change
//
{

(only use letter, post card, and penny stamps)

int minStamps;
Assert.pre(amount >= 0,"Reasonable amount of change.");
if (amount == 0) return 0;
if (answer[amount] != 0) return answer[amount];
// consider use of a penny stamp
minStamps = 1+stampCount(amount-1,answer);
// consider use of a post card stamp
if (amount >= CARD) {

int possible = 1+stampCount(amount-CARD,answer);
if (minStamps > possible) minStamps = possible;

}
// consider use of a letter stamp
if (amount >= LETTER) {

int possible = 1+stampCount(amount-LETTER,answer);
if (minStamps > possible) minStamps = possible;

}
answer[amount] = minStamps;
return minStamps;

}

When we call the method for the ﬁrst time, we allocate an array of sufﬁcient
size (amount+1 because arrays are indexed beginning at zero) and pass it as
answer in the protected two-parameter version of the method. If the answer
is not found in the array, it is computed using up to three recursive calls that
pass the array of previously computed answers. Just before returning, the newly
computed answer is placed in the appropriate slot. In this way, when solutions
are sought for this problem again, they can be retrieved without the overhead
of redundant computation.

When we seek the solution to the 70 cent problem, 146 calls are made to the
procedure. Only a few of these get past the ﬁrst few statements to potentially
make recursive calls. The combination of the power recursion and the efﬁciency
of dynamic programming yields elegant solutions to many seemingly difﬁcult

5.2 Self-Reference

problems.

101

Exercise 5.5 Explain why the dynamic programming approach to the problem
runs in linear time.

In the next section, we consider induction, a recursive proof technique. In-
duction is as elegant a means of proving theorems as recursion is for writing
programs.

5.2.2 Mathematical Induction

The accurate analysis of data structures often requires mathematical proof. An
effective proof technique that designers may apply to many computer science
problems is mathematical induction. The technique is, essentially, the construc-
tion of a recursive proof. Just as we can solve some problems elegantly using
recursion, some properties may be elegantly veriﬁed using induction.

A common template for proving statements by mathematical induction is as

follows:

1. Begin your proof with “We will prove this using induction on the size of

the problem.” This informs the reader of your approach.

2. Directly prove whatever base cases are necessary. Strive, whenever possi-
ble to keep the number of cases small and the proofs as simple as possible.

3. State the assumption that the observation holds for all values from the
base case, up to but not including the nth case. Sometimes this assump-
tion can be relaxed in simple inductive proofs.

4. Prove, from simpler cases, that the nth case also holds.

5. Claim that, by mathematical induction on n, the observation is true for all

cases more complex than the base case.

Individual proofs, of course, can deviate from this pattern, but most follow the
given outline.

As an initial example, we construct a formula for computing the sum of
integers between 0 and n ≥ 0 inclusively. Recall that this result was used in
Section 3.5 when we considered the cost of extending Vectors, and earlier, in
Section 5.1.2, when we analyzed buildVector2. Proof of this statement also
yields a constant-time method for implementing sum3.

Observation 5.1 Pn

i=0 i = n(n+1)

2

.

Proof: We prove this by induction. First, consider the simplest case, or base case.
If n = 0, then the sum is 0. The formula gives us 0(0+1)
2 = 0. The observation
appears to hold for the base case.

Now, suppose we know—for some reason—that our closed-form formula
holds for all values between 0 (our base case) and n − 1. This knowledge may

102

Design Fundamentals

help us solve a more complex problem, namely, the sum of integers between 0
and n. The sum

0 + 1 + 2 + · · · + (n − 1) + n

conveniently contains the sum of the ﬁrst n − 1 integers, so we rewrite it as

[0 + 1 + 2 + · · · + (n − 1)] + n

Because we have assumed that the sum of the natural numbers to n − 1 can be
computed by the formula, we may rewrite the sum as

(cid:21)

(cid:20) (n − 1)n
2

+ n

The terms of this expression may be simpliﬁed and reorganized:

(n − 1)n + 2n
2

=

n(n + 1)
2

Thus given only the knowledge that the formula worked for n − 1, we have
been able to extend it to n. It is not hard to convince yourself, then, that the
observation holds for any nonnegative value of n. Our base case was for n = 0,
so it must hold as well for n = 1. Since it holds for n = 1, it must hold for
n = 2. In fact, it holds for any value of n ≥ 0 by simply proving it holds for
values 0, 1, 2, . . . , n − 1 and then observing it can be extended to n.(cid:5)

The induction can be viewed as a recursively constructed proof (consider
Figure 5.6). Suppose we wish to see if our observation holds for n = 100. Our
method requires us to show it holds for n = 99. Given that, it is a simple matter
to extend the result to 100. Proving the result for n = 99, however, is almost2
as difﬁcult as it is for n = 100. We need to prove it for n = 98, and extend
that result. This process of developing the proof for 100 eventually unravels
into a recursive construction of a (very long) proof that demonstrates that the
observation holds for values 0 through 99, and then 100.

The whole process, like recursion, depends critically on the proof of appro-
priate base cases. In our proof of Observation 5.1, for example, we proved that
the observation held for n = 0. If we do not prove this simple case, then our
recursive construction of the proof for any value of n ≥ 0 does not terminate:
when we try to prove it holds for n = 0, we have no base case, and therefore
must prove it holds for n = −1, and in proving that, we prove that it holds for
−2, −3, . . . , ad inﬁnitum. The proof construction never terminates!

Our next example of proof by induction is a correctness proof . Our intent is
to show that a piece of code runs as advertised. In this case, we reinvestigate
sum3 from page 95:

99 cases left to
prove! Take one
down, pass it
around, 98
cases left to
prove! . . .

Recursion

2 It is important, of course, to base your inductive step on simpler problems—problems that take
you closer to your base case. If you avoid basing it on simpler cases, then the recursive proof will
never be completely constructed, and the induction will fail.

5.2 Self-Reference

103

Figure 5.6
proof. Compare with Figure 5.5.

The process of proof by induction simulates the recursive construction of a

public static int sum3(int n)
// pre: n >= 0
// post: compute the sum of 0..n
{

if (n < 1) return 0;
else return

sum3(

n-1

) + n;

// 1
// 2
// 3
// 4
// 5

}

(The code has been reformatted to allow discussion of portions of the computa-
tion.) As with our mathematical proofs, we state our result formally:

Observation 5.2 Given that n ≥ 0, the method sum3 computes the sum of the
integers 0 through n, inclusive.

Proof: Our proof is by induction, based on the parameter n. First, consider the
action of sum3 when it is passed the parameter 0. The if statement of line 1 is
true, and the program returns 0, the desired result.

We now consider n>0 and assume that the method computes the correct
result for all values less that n. We extend our proof of correctness to the pa-
rameter value of n. Since n is greater than 0, the if of line 1 fails, and the else
beginning on line 2 is considered. On line 4, the parameter is decremented,
and on line 3, the recursion takes place. By our assumption, this recursive

Proof for 0:* Trivial proof.Q.E.D.Proof for 100:* It works for 99.* Extend to 100.Q.E.D.Proof for 99:* Extend to 99.Q.E.D.* It works for 97.* Extend to 98.Q.E.D.* It works for 98.Proof for 98:104

Design Fundamentals

Figure 5.7 A group of n computer scientists composed of Alice and n − 1 others.

call returns the correct result—the sum of values between 0 and n-1, inclusive.
Line 5 adds in the ﬁnal value, and the entire result is returned. The program
works correctly for a parameter n greater than 0. By induction on n, the method
computes the correct result for all n>=0.(cid:5)

Proofs of correctness are important steps in the process of verifying that code
works as desired. Clearly, since induction and recursion have similar forms, the
application of inductive proof techniques to recursive methods often leads to
straightforward proofs. Even when iteration is used, however, induction can be
used to demonstrate assumptions made about loops, no matter the number of
iterations.

We state here an important result that gives us a closed-form expression for

computing the sum of powers of 2.

Observation 5.3 Pn

i=0 2i = 2n+1 − 1.

Exercise 5.6 Prove Observation 5.3.

There are, of course, ways that the inductive proof can go awry. Not proving
the appropriate base cases is the most common mistake and can lead to some
interesting results. Here we prove what few have suspected all along:

Observation 5.4 All computer scientists are good programmers.

Warning: bad

Proof: We prove the observation is true, using mathematical induction. First,
proof! we use traditional techniques (examinations, etc.) to demonstrate that Alice is

a good programmer.

Now, assume that our observation is true of any group of fewer than n com-
puter scientists. Let’s extend our result: select n computer scientists, including
Alice (see Figure 5.7). Clearly, the subgroup consisting of all computer scien-
tists that are “not Alice” is a group of n − 1 computer scientists. Our assumption
states that this group of n − 1 computer scientists is made up of good program-
mers. So Alice and all the other computer scientists are good programmers.
By induction on n, we have demonstrated that all computer scientists are good
programmers.(cid:5)

(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)Alicen-1 others5.2 Self-Reference

105

Figure 5.8 A group of n computer scientists, including Alice, Bob, and Carol.

This is a very interesting result, especially since it is not true. (Among other
things, some computer scientists do not program computers!) How, then, were
we successful in proving it? If you look carefully, our base case is Alice. The
assumption, on the other hand, is based on any group of n − 1 programmers.
Unfortunately, since our only solid proof of quality programming is Alice, and
non-Alice programmers cannot be reduced to cases involving Alice, our proof is
fatally ﬂawed.

Still, a slight reworking of the logic might make the proof of this observa-
tion possible. Since Alice is a computer scientist, we can attempt to prove the
observation by induction on groups of computer scientists that include Alice:

Proof: We prove the observation by induction. First, as our base case, consider Warning: bad
proof, take 2!
Alice. Alice is well known for being a good programmer. Now, assume that for
any group of fewer than n computer scientists that includes Alice, the members
are excellent programmers. Take n computer scientists, including Alice (see
Figure 5.8). Select a non-Alice programmer. Call him Bob. If we consider all
non-Bob computer scientists, we have a group of n − 1 computer scientists—
including Alice. By our assumption, they must all be good. What about Bob?
Select another non-Alice, non-Bob computer scientist from the group of n. Call
her Carol. Carol must be a good programmer, because she was a member of the
n − 1 non-Bob programmers. If we consider the n − 1 non-Carol programmers,
the group includes both Alice and Bob. Because it includes Alice, the non-
Carol programmers must all be good. Since Carol is a good programmer, then
all n must program well. By induction on n, all groups of computer scientists
that include Alice must be good programmers. Since the group of all computer
scientists is ﬁnite, and it includes Alice, the entire population must program
well. The observation holds!(cid:5)

This proof looks pretty solid—until you consider that in order for it to work,
you must be able to distinguish between Alice, Bob, and Carol. There are three
people. The proof of the three-person case depends directly on the observation
holding for just two people. But we have not considered the two-person case!
In fact, that is the hole in the argument. If we know of a bad programmer, Ted,
we can say nothing about the group consisting of Alice and Ted (see Figure 5.9).

(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)BobAliceCarol106

Design Fundamentals

Lesson:
it’s hard to ﬁnd
good
programmers.

Figure 5.9
programmer.

The proof does not hold for the simplest nontrivial case: Alice and any bad

As a result, we have a worrisome hole in the proof of the group consisting of
Alice, Bob, and Ted. In the end, the attempt at a complete proof unravels.

What have we learned from this discussion? For an inductive proof, the base
cases must be carefully enumerated and proved. When proving the inductive
step, the step must be made upon a proved foundation. If not, the entire state-
ment collapses. The subtlety of this difﬁculty should put us on alert: even the
most thoughtful proofs can go awry if the base case is not well considered.

We can now make a similar statement about recursion: it is important to
identify and correctly code the base cases you need. If you don’t, you run the
risk that your method will fail to stop or will compute the wrong answer. One
of the most difﬁcult debugging situations occurs when multiple base cases are
to be considered and only a few are actually programmed.

Our ﬁnal investigation considers the implementation of a Java method to

compute the following sequence of values:

0, 1, 1, 2, 3, 5, 8, 13, 21, . . .

These values are the ﬁrst of the sequence of Fibonacci numbers. Each value is
the sum of the two values that fall before it. We should be careful—especially
given our last discussion—that we have the base cases carefully considered. In
this particular case, we must specify two initial values: 0 and 1.

This sequence may be familiar to you. If it is, you may have seen the deﬁni-

tion of Fn, the nth value of the sequence as

Fn =

(cid:26) n

n = 0 or n = 1

Fn−2 + Fn−1 n > 1

The translation of this type of equation into Java is fairly straightforward. We
make the following attempt:

static public int fibo(int n)
// pre: n is a nonnegative integer
// post: result is the ith term from the sequence
//

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, . . .

Fibo

AliceTed5.2 Self-Reference

107

{

}

Assert.pre(n >= 0, "Index is nonnegative.");
// when n < 2, return n
if (n == 0) return 0;
else if (n == 1) return 1;
// complex, self-referential case:
else return fibo(n-2)+fibo(n-1);

// line 1
// line 2

// line 3

We now seek to prove that the recursive method computes and returns the nth
member of the sequence.
Proof: First, suppose n = 0: the method returns 0, on line 1. Next, suppose
that n = 1: the method returns 1, on line 2. So, for the two very simplest
cases, the method computes the correct values. Now, suppose that n > 1, and
furthermore, assume that fibo returns the correct value for all terms with index
less than n. Since n > 1, lines 1 and 2 have no effect. Instead, the method
resorts to using line 3 to compute the value. Since the method works for all
values less than n, it speciﬁcally computes the two previous terms—Fn−2 and
Fn−1—correctly. The sum of these two values (Fn) is therefore computed and
immediately returned on line 3. We have, then, by mathematical induction on
n proved that f ibo(n) computes Fn for all n ≥ 0.(cid:5)

Another approach to computing Fibonacci numbers, of course, would be to

use an iterative method:

static public int fibo2(int n)
// pre: n is a nonnegative integer
// post: result is the ith term from the sequence
//
{

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, . . .

Assert.pre(n >= 0, "Index is nonnegative.");
int a = 0;
int b = 1;
if (n == 0) return a;
if (n == 1) return b;
// for large values of n, iteratively compute sequence
int i=2,F;
do
{

// line 1
// line 2

a is the i-2nd member

// Assertion: b is the i-1st member of the sequence
//
F = a + b;
// Assertion: F is the ith member
// update previous two values:
a = b;
b = F;
i++;

// line 3

// line 4
// line 5
// line 6
// line 7
// line 8

} while (i <= n);
return F;

}

108

Design Fundamentals

To demonstrate that such a program works, we perform a step-wise analysis of
the method as it computes Fn.
Proof: Suppose n = 0. The condition in the if statement on line 1 is true and
the value a (0) is returned. If n = 1, the condition on line 1 is false, but the
if statement on line 2 is true and b (1) is returned. In both of these cases the
correct value is returned.

We now investigate the loop. We notice that when the loop starts (it is a do
loop, it must execute at least once if n > 1), a and b contain the values F0 and
F1, and i is 2. Thus, the loop invariant before line 3 holds on the ﬁrst iteration.
Now, assume that i ≥ 2 and the loop invariant before line 3 holds. The
effect of line 3 is to compute Fi from Fi−1 and Fi−2. The result is placed in F ,
and the loop invariant after line 3 is met. The remaining statements, on lines 4
through 6 result in Fi−2 in a and Fi−1 in b. If the condition on line 7 should be
true, we meet the loop invariant for the next iteration.

If the condition on line 7 should be false, then we note that this value of i is
the ﬁrst that is greater than n, so F = Fi−1 = Fn, and the result returned is the
correct result. (cid:5)

It is interesting to note that the initial values of the sequence are rather arbi-
trary, and that different natural phenomena related to Fibonacci numbers can
be modeled by sequences that begin with different initial values.

5.3 Properties of Design

This section is dedicated to two informal properties of design that are referenced
elsewhere within this text. The property of symmetry describes the predictability
of a design, while friction describes the difﬁculty of moving a data structure from
one state to another. Both terms extend the vocabulary of implementors when
discussing design decisions.

5.3.1 Symmetry

For the most part, our instruction of computers occurs through programs. As
a result, programs can be nonintuitive and hard to understand if they are not
designed with human-readability in mind. On the other hand, a well-designed
program can be used by a novice without a signiﬁcant learning curve. Systems
that are easy to use tend to survive longer.

The programmer, as a designer of a data structure, is responsible for deliv-
ering a usable implementation of an abstract data structure. For an implemen-
tation to be usable, it should provide access to the structure with methods that
are predictable and easy to use. The notion of predictability is particularly dif-
ﬁcult for designers of data structures to understand, and it is something often
overlooked by novice programmers.

When designing a system (here, a program or data structure) a useful princi-
ple is to make its interface symmetric. What is symmetry? Symmetry allows one

5.3 Properties of Design

109

to view a system from different points of view and see similarities. Mathemati-
cians would say that a system exhibits a symmetry if “it looks like itself under
a nontrivial transformation.” Given this, programmers consider asymmetries in
transformed programs to be early warning signs of errors in logic.

Consider the following method (you will see this as part of the swap proce-
dure of page 120). It exchanges two object references—data[i] and data[j].

int temp;
temp = data[i];
data[i] = data[j];
data[j] = temp;

Close inspection of this code demonstrates that it does what it claims to do.
Even if we stand back, not thinking so much about the actual workings of the
code, we can see that the code is pretty symmetric. For example, if we squint
our eyes and look at the code from the standpoint of variable data[i], we see
it as:

... = data[i];
data[i] = ...;

Here, data[i] is assigned to a variable, and a value is assigned to data[i]. We
see a similar pattern with data[j]:

... = data[j];
data[j] = ...;

While this is not direct proof that the code works, it is an indication that the
code is, in some way, “symmetric,” and that helps make the argument that it is
well designed.

Not everything we do is symmetric. If we consider the Association class,
for example, the key and value components of the Association are different.
The value, of course, has two associated methods, getValue and setValue.
The ﬁrst of the methods reads and returns a value, while the second method
consumes and sets a value. Everything is in balance, and so we are hopeful that
the design of the structure is correct. On the other hand, the key can only be
read: while there is a getKey method, there is no setKey. We have suggested
good reasons for doing this. As long as you can make a good argument for
asymmetry in design, the breaking of symmetry can be useful. Unreasoned
asymmetry, however, is a sign of poor and unpredictable design.

Here are various ways that one can look at a system to evaluate it for sym-

metry:

1. Compare methods that extend the structure with methods that trim the
structure. Do they have similar approaches? Are they similar in number?

2. Consider methods that read and write values. Can the input methods read
what is written by the output methods? Can the writing methods write all
values that can be read?

110

Design Fundamentals

Should 6= will.

3. Are procedures that consume parameters matched by functions that de-

liver values?

4. Can points of potential garbage collection be equally balanced by new in-

vocations?

5. In linked structures, does unlinking a value from the structure appear to

be the reverse of linking a new value into the structure?

When asymmetries are found, it is important to consider why they occur. Argu-
ments such as “I can’t imagine that anyone would need an opposite method!”
are usually unconvincing. Many methods are added to the structures, not be-
cause they are obviously necessary, but because there is no good argument
against them. Sometimes, of course, the language or underlying system forces
an asymmetry. In Java, for example, every Object has a toString method that
converts an internal representation of an object to a human-readable form, but
there’s no fromString required method that reads the value of an Object from
a String. There should be, but there isn’t.

5.3.2 Friction

One of the obvious beneﬁts of a data structure is that it provides a means of
storing information. The ease with which the structure accepts and provides in-
formation about its contents can often be determined by its interface. Likewise,
the difﬁculty of moving a data structure from one state to another determines,
in some way, its “stiffness” or the amount of friction the structure provides when
the state of the structure is to be modiﬁed.

One way that we might measure friction is to determine a sequence of logical
states for the structure, and then determine the number of operations that are
necessary to move the structure from each state to the next.
If the number
of operations is high, we imagine a certain degree of friction; if the operation
count is low, the structure moves forward with relative ease.

Often we see that the less space provided to the structure, the more friction
appears to be inherent in its structure. This friction can be good—it may make
it less possible to get our structure into states that are inconsistent with the
deﬁnition of the structure, or it may be bad—it may make it difﬁcult to get
something done.

5.4 Conclusions

Several formal concepts play an important role in modern data structure des-
ign—the use of big-O analysis to support claims of efﬁciency, the use of recur-
sion to develop concise but powerful structures, and the use of induction to
prove statements made about both data structures and algorithms. Mastery of
these concepts improves one’s approach to solving problems of data structure
design.

5.4 Conclusions

111

The purpose of big-O analysis is to demonstrate upper bounds on the growth
of functions that describe behavior of the structures we use. Since these are
upper bounds, the tightest bounds provide the most information. Still, it is
often not very difﬁcult to identify the fastest-growing component of a function—
analysis of that component is likely to lead to fairly tight bounds and useful
results.

Self-reference is a powerful concept. When used to develop methods, we call
this recursion. Recursion allows us to break down large problems into smaller
problems whose solutions can be brought together to solve the original prob-
lem. Interestingly, recursion is often a suitable substitute for loops as a means of
progressing through the problem solution, but compilers can often convert tail
recursive code back into loops, for better performance. All terminating recur-
sive methods involve at least one test that distinguishes the base case from the
recursive, and every recursive program must eventually make progress toward
a base case to construct a solution.

Mathematical induction provides a means of recursively generating proofs.
Perhaps more than most other proof mechanisms, mathematical induction is
a useful method for demonstrating a bound on a function, or the correct ter-
mination of a method. Since computers are not (yet) able to verify everyday
inductive proofs, it is important that they be constructed with appropriate care.
Knowing how to correctly base induction on special cases can be tricky and, as
we have recently seen, difﬁcult to verify.

In all these areas, practice makes perfect.

Self Check Problems

Solutions to these problems begin on page 444.

Suppose f (x) = x. What is its best growth rate, in big-O notation?
Suppose f (x) = 3x. What is its growth rate?
What is the growth rate of f (x) = x + 900?
How fast does f (x) grow if f (x) = x for odd integers and f (x) = 900

5.1
5.2
5.3
5.4
for even integers?

Evaluate and order the functions log2 x,
5.5
x = 2, 4, 16, and 64. For each value of x, which is largest?
What are three features of recursive programs?
5.6
The latest Harry Potter book may be read by as much as 75 percent of
5.7
the reading child population in the United States. Approximately how many
child-years of reading time does this represent?

x, x, 30x, x2, 2x, and x! at

√

Given an inﬁnite supply of 37 cent stamps, 21 cent stamps, and penny
5.8
stamps a postmaster returns a minimum number of stamps composed of c37(x),
c21(x), and c1(x) stamps for x dollars in change. What are the growth rates of
these functions?

112

Design Fundamentals

Problems

Solutions to the odd-numbered problems begin on page 462.
What is the time complexity associated with accessing a single value in
5.1
an array? The Vector class is clearly more complex than the array. What is the
time complexity of accessing an element with the get method?
What is the worst-case time complexity of the index-based remove code
5.2
in the Vector class? What is the best-case time complexity? (You may assume
the Vector does not get resized during this operation.)
5.3

What is the running time of the following method?

public static int reduce(int n)
{

int result = 0;
while (n > 1)
{

n = n/2;
result = result+1;

}
return result;

}

What is the time complexity of determining the length of an n-character
5.4
null-terminated string? What is the time complexity of determining the length
of an n-character counted string?
5.5

What is the running time of the following matrix multiplication method?

// square matrix multiplication
// m1, m2, and result are n by n arrays
for (int row = 0; row < n; row++)
{

for (int col = 0; col < n; col++)
{

int sum = 0;
for (int entry = 0; entry < n; entry++)
{

sum = sum + m1[row][entry]*m2[entry][col];

}
result[row][col] = sum;

}

}

5.6
bound. An alternative deﬁnition provides a lower bound for a function:

In Deﬁnition 5.1 we see what it means for a function to be an upper

Deﬁnition 5.2 A function f (n) is Ω(g(n)) (read “big-omega of g” or “at least
order g”), if and only if there exist two positive constants, c and n0, such that

for all n ≥ n0.

f (n) ≥ c · g(n)

5.4 Conclusions

113

What is a lower bound on the time it takes to remove a value from a Vector by
index?

What is a lower bound on adding a value to the end of a Vector? Does
5.7
it matter that sometimes we may have to spend time doubling the size of the
underlying array?

When discussing symmetry, we investigated a procedure that swapped
Is it possible to write a routine that swaps two

5.8
two values within an array.
integer values? If so, provide the code; if not, indicate why.

For subtle reasons String objects cannot be modiﬁed. Instead, Strings
5.9
are used as parameters to functions that build new Strings. Suppose that a is
an n-character String. What is the time complexity of performing a=a+"!"?

Read Problem 5.9. Suppose that a and b are n-character Strings. What

5.10
is the complexity of performing a=a+b?
5.11 What is the rate of growth (using big-O analysis) of the function f (n) =
n + log n? Justify your answer.

In this text, logarithms are assumed to be in base 2. Does it make a

5.12
difference, from a complexity viewpoint?
5.13 What is the rate of growth of the function 1
5.14 What is the rate of growth of the function sin n
5.15

n + 12? Justify your answer.
n ? Justify your answer.

Trick question: What is the rate of growth of tan n?
Suppose n integers between 1 and 366 are presented as input, and you
5.16
want to know if there are any duplicates. How would you solve this problem?
What is the rate of growth of the function T (n), describing the time it takes for
you to determine if there are duplicates? (Hint: Pick an appropriate n0.)
The ﬁrst element of a Syracuse sequence is a positive integer s0. The
5.17
value si (for i > 0) is deﬁned to be si−1/2 if si−1 is even, or 3si−1 + 1 if si−1
is odd. The sequence is ﬁnished when a 1 is encountered. Write a procedure to
print the Syracuse sequence for any integer s0. (It is not immediately obvious
that this method should always terminate.)

Rewrite the sqrt function of Section 2.1 as a recursive procedure.

5.18
5.19 Write a recursive procedure to draw a line segment between (x0, y0)
and (x1, y1) on a screen of pixels with integer coordinates. (Hint: The pixel
closest to the midpoint is not far off the line segment.)

5.20

Rewrite the reduce method of Problem 5.3 as a recursive method.

One day you notice that integer multiplication no longer works. Write
5.21
a recursive procedure to multiply two values a and b using only addition. What
is the complexity of this function?

5.22 Modify the “stamp change” problem of Section 5.2.1 to report the num-
ber of each type of stamp to be found in the minimum stamp change.
Prove that 5n − 4n − 1 is divisible by 16 for all n ≥ 0.
Prove Observation 5.3, that Pn

i=0 2i = 2n+1 − 1 for n ≥ 0.

5.24

5.23

114

Design Fundamentals

i=1 log i ≤ n log n.

5.25
5.26
5.27

5.28

i=0 ci = cn+1+(c−2)

Prove that a function nc is O(nd) for any d ≥ c.
Prove that Pn
i=1 2i = n(n + 1).
Prove that Pn
i=1 (2i − 1) = n2.
Show that for c ≥ 2 and n ≥ 0, Pn
Prove that Pn
Some artists seek asymmetry. Physicists tell us the universe doesn’t

5.29
5.30
always appear symmetric. Why are we unfazed?
5.31 With a colleague, implement a fresh version of Lists. First, agree on
the types and names of private ﬁelds. Then, going down the list of methods
required by the List interface, split methods to be implemented between you
by assigning every other method to your colleague. Bring the code together
and compile it. What types of bugs occur? Did you depend on your colleague’s
code?

− 1.

c−1

Consider the implementation of a Ratio data type. How does symmetry

5.32
appear in this implementation?

In the Vector class, we extend by doubling, but we never discuss re-

5.33
ducing by a similar technique. What is a good strategy?
5.34

Consider the following Java method:

static public int fido(int n)
// pre: n is a positive integer
// post: result is the nth term from the sequence
//
{

1, 3, 7, 15, 31, 63, 127, ...

int result = 1;
if (n > 1) result = 1+fido(n-1)+fido(n-1);
// assertion: the above if condition was tested
//
fido(n) times while computing result
return result;

}

a. What does it compute?

b. Prove or disprove the informal assertion following the if statement.

c. What is the time complexity of the method?

d. Why is fido an appropriate name for this method?

5.5 Laboratory: How Fast Is Java?

Objective. To develop an appreciation for the speed of basic Java operations
including assignment of value to variables, arrays, and Vectors.

Discussion. How long does it take to add two integers in Java? How long does
it take to assign a value to an entry in an array? The answers to these questions
depend heavily on the type of environment a programmer is using and yet play
an important role in evaluating the trade-offs in performance between different
implementations of data structures.

If we are interested in estimating the time associated with an operation,
it is difﬁcult to measure it accurately with clocks available on most modern
machines. If an operation takes 100 ns (nanoseconds, or billionths of a second),
10,000 of these operations can be performed within a single millisecond clock
tick. It is unlikely that we would see a change in the millisecond clock while the
operation is being performed.

One approach is to measure, say, the time it takes to perform a million of
these operations, and divide that portion of the time associated with the opera-
tion by a million. The result can be a very accurate measurement of the time it
takes to perform the operation. Several important things must be kept in mind:

• Different runs of the experiment can generate different times. This vari-
ation is unlikely to be due to signiﬁcant differences in the speed of the
operation, but instead to various interruptions that regularly occur when
a program is running. Instead of computing the average of the running
times, it is best to compute the minimum of the experiment’s elapsed
times. It’s unlikely that this is much of an underestimate!

• Never perform input or output while you are timing an experiment. These
operations are very expensive and variable. When reading or writing,
make sure these operations appear before or after the experiment being
timed.

• On modern systems there are many things happening concurrently with
your program. Clocks tick forward, printer queues manage printers, net-
work cards are accepting viruses. If you can keep your total experiment
time below, say, a tenth of a second, it is likely that you will eliminate
many of these distractions.

• The process of repeating an operation takes time. One of our tasks will be
to measure the time it takes to execute an empty for loop. The loop, of
course, is not really empty: it performs a test at the top of the loop and an
increment at the bottom. Failing to account for the overhead of a for loop
makes it impossible to measure any operation that is signiﬁcantly faster.

• Good compilers can recognize certain operations that can be performed
more efﬁciently in a different way. For example, traditional computers

116

Design Fundamentals

can assign a value of 0 much faster than the assignment of a value of 42.
If an experiment yields an unexpectedly short operation time, change the
Java to obscure any easy optimizations that may be performed. Don’t
forget to subtract the overhead of these obscuring operations!

Keeping a mindful eye on your experimental data will allow you to effectively
measure very, very short events accurate to nanoseconds. In one nanosecond,
light travels 11.80 inches!
Procedure. The ultimate goal of this experiment is a formally written lab report
presenting your results. Carefully design your experiment, and be prepared to
defend your approach. The data you collect here is experimental, and necessar-
ily involves error. To reduce the errors described above, perform multiple runs
of each experiment, and carefully document your ﬁndings. Your report should
include results from the following experiments:

1. A description of the machine you are using. Make sure you use this ma-

chine for all of your experiments.

2. Write a short program to measure the time that elapses, say, when an
empty for loop counts to one million. Print out the elapsed time, as well
as the per-iteration elapsed time. Adjust the number of loops so that the
total elapsed time falls between, say, one-hundredth and one-tenth of a
second.

Recall that we can measure times in nanoseconds (as accurately as possi-
ble, given your machine) using System.nanoTime():

int i, loops;
double speed;
loops = 10000000;
long start,stop,duration;

start = System.nanoTime();
for (i = 0; i < loops; i++)
{

// code to be timed goes here

}
stop = System.nanoTime();

duration = stop-start;
System.out.println("# Elapsed time: "+duration+"ns");
System.out.println("# Mean time: "+

(((double)duration)/loops)+
"nanoseconds");

3. Measure the time it takes to do a single integer assignment (e.g., i=42;).
Do not forget to subtract the time associated with executing the for loop.

4. Measure the time it takes to assign an integer to an array entry. Make sure

that the array has been allocated before starting the timing loop.

5.5 Laboratory: How Fast Is Java?

117

5. Measure the time it takes to assign a String reference to an array.

6. Measure the length of time it takes to assign a String to a Vector. (Note

that it is not possible to directly assign an int to a Vector class.)

7. Copy one Vector to another, manually, using set. Carefully watch the
elapsed time and do not include the time it takes to construct the two
Vectors! Measure the time it takes to perform the copy for Vectors of
different lengths. Does this appear to grow linearly?

Formally present your results in a write-up of your experiments.

Thought Questions. Consider the following questions as you complete the lab:

1. Your Java compiler and environment may have several switches that affect
the performance of your program. For example, some environments allow
the use of just-in-time (jit) compilers, that compile frequently used pieces
of code (like your timing loop) into machine-speciﬁc instructions that are
likely to execute faster. How does this affect your experiment?

2. How might you automatically guarantee that your total experiment time

lasts between, say, 10 and 100 milliseconds?

3. It is, of course, possible for a timer to underestimate the running time of
an instruction. For example, if you time a single assignment, it is certainly
possible to get an elapsed time of 0—an impossibility. To what extent
would a timing underestimate affect your results?

Notes:

Chapter 6

Sorting

Concepts:
. Natural sorting techniques
. Recursive sorting techniques
.
. Use of compareTo methods
.
Comparator techniques

Vector sorting

“Come along, children. Follow me.”
Before you could wink an eyelash
Jack, Knak, Lack, Mack,
Nack, Ouack, Pack, and Quack
fell into line, just as they had been taught.
—Robert McCloskey

COMPUTERS SPEND A CONSIDERABLE AMOUNT of their time keeping data in or-
der. When we view a directory or folder, the items are sorted by name or type
or modiﬁcation date. When we search the Web, the results are returned sorted
by “applicability.” At the end of the month, our checks come back from the bank
sorted by number, and our deposits are sorted by date. Clearly, in the grand
scheme of things, sorting is an important function of computers. Not surpris-
ingly, data structures can play a signiﬁcant role in making sorts run quickly. This
chapter begins an investigation of sorting methods.

6.1 Approaching the Problem

For the moment we assume that we will be sorting an unordered array of in-
tegers (see Figure 6.1a).1 The problem is to arrange the integers so that every
adjacent pair of values is in the correct order (see Figure 6.1b). A simple tech-
nique to sort the array is to pass through the array from left to right, swapping
adjacent values that are out of order (see Figure 6.2). The exchange of values
is accomplished with a utility method:

public static void swap(int data[], int i, int j)
// pre: 0 <= i,j < data.length
// post: data[i] and data[j] are exchanged
{

int temp;

BubbleSort

1 We focus on arrays of integers to maintain a simple approach. These techniques, of course, can
be applied to vectors of objects, provided that some relative comparison can be made between two
elements. This is discussed in Section 6.7.

120

Sorting

Figure 6.1

The relations between entries in unordered and sorted arrays of integers.

temp = data[i];
data[i] = data[j];
data[j] = temp;

}

After a single pass the largest value will end up “bubbling” up to the high-
indexed side of the array. The next pass will, at least, bubble up the next largest
value, and so forth. The sort—called bubble sort—must be ﬁnished after n − 1
passes. Here is how we might write bubble sort in Java:

public static void bubbleSort(int data[], int n)
// pre: 0 <= n <= data.length
// post: values in data[0..n-1] in ascending order
{

int numSorted = 0;
int index;
while (numSorted < n)
{

// number of values in order
// general index

// bubble a large element to higher array index
for (index = 1; index < n-numSorted; index++)
{

if (data[index-1] > data[index])
swap(data,index-1,index);

}
// at least one more value in place
numSorted++;

}

}

Observe that the only potentially time-consuming operations that occur in this
sort are comparisons and exchanges. While the cost of comparing integers is rel-

(b) Sorted1435842401234567891011301234567891011−10123344042435865402650−13(a) Unordered6.1 Approaching the Problem

121

Figure 6.2
Shaded values are in sorted order. A pass with no exchanges indicates sorted data.

The passes of bubble sort: hops indicate “bubbling up” of large values.

4024365−158130342421403430−15834246523400−1433458656565656565656565585858585858581421230−1404342431230−134042443434343434342424242422−1334044040404058584343424240406544444433333333333322222200000101−11−11−11−101−11−1BubbleDetectable finish122

Sorting

atively small, if each element of the array were to contain a long string (for ex-
ample, a DNA sequence) or a complex object (for example, a Library of Congress
entry), then the comparison of two values might be a computationally intensive
operation. Similarly, the cost of performing an exchange is to be avoided.2 We
can, therefore, restrict our attention to the number of comparison and exchange
operations that occur in sorts in order to adequately evaluate their performance.
In bubble sort each pass of the bubbling phase performs n − 1 comparisons
and as many as n − 1 exchanges. Thus the worst-case cost of performing bubble
sort is O((n−1)2) or O(n2) operations. In the best case, none of the comparisons
leads to an exchange. Even then, though, the algorithm has quadratic behavior.3
Most of us are inefﬁcient sorters. Anyone having to sort a deck of cards or a
stack of checks is familiar with the feeling that there must be a better way to do
this. As we shall see, there probably is: most common sorting techniques used in
day-to-day life run in O(n2) time, whereas the best single processor comparison-
based sorting techniques are expected to run in only O(n log n) time. (If multi-
ple processors are used, we can reduce this to O(log n) time, but that algorithm
is beyond the scope of this text.) We shall investigate two sorting techniques
that run in O(n2) time, on average, and two that run in O(n log n) time. In the
end we will attempt to understand what makes the successful sorts successful.

Our ﬁrst two sorting techniques are based on natural analogies.

6.2 Selection Sort

Children are perhaps the greatest advocates of selection sort. Every October,
Halloween candies are consumed from best to worst. Whether daily sampling
is limited or not, it is clear that choices of the next treat consumed are based
on “the next biggest piece” or “the next-most favorite,” and so on. Children
consume treats in decreasing order of acceptability. Similarly, when we select
plants from a greenhouse, check produce in the store, or pick strawberries from
the farm we seek the best items ﬁrst.

This selection process can be applied to an array of integers. Our goal is
to identify the index of the largest element of the array. We begin by assuming
that the ﬁrst element is the largest, and then form a competition among all the
remaining values. As we come across larger values, we update the index of the
current maximum value. In the end, the index must point to the largest value.
This code is idiomatic, so we isolate it here:

int index;
int max;
// determine maximum value in array

// general index
// index of largest value

SelectionSort

2 In languages like Java, where large objects are manipulated through references, the cost of an
exchange is usually fairly trivial. In many languages, however, the cost of exchanging large values
stored directly in the array is a real concern.
3 If, as we noted in Figure 6.2, we detected the lack of exchanges, bubble sort would run in O(n)
time on data that were already sorted. Still, the average case would be quadratic.

6.2 Selection Sort

123

max = 0;
for (index = 1; index < numUnsorted; index++)
{

if (data[max] < data[index]) max = index;

}

(Notice that the maximum is not updated unless a larger value is found.) Now,
consider where this maximum value would be found if the data were sorted:
it should be clear to the right, in the highest indexed location. This is easily
accomplished: we simply swap the last element of the unordered array with the
maximum. Once this swap is completed, we know that at least that one value is
in the correct location, and we logically reduce the size of the problem by one.
If we remove the n − 1 largest values in successive passes (see Figure 6.3), we
have selection sort. Here is how the entire method appears in Java:

public static void selectionSort(int data[], int n)
// pre: 0 <= n <= data.length
// post: values in data[0..n-1] are in ascending order
{

int numUnsorted = n;
int index;
int max;
while (numUnsorted > 0)
{

// general index
// index of largest value

// determine maximum value in array
max = 0;
for (index = 1; index < numUnsorted; index++)
{

if (data[max] < data[index]) max = index;

}
swap(data,max,numUnsorted-1);
numUnsorted--;

}

}

Selection sort potentially performs far fewer exchanges than bubble sort: se-
lection sort performs exactly one per pass, while bubble sort performs as many
as n − 1. Like bubble sort, however, selection sort demands O(n2) time for
comparisons.

Interestingly, the performance of selection sort is independent of the order
of the data: if the data are already sorted, it takes selection sort just as long to
sort as if the data were unsorted. We can improve on this behavior through a
slightly different analogy.

124

Sorting

Proﬁle of the passes of selection sort: shaded values are sorted. Circled
Figure 6.3
values are maximum among unsorted values and are moved to the low end of sorted
values on each pass.

4021433650−1583424406558434240434343656565656558585842210−1656565656558585858585843434343434342424242424240404040404044444333333322140404040−1−1−1−1−1222222221111111114343333333444440000005842423333330042−1−1−1−1−10303Select & exchange6.3 Insertion Sort

6.3 Insertion Sort

Card players, when collecting a hand, often consider cards one at a time, in-
serting each into its sorted location. If we consider the “hand” to be the sorted
portion of the array, and the “table” to be the unsorted portion, we develop a
new sorting technique called insertion sort.

In the following Java implementation of insertion sort, the sorted values are
kept in the low end of the array, and the unsorted values are found at the high
end (see Figure 6.4). The algorithm consists of several “passes” of inserting the
lowest-indexed unsorted value into the list of sorted values. Once this is done,
of course, the list of sorted values increases by one. This process continues until
each of the unsorted values has been incorporated into the sorted portion of the
array. Here is the code:

125

InsertionSort

public static void insertionSort(int data[], int n)
// pre: 0 <= n <= data.length
// post: values in data[0..n-1] are in ascending order
{

int numSorted = 1;
int index;
while (numSorted < n)
{

// number of values in place
// general index

// take the first unsorted value
int temp = data[numSorted];
// ...and insert it among the sorted:
for (index = numSorted; index > 0; index--)
{

if (temp < data[index-1])
{

data[index] = data[index-1];

} else {

break;

}

}
// reinsert value
data[index] = temp;
numSorted++;

}

}

A total of n − 1 passes are made over the array, with a new unsorted value
inserted each time. The value inserted may not be a new minimum or maximum
value. Indeed, if the array was initially unordered, the value will, on average,
end up near the middle of the previously sorted values. On random data the
running time of insertion sort is expected to be dominated by O(n2) compares
and data movements (most of the compares will lead to the movement of a data
value).

If the array is initially in order, one compare is needed at every pass to
determine that the value is already in the correct location. Thus, the inner loop

126

Sorting

Figure 6.4
values. Circled values are successively inserted into the hand.

Proﬁle of the passes of insertion sort: shaded values form a “hand” of sorted

12403650−1583424434024021433650−158342412403650−1583424431240436565656565656565444444442424242424233333585858−1−1−1005830−1−1−1−1−10000011111112222222333333333344040404040404043434343434343585858584242Insert1433650−15834246.4 Mergesort

127

is executed exactly once for each of n − 1 passes. The best-case running time
performance of the sort is therefore dominated by O(n) comparisons (there are
no movements of data within the array). Because of this characteristic, insertion
sort is often used when data are very nearly ordered (imagine sorting a phone
book after a month of new customers has been appended).

In contrast, if the array was previously in reverse order, the value must be
compared with every sorted value to ﬁnd the correct location. As the compar-
isons are made, the larger values are moved to the right to make room for the
new value. The result is that each of O(n2) compares leads to a data movement,
and the worst-case running time of the algorithm is O(n2).

Note that each of these sorts uses a linear number of data cells. Not every

sorting technique is able to live within this constraint.

6.4 Mergesort

Suppose that two friends are to sort an array of values. One approach might be
to divide the deck in half. Each person then sorts one of two half-decks. The
sorted deck is then easily constructed by combining the two sorted half-decks.
This careful interleaving of sorted values is called a merge.

It is straightforward to see that a merge takes at least O(n) time, because
every value has to be moved into the destination deck. Still, within n − 1 com-
parisons, the merge must be ﬁnished. Since each of the n − 1 comparisons
(and potential movements of data) takes at most constant time, the merge is no
worse than linear.

There are, of course, some tricky aspects to the merge operation—for exam-
ple, it is possible that all the cards in one half-deck are smaller than all the cards
in the other. Still, the performance of the following merge code is O(n):

private static void merge(int data[], int temp[],

int low, int middle, int high)

// pre: data[middle..high] are ascending
//
// post: data[low..high] contains all values in ascending order
{

temp[low..middle-1] are ascending

int ri = low; // result index
int ti = low; // temp index
int di = middle; // destination index
// while two lists are not empty merge smaller value
while (ti < middle && di <= high)
{

if (data[di] < temp[ti]) {

data[ri++] = data[di++]; // smaller is in high data

} else {

data[ri++] = temp[ti++]; // smaller is in temp

}

}
// possibly some values left in temp array

MergeSort

128

Sorting

while (ti < middle)
{

data[ri++] = temp[ti++];

}
// ...or some values left (in correct place) in data array

}

This code is fairly general, but a little tricky to understand (see Figure 6.5). We
assume that the data from the two lists are located in the two arrays—in the
lower half of the range in temp and in the upper half of the range in data (see
Figure 6.5a). The ﬁrst loop compares the ﬁrst remaining element of each list to
determine which should be copied over to the result list ﬁrst (Figure 6.5b). That
loop continues until one list is emptied (Figure 6.5c). If data is the emptied list,
the remainder of the temp list is transferred (Figure 6.5d). If the temp list was
emptied, the remainder of the data list is already located in the correct place!

Returning to our two friends, we note that before the two lists are merged
each of the two friends is faced with sorting half the cards. How should this be
done? If a deck contains fewer than two cards, it’s already sorted. Otherwise,
each person could recursively hand off half of his or her respective deck (now
one-fourth of the entire deck) to a new individual. Once these small sorts are
ﬁnished, the quarter decks are merged, ﬁnishing the sort of the half decks, and
the two half decks are merged to construct a completely sorted deck. Thus,
we might consider a new sort, called mergesort, that recursively splits, sorts,
and reconstructs, through merging, a deck of cards. The logical “phases” of
mergesort are depicted in Figure 6.6.

private static void mergeSortRecursive(int data[],
int temp[],
int low, int high)

// pre: 0 <= low <= high < data.length
// post: values in data[low..high] are in ascending order
{

int n = high-low+1;
int middle = low + n/2;
int i;

if (n < 2) return;
// move lower half of data into temporary storage
for (i = low; i < middle; i++)
{

temp[i] = data[i];

}
// sort lower half of array
mergeSortRecursive(temp,data,low,middle-1);
// sort upper half of array
mergeSortRecursive(data,temp,middle,high);
// merge halves together
merge(data,temp,low,middle,high);

}

6.4 Mergesort

129

Four stages of a merge of two six element lists (shaded entries are partic-
Figure 6.5
ipating values): (a) the initial location of data; (b) the merge of several values; (c) the
point at which a list is emptied; and (d) the ﬁnal result.

datadataditiriritididitiriritiditempdatatempdatatemptemp3456789101112130120123456789101112131214511125678910111213032345(a)(b)(d)(c)6789101112130125−15842034265314340−1012334404243586565−101233440424358584234654340−1012361283600130

Sorting

Figure 6.6
then recursively merged into ascending order.

Proﬁle of mergesort: values are recursively split into unsorted lists that are

Note that this sort requires a temporary array to perform the merging. This
temporary array is only used by a single merge at a time, so it is allocated once
and garbage collected after the sort. We hide this detail with a public wrapper
procedure that allocates the array and calls the recursive sort:

public static void mergeSort(int data[], int n)
// pre: 0 <= n <= data.length
// post: values in data[0..n-1] are in ascending order
{

mergeSortRecursive(data,new int[n],0,n-1);

}

Clearly, the depth of the splitting is determined by the number of times that n
can be divided in two and still have a value of 1 or greater: log2 n. At each
It
level of splitting, every value must be merged into its respective subarray.
follows that at each logical level, there are O(n) compares over all the merges.
Since there are log2 n levels, we have O(n · log n) units of work in performing a
mergesort.

Mergesort is a common technique for sorting large sets of data that do not
ﬁt completely in fast memory. Instead, the data are saved in temporary ﬁles that
are merged together. When the recursion splits the collection into subsets of a
manageable size, they can be sorted using other techniques, if desired.

Merge4021433650−15834244021433650−15834244021433650−15834244021433650−1583424365−15840433650−158326558326542134421212424404334421314340−103458655834342−1400420−1Split6.5 Quicksort

131

One of the unappealing aspects of mergesort is that it is difﬁcult to merge
two lists without signiﬁcant extra memory. If we could avoid the use of this extra
space without signiﬁcant increases in the number of comparisons or data move-
ments, then we would have an excellent sorting technique. Our next method
demonstrates an O(n log n) method that requires signiﬁcantly less space.

6.5 Quicksort

Since the process of sorting numbers consists of moving each value to its ul-
timate location in the sorted array, we might make some progress toward a
solution if we could move a single value to its ultimate location. This idea forms
the basis of a fast sorting technique called quicksort.

One way to ﬁnd the correct location of, say, the leftmost value—called a
pivot—in an unsorted array is to rearrange the values so that all the smaller
values appear to the left of the pivot, and all the larger values appear to the
right. One method of partitioning the data is shown here. It returns the ﬁnal
location for what was originally the leftmost value:

private static int partition(int data[], int left, int right)
// pre: left <= right
// post: data[left] placed in the correct (returned) location
{

while (true)
{

// move right "pointer" toward left
while (left < right && data[left] < data[right]) right--;
if (left < right) swap(data,left++,right);
else return left;
// move left pointer toward right
while (left < right && data[left] < data[right]) left++;
if (left < right) swap(data,left,right--);
else return right;

}

}

The indices left and right start at the two ends of the array (see Figure 6.7)
and move toward each other until they coincide. The pivot value, being leftmost
in the array, is indexed by left. Everything to the left of left is smaller than the
pivot, while everything to the right of right is larger. Each step of the main loop
compares the left and right values and, if they’re out of order, exchanges them.
Every time an exchange occurs the index (left or right) that references the
pivot value is alternated. In any case, the nonpivot variable is moved toward
the other. Since, at each step, left and right move one step closer to each
other, within n steps, left and right are equal, and they point to the current
location of the pivot value. Since only smaller values are to the left of the pivot,
and larger values are to the right, the pivot must be located in its ﬁnal location.
Values correctly located are shaded in Figure 6.8.

QuickSort

132

Sorting

The partitioning of an array’s values based on the (shaded) pivot value 40.
Figure 6.7
Snapshots depict the state of the data after the if statements of the partition method.

rightrightleftleft21433650−158342404213650−15834244043213650−158424433402130−15842443365402130424433655840−12130424433655840−121433650−1583424406.5 Quicksort

133

Proﬁle of quicksort: leftmost value (the circled pivot) is used to position
Figure 6.8
value in ﬁnal location (indicated by shaded) and partition array into relatively smaller
and larger values. Recursive application of partitioning leads to quicksort.

Because the pivot segregates the larger and smaller values, we know that
none of these values will appear on the opposite side of the pivot in the ﬁnal
arrangement. This suggests that we can reduce the sorting of a problem of size
n to two problems of size approximately n
2 . To ﬁnish the sort, we need only
recursively sort the values to the left and right of the pivot:

public static void quickSort(int data[], int n)
// post: the values in data[0..n-1] are in ascending order
{

quickSortRecursive(data,0,n-1);

}

private static void quickSortRecursive(int data[],int left,int right)
// pre: left <= right
// post: data[left..right] in ascending order
{

// the final location of the leftmost value

int pivot;
if (left >= right) return;
pivot = partition(data,left,right);
quickSortRecursive(data,left,pivot-1); /* 2 - sort small */
quickSortRecursive(data,pivot+1,right);/* 3 - sort large */
/* done! */

/* 1 - place pivot */

}

partition404358654058434265132043−11358424402433650−1−10−11323−104434242436558651332323233133242−101234342134

Sorting

In practice, of course, the splitting of the values is not always optimal (see the
placement of the value 4 in Figure 6.8), but a careful analysis suggests that even
with these “tough breaks” quicksort takes only O(n log n) time.

When either sorted or reverse-sorted data are to be sorted by quicksort, the
results are disappointing. This is because the pivot value selected (here, the
leftmost value) ﬁnds its ultimate location at one end of the array or the other.
This reduces the sort of n values to n − 1 values (and not n/2), and the sort
requires O(n) passes of an O(n) step partition. The result is an O(n2) sort.
Since nearly sorted data are fairly common, this result is to be avoided.

Notice that picking the leftmost value is not special. If, instead, we attempt
to ﬁnd the correct location for the middle value, then other arrangements of
data will cause the degenerate behavior. In short, for any speciﬁc or determin-
istic partitioning technique, a degenerate arrangement exists. The key to more
consistent performance, then, is a nondeterministic partitioning that correctly
places a value selected at random (see Problem 6.15). There is, of course, a
very unlikely chance that the data are in order and the positions selected in-
duce a degenerate behavior, but that chance is small and successive runs of
the sorting algorithm on the same data are exceedingly unlikely to exhibit the
same behavior. So, although the worst-case behavior is still O(n2), its expected
behavior is O(n log n).

Quicksort is an excellent sort when data are to be sorted with little extra
space. Because the speed of partitioning depends on the random access nature
of arrays or Vectors, quicksort is not suitable when not used with random access
structures. In these cases, however, other fast sorts are often possible.

6.6 Radix Sort

After investigating a number of algorithms that sort in O(n2) or O(n log n) time,
one might wonder if it is possible to sort in linear time. If the right conditions
hold, we can sort certain types of data in linear time. First, we must investigate
a pseudogame, 52 pickup!

Suppose we drop a deck of 52 cards on the ﬂoor, and we want to not only
pick them up, but we wish to sort them at the same time.
It might be most
natural to use an insertion sort: we keep a pile of sorted cards and, as we pick
up new cards, we insert them in the deck in the appropriate position. A more
efﬁcient approach makes use of the fact that we know what the sorted deck
looks like. We simply lay out the cards in a row, with each position in the row
reserved for the particular card. As we pick up a card, we place it in its reserved
location. In the end, all the cards are in their correct location and we collect
them from left to right.

Exercise 6.1 Explain why this sorting technique always takes O(n) time for a
deck of n cards.

Such an approach is the basis for a general sorting technique called bucket sort.
By quickly inspecting values (perhaps a word) we can approximately sort them

6.6 Radix Sort

135

into different buckets (perhaps based on the ﬁrst letter). In a subsequent pass
we can sort the values in each bucket with, perhaps a different sort. The buck-
ets of sorted values are then accumulated, carefully maintaining the order of
the buckets, and the result is completely sorted. Unfortunately, the worst-case
behavior of this sorting technique is determined by the performance of the al-
gorithm we use to sort each bucket of values.

Exercise 6.2 Suppose we have n values and m buckets and we use insertion sort
to perform the sort of each bucket. What is the worst-case time complexity of this
sort?

Such a technique can be used to sort integers, especially if we can partially sort
the values based on a single digit. For example, we might develop a support
function, digit, that, given a number n and a decimal place d, returns the
value of the digit in the particular decimal place. If d was 0, it would return the
units digit of n. Here is a recursive implementation:

public static int digit(int n, int d)
// pre: n >= 0 and d >= 0
// post: returns the value of the dth decimal place of n
//
{

where the units place has position 0

if (d == 0) return n % 10;
else return digit(n/10,d-1);

}

RadixSort

Here is the code for placing an array of integer values among 10 buckets, based
on the value of digit d. For example, if we have numbers between 1 and 52 and
we set d to 2, this code almost sorts the values based on their 10’s digit.

public static void bucketPass(int data[], int d)
// pre: data is an array of data values, and d >= 0
// post: data is sorted by the digit found in location d;
// if two values have the same digit in location d, their
// relative positions do not change; i.e., they are not swapped
{

int i,j;
int value;
// allocate some buckets
Vector<Vector<Integer>> bucket = new Vector<Vector<Integer>>(10);
bucket.setSize(10);
// allocate Vectors to hold values in each bucket
for (j = 0; j < 10; j++) bucket.set(j,new Vector<Integer>());
// distribute the data among buckets
int n = data.length;
for (i = 0; i < n; i++)
{

value = data[i];
// determine the d’th digit of value

136

Sorting

j = digit(value,d);
// add data value to end of vector; keeps values in order
bucket.get(j).add(value);

}
// collect data from buckets back into array
// collect in reverse order to unload Vectors
// in linear time
i = n;
for (j = 9; j >= 0; j--)
{

// unload all values in bucket j
while (!bucket.get(j).isEmpty())
{

i--;
value = bucket.get(j).remove();
data[i] = value;

}

}

}

We now have the tools to support a new sort, radix sort. The approach is to
use the bucketPass code to sort all the values based on the units place. Next,
all the values are sorted based on their 10’s digit. The process continues until
enough passes have been made to consider all the digits.
If it is known that
values are bounded above, then we can also bound the number of passes as
well. Here is the code to perform a radix sort of values under 1 million (six
passes):

public static void radixSort(int data[])
// pre: data is array of values; each is less than 1,000,000
// post: data in the array are sorted into increasing order
{

for (int i = 0; i < 6; i++)
{

bucketPass(data,i);

}

}

After the ﬁrst bucketPass, the values are ordered, based on their units digit. All
values that end in 0 are placed near the front of data (see Figure 6.9), all the
values that end in 9 appear near the end. Among those values that end in 0, the
values appear in the order they originally appeared in the array. In this regard,
we can say that bucketPass is a stable sorting technique. All other things being
equal, the values appear in their original order.

During the second pass, the values are sorted, based on their 10’s digit.
Again, if two values have the same 10’s digit, the relative order of the values is
maintained. That means, for example, that 140 will appear before 42, because
after the ﬁrst pass, the 140 appeared before the 42. The process continues, until

6.6 Radix Sort

137

The state of the data array between the six passes of radixSort. The
Figure 6.9
boundaries of the buckets are identiﬁed by vertical lines; bold lines indicate empty buck-
ets. Since, on every pass, paths of incoming values to a bucket do not cross, the sort
is stable. Notice that after three passes, the radixSort is ﬁnished. The same would be
true, no matter the number of values, as long as they were all under 1000.

00012StartFinishDigit 0Digit 1Digit 2Digit 3Digit 4Digit 634580132424652143365058342111114014003341114042435865123341142435865140123341142435865140123341142435865140012341142435865140433138

Sorting

all digits are considered. Here, six passes are performed, but only three are
necessary (see Problem 6.9).

There are several important things to remember about the construction of
this sort. First, bucketPass is stable. That condition is necessary if the work
of previous passes is not to be undone. Secondly, the sort is unlikely to work if
the passes are performed from the most signiﬁcant digit toward the units digit.
Finally, since the number of passes is independent of the size of the data array,
the speed of the entire sort is proportional to the speed of a single pass. Careful
design allows the bucketPass to be accomplished in O(n) time. We see, then,
that radixSort is a O(n) sorting method.

While, theoretically, radixSort can be accomplished in linear time, practi-
cally, the constant of proportionality associated with the bound is large com-
pared to the other sorts we have seen in this chapter. In practice, radixSort is
inefﬁcient compared to most other sorts.

6.7 Sorting Objects

Sorting arrays of integers is suitable for understanding the performance of vari-
ous sorts, but it is hardly a real-world problem. Frequently, the object that needs
to be sorted is an Object with many ﬁelds, only some of which are actually used
in making a comparison.

Let’s consider the problem of sorting the entries associated with an electronic
phone book. The ﬁrst step is to identify the structure of a single entry in the
phone book. Perhaps it has the following form:

PhoneBook

class PhoneEntry
{

String name;
String title;
int extension;
int room;
String building;

// person’s name
// person’s title
// telephone number
// number of room
// office building

public PhoneEntry(String n, String t, int e,

String b, int r)

// post: construct a new phone entry
{

...

}

public int compareTo(PhoneEntry other)
// pre: other is non-null
// post: returns integer representing relation between values
{

return this.extension - other.extension;

}

}

6.7 Sorting Objects

139

Figure 6.10 An array of phone entries for the 107th Congressional Delegation from
Oregon State, before and after sorting by telephone (shaded).

We have added the compareTo method to describe the relation between two
entries in the phone book (the shaded ﬁelds of Figure 6.10). The compareTo
method returns an integer that is less than, equal to, or greater than 0 when
this is logically less than, equal to, or greater than other. We can now modify
any of the sort techniques provided in the previous section to sort an array of
phone entries:

public static void insertionSort(PhoneEntry data[], int n)
// pre: n <= data.length
// post: values in data[0..n-1] are in ascending order
{

int numSorted = 1;
int index;
while (numSorted < n)
{

// number of values in place
// general index

// take the first unsorted value
PhoneEntry temp = data[numSorted];
// ...and insert it among the sorted:

Rep.DeFazio, Peter564162134RayburnHooley, DarleneRep.557111130LongworthRep.Walden, Greg567301404LongworthRep.LongworthWu, David508551023SenatorWyden, Ron45244516HartSmith, GordonSenator43753404RussellSmith, GordonSenator43753404RussellSenatorWyden, Ron45244516HartRep.LongworthWu, David508551023Rep.Blumenauer, Earl548811406LongworthHooley, DarleneRep.557111130LongworthRep.DeFazio, Peter564162134RayburnRep.Walden, Greg567301404LongworthData before sortingData after sorting by telephone01345620134562Rep.Blumenauer, Earl548811406Longworth140

Sorting

for (index = numSorted; index > 0; index--)
{

if (temp.compareTo(data[index-1]) < 0)
{

data[index] = data[index-1];

} else {

break;

}

}
// reinsert value
data[index] = temp;
numSorted++;

}

}

Careful review of this insertion sort routine shows that all the < operators have
been replaced by checks for negative compareTo values. The result is that the
phone entries in the array are ordered by increasing phone number.

If two or more people use the same extension, then the order of the resulting
entries depends on the stability of the sort. If the sort is stable, then the relative
order of the phone entries with identical extensions in the sorted array is the
same as their relative order in the unordered array. If the sort is not stable, no
guarantee can be made. To ensure that entries are, say, sorted in increasing
order by extension and, in case of shared phones, sorted by increasing name,
the following compareTo method might be used:

public int compareTo(PhoneEntry other)
// pre: other is non-null
// post: returns integer representing relation between values
{

if (this.extension != other.extension)

return this.extension - other.extension;
else return this.name.compareTo(other.name);

}

Correctly specifying the relation between two objects with the compareTo meth-
od can be difﬁcult when the objects cannot be totally ordered. Is it always possi-
ble that one athletic team is strictly less than another? Is it always the case that
one set contains another? No. These are examples of domains that are partially
ordered. Usually, however, most types may be totally ordered, and imagining
how one might sort a collection of objects forces a suitable relation between
any pair.

6.8 Ordering Objects Using Comparators

The deﬁnition of the compareTo method for an object should deﬁne, in a sense,
the natural ordering of the objects. So, for example, in the case of a phone

6.8 Ordering Objects Using Comparators

141

book, the entries would ideally be ordered based on the name associated with
the entry. Sometimes, however, the compareTo method does not provide the
ordering desired, or worse, the compareTo method has not been deﬁned for an
object. In these cases, the programmer turns to a simple method for speciﬁng an
external comparison method called a comparator. A comparator is an object that
contains a method that is capable of comparing two objects. Sorting methods,
then, can be developed to apply a comparator to two objects when a comparison
is to be performed. The beauty of this mechanism is that different comparators
can be applied to the same data to sort in different orders or on different keys.
In Java a comparator is any class that implements the java.util.Comparator
interface. This interface provides the following method:

package java.util;
public interface Comparator
{

public abstract int compare(Object a, Object b);
// pre: a and b are valid objects, likely of similar type
// post: returns a value less than, equal to, or greater than 0
if a is less than, equal to, or greater than b
//

Comparator

}

Like the compareTo method we have seen earlier, the compare method re-
turns an integer that identiﬁes the relationship between two values. Unlike
the compareTo method, however, the compare method is not associated with
the compared objects. As a result, the comparator is not privy to the implemen-
tation of the objects; it must perform the comparison based on information that
is gained from accessor methods.

As an example of the implementation of a Comparator, we consider the
implementation of a case-insensitive comparison of Strings, called Caseless-
Comparator. This comparison method converts both String objects to upper-
case and then performs the standard String comparison:

public class CaselessComparator implements java.util.Comparator<String>
{

public int compare(String a, String b)
// pre: a and b are valid Strings
// post: returns a value less than, equal to, or greater than 0
//
//
{

if a is less than, equal to, or greater than b, without
consideration of case

Caseless-
Comparator

String upperA = ((String)a).toUpperCase();
String upperB = ((String)b).toUpperCase();
return upperA.compareTo(upperB);

}

}

The result of the comparison is that strings that are spelled similarly in differ-
ent cases appear together. For example, if an array contains the words of the
children’s tongue twister:

142

Sorting

Fuzzy Wuzzy was a bear.
Fuzzy Wuzzy had no hair.
Fuzzy Wuzzy wasn’t fuzzy, wuzzy?

we would expect the words to be sorted into the following order:

a bear. Fuzzy Fuzzy Fuzzy fuzzy, had hair.
no was wasn’t Wuzzy Wuzzy Wuzzy wuzzy?

This should be compared with the standard ordering of String values, which
would generate the following output:

Fuzzy Fuzzy Fuzzy Wuzzy Wuzzy Wuzzy a bear.
fuzzy, had hair. no was wasn’t wuzzy?

To use a Comparator in a sorting technique, we need only replace the use
of compareTo methods with compare methods from a Comparator. Here, for
example, is an insertion sort that makes use of a Comparator to order the values
in an array of Objects:

CompInsSort

public static <T> void insertionSort(T data[], Comparator<T> c)
// pre: c compares objects found in data
// post: values in data[0..n-1] are in ascending order
{

int numSorted = 1;
int index;
int n = data.length;
while (numSorted < n)
{

// number of values in place
// general index
// length of the array

// take the first unsorted value
T temp = data[numSorted];
// ...and insert it among the sorted:
for (index = numSorted; index > 0; index--)
{

if (c.compare(temp,data[index-1]) < 0)
{

data[index] = data[index-1];

} else {

break;

}

}
// reinsert value
data[index] = temp;
numSorted++;

}

}

Note that in this description we don’t see the particulars of the types involved.
Instead, all data are manipulated as Objects, which are speciﬁcally manipulated
by the compare method of the provided Comparator.

6.9 Vector-Based Sorting

143

6.9 Vector-Based Sorting

We extend the phone book example one more time, by allowing the PhoneEntrys
to be stored in a Vector. There are, of course, good reasons to use Vector over
arrays, but there are some added complexities that should be considered. Here
is an alternative Java implementation of insertionSort that is dedicated to the
sorting of a Vector of PhoneEntrys:

protected void swap(int i, int j)
// pre: 0 <= i,j < this.size
// post: elements i and j are exchanged within the vector
{

PhoneEntry temp;
temp = get(i);
set(i,get(j));
set(j,temp);

}

public void insertionSort()
// post: values of vector are in ascending order
{

int numSorted = 0;
int index;
while (numSorted < size())
{

// number of values in place
// general index

// take the first unsorted value
PhoneEntry temp = (PhoneEntry)get(numSorted);
// ...and insert it among the sorted:
for (index = numSorted; index > 0; index--)
{

if (temp.compareTo((PhoneEntry)get(index-1)) < 0)
{

set(index,get(index-1));

PhoneBook

} else {

break;

}

}
// reinsert value
set(index,temp);
numSorted++;

}

}

Recall that, for Vectors, we use the get method to fetch a value and set to
store. Since any type of object may be referenced by a vector entry, we verify the
type expected when a value is retrieved from the vector. This is accomplished
If the type of the fetched value doesn’t match
through a parenthesized cast.
the type of the cast, the program throws a class cast exception. Here, we cast
the result of get in the compareTo method to indicate that we are comparing
PhoneEntrys.

144

Sorting

It is unfortunate that the insertionSort has to be specially coded for use

with the PhoneEntry objects.

Exercise 6.3 Write an insertionSort that uses a Comparator to sort a Vector
of objects.

6.10 Conclusions

Sorting is an important and common process on computers. In this chapter we
considered several sorting techniques with quadratic running times. Bubble sort
approaches the problem by checking and rechecking the relationships between
elements. Selection and insertion sorts are based on techniques that people
commonly use. Of these, insertion sort is most frequently used; it is easily
coded and provides excellent performance when data are nearly sorted.

Two recursive sorting techniques, mergesort and quicksort, use recursion
to achieve O(n log n) running times, which are optimal for comparison-based
techniques on single processors. Mergesort works well in a variety of situations,
but often requires signiﬁcant extra memory. Quicksort requires a random access
structure, but runs with little space overhead. Quicksort is not a stable sort
because it has the potential to swap two values with equivalent keys.

We have seen with radix sort, it is possible to have a linear sorting algorithm,
but it cannot be based on compares. Instead, the technique involves carefully
ordering values based on looking at portions of the key. The technique is, practi-
cally, not useful for general-purpose sorting, although for many years, punched
cards were efﬁciently sorted using precisely the method described here.

Sorting is, arguably, the most frequently executed algorithm on computers
today. When we consider the notion of an ordered structure, we will ﬁnd that
algorithms and structures work hand in hand to help keep data in the correct
order.

Self Check Problems

Solutions to these problems begin on page 445.
Why does it facilitate the swap method to have a temporary reference?
6.1
Cover the numbers below with your hand. Now, moving your hand to
6.2
the right, expose each number in turn. On a separate sheet of paper, keep the
list of values you have encounted in order. At the end you have sorted all of the
values. Which sorting technique are you using?
64

296

457

998

989

-95

3.1

12

21

39

Copy the above table onto a piece of scrap paper. Start a column of
6.3
numbers: write down the smallest table value you see into your column, cross-
ing it out of the table. Continue until you have considered each of the values.
What sorting technique are you using?

6.10 Conclusions

145

During spring cleaning, you decide to sort four months of checks re-
6.4
turned with your bank statements. You decide to sort each month separately
and go from there. Is this valid? If not, why. If it is, what happens next?

A postal employee approximately sorts mail into, say, 10 piles based
6.5
on the magnitude of the street number of each address, pile 1 has 1-10, pile
2 has 11-20, etc. The letters are then collected together by increasing pile
number. She then sorts them into a delivery crate with dividers labeled with
street names. The order of streets corresponds to the order they appear on her
mail route. What type of sort is she performing?

6.6

What is the purpose of the compareTo method?

Problems

Solutions to the odd-numbered problems begin on page 464.

Show that to exchange two integer values it is not strictly necessary to
6.1
use a third, temporary integer variable. (Hint: Use addition and/or subtrac-
tion.)

We demonstrated that, in the worst case, bubble sort performs O(n2)
6.2
operations. We assumed, however, that each pass performed approximately
O(n) operations. In fact, pass i performs as many as O(n − i) operations, for
1 ≤ i ≤ n − 1. Show that bubble sort still takes O(n2) time.
6.3
cases?

How does bubbleSort (as presented) perform in the best and average

On any pass of bubble sort, if no exchanges are made, then the rela-
6.4
tions between all the values are those desired, and the sort is done. Using this
information, how fast will bubble sort run in worst, best, and average cases?

6.5

How fast does selection sort run in the best, worst, and average cases?

How fast does insertion sort run in the best, worst, and average cases?

6.6
Give examples of best- and worst-case input for insertion sort.

Running an actual program, count the number of compares needed to
6.7
sort n values using insertion sort, where n varies (e.g., powers of 2). Plot your
data. Do the same thing for quicksort. Do the curves appear as theoretically
expected? Does insertion sort ever run faster than quicksort? If so, at what
point does it run slower?

6.8
more quickly without any increase in space. Is that true?

Comparing insertion sort to quicksort, it appears that quicksort sorts

At the end of the discussion on radix sort, we pointed out that the digit
6.9
sorting passes must occur from right to left. Give an example of an array of 5
two-digit values that do not sort properly if you perform the passes left to right.

In radix sort, it might be useful to terminate the sorting process when
6.10
numbers do not change position during a call to bucketPass. Should this mod-
iﬁcation be adopted or not?

146

Sorting

Using the millisecond timer, determine the length of time it takes to
6.11
perform an assignment of a nonzero value to an int. (Hint: It will take less
than a millisecond, so you will have to design several experiments that measure
thousands or millions of assignments; see the previous lab, on page 115, for
details.)

Running an actual program, and using the millisecond timer, System.-
6.12
currentTimeMillis, measure the length of time needed to sort arrays of data
of various sizes using a sort of your choice. Repeat the experiment but use
Vectors. Is there a difference? In either case, explain why. (Hint: You may
have to develop code along the lines of Problem 6.11.)

A sort is said to be stable if the order of equal values is maintained
6.13
throughout the sort. Bubble sort is stable, because whenever two equal val-
ues are compared, no exchange occurs. Which other sorts are stable (consider
insertion sort, selection sort, mergesort, and quicksort)?

The partition function of quicksort could be changed as follows: To
6.14
place the leftmost value in the correct location, count the number of values that
are strictly less than the leftmost value. The resulting number is the correct
index for the desired value. Exchange the leftmost value for the value at the
indexed location. With all other code left as it is, does this support a correctly
functioning quicksort? If not, explain why.
6.15 Modify the partition method used by quicksort so that the pivot is
randomly selected. (Hint: Before partitioning, consider placing the randomly
selected value at the left side of the array.)
6.16 Write a recursive selectionSort algorithm. (Hint: Each level of recur-
sion positions a value in the correct location.)
6.17 Write a recursive insertionSort algorithm.
Some of the best-performing sorts depend on the best-performing shuf-
6.18
ﬂes. A good shufﬂing technique rearranges data into any arrangement with
equal probability. Design the most efﬁcient shufﬂing mechanism you can, and
argue its quality. What is its performance?
6.19 Write a program called shuffleSort. It ﬁrst checks to see if the data are
in order. If they are, the sort is ﬁnished. If they aren’t, the data are shufﬂed and
the process repeats. What is the best-case running time? Is there a worst-case
running time? Why or why not? If each time the data were shufﬂed they were
arranged in a never-seen-before conﬁguration, would this change your answer?
6.20 Write a program to sort a list of unique integers between 0 and 1 million,
but only using 1000 32-bit integers of space. The integers are read from a ﬁle.

6.11 Laboratory: Sorting with Comparators

Objective. To gain experience with Java’s java.util.Comparator interface.
Discussion. In Chapter 6 we have seen a number of different sorting techniques.
Each of the techniques demonstrated was based on the ﬁxed, natural ordering
of values found in an array. In this lab we will modify the Vector class so that it
provides a method, sort, that can be used—with the help of a Comparator—to
order the elements of the Vector in any of a number of different ways.
Procedure. Develop an extension of structure.Vector, called MyVector, that
includes a new method, sort.

Here are some steps toward implementing this new class:

1. Create a new class, MyVector, which is declared to be an extension of the
structure.Vector class. You should write a default constructor for this
class that simply calls super();. This will force the structure.Vector
constructor to be called. This, in turn, will initialize the protected ﬁelds of
the Vector class.

2. Construct a new Vector method called sort. It should have the following

declaration:

public void sort(Comparator<T> c)
// pre: c is a valid comparator
// post: sorts this vector in order determined by c

This method uses a Comparator type object to actually perform a sort of
the values in MyVector. You may use any sort that you like.

3. Write an application that reads in a data ﬁle with several ﬁelds, and, de-
pending on the Comparator used, sorts and prints the data in different
orders.

Thought Questions. Consider the following questions as you complete the lab:

1. Suppose we write the following Comparator:

import structure5.*;
import java.util.Iterator;
import java.util.Comparator;
import java.util.Scanner;
public class RevComparator<T> implements Comparator<T>
{

protected Comparator<T> base;

public RevComparator(Comparator<T> baseCompare)
{

base = baseCompare;

148

Sorting

}

public int compare(T a, T b)
{

return -base.compare(a,b);

}

}

What happens when we construct:

MyVector<Integer> v = new MyVector<Integer>();
Scanner s = new Scanner(System.in);

while (s.hasNextInt())
{

v.add(s.nextInt());

}

Comparator<Integer> c = new RevComparator<Integer>(new IntegerComparator());
v.sort(c);

2. In our examples, here, a new Comparator is necessary for each sorting
order. How might it be possible to add state information (protected
data) to the Comparator to allow it to sort in a variety of different ways?
One might imagine, for example, a method called ascending that sets the
Comparator to sort into increasing order. The descending method would
set the Comparator to sort in reverse order.

Notes:

Chapter 7

A Design Method

Concepts:
.
Signatures
.
Interface design
. Abstract base classes

But, luckily, he kept his wits and his purple crayon.
—Crockett Johnson

THROUGHOUT THE REST of this book we consider a number of data structures–
ﬁrst from an abstract viewpoint and then as more detailed implementations. In
the process of considering these implementations, we will use a simple design
method that focuses on the staged development of interfaces and abstract base
classes. Here, we outline the design method. In the ﬁrst section we describe the
process. The remaining sections are dedicated to developing several examples.

7.1 The Interface-Based Approach

As we have seen in our discussion of abstract data types, the public aspect of
the data type—that part that users depend on—is almost entirely incorporated
in the interface. In Java the interface is a formal feature of the language. Inter-
faces allow the programmer of an abstract data type to specify the signatures of
each of the externally visible methods, as well as any constants that might be
associated with implementations.

The development and adherence to the interface is the most important part
of the development of an implementation. The process can be outlined as fol-
lows:

1. Design of the interface. The interface describes the common external fea-

tures of all implementations.

2. An abstract implementation. The abstract implementation describes the

common internal features of all implementations.

3. Extension of the abstract class. Each implementation suggests an indepen-
dent approach to writing code that extends the abstract implementation
and supports the required elements of the interface.

150

A Design Method

7.1.1 Design of the Interface

The designer ﬁrst considers, in as much detail as possible, an abstract data
structure’s various internal states. For example, if we are to consider a Vector
abstractly, we pull out a napkin, and draw an abstract Vector and we consider
the various effects that Vector operations might have on its structure. In our
napkin-as-design-tool strategy, we might ﬁnd ourselves using ellipses to suggest
that we expect the Vector to be unbounded in size; we might use outward
arrows to suggest the effect of accessor methods or methods that remove values;
and we might blot out old Vector values in favor of new values when mutators
are used. The designer must develop, from these free-ﬂowing abstract notions
of a structure, a collection of precise notions of how structures are accessed and
mutated. It is also important to understand which states of the abstract structure
are valid, and how the various methods ensure that the structure moves from
one valid state to the next.

Armed with this information, the designer develops the interface—the exter-
nal description of how users of the structure can interact with it. This consists
of

1. A list of constant (final static) values that help provide meaning to
values associated with the abstract structure. For example, any models of
an atom might provide constants for the masses of electrons, protons, and
neutrons. Each of these constants is declared within the atom interface
and becomes part of any implementation.

2. Implementation-independent publicly accessible methods that access or
modify data. For example, if we described an interface for a time-of-day
clock, methods for reading and setting the current time would be declared
part of the public interface. A method that made the clock “tick,” caus-
ing time to move forward, would not be declared as part of the interface
because it would not be a feature visible to the user. From the user’s stand-
point, the clock ticks on its own. How the clock ticks is not an abstract
feature of the clock.

3. If an interface appears to be a reﬁnement of another interface, it is com-
mon practice to have the one extend the other. This is useful if the new
objects should be usable wherever the previously declared values are used.

Once the interface has been deﬁned, it can be put to immediate use. Potential
users of the class can be asked to review it.
In fact, application code can be
written to make use of the new interface. Later, if particular implementations
are developed, the application code can be put to use. It is important to remem-
ber, however, that the development of the application and the implementation
of the data structure may both proceed at the same time.

7.1 The Interface-Based Approach

151

7.1.2 Development of an Abstract Implementation

Once the interface has been outlined, it is useful for the designer to consider
those functions of the structure that are implementation independent. These
pieces are gathered together in a partial or abstract implementation of the in-
terface called an abstract base class. Since some parts of the interface may not
be implemented in the abstract base class—perhaps they require a commitment
to a particular implementation approach—it is necessary for the class to be de-
clared abstract. Once the abstract class is implemented, it may be used as a
basis for developing extensions that describe particular implementations.

Some structures may have some built-in redundancy in their public methods.
An interface supporting trigonometric calculations might, for example, have a
method tan which computes its result from sin and cos. No matter how sin
and cos are actually implemented, the tan method can be implemented in this
manner. On the other hand, if a tan function can be computed in a more di-
rect manner—one not dependent on sin and cos—a particular implementation
might override the method outlined in the abstract base class.

A similar approach is taken in Vector-like classes that implement a backward-
compatible setElementAt method based on the more recently added set method.
Such a method might appear in an abstract base class as

public void setElementAt(E obj, int index)
// pre: 0 <= index && index < size()
// post: element value is changed to obj
{

set(index,obj);

}

Vector

Because such code cannot be included in any interface that might describe a
Vector, we place the code in the abstract class.

Implementors of other abstract objects may ﬁnd it useful to develop a com-
mon library of methods that support all implementations. These methods—
often declared privately—are only made available to the implementor of the
class.
If they are utility methods (like sin and sqrt) that are not associated
with a particular object, we also declare them static.

Thus the abstract base class provides the basis for all implementations of
an interface. This includes implementation-independent code that need not be
rewritten for each implementation. Even if an abstract base class is empty, it is
useful to declare the class to facilitate the implementation of implementation-
independent development later in the life of the data structure. It is frequently
the case, for example, that code that appears in several implementations is re-
moved from the speciﬁc implementations and shared from the abstract base
class. When implementations are actually developed, they extend the associ-
ated abstract base class.

152

A Design Method

7.1.3 Implementation

When the abstract base class is ﬁnished, it is then possible for one or more im-
plementations to be developed. Usually each of these implementations extends
the work that was started in the abstract base class. For example, a Fraction
interface might be implemented as the ratio of two values (as in the Ratio
class we saw in Chapter 1), or it might be implemented as a double.
In the
latter case, the double is converted to an approximate ratio of two integers,
the numerator and denominator. In both cases, it might be useful to declare an
AbstractFraction class that includes a greatest common divisor (gcd) method.
Such a method would be declared protected and static.

7.2 Example: Development of Generators

As an example, we develop several implementations of an object that gener-
ates a sequence of values—an object we will call a generator. Such an object
might generate randomly selected values, or it might generate, incrementally, a
sequence of primes.

We imagine the interface would include the following methods:

public interface Generator
{

Generator

public void reset();
// post: the generator is reset to the beginning of the sequence;
the current value can be immediately retrieved with get.
//

public int next();
// post: returns true iff more elements are to be generated.

public int get();
// post: returns the current value of the generator.

}

The Generator is constructed and the get method can be used to get its initial
value. When necessary, the next routine generates, one at a time, a sequence of
integer values. Each call to next returns the next value in the sequence, a value
that can be recalled using the get method. If necessary, the reset method can
be used to restart the sequence of generated values.

The next step is to generate an abstract base class that implements the inter-
face. For the AbstractGenerator we implement any of the methods that can
be implemented in a general manner. We choose, here, for example, to provide
a mechanism for the next method to save the current value:

abstract public class AbstractGenerator

Abstract-
Generator

{

implements Generator

protected int current;

// the last value saved

7.2 Example: Development of Generators

153

public AbstractGenerator(int initial)
// post: initialize the current value to initial
{

current = initial;

}

public AbstractGenerator()
// post: initialize the current value to zero
{

this(0);

}

protected int set(Integer next)
// post: sets the current value to next, and extends the sequence
{

int result = current;
current = next;
return result;

}

public int get()
// post: returns the current value of the sequence
{

return current;

}

public void reset()
// post: resets the Generator (by default, does nothing)
{
}

}

The current variable keeps track of a single integer—ideally the last value gen-
erated. This value is the value returned by the get method. A hidden method—
set—allows any implementation to set the value of current.
It is expected
that this is called by the next method of the particular implementation. By pro-
viding this code in the abstract base class, individual implementations needn’t
repeatedly reimplement this common code. By default, the reset method does
nothing. If a particular generator does not require a reset method, the default
method does nothing.

Here is a simple implementation of a Generator that generates a constant

value. The value is provided in the constructor:

public class ConstantGenerator extends AbstractGenerator
{

public ConstantGenerator(int c)
// pre: c is the value to be generated.
// post: constructs a generator that returns a constant value

Constant-
Generator

154

A Design Method

{

}

set(c);

public int next()
// post: returns the constant value
{

return get();

}

}

The set method of the AbstractGenerator is called from the constructor, record-
ing the constant value to be returned. This effectively implements the get
method—that code was provided in the AbstractGenerator class. The next
method simply returns the value available from the get method.

Another implementation of a Generator returns a sequence of prime num-
bers.
In this case, the constructor sets the current value of the generator to
2—the ﬁrst prime. The next method searches for the next prime and returns
that value after it has been saved. Here, the private set and the public get
methods from the AbstractGenerator class help to develop the state of the
Generator:

public class PrimeGenerator extends AbstractGenerator
{

PrimeGenerator

public PrimeGenerator()
// post: construct a generator that delivers primes starting at 2.
{

reset();

}

public void reset()
// post: reset the generator to return primes starting at 2
{

set(2);

}

public int next()
// post: generate the next prime
{

int f,n = get();
do
{

if (n == 2) n = 3;
else n += 2;

// check the next value
for (f = 2; f*f <= n; f++)
{

if (0 ==(n % f)) break;

7.3 Example: Playing Cards

155

}

} while (f*f <= n);
set(n);
return n;

}

}

Clearly, the reset method is responsible for restarting the sequence at 2. While
it would be possible for each Generator to keep track of its current value in its
own manner, placing that general code in the AbstractGenerator reduces the
overall cost of keeping track of this information for each of the many implemen-
tations.

Exercise 7.1 Implement a Generator that provides a stream of random integers.
After a call to reset, the random sequence is “rewound” to return the same se-
quence again. Different generators should probably generate different sequences.
(Hint: You may ﬁnd it useful to use the setSeed method of java.util.Random.)

7.3 Example: Playing Cards

Many games involve the use of playing cards. Not all decks of cards are the
same. For example, a deck of bridge cards has four suits with thirteen cards in
each suit. There are no jokers. Poker has a similar deck of cards, but various
games include special joker or wild cards. A pinochle deck has 48 cards con-
sisting of two copies of 9, jack, queen, king, 10, and ace in each of four suits.
The ranking of cards places 10 between king and ace. A baccarat deck is as in
bridge, except that face cards are worth nothing. Cribbage uses a standard deck
of cards, with aces low.

While there are many differences among these card decks, there are some
common features. In each, cards have a suit and a face (e.g., ace, king, 5). Most
decks assign a value or rank to each of the cards. In many games it is desirable
to be able to compare the relative values of two cards. With these features in
mind, we can develop the following Card interface:

public interface Card
{

public static final int ACE = 1;
public static final int JACK = 11;
public static final int QUEEN = 12;
public static final int KING = 13;
public static final int JOKER = 14;
public static final int CLUBS = 0;
public static final int DIAMONDS = 1;
public static final int HEARTS = 2;
public static final int SPADES = 3;
public int suit();

Card

156

A Design Method

// post: returns the suit of the card

public int face();
// post: returns the face of the card, e.g., ACE, 3, JACK

public boolean isWild();
// post: returns true iff this card is a wild card

public int value();
// post: return the point value of the card

public int compareTo(Object other);
// pre: other is valid Card
// post: returns int <,==,> 0 if this card is <,==,> other

public String toString();
// post: returns a printable version of this card

}

The card interface provides all the public methods that we need to have in our
card games, but it does not provide any hints at how the cards are implemented.
The interface also provides standard names for faces and suits that are passed
to and returned from the various card methods.

In the expectation that most card implementations are similar to a stan-
dard deck of cards, we provide an AbstractCard class that keeps track of an
integer—a card index—that may be changed with set or retrieved with get
(both are protected methods):

import java.util.Random;
public abstract class AbstractCard implements Card
{

protected int cardIndex;
protected static Random gen = new Random();

AbstractCard

public AbstractCard()
// post: constructs a random card in a standard deck
{

set(randomIndex(52));

}

protected static int randomIndex(int max)
// pre: max > 0
// post: returns a random number n, 0 <= n < max
{

return Math.abs(gen.nextInt()) % max;

}

protected void set(int index)
// post: this card has cardIndex index
{

7.3 Example: Playing Cards

157

cardIndex = index;

}

protected int get()
// post: returns this card’s card index
{

return cardIndex;

}

public int suit()
// post: returns the suit of the card
{

return cardIndex / 13;

}

public int face()
// post: returns the face of the card, e.g. ACE, 3, JACK
{

return (cardIndex % 13)+1;

}

public boolean isWild()
// post: returns true iff this card is a wild card
// (default is false)
{

return false;

}

public int value()
// post: return the point value of the card, Ace..King
{

return face();

}

public String toString()
// post: returns a printable version of this card
{

String cardName = "";
switch (face())
{

case ACE: cardName = "Ace"; break;
case JACK: cardName = "Jack"; break;
case QUEEN: cardName = "Queen"; break;
case KING: cardName = "King"; break;
default: cardName = cardName + face(); break;

}
switch (suit())
{

case HEARTS: cardName += " of Hearts"; break;
case DIAMONDS: cardName += " of Diamonds"; break;

158

A Design Method

case CLUBS: cardName += " of Clubs"; break;
case SPADES: cardName += " of Spades"; break;

}
return cardName;

}

}

Our abstract base class also provides a protected random number generator that
returns values from 0 to max-1. We make use of this in the default constructor
for a standard deck; it picks a random card from the usual 52. The cards are
indexed from the ace of clubs through the king of spades. Thus, the face and
suit methods must use division and modulo operators to split the card index
into the two constituent parts. By default, the value method returns the face of
the card as its value. This is likely to be different for different implementations
of cards, as the face values of cards in different games vary considerably.

We also provide a standard toString method that allows us to easily print
out a card. We do not provide a compareTo method because there are complex-
ities with comparing cards that cannot be predicted at this stage. For example,
in bridge, suits play a predominant role in comparing cards. In baccarat they do
not.

Since a poker deck is very similar to our standard implementation, we ﬁnd
the PokerCard implementation is very short. All that is important is that we
allow aces to have high values:

PokerCard

public class PokerCard extends AbstractCard
{

public PokerCard(int face, int suit)
// pre: face and suit have valid values
// post: constructs a card with the particular face value
{

set(suit*13+face-1);

}

public PokerCard()
// post: construct a random poker card.
{

// by default, calls the AbstractCard constructor

}

public int value()
// post: returns rank of card - aces are high
{

if (face() == ACE) return KING+1;
else return face();

}

public int compareTo(Object other)
// pre: other is valid PokerCard
// post: returns relationship between this card and other

7.3 Example: Playing Cards

159

{

}

}

PokerCard that = (PokerCard)other;
return value()-that.value();

Exercise 7.2 Write the value and compareTo methods for a pair of cards where
suits play an important role. Aces are high, and assume that suits are ranked clubs
(low), diamonds, hearts, and spades (high). Assume that face values are only
considered if the suits are the same; otherwise ranking of cards depends on their
suits alone.

The implementation of a pinochle card is particularly difﬁcult. We are inter-
ested in providing the standard interface for a pinochle card, but we are faced
with the fact that there are two copies each of the six cards 9, jack, queen, king,
10, and ace, in each of the four suits. Furthermore we assume that 10 has the
unusual ranking between king and ace. Here’s one approach:

public class PinochleCard extends AbstractCard
{

// cardIndex
// 0
// 1
// ...
// 10
// 11
// 12
// 13
// ...
// 47

face
9
9

ACE
ACE
9
9

ACE

suit
clubs
clubs (duplicate)

clubs
clubs (duplicate)
diamonds
diamonds (duplicate)

spades (duplicate)

public PinochleCard(int face, int suit, int copy)
// pre: face and suit have valid values
// post: constructs a card with the particular face value
{

if (face == ACE) face = KING+1;
set((suit*2+copy)*6+face-9);

}

public PinochleCard()
// post: construct a random Pinochle card.
{

set(randomIndex(48));

}

public int face()
// post: returns the face value of the card (9 thru Ace)
{

int result = get()%6 + 9;
if (result == 14) result = ACE;

PinochleCard

160

A Design Method

return result;

}

public int suit()
// post: returns the suit of the card (there are duplicates!)
{

// this is tricky; we divide by 12 cards (including duplicates)
// per suit, and again by 2 to remove the duplicate
return cardIndex / 12 / 2;

}

public int value()
// post: returns rank of card - aces are high
{

if (face() == ACE) return KING+2;
else if (face() == 10) return KING+1;
else return face();

}

public int compareTo(Object other)
// pre: other is valid PinochleCard
// post: returns relationship between this card and other
{

PinochleCard that = (PinochleCard)other;
return value()-that.value();

}

}

The difﬁculty is that there is more than one copy of a card. We choose to keep
track of the extra copy, in case we need to distinguish between them at some
point, but we treat duplicates the same in determining face value, suit, and
relative rankings.

7.4 Conclusions

Throughout the remainder of this book we will ﬁnd it useful to approach each
type of data structure ﬁrst in an abstract manner, and then provide the details of
various implementations. While each implementation tends to have a distinct
approach to supporting the abstract structure, many features are common to all
implementations. The basic interface, for example, is a shared concept of the
methods that are used to access the data structure. Other features—including
common private methods and shared utility methods—are provided in a basic
implementation called the abstract base class. This incomplete class serves as
a single point of extension for many implementations; the public and private
features of the abstract base class are shared (and possibly overridden) by the
varied approaches to solving the problem.

Chapter 8

Iterators

Concepts:
.
.
.
. Numeric iteration

Iterators
The AbstractIterator class
Vector iterators

One potato, two potato, three potato, four,
ﬁve potato, six potato, seven potato, more.
—A child’s iterator

PROGRAMS MOVE FROM ONE STATE TO ANOTHER. As we have seen, this “state” Ah! Interstate
is composed of the current value of user variables as well as some notion of
“where” the computer is executing the program. This chapter discusses enumer-
ations and iterators—objects that hide the complexities of maintaining the state
of a traversal of a data structure.

programs!

Consider a program that prints each of the values in a list. It is important
to maintain enough information to know exactly “where we are” at all times.
This might correspond to a reference to the current value. In other structures
it may be less clear how the state of a traversal is maintained. Iterators help us
hide these complexities. The careful design of these control structures involves,
as always, the development of a useful interface that avoids compromising the
iterator’s implementation or harming the object it traverses.

8.1 Java’s Enumeration Interface

Java deﬁnes an interface called an Enumeration that provides the user indirect,
iterative access to each of the elements of an associated data structure, exactly
once. The Enumeration is returned as the result of calling the elements method
of various container classes. Every Enumeration provides two methods:

public interface java.util.Enumeration
{

public abstract boolean hasMoreElements();
// post: returns true iff enumeration has outstanding elements

Enumeration

public abstract java.lang.Object nextElement();
// pre: hasMoreElements
// post: returns the next element to be visited in the traversal

}

162

Iterators

The hasMoreElements method returns true if there are unvisited elements of
the associated structure. When hasMoreElements returns false, the traversal
is ﬁnished and the Enumeration expires. To access an element of the under-
lying structure, nextElement must be called. This method does two things:
it returns a reference to the current element and then marks it visited. Typi-
cally hasMoreElements is the predicate of a while loop whose body processes
a single element using nextElement. Clearly, hasMoreElements is an impor-
tant method, as it provides a test to see if the precondition for the nextElement
method is met.

The following code prints out a catchy phrase using a Vector enumeration:

HelloWorld

public static void main(String args[])
{

// construct a vector containing two strings:
Vector<String> v = new Vector<String>();
v.add("Hello");
v.add("world!");

// construct an enumeration to view values of v
Enumeration i = (Enumeration)v.elements();
while (i.hasMoreElements())
{

// SILLY: v.add(1,"silly");
System.out.print(i.nextElement()+" ");

}
System.out.println();

}

When run, the following immortal words are printed:

Hello world!

There are some important caveats that come with the use of Java’s Enumera-
tion construct. First, it is important to avoid modifying the associated structure
while the Enumeration is active or live. Uncommenting the line marked SILLY
causes the following inﬁnite output to begin:

Hello silly silly silly silly silly silly

A silly virus
vector!

Inserting the string "silly" as the new second element of the Vector causes it
to expand each iteration of the loop, making it difﬁcult for the Enumeration to
detect the end of the Vector.

Principle 9 Never modify a data structure while an associated Enumeration is
live.

Modifying the structure behind an Enumeration can lead to unpredictable re-
sults. Clearly, if the designer has done a good job, the implementations of both

NNWSWSENEWSE8.2 The Iterator Interface

163

the Enumeration and its associated structure are hidden. Making assumptions
about their interaction can be dangerous.

Another subtle aspect of Enumerations is that they do not guarantee a par-
ticular traversal order. All that is known is that each element will be visited
exactly once before hasMoreElements becomes false. While we assume that
our ﬁrst example above will print out Hello world!, the opposite order may
also be possible.

Presently, we develop the concept of an iterator.

8.2 The Iterator Interface

An Iterator is similar to an Enumerator except that the Iterator traverses
an associated data structure in a predictable order. Since this is a behavior and
not necessarily a characteristic of its interface, it cannot be controlled or veriﬁed
by a Java compiler.
Instead, we must assume that developers of Iterators
will implement and document their structures in a manner consistent with the
following interface:

public interface java.util.Iterator
{

public abstract boolean hasNext();
// post: returns true if there is at least one more value to visit

Iterator

public abstract java.lang.Object next();
// pre: hasNext()
// post: returns the next value to be visited

}

While the Iterator is a feature built into the Java language, we will choose to
implement our own AbstractIterator class.

public abstract class AbstractIterator<E>

implements Enumeration<E>, Iterator<E>, Iterable<E>

{

public abstract void reset();
// pre: iterator may be initialized or even amid-traversal
// post: reset iterator to the beginning of the structure

Abstract-
Iterator

public abstract boolean hasNext();
// post: true iff the iterator has more elements to visit

public abstract E get();
// pre: there are more elements to be considered; hasNext()
// post: returns current value; ie. value next() will return

public abstract E next();
// pre: hasNext()
// post: returns current value, and then increments iterator

164

Iterators

public void remove()
// pre: hasNext() is true and get() has not been called
// post: the value has been removed from the structure
{

Assert.fail("Remove not implemented.");

}

final public boolean hasMoreElements()
// post: returns true iff there are more elements
{

return hasNext();

}

final public E nextElement()
// pre: hasNext()
// post: returns the current value and "increments" the iterator
{

return next();

}

final public Iterator<E> iterator()
// post: returns this iterator as a subject for a for-loop
{

return this;

}

}

This abstract base class not only meets the Iterator interface, but also im-
plements the Enumeration interface by recasting the Enumeration methods in
terms of Iterator methods. We also provide some important methods that are
not part of general Iterators: reset and get. The reset method reinitializes
the AbstractIterator for another traversal. The ability to traverse a structure
multiple times can be useful when an algorithm makes multiple passes through
a structure to perform a single logical operation. The same functionality can be
achieved by constructing a new AbstractIterator between passes. The get
method of the AbstractIterator retrieves a reference to the current element of
the traversal. The same reference will be returned by the call to next. Unlike
next, however, get does not push the traversal forward. This is useful when
the current value of an AbstractIterator is needed at a point logically distant
from the call to next.

The use of an AbstractIterator leads to the following idiomatic loop for

traversing a structure:

public static void main(String args[])
{

// construct a vector containing two strings:

HelloWorld

8.3 Example: Vector Iterators

165

Vector<String> v = new Vector<String>();
AbstractIterator<String> i;
v.add("Hello");
v.add("world!");

// construct an iterator to view values of v
for (i = (AbstractIterator<String>)v.iterator(); i.hasNext(); i.next())
{

System.out.print(i.get()+" ");

}
System.out.println();

}

The result is the expected Hello world!

In Java 5 any type that has a method iterator that returns an Iterator<T>
for traversing an object meets the requirements of the Iterable<T> interface.
These classes can make use of a new form of the for loop that simpliﬁes the
previous idiom to:

Vector<String> v = new Vector<String>();
...
for (String word : v)
{

System.out.print(word+" ");

}
System.out.println();

We will see this form of for loop used on many structure classes.

8.3 Example: Vector Iterators

For our ﬁrst example, we design an Iterator to traverse a Vector called, not
surprisingly, a VectorIterator. We do not expect the user to construct Vector-
Iterators directly—instead the Vector hides the construction and returns the
new structure as a generic Iterator, as was seen in the HelloWorld example.
Here is the iterator method:

public Iterator<E> iterator()
// post: returns an iterator allowing one to
//
{

view elements of vector

return new VectorIterator<E>(this);

}

When a Vector constructs an Iterator, it provides a reference to itself (this)
as a parameter. This reference is used by the VectorIterator to recall which
Vector it is traversing.

We now consider the interface for a VectorIterator:

Vector

Vector-
Iterator

166

Iterators

class VectorIterator<E> extends AbstractIterator<E>
{

public VectorIterator(Vector<E> v)
// post: constructs an initialized iterator associated with v

public void reset()
// post: the iterator is reset to the beginning of the traversal

public boolean hasNext()
// post: returns true if there is more structure to be traversed

public E get()
// pre: traversal has more elements
// post: returns the current value referenced by the iterator

public E next()
// pre: traversal has more elements
// post: increments the iterated traversal

}

As is usually the case, the nonconstructor methods of VectorIterator exactly
match those required by the Iterator interface. Here is how the VectorIter-
ator is constructed and initialized:

protected Vector<E> theVector;
protected int current;

public VectorIterator(Vector<E> v)
// post: constructs an initialized iterator associated with v
{

theVector = v;
reset();

}

public void reset()
// post: the iterator is reset to the beginning of the traversal
{

current = 0;

}

The constructor saves a reference to the associated Vector and calls reset. This
logically attaches the Iterator to the Vector and makes the ﬁrst element (if
one exists) current. Calling the reset method allows us to place all the resetting
code in one location.

To see if the traversal is ﬁnished, we invoke hasNext:

public boolean hasNext()
// post: returns true if there is more structure to be traversed
{

return current < theVector.size();

}

8.4 Example: Rethinking Generators

167

This routine simply checks to see if the current index is valid. If the index is less
than the size of the Vector, then it can be used to retrieve a current element
from the Vector. The two value-returning methods are get and next:

public E get()
// pre: traversal has more elements
// post: returns the current value referenced by the iterator
{

return theVector.get(current);

}

public E next()
// pre: traversal has more elements
// post: increments the iterated traversal
{

return theVector.get(current++);

}

The get method simply returns the current element. It may be called arbitrarily
many times without pushing the traversal along. The next method, on the other
hand, returns the same reference, but only after having incremented current.
The next value in the Vector (again, if there is one) becomes the current value.
Since all the Iterator methods have been implemented, Java will allow a
VectorIterator to be used anywhere an Iterator is required. In particular, it
can now be returned from the iterator method of the Vector class.

Observe that while the user cannot directly construct a VectorIterator (it
is a nonpublic class), the Vector can construct one on the user’s behalf. This
allows measured control over the agents that access data within the Vector.
Also, an Iterator is a Java interface. It is not possible to directly construct an
Iterator. We can, however, construct any class that implements the Iterator
interface and use that as we would any instance of an Iterator.

Since an AbstractIterator implements the Enumeration interface, we may
use the value returned by Vector’s iterator method as an Enumeration to
access the data contained within the Vector. Of course, treating the Vector-
Iterator as an Enumeration makes it difﬁcult to call the AbstractIterator
methods reset and get.

8.4 Example: Rethinking Generators

In Section 7.2 we discussed the construction of a class of objects that gener-
ated numeric values. These Generator objects are very similar to Abstract-
Iterators—they have next, get, and reset methods. They lack, however, a
hasNext method, mainly because of a lack of foresight, and because many se-
quences of integers are inﬁnite—their hasNext would, essentially, always return
true.

Generators are different from Iterators in another important way: Gen-
erators return the int type, while Iterators return Objects. Because of this,

168

Iterators

the Iterator interface is more general. Any Object, including Integer values,
may be returned from an Iterator.

In this section we experiment with the construction of a numeric iterator—a
Generator-like class that meets the Iterator interface. In particular, we are
interested in constructing an Iterator that generates prime factors of a speciﬁc
integer. The PFIterator accepts the integer to be factored as the sole parameter
on the constructor:

import structure5.AbstractIterator;
public class PFGenerator extends AbstractIterator<Integer>
{

PFGenerator

// the original number to be factored
protected int base;

public PFGenerator(int value)
// post: an iterator is constructed that factors numbers
{

base = value;
reset();

}

}

The process of determining the prime factor involves reducing the number by a
factor. Initially, the factor f starts at 2. It remains 2 as long as the reduced value
is even. At that point, all the prime factors of 2 have been determined, and we
next try 3. This process continues until the reduced value becomes 1.

Because we reduce the number at each step, we must keep a copy of the
original value to support the reset method. When the iterator is reset, the
original number is restored, and the current prime factor is set to 2.

// base, reduced by the prime factors discovered
protected int n;
// the current prime factor
protected int f;

public void reset()
// post: the iterator is reset to factoring the original value
{

n = base;
// initial guess at prime factor
f = 2;

}

If, at any point, the number n has not been reduced to 1, prime factors
remain undiscovered. When we need to ﬁnd the current prime factor, we ﬁrst
check to see if f divides n—if it does, then f is a factor. If it does not, we simply
increase f until it divides n. The next method is responsible for reducing n by a
factor of f.

8.4 Example: Rethinking Generators

169

public boolean hasNext()
// post: returns true iff there are more prime factors to be considered
{

return f <= n;

// there is a factor <= n

}

public Integer next()
// post: returns the current prime factor and "increments" the iterator
{

Integer result = get(); // factor to return
n /= f;
return result;

// reduce n by factor

}

public Integer get()
// pre: hasNext()
// post: returns the current prime factor
{

// make sure f is a factor of n
while (f <= n && n%f != 0) f++;
return f;

}

We can now write a program that uses the iterator to print out the prime
factors of the values presented on the command line of the Java program as it
is run:

public static void main(String[]args)
{

// for each of the command line arguments
for (int i = 0; i < args.length; i++)
{

// determine the value
int n = Integer.parseInt(args[i]);
PFGenerator g = new PFGenerator(n);
System.out.print(n+": ");
// and print the prime factors of n
while (g.hasNext()) System.out.print(g.next()+" ");
System.out.println();

}

}

For those programmers that prefer to use the hasMoreElements and next-
Element methods of the Enumeration interface, those methods are automat-
ically provided by the AbstractIterator base class, which PFGenerator ex-
tends.

Exercise 8.1 The 3n + 1 sequence is computed in the following manner. Given a
seed n, the next element of the sequence is 3n + 1 if n is odd, or n/2 if n is even.
This sequence of values stops whenever a 1 is encountered; this happens for all

170

Iterators

seeds ever tested. Write an Iterator that, given a seed, generates the sequence of
values that ends with 1.

8.5 Example: Filtering Iterators

We now consider the construction of a ﬁltering iterator. Instead of traversing
structures, a ﬁltering iterator traverses another iterator! As an example, we
construct an iterator that returns the unique values of a structure.

Before we consider the implementation, we demonstrate its use with a sim-
ple example. In the following code, suppose that data is a Vector of Strings,
some of which may be duplicates. For example, the Vector could represent the
text of the Gettysburg Address. The iterator method of data is used to con-
struct a VectorIterator. This is, in turn, used as a parameter to the construc-
tion of a UniqueFilter. Once constructed, the ﬁlter can be used as a standard
Iterator, but it only returns the ﬁrst instance of each String appearing in the
Vector:

Vector<String> data = new Vector<String>(1000);

...

UniqueFilter

AbstractIterator<String> dataIterator =

(AbstractIterator<String>)data.iterator();

AbstractIterator<String> ui = new UniqueFilter(dataIterator);
int count=0;

for (ui.reset(); ui.hasNext(); ui.next())
{

System.out.print(ui.get()+" ");
if (++count%7==0) System.out.println();

}
System.out.println();

The result of the program, when run on the Gettysburg Address, is the follow-
ing output, which helps increase the vocabulary of this textbook by nearly 139
words:

four score and seven years ago our
fathers brought forth on this continent a
new nation conceived in liberty dedicated to
the proposition that all men are created
equal now we engaged great civil war
testing whether or any so can long
endure met battlefield of have come dedicate
portion field as final resting place for
those who here gave their lives might
live it is altogether fitting proper should
do but larger sense cannot consecrate hallow
ground brave living dead struggled consecrated far

8.5 Example: Filtering Iterators

171

above poor power add detract world will
little note nor remember what say itcan
never forget they did us rather be
unfinished work which fought thus nobly advanced
task remaining before from these honored take
increased devotion cause last full measure highly
resolve shall not died vain under God
birth freedom government people by perish earth

Fans of compact writing will ﬁnd this unique.

The UniqueFilter provides the same interface as other iterators. Its con-

structor, however, takes a “base” Iterator as its parameter:

protected AbstractIterator<T> base; // slave iterator
protected List<T> observed; // list of previous values

public UniqueFilter(AbstractIterator<T> baseIterator)
// pre: baseIterator is a non-null iterator
// post: constructs unique-value filter
//
host iterator is reset
{

base = baseIterator;
reset();

}

public void reset()
// post: master and base iterators are reset
{

base.reset();
observed = new SinglyLinkedList<T>();

}

When the ﬁlter is reset using the reset method, the base iterator is reset as
well. We then construct an empty List of words previously observed. As the
ﬁlter progresses, words encountered are incorporated into the observed list.

The current value is fetched by the get method. It just passes the request
along to the base iterator. A similar technique is used with the hasNext method:

public boolean hasNext()
// post: returns true if there are more values available
//
{

from base stream

return base.hasNext();

}

public T get()
// pre: traversal has more elements
// post: returns the current value referenced by the iterator
{

172

Iterators

return base.get();

}

Finally, the substance of the iterator is found in the remaining method, next:

public T next()
// pre: traversal has more elements
// post: returns current value and increments the iterator
{

T current = base.next();
// record observation of current value
observed.add(current);
// now seek next new value
while (base.hasNext())
{

T possible = base.get();
if (!observed.contains(possible))
// new value found! leave
{
break;

} else {

// old value, continue
base.next();

}

}
return current;

}

Because this routine can only be called if there is a current value, we record the
current value in the observed list. The method then increments the base iterator
until a new, previously unobserved value is produced, or the base iterator runs
dry.

Some subtle details are worth noting here. First, while we have used a
VectorIterator on a Vector of Strings, the UniqueFilter can be applied, as
is, to any type of iterator and can deliver any type of value. All that is required
is that the base type support the equals method. Secondly, as the ﬁlter iterator
progresses, it forces the base iterator to progress, too. Because of this, two ﬁlters
are usually not applied to the same base iterator, and the base iterator should
never be modiﬁed while the ﬁlter is running.

8.6 Conclusions

We have seen that data structures can sometimes be used to control the way
programs focus on and access data. This is made very explicit with Java’s
Enumeration construct that facilitates visiting all the elements of a structure.

When we wish to traverse the elements of a data structure in a predeter-
mined order, we use an Iterator. The Iterator provides access to the ele-
ments of a structure using an interface that is similar to that of an Enumeration.

8.6 Conclusions

173

The abstract base class AbstractIterator implements both the Iterator and
Enumeration interfaces, and provides two new methods—get and reset—as
well. We have also seen that there are weaknesses in the concept of both of
these constructs, because they surrender some of the data hiding and access
controls that are provided by the associated structure. Careful use of these con-
trolling structures, however, can yield useful tools to make traversal of struc-
tures simpler.

Self Check Problems

Solutions to these problems begin on page 445.

8.1
using e to print all the values of the data structure.

Suppose e is an Enumeration over some data structure. Write a loop

Suppose i is an Iterator over some data structure. Write a loop using

8.2
i to print all the values of the data structure.

Suppose that v is a Vector of Integer values. Write a loop that will use

8.3
an Iterator to print those Integer values that are even.

It is possible to write down the integers 1 through 15 in an order such
8.4
that each adjacent pair of integers sums to a perfect square. Write a loop that
prints Perfect! only if the adjacent Integer values generated by the Iterator
g sum to perfect squares. (You needn’t verify the number or range of values.)

Problems

Solutions to the odd-numbered problems begin on page 467.

Since the get method is available to the AbstractIterator, the next
8.1
method does not appear to need to return a value. Why does our implementa-
tion return the value?

Write an Iterator that works on Strings. Each value returned should

8.2
be an object of type Character.

8.3
How close is it to the Generator implementation of Section 7.2?

Write an Iterator that returns a stream of Integers that are prime.

Write a ﬁltering iterator, ReverseIterator, that reverses the stream of
8.4
values produced by another Iterator. You may assume that the base Iterator
will eventually have no more elements, but you may not bound the number.

Write a ﬁltering iterator, OrderedIterator, that sorts the stream of
8.5
values produced by another Iterator. You may assume that the base Iterator
will eventually have no more elements, but you may not bound the number.

Write a ﬁltering iterator, ShuffleIterator, that shufﬂes the stream of
8.6
values produced by another Iterator. You may assume that the base Iterator
will eventually have no more elements, but you may not bound the number.

174

Iterators

Write a ﬁltering iterator that takes a base iterator and an Object (called
8.7
predicate) with a static select method deﬁned. This iterator passes along
only those values that generate true when passed to the select method of the
predicate Object.

8.7 Laboratory: The Two-Towers Problem

Objective. To investigate a difﬁcult problem using Iterators.

Discussion. Suppose that we are given n uniquely sized cubic blocks and that
each block has a face area between 1 and n. Build two towers by stacking these
blocks. How close can we get the heights of the two towers? The following
two towers built by stacking 15 blocks, for example, differ in height by only 129
millions of an inch (each unit is one-tenth of an inch):

Still, this stacking is only the second-best solution! To ﬁnd the best stacking, we
could consider all the possible conﬁgurations.

We do know one thing: the total height of the two towers is computed by

summing the heights of all the blocks:

n
X

√

i

h =

i=1

If we consider all the subsets of the n blocks, we can think of the subset as the
set of blocks that make up, say, the left tower. We need only keep track of that
subset that comes closest to h/2 without exceeding it.

In this lab, we will represent a set of n distinct objects by a Vector, and we

will construct an Iterator that returns each of the 2n subsets.
Procedure. The trick to understanding how to generate a subset of n values
from a Vector is to ﬁrst consider how to generate a subset of indices of elements
from 0 to n − 1. Once this simpler problem is solved, we can use the indices to
help us build a Vector (or subset) of values identiﬁed by the indices.

There are exactly 2n subsets of values 0 to n−1. We can see this by imagining
that a coin is tossed n times—once for each value—and the value is added to
the subset if the coin ﬂip shows a head. Since there are 2 × 2 × · · · × 2 = 2n
different sequences of coin tosses, there are 2n different sets.

We can also think of the coin tosses as determining the place values for n
different digits in a binary number. The 2n different sequences generate binary
numbers in the range 0 through 2n − 1. Given this, we can see a line of attack:

151410976415131211832176

Iterators

count from 0 to 2n−1 and use the binary digits (bits) of the number to determine
which of the original values of the Vector are to be included in a subset.

Computer scientists work with binary numbers frequently, so there are a

number of useful things to remember:

• An int type is represented by 32 bits. A long is represented by 64 bits. For
maximum ﬂexibility, it would be useful to use long integers to represent
sets of up to 64 elements.

• The arithmetic shift operator (<<) can be used to quickly compute powers
of 2. The value 2i can be computed by shifting a unit bit (1) i places to the
left. In Java we write this 1<<i. This works only for nonnegative, integral
powers. (For long integers, use 1L<<i.)

• The bitwise and of two integers can be used to determine the value of
a single bit in a number’s binary representation. To retrieve bit i of an
integer m we need only compute m & (1<<i).

Armed with this information, the process of generating subsets is fairly straight-
forward. One line of attack is the following:

1. Construct a new extension to the AbstractIterator class. (By extending
the AbstractIterator we support both the Iterator and Enumeration
interfaces.) This new class should have a constructor that takes a Vector
as its sole argument. Subsets of this Vector will be returned as the
Iterator progresses.

2. Internally, a long value is used to represent the current subset. This value
increases from 0 (the empty set) to 2n − 1 (the entire set of values) as the
Iterator progresses. Write a reset method that resets the subset counter
to 0.

3. Write a hasNext method that returns true if the current value is a reason-

able representation of a subset.

4. Write a get method that returns a new Vector of values that are part of
the current subset. If bit i of the current counter is 1, element i of the
Vector is included in the resulting subset Vector.

5. Write a next method. Remember it returns the current subset before in-

crementing the counter.

6. For an Iterator you would normally have to write a remove method. If
you extend the AbstractIterator class, this method is provided and will
do nothing (this is reasonable).

You can now test your new SubsetIterator by having it print all the subsets
of a Vector of values. Remember to keep the Vector small. If the original values
are all distinct, the subsets should all have different values.

8.7 Laboratory: The Two-Towers Problem

177

1,

√

√

√

2,. . . ,

To solve the two-towers problem, write a main method that inserts the values
n as Double objects into a Vector. A SubsetIterator is then
used to construct 2n subsets of these values. The values of each subset are
summed, and the sum that comes closest to, but does not exceed, the value
h/2 is remembered. After all the subsets have been considered, print the best
solution.
Thought Questions. Consider the following questions as you complete the lab:

1. What is the best solution to the 15-block problem?

2. This method of exhaustively checking the subsets of blocks will not work
for very large problems. Consider, for example, the problem with 50
blocks: there are 250 different subsets. One approach is to repeatedly
pick and evaluate random subsets of blocks (stop the computation after
1 second of elapsed time, printing the best subset found). How would you
implement randomSubset, a new SubsetIterator method that returns a
random subset?

Notes:

Chapter 9

Lists

Concepts:
.
The list abstraction
.
Singly linked lists
. Doubly linked lists
. Circular lists
. Vectors as lists

He’s makin’ a list
and checkin’ it twice!
—Haven Gillespie

IMAGINE YOU ARE TO WRITE A ROBUST PROGRAM to handle varying amounts of
data. An inventory program is a classic example. The same inventory program
might be used to keep track of either tens of items, or millions. To support such
applications, Vectors and arrays are not ideal. As they reach their capacity they
must be expanded. Either this happens manually, as with arrays, or automati-
cally, as with Vectors. In either case the penalty for growing the structure is the
same over the growth of the structure. With a Vector, for example, when the
structure must double in size, the cost of adding an element is proportional to
the size of the Vector.

In this chapter, we develop the concept of a linked list. A linked list is a
dynamic structure that grows and shrinks exactly when necessary and whose
elements may be added in constant time. There is some cost for this dynamic
behavior, however. As with Vectors and arrays, each of the elements of a linked-
list has an associated index, but the elements of many linked list implementa-
tions cannot be efﬁciently accessed out of order or accessed randomly. Despite
this one inefﬁciency, linked lists provide an important building block for the
design of many effective data structures.

An analogy for linked lists is a child’s string of snap-together beads. As we
grow the string of beads, we attach and detach new beads on either the front
(head) or rear (tail). Since there are two modifying operations that we can per-
form (add or remove) and two ends (at the location of the ﬁrst or last element)
there are four operations that change the length of the structure at the end.

We may also wish to perform operations on the internal portion of the list.
For example, we may want to test for inclusion (Is there a red bead?) or extract
an element (Remove a red bead!). These operations require a traversal of the
linked list from one of the two ends.

If you have
never seen
these, visit your
niece.

180

Lists

Now, let’s see what the Java description of a list looks like:

List

public interface List<E> extends Structure<E>
{

public int size();
// post: returns number of elements in list

public boolean isEmpty();
// post: returns true iff list has no elements

public void clear();
// post: empties list

public void addFirst(E value);
// post: value is added to beginning of list

public void addLast(E value);
// post: value is added to end of list

public E getFirst();
// pre: list is not empty
// post: returns first value in list

public E getLast();
// pre: list is not empty
// post: returns last value in list

public E removeFirst();
// pre: list is not empty
// post: removes first value from list

public E removeLast();
// pre: list is not empty
// post: removes last value from list

public E remove(E value);
// post: removes and returns element equal to value
//

otherwise returns null

public void add(E value);
// post: value is added to tail of list

public E remove();
// pre: list has at least one element
// post: removes last value found in list

public E get();
// pre: list has at least one element
// post: returns last value found in list

public boolean contains(E value);
// pre: value is not null
// post: returns true iff list contains an object equal to value

181

public int indexOf(E value);
// pre: value is not null
// post: returns (0-origin) index of value,
//

or -1 if value is not found

public int lastIndexOf(E value);
// pre: value is not null
// post: returns (0-origin) index of value,
//

or -1 if value is not found

public E get(int i);
// pre: 0 <= i < size()
// post: returns object found at that location

public E set(int i, E o);
// pre: 0 <= i < size()
// post: sets ith entry of list to value o;
//

returns old value

public void add(int i, E o);
// pre: 0 <= i <= size()
// post: adds ith entry of list to value o

public E remove(int i);
// pre: 0 <= i < size()
// post: removes and returns object found at that location

public Iterator<E> iterator();
// post: returns an iterator allowing
//

ordered traversal of elements in list

}

Again, because this structure is described as an interface (as opposed to a
class) Java understands this to be a contract describing the methods that are
required of lists. We might think of an interface as being a “structural precondi-
tion” describing the outward appearance of any “listlike” class. If we write our
code in terms of this interface, we may only invoke methods speciﬁed within
the contract.

Note that the List interface is an extension of the Structure interface that
we have seen earlier, in Section 1.8. Thus, every List is also a Structure—a
structure that supports operations like add and remove as well as other size-
related methods. We will see, over the course of this text, several abstract types
that may serve as Structures.

The interface, along with pre- and postconditions, makes many of the imple-
mentation-independent decisions about the semantics of associated structures.

182

Lists

When we develop speciﬁc implementations, we determine the implementation-
speciﬁc features of the structure, including its performance. When we compare
speciﬁc implementations, we compare their performance in terms of space and
time. Often, performance can be used to help us select among different imple-
mentations for a speciﬁc use.

9.1 Example: A Unique Program

As an example of how we might use lists, we write a program that writes out
the input with duplicate lines removed. The approach is to store each of the
unique lines in a structure (lines) as they are printed out. When new lines are
read in, they are compared against the existing list of unique, printed lines. If
the current line (current) is not in the list, it is added. If the current line is in
the list, it is ignored.

Unique

public static void main(String[] args)
{

// input is read from System.in
Scanner s = new Scanner(System.in);
String current;
// list of unique lines
List<String> lines = new SinglyLinkedList<String>();
// read a list of possibly duplicated lines
while (s.hasNextLine()) {

// current line

current = s.nextLine();
// check to see if we need to add it
if (!lines.contains(current)) {

System.out.println(current);
lines.add(current);

}

}

}

In this example we actually construct a particular type of list, a SinglyLinked-
List. The details of that implementation, discussed in the next section, are
not important to us because lines is declared to be a generic interface, a List.
Accessing data through the lines variable, we are only allowed to invoke meth-
ods found in the List interface. On the other hand, if we are able to cast our
algorithms in terms of Lists, any implementation of a List will support our
program.

When given input

madam
I’m
Adam!
...
Adam!

9.2 Example: Free Lists

183

I’m
Ada!
...
mad
am I...
madam

the program generates the following output:

madam
I’m
Adam!
...
Ada!
mad
am I...

Because there is no practical limit (other than the amount of memory available)
on the length of a list, there is no practical limit on the size of the input that
can be fed to the program. The list interface does not provide any hint of how
the list is actually implemented, so it is difﬁcult to estimate the performance of
the program. It is likely, however, that the contains method—which is likely
to have to consider every existing element of the list—and the add method—
which might have to pass over every element to ﬁnd its correct position—will
govern the complexity of the management of this List. As we consider imple-
mentations of Lists, we should keep the performance of programs like Unique
in mind.

9.2 Example: Free Lists

In situations where a pool of resources is to be managed, it is often convenient
to allocate a large number and keep track of those that have not been allocated.
This technique is often used to allocate chunks of physical memory that might
eventually be allocated to individual applications or printers from a pool that
might be used to service a particular type of printing request.

The following application maintains rental contracts for a small parking lot.

We maintain each parking space using a simple class, Space:

class Space
{

// structure describing parking space
public final static int COMPACT = 0; // small space
public final static int MINIVAN = 1; // medium space
public final static int TRUCK = 2;
protected int number;
protected int size;
public Space(int n, int s)
// post: construct parking space #n, size s
{

// address in parking lot
// size of space

// large space

ParkingLot

184

Lists

number = n;
size = s;

}
public boolean equals(Object other)
// pre: other is not null
// post: true iff spaces are equivalent size
{

Space that = (Space)other;
return this.size == that.size;

}

}

The lot consists of 10 spaces of various sizes: one large, six medium, and three
small. Renters may rent a space if one of appropriate size can be found on
the free list. The equals method of the Space class determines an appropriate
match. The rented list maintains Associations between names and space de-
scriptions. The following code initializes the free list so that it contains all the
parking spaces, while the rented list is initially empty:

List<Space> free = new SinglyLinkedList<Space>();
List<Association<String,Space>> rented =

// available

new SinglyLinkedList<Association<String,Space>>(); // rented spaces

for (int number = 0; number < 10; number++)
{

if (number < 3) // three small spaces

free.add(new Space(number,Space.COMPACT));

else if (number < 9) // six medium spaces

free.add(new Space(number,Space.MINIVAN));

else // one large space

free.add(new Space(number,Space.TRUCK));

}

The main loop of our program reads in commands from the keyboard—either
rent or return:

Scanner s = new Scanner(System.in);
while (s.hasNext())
{

String command = s.next(); // rent/return

...

}
System.out.println(free.size()+" slots remain available.");

Within the loop, when the rent command is entered, it is followed by the size
of the space needed and the name of the renter. This information is used to
construct a contract:

Space location;
if (command.equals("rent"))
{

// attempt to rent a parking space

9.2 Example: Free Lists

185

String size = s.next();
Space request;
if (size.equals("small"))

request = new Space(0,Space.COMPACT);

else if (size.equals("medium"))

request = new Space(0,Space.MINIVAN);

else request = new Space(0,Space.TRUCK);
// check free list for appropriate-sized space
if (free.contains(request))
// a space is available
{
location = free.remove(request);
String renter = s.next(); // to whom?
// link renter with space description
rented.add(new Association<String,Space>(renter,location));
System.out.println("Space "+location.number+" rented.");

} else {

System.out.println("No space available. Sorry.");

}

}

Notice that when the contains method is called on a List, a dummy element is
constructed to specify the type of object sought. When the dummy item is used
in the remove command, the actual item removed is returned. This allows us to
maintain a single copy of the object that describes a single parking space.

When the spaces are returned, they are returned by name. The contract is

looked up and the associated space is returned to the free list:

Space location;
if (command.equals("return")){

String renter = s.next(); // from whom?
// template for finding "rental contract"
Association<String,Space> query = new Association<String,Space>(renter);
if (rented.contains(query))
{

// contract found
Association<String,Space> contract =

rented.remove(query);

location = contract.getValue(); // where?
free.add(location); // put in free list
System.out.println("Space "+location.number+" is now free.");

} else {

System.out.println("No space rented to "+renter);

}

}

Here is a run of the program:

rent small Alice

Space 0 rented.

rent large Bob

Space 9 rented.

186

Lists

rent small Carol

Space 1 rented.
return Alice

Space 0 is now free.

return David

No space rented to David

rent small David

Space 2 rented.

rent small Eva

Space 0 rented.

quit

6 slots remain available.

Notice that when Alice’s space is returned, it is not immediately reused because
the free list contains other small, free spaces. The use of addLast instead of
addFirst (or the equivalent method, add) would change the reallocation policy
of the parking lot.

We now consider an abstract base class implementation of the List inter-

face.

9.3 Partial Implementation: Abstract Lists

Although we don’t have in mind any particular implementation, there are some
pieces of code that may be written, given the little experience we already have
with the use of Lists.

For example, we realize that it is useful to have a number of synonym meth-
ods for common operations that we perform on Lists. We have seen, for ex-
ample, that the add method is another way of indicating we want to add a new
value to one end of the List. Similarly, the parameterless remove method per-
forms a removeLast. In turn removeLast is simply a shorthand for removing
the value found at location size()-1.

public abstract class AbstractList<E>

extends AbstractStructure<E> implements List<E>

{

AbstractList

public AbstractList()
// post: does nothing
{
}

public boolean isEmpty()
// post: returns true iff list has no elements
{

return size() == 0;

}

public void addFirst(E value)
// post: value is added to beginning of list

9.3 Partial Implementation: Abstract Lists

187

{

}

add(0,value);

public void addLast(E value)
// post: value is added to end of list
{

add(size(),value);

}

public E getFirst()
// pre: list is not empty
// post: returns first value in list
{

return get(0);

}

public E getLast()
// pre: list is not empty
// post: returns last value in list
{

return get(size()-1);

}

public E removeFirst()
// pre: list is not empty
// post: removes first value from list
{

return remove(0);

}

public E removeLast()
// pre: list is not empty
// post: removes last value from list
{

return remove(size()-1);

}

public void add(E value)
// post: value is added to tail of list
{

addLast(value);

}

public E remove()
// pre: list has at least one element
// post: removes last value found in list
{

return removeLast();

}

188

Lists

public E get()
// pre: list has at least one element
// post: returns last value found in list
{

return getLast();

}

public boolean contains(E value)
// pre: value is not null
// post: returns true iff list contains an object equal to value
{

return -1 != indexOf(value);

}

}

Position-independent operations, like contains, can be written in an implemen-
tation-independent manner. To see if a value is contained in a List we could
If the value returned
simply determine its index with the indexOf method.
is −1, it was not in the list, otherwise the list contains the value. This approach
to the implementation does not reduce the cost of performing the contains
operation, but it does reduce the cost of implementing the contains operation:
once the indexOf method is written, the contains method will be complete.
When we expect that there will be multiple implementations of a class, sup-
porting the implementations in the abstract base class can be cost effective. If
improvements can be made on the generic code, each implementation has the
option of providing an alternative version of the method.

Notice that we provide a parameterless constructor for AbstractList ob-
jects. Since the class is declared abstract, the constructor does not seem nec-
essary. If, however, we write an implementation that extends the AbstractList
class, the constructors for the implementation implicitly call the parameterless
constructor for the AbstractList class. That constructor would be responsible
for initializing any data associated with the AbstractList portion of the imple-
mentation. In the examples of the last chapter, we saw the AbstractGenerator
initialized the current variable. Even if there is no class-speciﬁc data—as is
true with the AbstractList class—it is good to get in the habit of writing these
simple constructors.

We now consider a number of implementations of the List type. Each of
these implementations is an extension of the AbstractList class. Some inherit
the methods provided, while others override the selected method deﬁnitions to
provide more efﬁcient implementation.

9.4 Implementation: Singly Linked Lists

Dynamic memory is allocated using the new operator. Java programmers are
accustomed to using the new operator whenever classes or arrays are to be allo-
cated. The value returned from the new operator is a reference to the new object.

9.4 Implementation: Singly Linked Lists

189

Figure 9.1
of a class (right).

Pictures of a null reference (left) and a non-null reference to an instance

Thus, whenever we declare an instance of a class, we are actually declaring a
reference to one of those objects. Assignment of references provides multiple
variables with access to a single, shared instance of an object.

An instance of a class is like a helium-ﬁlled balloon. The balloon is the object
being allocated. The string on the balloon is a convenient handle that we can
use to hold onto with a hand. Anything that holds onto the string is a reference.
Assignment of references is similar to asking another hand to “hold the balloon
I’m holding.” To not reference anything (to let go of the balloon) we can assign
the reference the value null. If nothing references the balloon, then it ﬂoats
away and we can no longer get access to the instance. When memory is not
referenced in any way, it is recycled automatically by a garbage collector.

Principle 10 When manipulating references, draw pictures.

In this text, we will draw references as arrows pointing to their respective ob-
jects (Figure 9.1). When a reference is not referencing anything, we draw it as
a dot. Since references can only be in one of two states—pointing to nothing or
pointing to an object—these are the only pictures we will ever draw.

One approach to keeping track of arbitrarily large collections of objects is to
use a singly linked list to dynamically allocate each chunk of memory “on the
ﬂy.” As the chunks of memory are allocated, they are linked together to form First garbage,
the entire structure. This is accomplished by packaging with each user object a
reference to the next object in the chain. Thus, a list of 10 items contains 10
elements, each of which contains a value as well as another element reference.
Each element references the next, and the ﬁnal element does not reference
anything: it is assigned null (see Figure 9.2). Here, an implementation of a
Node contains an additional reference, nextElement:

now ﬂies!

public class Node<E>
{

protected E data; // value stored in this element
protected Node<E> nextElement; // ref to next

Node

ABNNWSWSENEWSE190

Lists

public Node(E v, Node<E> next)
// pre: v is a value, next is a reference to remainder of list
// post: an element is constructed as the new head of list
{

data = v;
nextElement = next;

}

public Node(E v)
// post: constructs a new tail of a list with value v
{

this(v,null);

}

public Node<E> next()
// post: returns reference to next value in list
{

return nextElement;

}

public void setNext(Node<E> next)
// post: sets reference to new next value
{

nextElement = next;

}

public E value()
// post: returns value associated with this element
{

return data;

}

public void setValue(E value)
// post: sets value associated with this element
{

data = value;

}

}

When a list element is constructed, the value provided is stored away in the ob-
ject. Here, nextElement is a reference to the next element in the list. We access
the nextElement and data ﬁelds through public methods to avoid accessing
protected ﬁelds. Notice that, for the ﬁrst time, we see a self-referential data
structure: the Node object has a reference to a Node. This is a feature common
to structures whose size can increase dynamically. This class is declared public
so that anyone can construct Nodes.

We now construct a new class that implements the List interface by extend-
ing the AbstractList base class. For that relation to be complete, it is necessary
to provide a complete implementation of each of the methods promised by the

9.4 Implementation: Singly Linked Lists

191

Figure 9.2 A nonempty singly linked list.

Figure 9.3 An empty singly linked list.

interface. Failure to implement any of the methods leaves the implementation
incomplete, leaving the class abstract.

Our approach will be to maintain, in head, a reference to the ﬁrst element
of the list in a protected ﬁeld (Figure 9.2). This initial element references the
second element, and so on. The ﬁnal element has a null-valued next reference.
If there are no elements, head contains a null reference (Figure 9.3). We also
maintain an integer that keeps track of the number of elements in the list. First,
as with all classes, we need to specify protected data and a constructor:

protected int count;
protected Node<E> head; // ref. to first element

// list size

public SinglyLinkedList()
// post: generates an empty list
{

head = null;
count = 0;

}

SinglyLinked-
List

This code sets the head reference to null and the count ﬁeld to 0. Notice that,
by the end of the constructor, the list is in a consistent state.

Principle 11 Every public method of an object should leave the object in a consis-
tent state.

headcountLifeMarson3headcount0NNWSWSENEWSE192

Lists

What constitutes a “consistent state” depends on the particular structure, but
in most cases the concept is reasonably clear.
In the SinglyLinkedList, the
constructor constructs a list that is empty.

The size-oriented methods are simply written in terms of the count identi-

ﬁer. The size method returns the number of elements in the list.

public int size()
// post: returns number of elements in list
{

return count;

}

Recall that the isEmpty method described in the AbstractList class simply
returns whether or not the size method would return 0. There’s a great advan-
tage to calling the size method to implement isEmpty: if we ever change the
implementation, we need only change the implementation of size.

Both of these methods could avoid referencing the count ﬁeld, by travers-
ing each of the next references.
In this alternative code we use the analogy
of a ﬁnger referencing each of the elements in the list. Every time the ﬁnger
references a new element, we increment a counter. The result is the number of
elements. This time-consuming process is equivalent to constructing the infor-
mation stored explicitly in the count ﬁeld.

public int size()
// post: returns number of elements in list
{

// number of elements we’ve seen in list
int elementCount = 0;
// reference to potential first element
Node<E> finger = head;

while (finger != null) {

// finger references a new element, count it
elementCount++;
// reference possible next element
finger = finger.next();

}
return elementCount;

}

Note that isEmpty does not need to change.1 It is early veriﬁcation that the
interface for size helps to hide the implementation.

The decision between the two implementations has little impact on the user
of the class, as long as both implementations meet the postconditions. Since
the user is insulated from the details of the implementation, the decision can
be made even after applications have been written. If, for example, an environ-
ment is memory-poor, it might be wise to avoid the use of the count ﬁeld and

1 In either case, the method isEmpty could be written more efﬁciently, checking a null head
reference.

9.4 Implementation: Singly Linked Lists

193

Figure 9.4 A singly linked list before and after the call to addFirst. Shaded value is
added to the list. The removeFirst method reverses this process and returns value.

instead traverse the list to determine the number of elements by counting them.
If, however, a machine is slow but memory-rich, then the ﬁrst implementation
would be preferred. Both implementations could be made available, with the
user selecting the appropriate design, based on broad guidelines (e.g., memory
versus speed). If this trade-off does not appear dramatic, you might consider
Problem 9.10. We also discuss space–time trade-offs in more detail in Chap-
ter 10.

Let us now consider the implementation of the methods that manipulate
items at the head of the list (see Figure 9.4). First, to add an element at the
head of the list, we simply need to create a new Node that has the appropriate
value and references the very ﬁrst element of the list (currently, head). The head
of the new list is simply a reference to the new element. Finally, we modify the
count variable to reﬂect an increase in the number of elements.

public void addFirst(E value)
// post: value is added to beginning of list
{

// note order that things happen:
// head is parameter, then assigned
head = new Node<E>(value, head);
count++;

}

Removing a value should simply perform the reverse process. We copy the
reference2 to a temporary variable where it can be held for return, and then we
simply move the head reference down the list. Once completed, the value is
returned.

2 Remember: The assignment operator does not copy the value, just the reference. If you want a
reference to a new element, you should use the new operator and explicitly create a new object to
be referenced.

3LifeonMarsLifeonMarsYes!4194

Lists

public E removeFirst()
// pre: list is not empty
// post: removes and returns value from beginning of list
{

Node<E> temp = head;
head = head.next(); // move head down list
count--;
return temp.value();

}

Notice that removeFirst returns a value. Why not? Since addFirst “absorbs” a
value, removeFirst should do the reverse and “emit” one. Typically, the caller
will not dispose of the value, but re-insert it into another data structure. Of
course, if the value is not desired, the user can avoid assigning it a variable,
and it will be garbage-collected at some later time. Since we think of these
two operations as being inverses of each other, it is only natural to have them
balance the consumption of objects in this way.

Principle 12 Symmetry is good.

One interesting exception to Principle 12 only occurs in languages like Java,
where a garbage collector manages the recycling of dynamic memory. Clearly,
addFirst must construct a new element to hold the value for the list. On the
other hand, removeFirst does not explicitly get rid of the element. This is
because after removeFirst is ﬁnished, there are no references to the element
that was just removed. Since there are no references to the object, the garbage
collector can be assured that the object can be recycled. All of this makes the
programmer a little more lax about thinking about when memory has been
logically freed. In languages without garbage collection, a “dispose” operation
must be called for any object allocated by a new command. Forgetting to dispose
of your garbage properly can be a rude shock, causing your program to run out
of precious memory. We call this a memory leak. Java avoids all of this by
collecting your garbage for you.

There’s one more method that we provide for the sake of completeness:
getFirst.
It is a nondestructive method that returns a reference to the ﬁrst
value in the list; the list is not modiﬁed by this method; we just get access to
the data:

public E getFirst()
// pre: list is not empty
// post: returns first value in list
{

return head.value();

}

Next, we must write the methods that manipulate the tail of the list (see
Figure 9.5). While the interface makes these methods appear similar to those
that manipulate the head of the list, our implementation has a natural bias
against tail-oriented methods. Access through a single reference to the head

NNWSWSENEWSE9.4 Implementation: Singly Linked Lists

195

The process of adding a new value (shaded) to the tail of a list. The finger
Figure 9.5
reference keeps track of progress while searching for the element whose reference must
be modiﬁed.

of the list makes it difﬁcult to get to the end of a long singly linked list. More
“energy” will have to be put into manipulating items at the tail of the list.
Let’s see how these methods are implemented:

public void addLast(E value)
// post: adds value to end of list
{

// location for new value
Node<E> temp = new Node<E>(value,null);
if (head != null)
{

// pointer to possible tail
Node<E> finger = head;
while (finger.next() != null)
{

finger = finger.next();

}
finger.setNext(temp);

} else head = temp;

LifeonMarsfinger33LifeonMarsLifeonMarsfingergone!(a)(b)(c)4(d)3LifeonMarsfinger196

Lists

count++;

}

public E removeLast()
// pre: list is not empty
// post: removes last value from list
{

Node<E> finger = head;
Node<E> previous = null;
Assert.pre(head != null,"List is not empty.");
while (finger.next() != null) // find end of list
{

previous = finger;
finger = finger.next();

}
// finger is null, or points to end of list
if (previous == null)
{

// has exactly one element
head = null;

}
else
{

// pointer to last element is reset
previous.setNext(null);

}
count--;
return finger.value();

}

Each of these (complex) methods uses the ﬁnger-based list traversal tech-
nique. We reference each element of the list, starting at the top and moving
downward, until we ﬁnally reach the tail. At that point we have constructed
the desired reference to the end of the list, and we continue as we would have
in the head-manipulating methods. We have to be aware of one slight problem
that concerns the very simplest case—when the list is empty.
If there are no
elements, then finger never becomes non-null, and we have to write special
code to manipulate the head reference.

To support the add and remove methods of the Structure (and thus List)
interface, we had them call the addLast and removeLast methods, respectively.
Given their expense, there might be a good argument to have them manipulate
values at the head of the list, but that leads to an inconsistency with other
potential implementations. The correct choice in design is not always obvious.
Several methods potentially work in the context of the middle of lists—
including contains and remove. Here, the code becomes particularly tricky
because we cannot depend on lists having any values, and, for remove, we must
carefully handle the boundary cases—when the elements are the ﬁrst or last
elements of the list. Errors in code usually occur at these difﬁcult points, so it is
important to make sure they are tested.

9.4 Implementation: Singly Linked Lists

197

Principle 13 Test the boundaries of your structures and methods.

Here is the code for these methods:

public boolean contains(E value)
// pre: value is not null
// post: returns true iff value is found in list
{

Node<E> finger = head;
while (finger != null &&

!finger.value().equals(value))

{

finger = finger.next();

}
return finger != null;

}

public E remove(E value)
// pre: value is not null
// post: removes first element with matching value, if any
{

Node<E> finger = head;
Node<E> previous = null;
while (finger != null &&

!finger.value().equals(value))

{

previous = finger;
finger = finger.next();

}
// finger points to target value
if (finger != null) {

// we found element to remove
if (previous == null) // it is first
{

head = finger.next();

} else {

// it’s not first

previous.setNext(finger.next());

}
count--;
return finger.value();

}
// didn’t find it, return null
return null;

}

In the contains method we call the value’s equals method to test to see if the
values are logically equal. Comparing the values with the == operator checks to
see if the references are the same (i.e., that they are, in fact, the same object).
We are interested in ﬁnding a logically equal object, so we invoke the object’s
equals method.

NNWSWSENEWSE198

Lists

Figure 9.6
head of the list, (b) in the middle, (c) at the tail, or (d) not present.

The relation between finger and previous. The target element is (a) the

previousfingerLifeonMarspreviousfingerLifeonMars(b)(a)(c)(d)previousfingerLifeonMarspreviousfingerLifeonMars9.4 Implementation: Singly Linked Lists

199

Some fancy reference manipulation is needed in any routine that removes
an element from the list. When we ﬁnd the target value, the finger variable
has moved too far down to help with removing the element. By the time finger
references the element holding the target value, we lose the reference to the pre-
vious element—precisely the element that needs to have its next reference reset
when the value is removed. To avoid this difﬁculty, we keep another reference,
local to the particular method, that is either null or references the element just
before finger. When (and if) we ﬁnd a value to be removed, the element to
be ﬁxed is referenced by previous (Figure 9.6). Of course, if previous is null,
we must be removing the ﬁrst element, and we update the head reference. All
of this can be very difﬁcult to write correctly, which is another good reason to
write it carefully once and reuse the code whenever possible (see Principle 2,
Free the future: Reuse code).

One ﬁnal method with subtle behavior is the clear method. This removes
In Java, this is accomplished by clearing the

all the elements from the list.
reference to the head and adjusting the list size:

public void clear()
// post: removes all elements from list
{

head = null;
count = 0;

}

All that happens is that head stops referencing the list. Instead, it is explicitly
made to reference nothing. What happens to the elements of the list? When
the garbage collector comes along, it notices that the ﬁrst element of the former
list is not referenced by anything—after all it was only referenced by head be-
fore. So, the garbage collector collects that ﬁrst element as garbage. It is pretty
easy to see that if anything is referenced only by garbage, it is garbage. Thus,
the second element (as well as the value referenced by the ﬁrst element) will
be marked as garbage, and so forth. This cascading identiﬁcation of garbage
elements is responsible for recycling all the elements of the list and, potentially,
the Objects they reference. (If the list-referenced objects are referenced outside
of the list, they may not be garbage after all!)

We have left to this point the implementation of general methods for sup-
porting indexed versions of add and remove. These routines insert and remove
values found at particular offsets from the beginning of this list. Careful in-
spection of the AbstractList class shows that we have chosen to implement
addFirst and similar procedures in terms of the generic add and remove rou-
tines. We have, however, already seen quite efﬁcient implementations of these
routines. Instead, we choose to make use of the end-based routines to handle
special cases of the generic problem.

Here, we approach the adding of a value to the middle of a list. An index
is passed with a value and indicates the desired index of the value in the aug-
mented list. A ﬁnger keeps track of our progress in ﬁnding the correct location.

You are what
references you.

200

Lists

public void add(int i, E o)
// pre: 0 <= i <= size()
// post: adds ith entry of list to value o
{

Assert.pre((0 <= i)

&& (i <= size()),

"Index in range.");

if (i == size()) {
addLast(o);
} else if (i == 0) {
addFirst(o);

} else {

Node<E> previous = null;
Node<E> finger = head;
// search for ith position, or end of list
while (i > 0)
{

previous = finger;
finger = finger.next();
i--;

}
// create new value to insert in correct position
Node<E> current =

new Node<E>(o,finger);

count++;
// make previous value point to new value
previous.setNext(current);

}

}

Some thought demonstrates that the general code can be considerably simpli-
ﬁed if the boundary cases (adding near the ends) can be handled directly. By
handling the head and tail cases we can be sure that the new value will be in-
serted in a location that has a non-null previous value, as well as a non-null
next value. The loop is simpler, then, and the routine runs considerably faster.

A similar approach is used in the indexed remove routine:

public E remove(int i)
// pre: 0 <= i < size()
// post: removes and returns object found at that location
{

Assert.pre((0 <= i) && (i < size()),

"Index in range.");

if (i == 0) return removeFirst();
else if (i == size()-1) return removeLast();
Node<E> previous = null;
Node<E> finger = head;
// search for value indexed, keep track of previous
while (i > 0)
{

previous = finger;

9.5 Implementation: Doubly Linked Lists

201

finger = finger.next();
i--;

}
// in list, somewhere in middle
previous.setNext(finger.next());
count--;
// finger’s value is old value, return it
return finger.value();

}

Exercise 9.1 Implement the indexed set and get routines. You may assume the
existence of setFirst, setLast, getFirst, and getLast.

We now consider another implementation of the list interface that makes

use of two references per element.

Swoon!

9.5 Implementation: Doubly Linked Lists

In Section 9.4, we saw indications that some operations can take more “energy”
to perform than others, and expending energy takes time. Operations such as
modifying the tail of a singly linked list can take signiﬁcantly longer than those
that modify the head. If we, as users of lists, expect to modify the tail of the list
frequently, we might be willing to make our code more complex, or use more
space to store our data structure if we could be assured of signiﬁcant reductions
in time spent manipulating the list.

We now consider an implementation of a doubly linked list.

In a doubly
linked list, each element points not only to the next element in the list, but also
to the previous element (see Figure 9.7). The ﬁrst and last elements, of course,
have null previousElement and nextElement references, respectively.

In addition to maintaining a second reference within each element, we will
also consider the addition of a reference to the tail of the list (see Figure 9.8).
This one reference provides us direct access to the end of the list and has the
potential to improve the addLast and removeLast methods.

A cursory glance at the resulting data structure identiﬁes that it is more
symmetric with respect to the head and tail of the list. Writing the tail-related
methods can be accomplished by a simple rewriting of the head-related meth-
ods. Symmetry is a powerful concept in the design of complex structures; if
something is asymmetric, you should step back and ask yourself why.

Principle 14 Question asymmetry.

We begin by constructing a DoublyLinkedNode structure that parallels the
Node. The major difference is the addition of the previous reference that refers
to the element that occurs immediately before this element in the doubly linked
list. One side effect of doubling the number of references is that we duplicate
some of the information.

NNWSWSENEWSE202

Lists

Figure 9.7 A nonempty doubly linked list.

Figure 9.8 An empty doubly linked list.

Figure 9.9 Rhonda’s next reference duplicates Rhoda’s previous reference.

headcounttail3RhodavaluenpRhondaRhoryheadcount0tailRhodaRhonda9.5 Implementation: Doubly Linked Lists

203

If we look at two adjacent elements Rhonda and Rhoda in a doubly linked
list, their mutual adjacency is recorded in two references (Figure 9.9): Rhonda’s
nextElement reference refers to Rhoda, while Rhoda’s previousElement refer-
ence refers to Rhonda.Whenever one of the references is modiﬁed, the other
must be modiﬁed also. When we construct a new DoublyLinkedNode, we set
both the nextElement and previousElement references. If either is non-null,
a reference in the newly adjacent structure must be updated.
If we fail to do
this, the data structure is left in an inconsistent state.

Here’s the code:

Say that twice!

protected E data;
protected DoublyLinkedNode<E> nextElement;
protected DoublyLinkedNode<E> previousElement;

public DoublyLinkedNode(E v,

DoublyLinkedNode<E> next,
DoublyLinkedNode<E> previous)

{

}

data = v;
nextElement = next;
if (nextElement != null)

nextElement.previousElement = this;

previousElement = previous;
if (previousElement != null)

previousElement.nextElement = this;

public DoublyLinkedNode(E v)
// post: constructs a single element
{

this(v,null,null);

}

Now we construct the class describing the doubly linked list, proper. As with
any implementation of the list interface, it is necessary for our new Doubly-
LinkedList to provide code for each method not addressed in the AbstractList
class. The constructor simply sets the head and tail references to null and the
count to 0—the state identifying an empty list:

protected int count;
protected DoublyLinkedNode<E> head;
protected DoublyLinkedNode<E> tail;

public DoublyLinkedList()
// post: constructs an empty list
{

head = null;
tail = null;
count = 0;

}

DoublyLinked-
Node

DoublyLinked-
List

204

Lists

Many of the fast methods of SinglyLinkedLists, like addFirst, require

only minor modiﬁcations to maintain the extra references.

public void addFirst(E value)
// pre: value is not null
// post: adds element to head of list
{

// construct a new element, making it head
head = new DoublyLinkedNode<E>(value, head, null);
// fix tail, if necessary
if (tail == null) tail = head;
count++;

}

The payoff for all our extra references comes when we implement methods like
those modifying the tail of the list:

public void addLast(E value)
// pre: value is not null
// post: adds new value to tail of list
{

// construct new element
tail = new DoublyLinkedNode<E>(value, null, tail);
// fix up head
if (head == null) head = tail;
count++;

}

public E removeLast()
// pre: list is not empty
// post: removes value from tail of list
{

Assert.pre(!isEmpty(),"List is not empty.");
DoublyLinkedNode<E> temp = tail;
tail = tail.previous();
if (tail == null) {
head = null;

} else {

tail.setNext(null);

}
count--;
return temp.value();

}

Here, it is easy to see that head- and tail-based methods are textually similar,
making it easier to verify that they are written correctly. Special care needs to be
taken when these procedures handle a list that newly becomes either empty or
not empty. In these cases, both the head and tail references must be modiﬁed
to maintain a consistent view of the list. Some people consider the careful
manipulation of these references so time-consuming and error-prone that they
dedicate an unused element that permanently resides at the head of the list. It

9.5 Implementation: Doubly Linked Lists

205

is never seen or modiﬁed by the user, and it can simplify the code. Here, for
example, are the addLast and removeLast methods for this type of list:

public void addLast(E value)
{

// construct new element
tail = new DoublyLinkedNode<E>(value, null, tail);
count++;

}

public E removeLast()
{

Assert.pre(!isEmpty(),"List is not empty.");
DoublyLinkedNode<E> temp = tail;
tail = tail.previous();
tail.setNext(null);
count--;
return temp.value();

}

The reserved-element technique increases the amount of space necessary to
store a DoublyLinkedList by the size of a single element. The choice is left
to the implementor and is another example of a time–space trade-off.

Returning to our original implementation, we note that remove is simpliﬁed

by the addition of the previous reference:

public E remove(E value)
// pre: value is not null.
// post: first element matching value is removed from list
{

List can be empty

DoublyLinkedNode<E> finger = head;
while (finger != null &&

!finger.value().equals(value))

{

finger = finger.next();

}
if (finger != null)
{

// fix next field of element above
if (finger.previous() != null)
{

finger.previous().setNext(finger.next());

} else {

head = finger.next();

}
// fix previous field of element below
if (finger.next() != null)
{

finger.next().setPrevious(finger.previous());

} else {

206

Lists

tail = finger.previous();

}
count--;
return finger.value();

// fewer elements

}
return null;

}

Because every element keeps track of its previous element, there is no difﬁculty
in ﬁnding it from the element that is to be removed. Of course, once the removal
is to be done, several references need to be updated, and they must be assigned
carefully to avoid problems when removing the ﬁrst or last value of a list.

The List interface requires the implementation of two index-based methods
called indexOf and lastIndexOf. These routines return the index associated
with the ﬁrst (or last) element that is equivalent to a particular value. The
indexOf method is similar to the implementation of contains, but it returns
the index of the element, instead of the element itself. For DoublyLinkedLists,
the lastIndexOf method performs the same search, but starts at the tail of the
list. It is, essentially, the mirror image of an indexOf method.

public int lastIndexOf(E value)
// pre: value is not null
// post: returns the (0-origin) index of value,
//
{

or -1 if value is not found

int i = size()-1;
DoublyLinkedNode<E> finger = tail;
// search for last matching value, result is desired index
while (finger != null && !finger.value().equals(value))
{

finger = finger.previous();
i--;

}
if (finger == null)
{

// value not found, return indicator
return -1;

} else {

// value found, return index
return i;

}

}

9.6 Implementation: Circularly Linked Lists

Careful inspection of the singly linked list implementation identiﬁes one seem-
ingly unnecessary piece of data: the ﬁnal reference of the list. This reference is
always null, but takes up as much space as any varying reference. At the same
time, we were motivated to add a tail reference in the doubly linked list to help

9.6 Implementation: Circularly Linked Lists

207

The tail wags
the dog.

CircularList

Figure 9.10 A nonempty circularly linked list.

us access either end of the list with equal ease. Perhaps we could use the last
reference as the extra reference we need to keep track of one of the ends!

Here’s the technique: Instead of keeping track of both a head and a tail
reference, we explicitly keep only the reference to the tail. Since this element
would normally have a null reference, we use that reference to refer, implicitly,
to the head (see Figure 9.10). This implementation marries the speed of the
DoublyLinkedList with the space needed by the SinglyLinkedList. In fact,
we are able to make use of the Node class as the basis for our implementation.
To build an empty list we initialize the tail to null and the count to 0:

protected Node<E> tail;
protected int count;

public CircularList()
// pre: constructs a new circular list
{

tail = null;
count = 0;

}

Whenever access to the head of the list is necessary, we use tail.next(),
instead.3 Thus, methods that manipulate the head of the list are only slight
modiﬁcations of the implementations we have seen before for singly and doubly
linked lists. Here is how we add a value to the head of the list:

public void addFirst(E value)
// pre: value non-null
// post: adds element to head of list
{

3 This longhand even works in the case when there is exactly one element, since its next reference
points to itself.

count3tail"head"muffinsdonutsbagels208

Lists

Node<E> temp = new Node<E>(value);
if (tail == null) { // first value added

tail = temp;
tail.setNext(tail);

} else { // element exists in list
temp.setNext(tail.next());
tail.setNext(temp);

}
count++;

}

Now, to add an element to the end of the list, we ﬁrst add it to the head, and
then “rotate” the list by moving the tail down the list. The overall effect is to
have added the element to the tail!

public void addLast(E value)
// pre: value non-null
// post: adds element to tail of list
{

// new entry:
addFirst(value);
tail = tail.next();

}

The “recycling” of the tail reference as a new head reference does not solve
all our problems. Careful thought will demonstrate that the removal of a value
from the tail of the list remains a difﬁcult problem. Because we only have
access to the tail of the list, and not the value that precedes it, it is difﬁcult
to remove the ﬁnal value. To accomplish this, we must iterate through the
structure, looking for an element that refers to the same element as the tail
reference.

public E removeLast()
// pre: !isEmpty()
// post: returns and removes value from tail of list
{

Assert.pre(!isEmpty(),"list is not empty.");
Node<E> finger = tail;
while (finger.next() != tail) {
finger = finger.next();

}
// finger now points to second-to-last value
Node<E> temp = tail;
if (finger == tail)
{

tail = null;

} else {

finger.setNext(tail.next());
tail = finger;

}

9.7 Implementation: Vectors

209

count--;
return temp.value();

}

There are two approaches to improving the performance of this operation. First,
we could reconsider the previous links of the doubly linked list. There’s not
much advantage to doing this, and if we did, we could then keep the head
reference instead of the tail reference. The second technique is to point instead
to the element before the tail; that is the subject of Problem 9.11.

9.7 Implementation: Vectors

Careful inspection of the List interface makes it clear that the Vector class
actually implements the List interface. Thus, we can augment the Vector
deﬁnition with the phrase implements List.

With such varied implementations, it is important to identify the situations
where each of the particular implementations is most efﬁcient. As we had noted
before, the Vector is a good random access data structure. Elements in the
middle of the Vector can be accessed with little overhead. On the other hand,
the operations of adding or removing a value from the front of the Vector are
potentially inefﬁcient, since a large number of values must be moved in each
case.

In contrast, the dynamically allocated lists manipulate the head of the list
quite efﬁciently, but do not allow the random access of the structure without a
signiﬁcant cost. When the tail of a dynamically allocated list must be accessed
quickly, the DoublyLinkedList or CircularList classes should be used.

Exercise 9.2 In Exercise 6.3 (see page 144) we wrote a version of insertionSort
that sorts Vectors. Follow up on that work by making whatever modiﬁcation
would be necessary to have the insertionSort work on any type of List.

9.8 List Iterators

The observant reader will note that all classes that implement the Structure
class (see page 24) are required to provide an iterator method. Since the
List interface extends the Structure interface, all Lists are required to im-
plement an iterator method. We sketch the details of an Iterator over
Implementations of other List-based iterators are
SinglyLinkedLists here.
similar.

When implementing the VectorIterator it may be desirable to use only
methods available through the Vector’s public interface to access the Vector’s
data. Considering the List interface—an interface biased toward manipulating
the ends of the structure—it is not clear how a traversal might be accomplished
without disturbing the underlying List. Since several Iterators may be active
on a single List at a time, it is important not to disturb the host structure.

210

Lists

As a result, efﬁcient implementations of ListIterators must make use of the
protected ﬁelds of the List object.

The SinglyLinkedListIterator implements all the standard Iterator meth-

ods. To maintain its positioning within the List, the iterator maintains two
references: the head of the associated list and a reference to the current node.
The constructor and initialization methods appear as follows:

protected Node<E> current;
protected Node<E> head;

SinglyLinked-
ListIterator

public SinglyLinkedListIterator(Node<E> t)
// post: returns an iterator that traverses a linked list
{

head = t;
reset();

}

public void reset()
// post: iterator is reset to beginning of traversal
{

current = head;

}

When called by the SinglyLinkedList’s iterator method, the protected head
reference is passed along. The constructor caches away this value for use in
reset. The reset routine is then responsible for initializing current to the
value of head. The Iterator is able to refer to the Nodes because both structures
are in the same package.

The value-returning routines visit each element and “increment” the current

reference by following the next reference:

protected Node<E> current;
protected Node<E> head;

public boolean hasNext()
// post: returns true if there is more structure to be viewed:
//
{

i.e., if value (next) can return a useful value.

return current != null;

}

public E next()
// pre: traversal has more elements
// post: returns current value and increments iterator
{

E temp = current.value();
current = current.next();
return temp;

}

9.9 Conclusions

211

The traversal is ﬁnished when the current reference “falls off” the end of the
List and becomes null.

Observe that the Iterator is able to develop references to values that are
not accessible through the public interface of the underlying List structure.
While it is of obvious utility to access the middle elements of the List, these
references could be used to modify the associated List structure. If the objects
referred to through the Iterator are modiﬁed, this underlying structure could
become corrupted. One solution to the problem is to return copies or clones of
the current object, but then the references returned are not really part of the
List. The best advice is to think of the values returned by the Iterator as
read-only.

Principle 15 Assume that values returned by iterators are read-only.

9.9 Conclusions

In this chapter we have developed the notion of a list and three different im-
plementations. One of the features of the list is that as each of the elements is
added to the list, the structure is expanded dynamically, using dynamic mem-
ory. To aid in keeping track of an arbitrarily large number of chunks of dynamic
memory, we allocate, with each chunk, at least one reference for keeping track
of logically nearby memory.

Although the description of the interface for lists is quite detailed, none of
the details of any particular implementation show through the interface. This
approach to designing data structures makes it less possible for applications to
depend on the peculiarities of any particular implementation, making it more
likely that implementations can be improved without having to reconsider indi-
vidual applications.

Finally, as we investigated each of the three implementations, it became
clear that there were certain basic trade-offs in good data structure design. In-
creased speed is often matched by an increased need for space, and an increase
in complexity makes the code less maintainable. We discuss these trade-offs in
more detail in upcoming chapters.

NNWSWSENEWSE212

Lists

Self Check Problems

Solutions to these problems begin on page 446.

9.1
Vector implementation?

What are the essential distinctions between the List types and the

Why do most List implementations make use of one or more references

9.2
for each stored value?

9.3

How do we know if a structure qualiﬁes as a List?

If class C extends the SinglyLinkedList class, is it a SinglyLinkedList?

9.4
Is it a List? Is it an AbstractList? Is it a DoublyLinkedList?

The DoublyLinkedList class has elements with two pointers, while the
9.5
SinglyLinkedList class has elements with one pointer. Is DoublyLinkedList
a SinglyLinkedList with additional information?

9.6

9.7

Why do we have a tail reference in the DoublyLinkedList?

Why don’t we have a tail reference in the SinglyLinkedList?

9.8
might we use it, in any case?

The ListVector implementation of a List is potentially slow? Why

9.9
references. Why?

The AbstractList class does not make use of any element types or

9.10
does it get added?

If you use the add method to add an element to a List, to which end

The get and set methods take an integer index. Which element of the

9.11
list is referred to by index 1?

Problems

Solutions to the odd-numbered problems begin on page 471.

When considering a data structure it is important to see how it works in
9.1
the boundary cases. Given an empty List, which methods may be called without
violating preconditions?

9.2
types we have seen in this chapter.

Compare the implementation of getLast for each of the three List

From within Java programs, you may access information on the Web
9.3
using URL’s (uniform resource locators). Programmers at MindSlave software
(working on their new NetPotato browser) would like to keep track of a poten-
tially large bookmark list of frequently visited URL’s. It would be most useful
if they had arbitrary access to the values saved within the list.
Is a List an
appropriate data structure? (Hint: If not, why?)

Write a List method, equals, that returns true exactly when the el-
Ideally, your implementation should

9.4
ements of two lists are pair-wise equal.
work for any List implementation, without change.

9.9 Conclusions

213

Design a method that inserts an object into the middle of a CircularList.

Write a method of CircularList, called reverse, that reverses the

Write a method of DoublyLinkedList, called reverse, that reverses the

Write a method of SinglyLinkedList, called reverse, that reverses the
9.5
order of the elements in the list. This method should be destructive—it should
modify the list upon which it acts.
9.6
order of the elements in the list. This method should be destructive.
9.7
order of the element in the list. This method should be destructive.
Each of the n references in a singly linked list are needed if we wish
9.8
In a doubly linked list, are each of the addi-
to remove the ﬁnal element.
tional n previous references necessary if we want to remove the tail of the
list in constant time? (Hint: What would happen if we mixed Nodes and
DoublyLinkedNodes?)
9.9
9.10 Which implementation of the size and isEmpty methods would you
use if you had the potential for a million-element list. (Consider the problem
of keeping track of the alumni for the University of Michigan.) How would you
choose if you had the potential for a million small lists. (Consider the problem
of keeping track of the dependents for each of a million income-tax returns.)
One way to make all the circular list operations run quickly is to keep
9.11
track of the element that points to the last element in the list. If we call this
penultimate, then the tail is referenced by penultimate.next, and the head
by penultimate.next.next. What are the disadvantages of this?
Suppose we read n integers 1, 2, . . . , n from the input, in order. Flipping
9.12
a coin, we add each new value to either the head or tail of the list. Does this
shufﬂe the data? (Hint: See Problem 6.18.)
9.13 Measure the performance of addFirst, remove(Object), and remove-
Last for each of the three implementations (you may include Vectors, if you
wish). Which implementations perform best for small lists? Which implemen-
tations perform best for large lists?
Consider the implementation of an insertionSort that works on Lists.
9.14
(See Exercises 6.3 and 9.2.) What is the worst-case performance of this sort?
Be careful.
9.15
Lists. (Hint: A wrapper may be useful.)
9.16
LinkedLists.
Suppose the add of the Unique program is replaced by addFirst and the
9.17
program is run on (for example) the ﬁrst chapter of Mark Twain’s Tom Sawyer.
Why does the modiﬁed program run as much as 25 percent slower than the
program using the add (i.e., addLast) method? (Hint: Mark Twain didn’t write
randomly.)
9.18
Lists.

Describe an implementation for an iterator associated with Circular-

Implement a recursive version of the size method for SinglyLinked-

Implement a recursive version of the contains method for Singly-

9.10 Laboratory: Lists with Dummy Nodes

Objective. To gain experience implementing List-like objects.

Discussion. Anyone attempting to understand the workings of a doubly linked
list understands that it is potentially difﬁcult to keep track of the references. One
of the problems with writing code associated with linked structures is that there
are frequently boundary cases. These are special cases that must be handled
carefully because the “common” path through the code makes an assumption
that does not hold in the special case.

Take, for example, the addFirst method for DoublyLinkedLists:

public void addFirst(E value)
// pre: value is not null
// post: adds element to head of list
{

// construct a new element, making it head
head = new DoublyLinkedNode<E>(value, head, null);
// fix tail, if necessary
if (tail == null) tail = head;
count++;

}

The presence of the if statement suggests that sometimes the code must reas-
sign the value of the tail reference. Indeed, if the list is empty, the ﬁrst element
must give an initial non-null value to tail. Keeping track of the various special
cases associated with a structure can be very time consuming and error-prone.
One way that the complexity of the code can be reduced is to introduce
dummy nodes. Usually, there is one dummy node associated with each external
reference associated with the structure. In the DoublyLinkedList, for example,
we have two references (head and tail); both will refer to a dedicated dummy
node:

RhodaRhoryRhondaheadcounttail3valuepn216

Lists

LinkedList

These nodes appear to the code to be normal elements of the list. In fact, they
do not hold any useful data. They are completely hidden by the abstraction of
the data structure. They are transparent.

Because most of the boundary cases are associated with maintaining the
correct values of external references and because these external references are
now “hidden” behind their respective dummy nodes, most of the method code
is simpliﬁed. This comes at some cost: the dummy nodes take a small amount
of space, and they must be explicitly stepped over if we work at either end of
the list. On the other hand, the total amount of code to be written is likely
to be reduced, and the running time of many methods decreases if the special
condition testing would have been expensive.

Procedure. In this lab we will extend the DoublyLinkedList, building a new
class, LinkedList, that makes use of two dummy nodes: one at the head of the
list, and one at the end.

You should begin taking a copy of the LinkedList.java starter ﬁle. This
ﬁle simply declares LinkedList to be an extension of the structure package’s
DoublyLinkedList class. The code associated with each of the existing methods
is similar to the code from DoublyLinkedList. You should replace that code
with working code that makes use of two dummy nodes:

1. First, recall that the three-parameter constructor for DoublyLinkedList-
Elements takes a value and two references—the nodes that are to be next
and previous to this new node. That constructor will also update the
next and previous nodes to point to the newly constructed node. You
may ﬁnd it useful to use the one-parameter constructor, which builds a
node with null next and previous references.

2. Replace the constructor for the LinkedList. Instead of constructing head
and tail references that are null, you should construct two dummy
nodes; one node is referred to by head and the other by tail. These
dummy nodes should point to each other in the natural way. Because these
dummy nodes replace the null references of the DoublyLinkedList class,
we will not see any need for null values in the rest of the code. Amen.

3. Check and make necessary modiﬁcations to size, isEmpty, and clear.

4. Now, construct two important protected methods. The method insert-
After takes a value and a reference to a node, previous. It inserts a new
node with the value value that directly follows previous. It should be
declared protected because we are not interested in making it a formal
feature of the class. The other method, remove, is given a reference to a
node. It should unlink the node from the linked list and return the value
stored in the node. You should, of course, assume that the node removed
is not one of the dummy nodes. These methods should be simple with no
if statements.

9.10 Laboratory: Lists with Dummy Nodes

217

5. Using insertAfter and remove, replace the code for addFirst, addLast,
getFirst, getLast, removeFirst, and removeLast. These methods should
be very simple (perhaps one line each), with no if statements.

6. Next, replace the code for the indexed versions of methods add, remove,
get, and set. Each of these should make use of methods you have already
written. They should work without any special if statements.

7. Finally, replace the versions of methods indexOf, lastIndexOf, and contains
(which can be written using indexOf), and the remove method that takes
an object. Each of these searches for the location of a value in the list
and then performs an action. You will ﬁnd that each of these methods is
simpliﬁed, making no reference to the null reference.

Thought Questions. Consider the following questions as you complete the lab:

1. The three-parameter constructor for DoublyLinkedNodes makes use of
two if statements. Suppose that you replace the calls to this construc-
tor with the one-parameter constructor and manually use setNext and
setPrevious to set the appropriate references. The if statements disap-
pear. Why?

2. The contains method can be written making use of the indexOf method,

but not the other way around. Why?

3. Notice that we could have replaced the method insertAfter with a sim-
ilar method, insertBefore. This method inserts a new value before the
indicated node. Some changes would have to be made to your code. There
does not appear, however, to be a choice between versions of remove. Why
is this the case? (Hint: Do you ever pass a dummy node to remove?)

4. Even though we don’t need to have the special cases in, for example, the
indexed version of add, it is desirable to handle one or more cases in a
special way. What are the cases, and why is it desirable?

5. Which ﬁle is bigger: your ﬁnal result source or the original?

Notes:

Chapter 10

Linear Structures

Concepts:
.
Stacks
. Queues

“Rule Forty-two.
All persons more than a mile high to leave the court.”. . .
“Well, I shan’t go,” said Alice; “Besides
that’s not a regular rule: you just invented it now.”
“It’s the oldest rule in the book,” said the King.
“Then it ought to be Number One,” said Alice.
—Charles Lutwidge Dodgson

THE STATE OF SOME STRUCTURES REFLECTS THEIR HISTORY. Many systems—for
example, a line at a ticket booth or, as Alice presumes, the King’s book of rules—
modify their state using two pivotal operations: add and remove. As with most
data structures, when values are added and removed the structure grows and
shrinks. Linear structures, however, shrink in a predetermined manner: values
are removed in an order based only on the order they were added. All linear
structures abide by a very simple interface:

public interface Linear<E> extends Structure<E>
{

public void add(E value);
// pre: value is non-null
// post: the value is added to the collection,
//

the consistent replacement policy is not specified

Linear

public E get();
// pre: structure is not empty
// post: returns reference to next object to be removed

public E remove();
// pre: structure is not empty
// post: removes an object from store

public int size();
// post: returns the number of elements in the structure

public boolean empty();
// post: returns true if and only if the linear structure is empty

}

For each structure the add and remove methods are used to insert new values
into the linear structure and remove those values later. A few utility routines are

220

Linear Structures

also provided: get retrieves a copy of the value that would be removed next,
while size returns the size of the structure and empty returns whether or not
the linear structure is empty.

One thing that is important to note is that the empty method seems to pro-
vide the same features as the isEmpty method we have seen in previous struc-
tures. This is a common feature of ongoing data structure design—as structures
evolve, aliases for methods arise. In this case, many native classes of Java ac-
tually have an empty method. We provide that for compatibility; we implement
it in our abstract implementation of the Linear interface. The empty method
simply calls the isEmpty method required by the Structure interface and ulti-
mately coded by the particular Linear implementation.

abstract public class AbstractLinear<E> extends AbstractStructure<E>

Abstract-
Linear

{

implements Linear<E>

public boolean empty()
// post: return true iff the linear structure is empty
{

return isEmpty();

}

public E remove(E o)
// pre: value is non-null
// post: value is removed from linear structure, if it was there
{

Assert.fail("Method not implemented.");
// never reaches this statement:
return null;

}

}

Since our Linear interface extends our notion of Structure, so we will also
In particular, the
be required to implement the methods of that interface.
remove(Object) method is required. We use the AbstractLinear implemen-
tation to provide a default version of the remove method that indicates it is not
yet implemented. The beneﬁts of making Linear classes instances of Structure
outweigh the perfect ﬁt of the Structure interface.

Since AbstractLinear extends AbstractStructure, any features of the
AbstractStructure implementation are enjoyed by the AbstractLinear in-
terface as well.

We are now ready to implement several implementations of the Linear in-
terface. The two that we will look at carefully in this chapter, stacks and queues,
are the most common—and most widely useful—examples of linear data struc-
tures we will encounter.

10.1 Stacks

10.1 Stacks

221

Our ﬁrst linear structure is a stack. A stack is a collection of items that exhibit
the behavior that the last item in is the ﬁrst item out.
It is a LIFO (“lie-foe”)
structure. The add method pushes an item onto the stack, while remove pops
off the item that was pushed on most recently. We will provide the traditionally
named methods push and pop as alternatives for add and remove, respectively.
We will use these stack-speciﬁc terms when we wish to emphasize the LIFO
quality. A nondestructive operation, get, returns the top element of the Stack—
the element that would be returned next. Since it is meaningless to remove or
get a Stack that is empty, it is important to have access to the size methods
(empty and size). Here is the interface that deﬁnes what it means to be a
Stack:

public interface Stack<E> extends Linear<E>
{

public void add(E item);
// post: item is added to stack
//

will be popped next if no intervening add

public void push(E item);
// post: item is added to stack
//

will be popped next if no intervening push

Stack

public E remove();
// pre: stack is not empty
// post: most recently added item is removed and returned

public E pop();
// pre: stack is not empty
// post: most recently pushed item is removed and returned

public E get();
// pre: stack is not empty
// post: top value (next to be popped) is returned

public E getFirst();
// pre: stack is not empty
// post: top value (next to be popped) is returned

public E peek();
// pre: stack is not empty
// post: top value (next to be popped) is returned

public boolean empty();
// post: returns true if and only if the stack is empty

public int size();
// post: returns the number of elements in the stack

222

Linear Structures

QuickSort

QuickSort

}

To maintain applications that are consistent with Java’s java.util.Stack the
alternative operations of push and pop may be preferred in some of our discus-
sions.

10.1.1 Example: Simulating Recursion

Earlier we mentioned that tail recursive methods could be transformed to use
loops. More complex recursive methods—especially those that perform mul-
tiple recursive calls—can be more complex to implement. In this section, we
focus on the implementation of an iterative version of the quicksort sorting al-
gorithm. Before we turn to the implementation, we will investigate the calling
mechanisms of languages like Java.

Data available to well-designed methods come mainly from two locations:
the method’s parameters and its local variables. These values help to deﬁne
the method’s current state. In addition to the explicit values, implicit parame-
ters also play a role. Let us consider the recursive version of quicksort we saw
earlier:

private static void quickSortRecursive(int data[],int left,int right)
// pre: left <= right
// post: data[left..right] in ascending order
{

// the final location of the leftmost value

int pivot;
if (left >= right) return;
pivot = partition(data,left,right);
quickSortRecursive(data,left,pivot-1); /* 2 - sort small */
quickSortRecursive(data,pivot+1,right);/* 3 - sort large */
/* done! */

/* 1 - place pivot */

}

The ﬂow of control in most methods is from top to bottom. If the machine stops
execution, it is found to be executing one of the statements. In our recursive
quicksort, there are three main points of execution: (1) before the ﬁrst recursive
call, (2) before the second recursive call, and (3) just before the return. To focus
on what needs to be accomplished, the computer keeps a special reference to
the code, called a program counter. For the purposes of our exercise, we will
assume the program counter takes on the value 1, 2, or 3, depending on its
location within the routine.

These values—the parameters, the local variables, and the program coun-
ter—reside in a structure called a call frame. Whenever a method is called,
a new call frame is constructed and ﬁlled out with appropriate values. Be-
cause many methods may be active at the same time (methods can call each
other), there are usually several active call frames. These frames are main-
tained in a call stack. Since the ﬁrst method to return is the last method called,
a stack seems appropriate. Figure 10.1 describes the relationship between the
call frames of the call stack and a partial run of quicksort.

10.1 Stacks

223

Figure 10.1 A partially complete quicksort and its call stack. The quicksort routine
is currently partitioning the three-element subarray containing 2, 3, and 3. The curve
indicates the progress of recursion in quicksort to this point.

4058434265132043−1003011151551212lowhighpivot1358424402433650−1−10−11323−104133232313320123456789101167PC−1014406225Quicksort progressCall stack baseCall frame 0Call frame 1Call frame 2Call frame 3Call frame 4Call stack top224

Linear Structures

Our approach is to construct a stack of frames that describes the progress of
the various “virtually recursive” calls to quicksort. Each frame must maintain
enough information to simulate the actual call frame of the recursive routine:

class callFrame
{

int pivot;// location of pivot
int low;
int high; // right index
int PC; // next statement (see numbering in recursive code)

// left index

public callFrame(int l, int h)
// post: generate new call frame with low and high as passed
{

low = l; high = h; PC = 1;

}

}

Just as with real call frames, it’s possible for variables to be uninitialized (e.g.,
pivot)! We can consider an iterative version of quicksort:

public static void quickSortIterative(int data[], int n)
// pre: n <= data.length
// post: data[0..n-1] in ascending order
{

Stack<callFrame> callStack = new StackList<callFrame>();
callStack.push(new callFrame(0,n-1));
while (!callStack.isEmpty())
{

// some "virtual" method outstanding
callFrame curr = callStack.get();
if (curr.PC == 1) { // partition and sort lower

// return if trivial
if (curr.low >= curr.high) { callStack.pop(); continue; }
// place the pivot at the correct location
curr.pivot = partition(data,curr.low,curr.high);
curr.PC++;
// sort the smaller values...
callStack.push(new callFrame(curr.low,curr.pivot-1));

} else if (curr.PC == 2) { // sort upper

curr.PC++;
// sort the larger values....
callStack.push(new callFrame(curr.pivot+1,curr.high));

} else { callStack.pop(); continue; } // return

}

}

We begin by creating a new stack initialized with a callFrame that simulates
ﬁrst call to the recursive routine. The low and high variables of the frame are
initialized to 0 and n − 1 respectively, and we are about to execute statement 1.
As long as there is a call frame on the stack, some invocation of quicksort is

10.1 Stacks

225

still executing. Looking at the top frame, we conditionally execute the code
associated with each statement number. For example, statement 1 returns (by
popping off the top call frame) if low and high suggest a trivial sort. Each vari-
able is preﬁxed by curr because the local variables reside within the call frame,
curr. When we would execute a recursive call, we instead increment the “pro-
gram counter” and push a new frame on the stack with appropriate initialization
of local variables. Because each local variable appears in several places on the
call stack, recursive procedures appear to have fewer local variables. The con-
version of recursion to iteration, then, requires more explicit handling of local
storage during execution.

The sort continues until all the call frames are popped off. This happens
when each method reaches statement 3, that is, when the recursive quicksort
would have returned.

We now discuss two implementations of Stacks: one based on a Vector and

one based on a List.

10.1.2 Vector-Based Stacks

Let us consider a traditional stack-based analogy: the storage of trays in a fast-
food restaurant. At the beginning of a meal, you are given a tray from a stack.
The process of removing a tray is the popping of a tray off the stack. When trays
are returned, they are pushed back on the stack.

Now, assume that the side of the tray holder is marked in a rulerlike fashion,
perhaps to measure the number of trays stacked. Squinting one’s eyes, this looks
like a sideways vector:

Following our analogy, the implementation of a Stack using a Vector can be
accomplished by aligning the “top” of the Stack with the “end” of the Vector
(see Figure 10.2). We provide two constructors, including one that provides the
Vector with the initial capacity:

protected Vector<E> data;

public StackVector()
// post: an empty stack is created
{

data = new Vector<E>();

}

StackVector

Tray Stacker 1009226

Linear Structures

Figure 10.2 A Vector-based stack containing n elements. The top of the stack is
implied by the length of the underlying vector. The arrow demonstrates the direction of
growth.

public StackVector(int size)
// post: an empty stack with initial capacity of size is created
{

data = new Vector<E>(size);

}

To add elements on the Stack, we simply use the add method of the Vector.
When an element is to be removed from the Stack, we carefully remove the last
element, returning its value.

public void add(E item)
// post: item is added to stack
//
{

will be popped next if no intervening add

data.add(item);

}

public E remove()
// pre: stack is not empty
// post: most recently added item is removed and returned
{

return data.remove(size()-1);

}

The add method appends the element to the end of the Vector, extending it if
necessary. Notice that if the vector has n elements, this element is written to slot
n, increasing the number of elements to n + 1. Removing an element reverses
this process: it removes and returns the last element. (Note the invocation of
our principle of symmetry, here. We will depend on this notion in our design
of a number of add and remove method pairs.) The get method is like remove,

−1n1234501234data.size() == ndataTop of stackFirst free elementn10.1 Stacks

227

except that the stack is not modiﬁed. The size of the stack is easily determined
by requesting the same information from the underlying Vector. Of course
when the Vector is empty, so is the Stack it supports.1

public boolean isEmpty()
// post: returns true if and only if the stack is empty
{

return size() == 0;

}

public int size()
// post: returns the number of elements in stack
{

return data.size();

}

public void clear()
// post: removes all elements from stack
{

data.clear();

}

The clear method is required because the Stack indirectly extends the Struc-
ture interface.

10.1.3 List-Based Stacks

Only the top “end” of the stack ever gets modiﬁed. It is reasonable, then, to seek
an efﬁcient implementation of a Stack using a SinglyLinkedList. Because our
SinglyLinkedList manipulates its head more efﬁciently than its tail, we align
the Stack top with the head of the SinglyLinkedList (see Figure 10.3).

The add method simply performs an addFirst, and the remove operation
performs removeFirst. Since we have implemented the list’s remove operation
so that it returns the value removed, the value can be passed along through the
stack’s remove method.

public void add(E value)
// post: adds an element to stack;
//
{

will be next element popped if no intervening push

StackList

data.addFirst(value);

}

public E remove()
// pre: stack is not empty

1 Again, the class java.util.Stack has an empty method that is analogous to our isEmpty
method. We prefer to use isEmpty for consistency.

228

Linear Structures

Figure 10.3 A stack implemented using a singly-linked list. The arrow demonstrates
the direction of stack growth.

// post: removes and returns the top element from stack
{

return data.removeFirst();

}

The remaining methods are simply the obvious wrappers for similar methods
from the SinglyLinkedList. Because the stack operations are trivial represen-
tations of the linked-list operations, their complexities are all the same as the
corresponding operations found in linked lists—each takes constant time.

It should be noted that any structure implementing the List interface would
be sufﬁcient for implementing a Stack. The distinction, however, between the
various lists we have seen presented here has focused on providing quick access
to the tail of the list. Since stacks do not require access to the tail, the alternative
implementations of list structure do not provide any beneﬁt. Thus we use the
SinglyLinkedList implementation.

10.1.4 Comparisons

Clearly, stacks are easily implemented using the structures we have seen so far.
In fact, the complexities of those operations make it difﬁcult to decide which of
the classes is better. How can we make the decision?

Let’s consider each a little more carefully. First, in terms of time, both under-
lying structures provide efﬁcient implementations. In addition, both structures
provide (virtually) unlimited extension of the structure. Their difference stems
from a difference in approach for expanding the structure. In the case of the
vector, the structure is responsible for deciding when the structure is extended.

head5432110.2 Queues

229

Because the structure grows by doubling, the reallocation of memory occurs in-
creasingly less often, but when that reallocation occurs, it takes an increasingly
long time. So, while the amortized cost of dynamically extending vectors is con-
stant time per element, the incremental cost of extending the vector either is
zero (if no extension actually occurs) or is occasionally proportional to the size
of the vector. Some applications may not be tolerant of this great variation in the
cost of extending the structure. In those cases the StackList implementation
should be considered.

The constant incremental overhead of expanding the StackList structure,
however, comes at a price. Since each element of the list structure requires a
reference to the list that follows, there is a potentially signiﬁcant overhead in
terms of space. If the items stored in the stack are roughly the size of a reference,
the overhead is signiﬁcant. If, however, the size of a reference is insigniﬁcant
when compared to the size of the elements stored, the increase in space used
may be reasonable.

10.2 Queues

Most of us have participated in queues—at movie theaters, toll booths, or ice
cream shops, or while waiting for a communal bathroom in a large family! A
queue, like a stack, is an ordered collection of elements with tightly controlled
access to the structure. Unlike a stack, however, the ﬁrst item in is the ﬁrst
item out. We call it a FIFO (“ﬁe-foe”) structure. FIFO’s are useful because they
maintain the order of the data that run through them.

The primary operations of queues are to enqueue and dequeue elements.
Again, to support the Linear interface, we supply the add and remove meth-
ods as alternatives. Elements are added at the tail of the structure, where they
then pass through the structure, eventually reaching the head where they are re-
moved. The interface provides a number of other features we have seen already:

public interface Queue<E> extends Linear<E>
{

public void add(E value);
// post: the value is added to the tail of the structure

public void enqueue(E value);
// post: the value is added to the tail of the structure

Queue

public E remove();
// pre: the queue is not empty
// post: the head of the queue is removed and returned

public E dequeue();
// pre: the queue is not empty
// post: the head of the queue is removed and returned

public E getFirst();

230

Linear Structures

// pre: the queue is not empty
// post: the element at the head of the queue is returned

public E get();
// pre: the queue is not empty
// post: the element at the head of the queue is returned

public E peek();
// pre: the queue is not empty
// post: the element at the head of the queue is returned

public boolean empty();
// post: returns true if and only if the queue is empty

public int size();
// post: returns the number of elements in the queue

}

As with the Stack deﬁnition, the Queue interface describes necessary char-
acteristics of a class, but not the code to implement it. We use, instead, the
AbstractQueue class to provide any code that might be of general use in imple-
menting queues. Here, for example, we provide the various traditional aliases
for the standard operations add and remove required by the Linear interface:

public abstract class AbstractQueue<E>

extends AbstractLinear<E> implements Queue<E>

AbstractQueue

{

public void enqueue(E item)
// post: the value is added to the tail of the structure
{

add(item);

}

public E dequeue()
// pre: the queue is not empty
// post: the head of the queue is removed and returned
{

return remove();

}

public E getFirst()
// pre: the queue is not empty
// post: the element at the head of the queue is returned
{

return get();

}

public E peek()
// pre: the queue is not empty
// post: the element at the head of the queue is returned

231

CoinPuzzle

10.2 Queues

{

}

}

return get();

We will use this abstract class as the basis for the various implementations of
queues that we see in this chapter.

10.2.1 Example: Solving a Coin Puzzle

As an example of an application of queues, we consider an interesting coin
puzzle (see Figure 10.4). A dime, penny, nickel, and quarter are arranged in
decreasing size in each of the leftmost four squares of a ﬁve-square board. The
object is to reverse the order of the coins and to leave them in the rightmost
four slots, in the least number of moves. In each move a single coin moves to
the left or right. Coins may be stacked, but only the top coin is allowed to move.
When a coin moves, it may not land off the board or on a smaller coin. A typical
intermediate legal position is shown in the middle of Figure 10.4. From this
point, the nickel may move right and the dime may move in either direction.

We begin by deﬁning a “board state.” This object keeps track of the posi-
tions of each of the four coins, as well as the series of board states that lead
to this state from the start. Its implementation is an interesting problem (see
Problem 1.14). We outline its interface here:

class State
{

public static final int DIME=0;
public static final int PENNY=1;
public static final int NICKEL=2;
public static final int QUARTER=3;
public static final int LEFT = -1; // directions
public static final int RIGHT = 1;

// coins

public State()
// post: construct initial layout of coins

public State(State prior)
// pre: prior is a non-null state
// post: constructs a copy of that state to be successor state

public boolean done()
// post: returns true if state is the finish state

public boolean validMove(int coin, int direction)
// pre: State.DIME <= coin <= State.QUARTER
//
// post: returns true if coin can be moved in desired direction

direction = State.left or State.right

232

Linear Structures

Figure 10.4 A four-coin puzzle. Top and bottom orientations depict the start and ﬁnish
positions. A typical legal position is shown in the middle.

public State move(int coin, int direction)
// pre: coin and direction describe valid move
// post: coin is moved in that direction

public int printMoves()
// post: print moves up to and including this point

public int id()
// post: construct an integer representing state
//

states of coins are equal iff id’s are equal

}

The parameterless constructor builds a board in the initial state, while the one-
parameter constructor generates a new state to follow prior. The done method
checks for a state that is in the ﬁnal position. Coins (identiﬁed by constants
such as State.PENNY) are moved in different directions (e.g., State.LEFT), but
only if the move is valid. Once the ﬁnal state is found, the intermediate states
leading to a particular board position are printed with printMoves.

Once the State class has been deﬁned, it is a fairly simple matter to solve

the puzzle:

public static void main(String args[])
{

Queue<State> pool = new QueueList<State>();
State board = new State();

25255551LegalFinish11Start40123401231041010012310.2 Queues

233

BitSet seen = new BitSet(5*5*5*5);
pool.add(board);
while (!pool.isEmpty())
{

board = (State)pool.remove();
if (board.done()) break;
int moveCode = board.id();
if (seen.contains(moveCode)) continue;
seen.add(moveCode);
for (int coin = State.DIME; coin <= State.QUARTER; coin++)
{

if (board.validMove(coin,State.LEFT))

pool.add(board.move(coin,State.LEFT));

if (board.validMove(coin,State.RIGHT))

pool.add(board.move(coin,State.RIGHT));

}

}
board.printMoves();

}

We begin by keeping a pool of potentially unvisited board states. This queue
initially includes the single starting board position. At each stage of the loop an
unvisited board position is checked. If it is not the ﬁnish position, the boards
generated by the legal moves from the current position are added to the state
pool. Processing continues until the ﬁnal position is encountered. At that point,
the intermediate positions are printed out.

Because the pool is a FIFO, each unvisited position near the head of the
queue must be at least as close to the initial position as those found near the
end. Thus, when the ﬁnal position is found, the distance from the start position
(in moves) is a minimum!

A subtle point in this code is the use of the id function and the BitSet.2
The id function returns a small integer (between 0 and 55 − 1) that uniquely
identiﬁes the state of the board. These integers are kept in the set. If, in the
future, a board position with a previously encountered state number is found,
the state can safely be ignored. Without this optimization, it becomes difﬁcult
to avoid processing previously visited positions hundreds or thousands of times
before a solution is found. We will leave it to the reader to ﬁnd the solution
either manually or automatically. (The fastest solution involves 22 moves.)

We now investigate three different implementations of queues, based on
Lists, Vectors, and arrays. Each has its merits and drawbacks, so we consider
them carefully.

2 BitSets are part of the java.util package, and we provide public implementations of these and
other sets in the structure package. They are not discussed formally within this text.

234

Linear Structures

10.2.2 List-Based Queues

Drawing upon our various analogies with real life, it is clear that there are two
points of interest in a queue: the head and the tail. This two-ended view of
queues makes the list a natural structure to support the queue. In this imple-
mentation, the head and tail of the queue correspond to the head and tail of the
list. The major difference is that a queue is restricted to peeking at and remov-
ing values from the head and appending elements to the tail. Lists, on the other
hand, allow adding and removing values from both ends.

This discussion leads us to the following protected ﬁeld within the QueueList

structure:

protected List<E> data;

QueueList

When we consider the constructor, however, we are forced to consider more
carefully which implementation of a list is actually most suitable. While for the
stack the SinglyLinkedList was ideal, that implementation only provides efﬁ-
cient manipulation of the head of the list. To get fast access to both ends of the
list, we are forced to consider either the DoublyLinkedList (see Figure 10.5)
or the CircularList. Either would be time-efﬁcient here, but we choose the
more compact CircularList. We now consider our constructor:

public QueueList()
// post: constructs a new, empty queue
{

data = new CircularList<E>();

}

Notice that, because the List structure has unbounded size, the Queue structure
built atop it is also unbounded. This is a nice feature, since many applications
of queues have no easily estimated upper bound.

To add a value to the queue, we simply add an element to the tail of the list.
Removing a value from the list provides what we need for the remove operation.
(Note, once again, that the decision to have remove operations symmetric with
the add operations was the right decision.)

public void add(E value)
// post: the value is added to the tail of the structure
{

data.addLast(value);

}

public E remove()
// pre: the queue is not empty
// post: the head of the queue is removed and returned
{

return data.removeFirst();

}

10.2 Queues

235

The needs of the remove operation are satisﬁed by the removeFirst operation
on a List, so very little code needs to be written. The simplicity of these meth-
ods demonstrates how code reuse can make the implementation of structures
less difﬁcult. This is particularly dramatic when we consider the implementa-
tion of the size-related methods of the QueueList structure:

public E get()
// pre: the queue is not empty
// post: the element at the head of the queue is returned
{

return data.getFirst();

}

public int size()
// post: returns the number of elements in the queue
{

return data.size();

}

public void clear()
// post: removes all elements from the queue
{

data.clear();

}

public boolean isEmpty()
// post: returns true iff the queue is empty
{

return data.isEmpty();

}

Because the QueueList is a rewrapping of the List structure, the complexity
of each of the methods corresponds to the complexity of the underlying List
implementation. For CircularLists, most of the operations can be performed
in constant time.

We now consider two implementations that do not use dynamically linked
structures for their underpinnings. While these structures have drawbacks, they
are useful when space is a concern or when the size of the queue can be bounded
above.

10.2.3 Vector-Based Queues

The lure of implementing structures through code reuse can lead to perfor-
mance problems if the underlying structure is not well considered. Thus, while
we have advocated hiding the unnecessary details of the implementation within
the object, it is important to have a sense of the complexity of the object’s meth-
ods, if the object is the basis for supporting larger structures.

Principle 16 Understand the complexity of the structures you use.

NNWSWSENEWSE236

Linear Structures

Figure 10.5 A Queue implemented using a List. The head of the List corresponds to
the head of the Queue.

Figure 10.6 A Queue implemented atop a Vector. As elements are added (enqueued),
the Queue expands the Vector. As elements are removed, the leftmost element is re-
moved, shrinking the Vector.

We will return to this issue a number of times. Careful design of data structures
sometimes involves careful reconsideration of the performance of structures.

Now, let’s consider an implementation of Queues using Vectors. Recall that
a Vector is a linear structure whose values can be randomly accessed and mod-
iﬁed. We will, again, reconsider the Vector as a two-ended structure in our
attempt to implement a queue (see Figure 10.6). The head of the queue will be
found in the ﬁrst location of the Vector, and the tail of the queue will be found
in the last. (The index associated with each element will, essentially, enumerate
the order in which they would be removed in the future.)

The constructor creates an empty QueueVector by creating an empty Vec-
tor. When an upper bound is provided, the second QueueVector constructor

tail3head5124m−1023n−1data1123mm−1Tail of queueFirst free elementHead of queue10.2 Queues

237

QueueVector

passes that information along to the Vector:

protected Vector<E> data;

public QueueVector()
// post: constructs an empty queue
{

data = new Vector<E>();

}

public QueueVector(int size)
// post: constructs an empty queue of appropriate size
{

data = new Vector<E>(size);

}

Adding an element to the end of the queue is as simple as adding an element
to the end of the Vector. This is accomplished through the Vector method
addElement.

public void add(E value)
// post: the value is added to the tail of the structure
{

data.add(value);

}

As with the remove method of the StackVector class, remove the ﬁrst ele-

ment of the Vector, and return the returned value.

public E remove()
// pre: the queue is not empty
// post: the head of the queue is removed and returned
{

return data.remove(0);

}

As usual, the remaining methods rewrap similar Vector methods in the ex-
pected way. Because of the restricted nature of this linear structure, we take
care not to publicize any features of the Vector class that violate the basic re-
strictions of a Linear interface.

When considering the complexity of these methods, it is important to keep
in mind the underlying complexities of the Vector. For example, adding a value
to the “end” of the Vector can be accomplished, on average, in constant time.3
That method, add, is so special in its simplicity that it was distinguished from
the general add(int) method: the latter operation has time complexity that
is expected to be O(n), where n is the length of the Vector. The worst-case

3 It can vary considerably if the Vector requires reallocation of memory, but the average time (as
we saw in Section 3.5) is still constant time.

238

Linear Structures

behavior, when a value is added at the front of the Vector, is O(n). All the
existing elements must be moved to the right to make room for the new value.
A similar situation occurs when we consider the removal of values from a
Vector. Unfortunately, the case we need to use—removing an element from
the beginning of a Vector—is precisely the worst case. It requires removing the
value and then shifting n − 1 elements to the left, one slot. That O(n) behavior
slows down our remove operation, probably by an unacceptable amount. For
example, the process of adding n elements to a queue and then dequeuing them
takes about O(n2) time, which may not be acceptable.

Even though the implementation seemed straightforward, we pay a signif-
icant price for this code reuse.
If we could remove a value from the Vector
without having to move the other elements, the complexity of these operations
could be simpliﬁed. We consider that approach in our implementation of the
Queue interface using arrays of objects.

10.2.4 Array-Based Queues

If we can determine an upper bound for the size of the queue we need, we can
gain some efﬁciency because we need not worry so much about managing the
memory we use. Keyboard hardware, for example, often implements a queue
of keystrokes that are to be shipped off to the attached processor. The size of
that queue can be easily bounded above by, say, several hundred keystrokes.
take note! Notice, by the way, that this does not limit the number of elements that can run
through the queue, only the number of values that can be resident within the
queue simultaneously.

Fast typists,

Once the array has been allocated, we simply place enqueued values in
successive locations within the array, starting at location 0. The head of the
queue—the location containing the oldest element—is initially found at loca-
tion 0. As elements are removed, we return the value stored at the head of the
queue, and move the head toward the right in the array. All the operations must
explicitly store and maintain the size of the queue in a counter. This counter
should be a nonnegative number less than the length of the array.

One potential problem occurs when a value is removed from the very end
of the array. Under normal circumstances, the head of the queue would move
to the right, but we now need to have it wrap around. One common solution is
to use modular arithmetic. When the head moves too far to the right, its value,
head, is no longer less than the length of the array, data.length. After moving
the head to the right (by adding 1 to head), we simply compute the remainder
when dividing by data.length. This always returns a valid index for the array,
data, and it indexes the value just to the right, in wrap-around fashion (see
Figure 10.7). It should be noted, at this point, that remainder computation is
reasonable for positive values; if head were ever to become negative, one must
take care to check to see if a negative remainder might occur. If that appears
to be a possibility, simply adding data.length to the value before remainder
computation ﬁxes the problem in a portable way.

10.2 Queues

239

Figure 10.7 A Queue implemented atop an array. As elements are added (enqueued),
the queue wraps from the back to the front of the array.

Our QueueArray implementation, then, requires three values: an array of
objects that we allocate once and never expand, the index of the head of the
queue, and the number of elements currently stored within the queue.

QueueArray

protected Object data[]; // an array of the data
protected int head; // next dequeue-able value
protected int count; // current size of queue

Given an upper bound, the constructor allocates the array and sets the head
and count variables to 0. The initial value of head could, actually, be any value
between 0 and size-1, but we use 0, for neatness.

public QueueArray(int size)
// post: create a queue capable of holding at most size values
{

42 isn’t the
ultimate
answer, then?

data = new Object[size];
head = 0;
count = 0;

}

The add and remove methods appear more complex than their QueueVector
counterparts, but review of the implementation of the Vector class proves their
similarity. The add method adds a value at the logical end of the queue—the
ﬁrst free slot in the array—in wrap-around fashion. That location is count slots
to the right of the head of the queue and is computed using modular arithmetic;
remove is similar, but carefully moves the head to the right. The reader should
verify that, in a valid QueueArray, the value of head will never pass the value
of tail.

public void add(E value)
// pre: the queue is not full
// post: the value is added to the tail of the structure
{

Assert.pre(!isFull(),"Queue is not full.");

−10123412345dataTail of queueFirst free elementHead of queuen240

Linear Structures

int tail = (head + count) % data.length;
data[tail] = value;
count++;

}

public E remove()
// pre: the queue is not empty
// post: the head of the queue is removed and returned
{

Assert.pre(!isEmpty(),"The queue is not empty.");
E value = (E)data[head];
head = (head + 1) % data.length;
count--;
return value;

}

The get method just provides quick access to the head entry of the array.

public E get()
// pre: the queue is not empty
// post: the element at the head of the queue is returned
{

Assert.pre(!isEmpty(),"The queue is not empty.");
return (E)data[head];

}

Because we are directly using arrays, we do not have the luxury of previously
constructed size-oriented methods, so we are forced to implement these directly.
Again, the cost of an efﬁcient implementation can mean less code reuse—or
increased original code and the potential for error.

public int size()
// post: returns the number of elements in the queue
{

return count;

}

public void clear()
// post: removes all elements from the queue
{

// we could remove all the elements from the queue
count = 0;
head = 0;

}

public boolean isFull()
// post: returns true if the queue is at its capacity
{

return count == data.length;

}

10.2 Queues

241

public boolean isEmpty()
// post: returns true iff the queue is empty
{

return count == 0;

}

Alternative Array Implementations

One aesthetic drawback of the implementation of queues described is that sym-
metric operations—add and remove—do not lead to symmetric code. There are
several reasons for this, but we might uncover some of the details by reconsid-
ering the implementations.

Two variables, head and count, are responsible for encoding (in an asym-
metric way) the information that might be more symmetrically encoded using
two variables, say head and tail. For the sake of argument, suppose that head
points to the oldest value in the queue, while tail points to the oldest value
not in the queue—the value to the right of the last value in the queue. A simple
test to see if the queue is empty is to check to see if head == tail. Strangely
enough this also appears to be a test to see if the queue is full! Since a queue
should never be simultaneously empty and full, this problem must be addressed.
Suppose the array has length l. Then both head and tail have a range of
0 to l − 1. They take on l values apiece. Now consider all queues with head
values stored beginning at location 0. The head is ﬁxed at 0, but tail may take
on any of the l values between 0 and l − 1, inclusive. The queues represented,
however, have l + 1 potential states: an empty queue, a queue with one value,
a queue with two values, up to a queue with l values. Because there are l + 1
queues, they cannot be adequately represented by a pair of variables that can
support only l different states. There are several solutions to this conﬂict that
we outline momentarily and discuss in problems at the end of the chapter.

1. A boolean variable, queueEmpty, could be added to distinguish between
the two states where head and tail are identical. Code written using this
technique is clean and symmetric. The disadvantage of this technique is
that more code must be written to maintain the extra variable.

2. An array element logically to the left of the head of the queue can be
reserved. The queue is full if there are l − 1 values within the queue.
Since it would not be possible to add a value when the queue is full, the
tail and head variables would never “become equal” through expansion.
The only signiﬁcant difference is the allocation of an extra reserved cell,
and a change in the isFull method to see if the tail is just to the logical
left of the head. The disadvantage of this technique is that it requires the
allocation of another array element. When objects are large, this cost may
be prohibitively expensive. (Again, in Java, an Object is a small reference
In other
to the actual memory used, so the cost is fairly insigniﬁcant.

242

Linear Structures

languages, where instances are stored directly, and not as references, this
cost may be prohibitively expensive.)

Our actual implementation, of course, provides a third solution to the prob-
lem. While we like our solution, data abstraction allows us to hide any changes
we might make if we change our minds.

10.3 Example: Solving Mazes

To demonstrate the utility of stacks and queues, we consider the automated
solution of a maze. A maze is simply a matrix of cells that are adjacent to one
another. The user begins in a special start cell and seeks a path of adjacent cells
that lead to the ﬁnish cell.

One general approach to solving a maze (or any other search problem) is
to consider unvisited cells as potential tasks. Each unvisited cell represents the
task of ﬁnding a path from that cell to the ﬁnish. Some cells, of course, may not
be reachable from the start position, so they are tasks we seek to avoid. Other
cells lead us on trails from the start to ﬁnish. Viewed in this manner, we can use
the linear structure to help us solve the problem by keeping track of outstanding
tasks—unvisited cells adjacent to visited cells.

In the following program, we make use of two abstract classes, Position
and Maze. A Position is used to identify a unique location within a maze. Any
Position can be transformed into another Position by asking it for an adjacent
position to the north, south, east, or west. Here is the class:

class Position
{

MazeRunner

public Position north()
// post: returns position above

public Position south()
// post: returns position below

public Position east()
// post: returns position to right

public Position west()
// post: returns position to left

public boolean equals(Object other)
// post: returns true iff objects represent same position

}

The interface for the Maze class reads a maze description from a ﬁle and gen-
erates the appropriate adjacency of cells. We ignore the implementation that is
not important to us at this point (it is available online):

10.3 Example: Solving Mazes

243

class Maze
{

public Maze(String filename)
// pre: filename is the name of a maze file. # is a wall.
//
’s’ marks the start, ’f’ marks the finish.
// post: reads and constructs maze from file <filename>

public void visit(Position p)
// pre: p is a position within the maze
// post: cell at position p is set to visited

public boolean isVisited(Position p)
// pre: p is a position within the maze
// pos: returns true if the position has been visited

public Position start()
// post: returns start position

public Position finish()
// post: returns finish position

public boolean isClear(Position p)
// post: returns true iff p is a clear location within the maze

}

Now, once we have the structures supporting the construction of mazes, we can
use a Linear structure to organize the search for a solution in a Maze:

public static void main(String[] arguments)
{

Maze m = new Maze(arguments[0]); // the maze
Position goal = m.finish(); // where the finish is
Position square = null; // the current position
// a linear structure to manage search
Linear<Position> todo = new StackList<Position>();

// begin by priming the queue(stack) w/starting position
todo.add(m.start());
while (!todo.isEmpty()) // while we haven’t finished exploring
{

// take the top position from the stack and check for finish
square = todo.remove();
if (m.isVisited(square)) continue; // been here before
if (square.equals(goal)) {

System.out.println(m); // print solution
break;

}
// not finished.
// visit this location, and add neighbors to pool
m.visit(square);

244

Linear Structures

#

#f

####################
#s#
#
# ####### #### # # #
# ### #
#
#
##### ### #
#
####### ##
# #
#
# # ### #
#
#
#
# # ## #
# # #
#
#
#
#
#
####################

#

####################
#s#
#f...#...#
#.####### ####.#.#.#
#.........#..#.###.#
##### ###.#........#
# #...#######.##
#
# #.### #...#..#
#
# #.#...#.#.##.#
#
#
#...#...#....#
####################

Figure 10.8 A classic maze and its solution found using a stack. Dots indicate locations
in the maze visited during the solution process.

if (m.isClear(square.north())) todo.add(square.north());
if (m.isClear(square.west())) todo.add(square.west());
if (m.isClear(square.south())) todo.add(square.south());
if (m.isClear(square.east())) todo.add(square.east());

}

}

We begin by placing the start position on the stack. If, ultimately, the stack is
emptied, a solution is impossible. If the stack is not empty, the top position is
removed and considered, if not visited. If an unvisited cell is not the ﬁnish, the
cell is marked visited, and open neighboring cells are added to the stack.

Notice that since the underlying data structure is a Stack, the order in which
the neighboring positions are pushed on the stack is the reverse of the order in
which they will be considered. The result is that the program prefers to head
east before any other direction. This can be seen as it gets distracted by going
east at the right border of the maze of Figure 10.8. Because stacks are LIFO
structures, the search for a solution prefers to deepen the search rather than
investigate alternatives. If a queue was used as the linear structure, the search
would expand along a frontier of cells that are equidistant from the start. The
solution found, then, would be the most direct route from start to ﬁnish, just as
in the coin puzzle.

10.4 Conclusions

In this chapter we have investigated two important linear structures: the Stack
and the Queue. Each implements add and remove operations. Traditional im-
plementations of Stacks refer to these operations as push and pop, while tradi-
tional Queue methods are called enqueue and dequeue. Since these structures
are often used to solve similar problems (e.g., search problems), they share a
common Linear interface.

No:
Go west
young Maze!

10.4 Conclusions

245

There are many different ways to implement each of these linear structures,
and we have investigated a number—including implementations using arrays.
Because of the trade-offs between speed and versatility, each implementation
has its own particular strengths. Still, for many applications where performance
is less important, we can select an implementation and use it without great
concern because a common interface allows us to freely swap implementations.
We have seen a number of examples that use Linear structures to solve com-
plex problems. Since stacks are used to maintain the state of executing methods,
we have seen that recursive programs can be converted to iterative programs
that maintain an explicit stack. Two explicit search problems—the coin puzzle
and the maze—have an obvious relation. Because the coin puzzle searches for a
short solution, we use a queue to maintain the pool of goal candidates. For the
maze, we chose a stack, but a queue is often just as effective. The coin puzzle
can be thought of as a maze whose rules determine the location of the barriers
between board positions.

Self Check Problems

Solutions to these problems begin on page 447.

10.1

10.2

Is a Stack a Linear? Is it a List? An AbstractLinear? A Queue?

Is a Stack a List? Is a StackList a List? Is a StackList a Stack?

10.3 Why might we use generic Queues in our code, instead of QueueLists?

10.4

10.5

Is it possible to construct a new Queue directly?

If you are in a line to wash your car, are you in a queue or a stack?

Suppose you surf to a page on the Web about gardening. From there,
10.6
you surf to a page about ﬂowers. From there, you surf to a ﬂower seed distribu-
tor. When you push the “go back” button, you return to the ﬂower page. What
structure is your history stored in?

10.7

In a breadth-ﬁrst search, what is special about the ﬁrst solution found?

You are in a garden maze and you are optimistically racing to the center.
10.8
Are you more likely to use a stack-based depth-ﬁrst search, or a queue-based
breadth-ﬁrst search?

10.9 Why do we use modular arithmetic in the QueueArray implementation?

Problems

Solutions to the odd-numbered problems begin on page 474.

Suppose we push each of the integers 1, 2, . . . , n, in order, on a stack,

10.1
and then perform m ≤ n pop operations. What is the ﬁnal state of the stack?

Suppose we enqueue each of the integers 1, 2, . . . , n, in order, into a
10.2
queue, and then perform m ≤ n dequeue operations. What is the ﬁnal state of
the queue?

246

Linear Structures

Suppose you wish to copy a queue into another, preserving the order of

Suppose you wish to ﬁll a stack with a copy of another, maintaining the
10.3
order of elements. Using only Stack operations, describe how this would be
done. How many additional stacks are necessary?
Suppose you wish to reverse the order of elements of a stack. Using
10.4
only Stack operations, describe how this would be done. Assuming you place
the result in the original stack, how many additional stacks are necessary?
10.5
elements. Using only Queue operations, describe how this would be done.
In the discussion of radix sort (see Section 6.6) bucketPass sorted in-
10.6
teger values based on a digit of the number. It was important that the sort was
stable—that values with similar digits remained in their original relative or-
der. Unfortunately, our implementation used Vectors, and to have bucketPass
work in O(n) time, it was important to add and remove values from the end of
the Vector. It was also necessary to unload the buckets in reverse order. Is it
possible to clean this code up using a Stack or a Queue? One of these two will
allow us to unload the buckets into the data array in increasing order. Does this
improved version run as quickly (in terms of big-O)?
Suppose you wish to reverse the order of elements of a queue. Using
10.7
only Queue operations, describe how this would be done. (Hint: While you can’t
use a stack, you can use something similar.)
Over time, the elements 1, 2, and 3 are pushed onto the stack, among
10.8
others, in that order. What sequence(s) of popping the elements 1, 2 and 3 off
the stack are not possible?
Generalize the solution to Problem 10.8. If elements 1, 2, 3, . . . , n are
10.9
pushed onto a stack in that order, what sequences of popping the elements off
the stack are not permissible?
10.10 Over time, the elements 1, 2, and 3 are added to a queue, in that order.
What sequence(s) of removing the elements from the queue is impossible?
10.11 Generalize the solution to Problem 10.10. If elements 1, 2, 3, . . . , n are
added to a queue in that order, what sequences of removing the elements are
not permissible?
10.12 It is conceivable that one linear structure is more general than another.
(a) Is it possible to implement a Queue using a Stack? What is the complexity
of each of the Queue operations? (b) Is it possible to implement a Stack using
a Queue? What are the complexities of the various Stack methods?
10.13 Describe how we might efﬁciently implement a Queue as a pair of
Stacks, called a “stack pair.” (Hint: Think of one of the stacks as the head
of the queue and the other as the tail.)
10.14 The implementation of QueueLists makes use of a CircularList. Im-
plement QueueLists in a manner that is efﬁcient in time and space using Node
with a head and tail reference.
10.15 Burger Death needs to keep track of orders placed at the drive-up win-
dow. Design a data structure to support their ordering system.

10.5 Laboratory: A Stack-Based Language

Objective. To implement a PostScript-based calculator.

Discussion. In this lab we will investigate a small portion of a stack-based lan-
guage called PostScript. You will probably recognize that PostScript is a ﬁle
format often used with printers. In fact, the ﬁle you send to your printer is a
program that instructs your printer to draw the appropriate output. PostScript
is stack-based: integral to the language is an operand stack. Each operation
that is executed pops its operands from the stack and pushes on a result. There
are other notable examples of stack-based languages, including forth, a lan-
guage commonly used by astronomers to program telescopes. If you have an
older Hewlett-Packard calculator, it likely uses a stack-based input mechanism
to perform calculations.

We will implement a few of the math operators available in PostScript.
To see how PostScript works, you can run a PostScript simulator. (A good
simulator for PostScript is the freely available ghostscript utility. It is available
from www.gnu.org.) If you have a simulator handy, you might try the following
example inputs. (To exit a PostScript simulator, type quit.)

1. The following program computes 1 + 1:

1 1 add pstack

Every item you type in is a token. Tokens include numbers, booleans,
or symbols. Here, we’ve typed in two numeric tokens, followed by two
symbolic tokens. Each number is pushed on the internal stack of operands.
When the add token is encountered, it causes PostScript to pop off two
values and add them together. The result is pushed back on the stack.
(Other mathematical operations include sub, mul, and div.) The pstack
command causes the entire stack to be printed to the console.

2. Provided the stack contains at least one value, the pop operator can be
used to remove it. Thus, the following computes 2 and prints nothing:

1 1 add pop pstack

3. The following “program” computes 1 + 3 ∗ 4:

1 3 4 mul add pstack

The result computed here, 13, is different than what is computed by the
following program:

1 3 add 4 mul pstack

In the latter case the addition is performed ﬁrst, computing 16.

248

Linear Structures

4. Some operations simply move values about. You can duplicate values—

the following squares the number 10.1:

10.1 dup mul pstack pop

The exch operator to exchange two values, computing 1 − 3:

3 1 exch sub pstack pop

5. Comparison operations compute logical values:

1 2 eq pstack pop

tests for equality of 1 and 2, and leaves false on the stack. The program

1 1 eq pstack pop

yields a value of true.

6. Symbols are deﬁned using the def operation. To deﬁne a symbolic value
we specify a “quoted” symbol (preceded by a slash) and the value, all
followed by the operator def:

/pi 3.141592653 def

Once we deﬁne a symbol, we can use it in computations:

/radius 1.6 def
pi radius dup mul mul pstack pop

computes and prints the area of a circle with radius 1.6. After the pop, the
stack is empty.

Procedure. Write a program that simulates the behavior of this small subset of
PostScript. To help you accomplish this, we’ve created three classes that you
will ﬁnd useful:

• Token. An immutable (constant) object that contains a double, boolean,
or symbol. Different constructors allow you to construct different Token
values. The class also provides methods to determine the type and value
of a token.

• Reader. A class that allows you to read Tokens from an input stream. The

typical use of a reader is as follows:

Reader r = new Reader();
Token t;
while (r.hasNext())
{

t = (Token)r.next();
if (t.isSymbol() && // only if symbol:

t.getSymbol().equals("quit")) break;

// process token

}

Token

Reader

10.5 Laboratory: A Stack-Based Language

249

This is actually our ﬁrst use of an Iterator. It always returns an Object
of type Token.

• SymbolTable. An object that allows you to keep track of String–Token
associations. Here is an example of how to save and recall the value of π:

SymbolTable table = new SymbolTable();
// sometime later:
table.add("pi",new Token(3.141592653));
// sometime even later:
if (table.contains("pi"))
{

Token token = table.get("pi");
System.out.println(token.getNumber());

}

SymbolTable

You should familiarize yourself with these classes before you launch into writing
your interpreter.

To complete your project, you should implement the PostScript commands
pstack, add, sub, mul, div, dup, exch, eq, ne, def, pop, quit. Also implement
the nonstandard PostScript command ptable that prints the symbol table.
Thought Questions. Consider the following questions as you complete the lab:

1. If we are performing an eq operation, is it necessary to assume that the

values on the top of the stack are, say, numbers?

2. The pstack operation should print the contents of the operand stack with-
out destroying it. What is the most elegant way of doing this? (There are
many choices.)

3. PostScript also has a notion of a procedure. A procedure is a series of
Tokens surrounded by braces (e.g., { 2 add }). The Token class reads
procedures and stores the procedure’s Tokens in a List. The Reader
class has a constructor that takes a List as a parameter and returns a
Reader that iteratively returns Tokens from its list. Can you augment your
PostScript interpreter to handle the deﬁnition of functions like area, be-
low?

/pi 3.141592653 def
/area { dup mul pi mul } def
1.6 area
9 area pstack
quit

Such a PostScript program deﬁnes a new procedure called area that com-
putes πr2 where r is the value found on the top of the stack when the
procedure is called. The result of running this code would be

254.469004893
8.042477191680002

250

Linear Structures

4. How might you implement the if operator? The if operator takes a
boolean and a token (usually a procedure) and executes the token if the
boolean is true. This would allow the deﬁnition of the absolute value
function (given a less than operator, lt):

/abs { dup 0 lt { -1 mul } if } def
3 abs
-3 abs
eq pstack

The result is true.

5. What does the following do?

/count { dup 1 ne { dup 1 sub count } if } def
10 count pstack

Notes:

10.6 Laboratory: The Web Crawler

Objective. To crawl over web pages in a breadth-ﬁrst manner.
Discussion. Web crawling devices are a fact of life. These programs automati-
cally venture out on the Web and internalize documents. Their actions are based
on the links between pages. The data structures involved in keeping track of the
progress of an avid web crawler are quite complex: imagine, for example, how
difﬁcult it must be for such a device to keep from chasing loops of references
between pages.

In this lab we will build a web crawler that determines the distance from
one page to another. If page A references page B, the distance from A to B is
1. Notice that page B may not have any links on it, so the distance from B to A
may not be deﬁned.

Here is an approach to determining the distance from page A to arbitrary

page B:

• Start a list of pages to consider. This list has two columns: the page, and
its distance from A. We can, for example, put page A on this list, and
assign it a distance of zero. If we ever see page B on this list, the problem
is solved: just print out the distance associated with B.

• Remove the ﬁrst page on our list: call it page X with distance d from page
A. If X has the same URL as B, B must be distance d from A. Otherwise,
consider any link off of page X: either it points to a page we’ve already
seen on our list (it has a distance d or less), or it is a new page. If it’s
a new page we haven’t encountered, add it to the end of our list and
associate with it a distance d + 1—it’s a link farther from A than page X.
We consider all the links off of page X before considering a new page from
our list.

If the list is a FIFO, this process is a breadth-ﬁrst traversal, and the distance
associated with B is the shortest distance possible. We can essentially think of
the Web as a large maze that we’re exploring.
Procedure. Necessary for this lab is a class HTML. It deﬁnes a reference to a
textual (HTML) web page. The constructor takes a URL that identiﬁes which
page you’re interested in:

HTML page = new HTML("http://www.yahoo.com");

Only pages that appear to have valid HTML code can actually be inspected
(other types of pages will appear empty). Once the reference is made to the
page, you can get its content with the method content:

HTML

System.out.println(page.content());

Two methods allow you to get the URL’s associated with each link on a page:
hasNext and nextURL. The hasNext method returns true if there are more links
you have not yet considered. The nextURL method returns a URL (a String)
that is pointed to by this page. Here’s a typical loop that prints all the links
associated with a page:

252

Linear Structures

int i = 0;
while (page.hasNext())
{

System.out.println(i+": "+page.nextURL());
i++;

}

For the sake of speed, the HTML method downloads 10K of information. For
simple pages, this covers at least the ﬁrst visible page on the screen, and it might
be argued that the most important links to other pages probably appear within
this short start (many crawlers, for example, limit their investigations to the
ﬁrst part of a document). You can change the size of the document considered
in the constructor:

HTML page = new HTML("http://www.yahoo.com",20*1024);

You should probably keep this within a reasonable range, to limit the total im-
pact on memory.

Write a program that will tell us the maximum number of links (found on
the ﬁrst page of a document) that are necessary to get from your home page to
any other page within your personal web. You can identify these pages because
they all begin with the same preﬁx. We might think of this as a crude estimate
of the “depth” or “diameter” of a site.
Thought Questions. Consider the following questions as you complete the lab:

1. How do your results change when you change the buffer size for the page
to 2K? 50K? Under what conditions would a large buffer change cause the
diameter of a Web to decrease? Under what conditions would this change
cause the diameter of a Web to increase?

Notes:

Chapter 11

Ordered Structures

Concepts:
.
.
.
.
.

The java.lang.Comparable interface
The java.util.Comparator
The OrderedStructure interface
The OrderedVector
The OrderedList

“Make no mistake about it.
A miracle has happened. . .
we have no ordinary pig.”
“Well,” said Mrs. Zuckerman,
“it seems to me you’re a little off.
It seems to me we have
no ordinary spider.”
—Elwyn Brooks White

WE HAVE MADE NO ASSUMPTIONS about the type of data we store within our
structures—so far. Instead, we have assumed only that the data referenced are
a subclass of the type Object. Recall that all classes are subtypes of Object
in Java, so this is hardly a constraint. Data structures serve a purpose, often
helping us perform tasks more complex than “just holding data.” For example,
we used the Stack and Queue classes to guide a search through search space in
the previous chapter.

One important use of data structures is to help keep data in order—the
smallest value in the structure might be stored close to the front, while the
largest value would be stored close to the rear. Once a structure is ordered it
becomes potentially useful as a mechanism for sorting: we simply insert our
possibly unordered data into the structure and then extract the values in order.
To do this, however, it is necessary to compare data values to see if they are in the
correct order. In this chapter we will discuss approaches to the various problems
associated with maintaining ordered structures. First we review material we
ﬁrst encountered when we considered sorting.

11.1 Comparable Objects Revisited

In languages like C++ it is possible to override the comparison operators (<, >,
==, etc.). When two objects are compared using these operators, a user-written
method is called. Java does not support overriding of built-in operators. Thus,
it is useful to come up with a convention for supporting comparable data.

First, let’s look closely at the interface for Java’s Object. Since every class
inherits and extends the interface of the Object class, each of its methods may
be applied to any class. For example, the equals method allows us to check

254

Ordered Structures

if an Object is logically equal to another Object. In contrast, the == operator
compares two references to see if they refer to the same instance of an object.

By default, the equals function returns true whenever two references point
to exactly the same object. This is often not the correct comparison—often we
wish to have different instances of the same type be equal—so the class designer
should consider rewriting it as a class-speciﬁc method.

For our purposes, we wish to require of comparable classes a method that
determines the relative order of objects. How do we require this? Through
an interface! Since an interface is a contract, we simply wrap the compareTo
method in a language-wide interface, java.lang.Comparable:

public interface Comparable<T>
{

public int compareTo(T that);

Comparable

}

This is pretty simple: When we want to compare two objects of type T, we
simply call the compareTo method of one on another. Now, if we require that
an object be a Comparable object, then we know that it may be compared to
similarly typed data using the compareTo method.

11.1.1 Example: Comparable Ratios

Common types, such as Integers and Strings, include a compareTo method.
In this section we add methods to make the Ratio class comparable. Recall that
a Ratio has the following interface:

public class Ratio

implements Comparable<Ratio>

{

Ratio

public Ratio(int top, int bottom)
// pre: bottom != 0
// post: constructs a ratio equivalent to top::bottom

public int getNumerator()
// post: return the numerator of the fraction

public int getDenominator()
// post: return the denominator of the fraction

public double getValue()
// post: return the double equivalent of the ratio

public Ratio add(Ratio other)
// pre: other is nonnull
// post: return new fraction--the sum of this and other

public String toString()

11.1 Comparable Objects Revisited

255

// post: returns a string that represents this fraction.

public int compareTo(Ratio that)
// pre: other is non-null and type Ratio
// post: returns value <, ==, > 0 if this value is <, ==, > that

public boolean equals(Object that)
// pre: that is type Ratio
// post: returns true iff this ratio is the same as that ratio

}

A Ratio is constructed by passing it a pair of integers. These integers are cached
away—we cannot tell how—where they can later be used to compare their ratio
with another Ratio. The protected data and the constructor that initializes them
appear as follows:

protected int numerator;
protected int denominator; // denominator of ratio

// numerator of ratio

public Ratio(int top, int bottom)
// pre: bottom != 0
// post: constructs a ratio equivalent to top::bottom
{

numerator = top;
denominator = bottom;
reduce();

}

We can see, now, that this class has a pair of protected ints to hold the
values. Let us turn to the compareTo method. The Comparable interface for a
type Ratio declares the compareTo method to take a Ratio parameter, so we
expect the parameter to be a Ratio; it’s not useful to compare Ratio types to
non-Ratio types. Implementation is fairly straightforward:

public int compareTo(Ratio that)
// pre: other is non-null and type Ratio
// post: returns value <, ==, > 0 if this value is <, ==, > that
{

return this.getNumerator()*that.getDenominator()-
that.getNumerator()*this.getDenominator();

}

The method checks the order of two ratios: the values stored within this ratio
are compared to the values stored in that ratio and an integer is returned. The
relationship of this integer to zero reprepresents the relationship between this
and that. The magnitude of the result is unimportant (other than being zero or
non-zero).

We can now consider the equals method:

256

Ordered Structures

public boolean equals(Object that)
// pre: that is type Ratio
// post: returns true iff this ratio is the same as that ratio
{

return compareTo((Ratio)that) == 0;

}

Conveniently, the equals method can be cast in terms of the compareTo method.
For the Ratio class, the compareTo method is not much more expensive to com-
pute than the equals, so this “handing off” of the work does not cost much.
For more complex classes, the compareTo method may be so expensive that a
consistent equals method can be constructed using independently considered
code. In either case, it is important that equals return true exactly when the
compareTo method returns 0.

Note also that the parameter to the equals method is declared as an Object.
If it is not, then the programmer is writing a new method, rather than overrid-
ing the default method inherited from the Object class. Because compareTo
compares Ratio types, we must cast the type of that to be a Ratio. If that is a
Ratio (or a subclass), the compare will work. If not, then a cast error will occur
at this point. Calling compareTo is the correct action here, since equal Ratios
may appear in different object instances. Direct comparison of references is not
appropriate. Failure to correctly implement the equals (or compareTo) method
can lead to very subtle logical errors.

Principle 17 Declare parameters of overriding methods with the most general
types possible.

To reiterate, failure to correctly declare these methods as generally as possible
makes it less likely that Java will call the correct method.

11.1.2 Example: Comparable Associations

Let us return now to the idea of an Association. An Association is a key-value
pair, bound together in a single class. For the same reasons that it is sometimes
nice to be able to compare integers, it is often useful to compare Associations.
Recall that when we constructed an Association we took great care in deﬁning
the equals operator to work on just the key ﬁeld of the Association. Similarly,
when we extend the concept of an Association to its Comparable equivalent,
we will have to be just as careful in constructing the compareTo method.

Unlike the Ratio class, the ComparableAssociation can be declared an
extension of the Association class. The outline of this extension appears as
follows:

public class ComparableAssociation<K extends Comparable<K>,V>

extends Association<K,V>
implements Comparable<ComparableAssociation<K,V>>

{

Comparable-
Association

NNWSWSENEWSE11.1 Comparable Objects Revisited

257

public ComparableAssociation(K key)
// pre: key is non-null
// post: constructs comparable association with null value

public ComparableAssociation(K key, V value)
// pre: key is non-null
// post: constructs association between a comparable key and a value

public int compareTo(ComparableAssociation<K,V> that)
// pre: other is non-null ComparableAssociation
// post: returns integer representing relation between values

}

Notice that there are very few methods. Since ComparableAssociation is
an extension of the Association class, all the methods written for Association
are available for use with ComparableAssociations. The only additions are
those shown here. Because one of the additional methods is the compareTo
method, it meets the speciﬁcation of what it means to be Comparable; thus we
claim it implements the Comparable interface.

Let’s look carefully at the implementation. As with the Association class,
there are two constructors for ComparableAssociations. The ﬁrst construc-
tor initializes the key and sets the value reference to null, while the second
initializes both key and value:

public ComparableAssociation(K key)
// pre: key is non-null
// post: constructs comparable association with null value
{

this(key,null);

}

public ComparableAssociation(K key, V value)
// pre: key is non-null
// post: constructs association between a comparable key and a value
{

super(key,value);

}

Remember that there are two special methods available to constructors: this
and super. The this method calls another constructor with a different set of
parameters (if the parameters are not different, the constructor could be recur-
sive!). We write one very general purpose constructor, and any special-purpose
constructors call the general constructor with reconsidered parameter values.
The super method is a means of calling the constructor for the superclass—the
class we are extending—Association. The second constructor simply calls the
constructor for the superclass. The ﬁrst constructor calls the second constructor
(which, in turn, calls the superclass’s constructor) with a null value ﬁeld. All
of this is necessary to be able to construct ComparableAssociations using the
Association’s constructors.

258

Ordered Structures

Now, the compareTo method requires some care:

public int compareTo(ComparableAssociation<K,V> that)
// pre: other is non-null ComparableAssociation
// post: returns integer representing relation between values
{

return this.getKey().compareTo(that.getKey());

}

Because the compareTo method must implement the Comparable interface be-
tween ComparableAssociations, its parameter is an ComparableAssociation.
Careful thought suggests that the relationship between the associations is com-
pletely determined by the relationship between their respective keys.

Since ComparableAssociations are associations with comparable keys, we
know that the key within the association has a compareTo method. Java would
not able to deduce this on its own, so we must give it hints, by declaring the
type parameter K to be any type that implements Comparable<K>. The dec-
laration, here, is essentially a catalyst to get Java to verify that a referenced
object has certain type characteristics. In any case, we get access to both keys
through independent variables. These allow us to make the comparison by
calling the compareTo method on the comparable objects. Very little logic is
directly encoded in these routines; we mostly make use of the prewritten code
to accomplish what we need.

In the next few sections we consider features that we can provide to existing
data structures, provided that the underlying data are known to be comparable.

11.2 Keeping Structures Ordered

We can make use of the natural ordering of classes suggested by the compareTo
method to organize our structure. Keeping data in order, however, places signif-
icant constraints on the type of operations that should be allowed. If a compa-
rable value is added to a structure that orders its elements, the relative position
of the new value is determined by the data, not the structure. Since this place-
ment decision is predetermined, ordered structures have little ﬂexibility in their
It is not possible, for example, to insert data at random locations.
interface.
While simpler to use, these operations also tend to be more costly than their
unordered counterparts. Typically, the increased energy required is the result of
an increase in the friction associated with decisions needed to accomplish add
and remove.

The implementation of the various structures we see in the remainder of this
chapter leads to simpler algorithms for sorting, as we will see in Section 11.2.3.

11.2.1 The OrderedStructure Interface

Recall that a Structure is any traversable structure that allows us to add and
remove elements and perform membership checks (see Section 1.8). Since the

11.2 Keeping Structures Ordered

259

Ordered-
Structure

The emperor
wears no
clothes!

OrderedVector

Structure interface also requires the usual size-related methods (e.g., size,
isEmpty, clear), none of these methods actually requires that the data within
the structure be kept in order. To ensure that the structures we create order their
data (according to their native ordering), we make them abide by an extended
interface—an OrderedStructure:

public interface OrderedStructure<K extends Comparable<K>>

extends Structure<K>

{
}

Amazingly enough we have accomplished something for almost nothing! Actu-
ally, what is happening is that we are using the type to store the fact that the
data can be kept in sorted order. Encoded in this, of course, is that we are
working with Comparable values of generic type K.

11.2.2 The Ordered Vector and Binary Search

We can now consider the implementation of an ordered Vector of values. Since
it implements an OrderedStructure, we know that the order in which elements
are added does not directly determine the order in which they are ultimately
removed.
Instead, when elements are added to an OrderedVector, they are
kept ascending in their natural order.

Constructing an ordered Vector requires little more than allocating the un-

derlying vector:

public class OrderedVector<E extends Comparable<E>>

extends AbstractStructure<E>
implements OrderedStructure<E>

{

}

public OrderedVector()
// post: constructs an empty, ordered vector
{

data = new Vector<E>();

}

Rather obviously, if there are no elements in the underlying Vector, then all of
the elements are in order. Initially, at least, the structure is in a consistent state.
We must always be mindful of consistency.

Because ﬁnding the correct location for a value is important to both adding
and removing values, we focus on the development of an appropriate search
technique for OrderedVectors. This process is much like looking up a word
in a dictionary, or a name in a phone book (see Figure 11.1). First we look at
the value halfway through the Vector and determine if the value for which we
are looking is bigger or smaller than this median. If it is smaller, we restart our
search with the left half of the structure. If it is bigger, we restart our search
with the right half of the Vector. Whenever we consider a section of the Vector

260

Ordered Structures

consisting of a single element, the search can be terminated, with the success
of the search dependent on whether or not the indicated element contains the
desired value. This approach is called binary search.

We present here the code for determining the index of a value in an Or-
deredVector. Be aware that if the value is not in the Vector, the routine returns
the ideal location to insert the value. This may be a location that is outside the
Vector.

protected int locate(E target)
{

Comparable<E> midValue;
int low = 0; // lowest possible location
int high = data.size(); // highest possible location
int mid = (low + high)/2; // low <= mid <= high
// mid == high iff low == high
while (low < high) {

// get median value
midValue = data.get(mid);
// determine on which side median resides:
if (midValue.compareTo(target) < 0) {

low = mid+1;

} else {

high = mid;

}
// low <= high
// recompute median index
mid = (low+high)/2;

}
return low;

}

For each iteration through the loop, low and high determine the bounds
of the Vector currently being searched. mid is computed to be the middle
element (if there are an even number of elements being considered, it is the
leftmost of the two middle elements). This middle element is compared with
the parameter, and the bounds are adjusted to further constrain the search.
Since the portion of the Vector participating in the search is roughly halved
each time, the total number of times around the loop is approximately O(log n).
This is a considerable improvement over the implementation of the indexOf
method for Vectors of arbitrary elements—that routine is linear in the size of
the structure.

Notice that locate is declared as a protected member of the class. This
makes it impossible for a user to call directly, and makes it more difﬁcult for a
user to write code that depends on the underlying implementation. To convince
yourself of the utility of this, both OrderedStructures of this chapter have ex-
actly the same interface (so these two data types can be interchanged), but they
are completely different structures. If the locate method were made public,
then code could be written that makes use of this Vector-speciﬁc method, and
it would be impossible to switch implementations.

11.2 Keeping Structures Ordered

261

Finding the correct location for a comparable value in an ordered array.
Figure 11.1
The top search ﬁnds a value in the array; the bottom search fails to ﬁnd the value, but
ﬁnds the correct point of insertion. The shaded area is not part of the Vector during
search.

13-10236543458403-10242465401425813-10234654043424313-10233654043425813-105834654043425813-1023465404342413-10234654043425823-10234654043425813-10234654043585813-10234654043425813-10234654043425814201234567891011midlowhigh12highlowmid0123456789101140midlowhigh12lowhighmidvaluevalue9393262

Ordered Structures

Implementation of the locate method makes most of the nontrivial Or-
deredVector methods more straightforward. The add operator simply adds an
element to the Vector in the position indicated by the locate operator:

public void add(E value)
// pre: value is non-null
// post: inserts value, leaves vector in order
{

int position = locate(value);
data.add(position,value);

}

It is interesting to note that the cost of looking up the value is O(log n), but
the insertElementAt for relatively “small” values can take O(n) time to insert.
Thus, the worst-case (and expected—see Problem 11.6) time complexity of the
add operation is O(n), linear in the size of the structure. In reality, for large
Vectors, the time required to ﬁnd the index of a value using the OrderedVector
method is signiﬁcantly reduced over the time required using the underlying
Vector method. If the cost of comparing two objects exceeds the cost of assign-
ing one object to another, the use of binary search can be expected to reduce
the cost of the add operation by as much as a factor of 2.

Both contains and remove can also make use of the locate operator. First,

we consider testing to see if an element is contained by the OrderedVector:

public boolean contains(E value)
// pre: value is non-null
// post: returns true if the value is in the vector
{

int position = locate(value);
return (position < size()) &&

data.get(position).equals(value);

}

We simply attempt to ﬁnd the item in the Vector, and if the location returned
contains the value we desire, we return true; otherwise we return false. Since
locate takes O(log n) time and the check for element equality is constant, the
total complexity of the operation is O(log n). The Vector version of the same
operation is O(n) time. This is a considerable improvement.

The return statement, you will note, returns the result of a logical and (&&)
operator. This is a short-circuiting operator: if, after evaluating the left half of
the expression, the ultimate value of the expression is known to be false, then
the second expression is not evaluated. That behavior is used here to avoid
calling the get operator with a position that might exceed the size of the
structure, that is, the length of the Vector. This is a feature of many languages,
but a potential trap if you ever consider reordering your boolean expressions.

Removing a value from an OrderedVector involves ﬁnding it within the

Vector and then explicitly extracting it from the structure:

11.2 Keeping Structures Ordered

263

public E remove(E value)
// pre: value is non-null
// post: removes one instance of value, if found in vector
{

if (contains(value)) {

// we know value is pointed to by location
int position = locate(value);
// since vector contains value, position < size()
// keep track of the value for return
E target = data.get(position);
// remove the value from the underlying vector
data.remove(position);
return target;

}
return null;

}

Like add, the operation has complexity O(n). But it executes faster than its
Vector equivalent, removeElement.

Note that by keeping the elements sorted, we have made adding and re-
moving an element from the OrderedVector relatively symmetric: the expected
complexity of each method is O(n). Yet, in the underlying Vector, an addElement
operation takes constant time, while the removeElement operation takes O(n)
time.

Extracting values in order from an OrderedStructure is accomplished by an
iterator returned from the elements method. Because the elements are stored
in the correct order in the Vector, the method need only return the value of the
Vector’s iterator method:

public Iterator<E> iterator()
{

return data.iterator();

}

The ease of implementing this particular method reassures us that our layout
of values within the vector (in ascending order) is appropriate. The rest of the
OrderedVector operators repackage similar operators from the Vector class:

public boolean isEmpty()
// post: returns true if the OrderedVector is empty
{

return data.size() == 0;

}

public void clear()
// post: vector is emptied
{

data.setSize(0);

}

264

Ordered Structures

public int size()
// post: returns the number of elements in vector
{

return data.size();

}

This “repackaging” brings up a point: Why is it necessary? If one were to,
instead, consider the OrderedVector to be an extension of the Vector class,
much of this repackaging would be unnecessary, because each of the repackaged
methods could be inherited, and those—like add, contains, and remove—that
required substantial reconsideration could be rewritten overriding the methods
provided in the underlying Vector class.

That’s all true! There’s one substantial drawback, however, that is uncov-
ered by asking a simple question: Is an OrderedVector suitably used wherever
a Vector is used? The answer is: No! Consider the following hypothetical code
that allocates an OrderedVector for use as a Vector:

static void main(String args[])
{

OrderedVector<String> v = new OrderedVector<String>();

v.add("Michael’s Pizza");
v.add(1,"Cozy Pizza");
v.add(0,"Hot Tomatoes Pizza");;

}

First, the add methods are not methods for OrderedVectors. Assuming this
could be done, the semantics become problematic. We are inserting elements
at speciﬁc locations within a Vector, but it is really an OrderedVector. The
values inserted violate the ordering of elements and the postconditions of the
add method of the OrderedVector.

We now consider a simple application of OrderedStructures—sorting.

11.2.3 Example: Sorting Revisited

Now that we have seen the implementation of an OrderedStructure, we can
use these structures to sort comparable values. (If values are not comparable,
it is hard to see how they might be sorted, but we will see an approach in
Section 11.2.4.) Here is a program to sort integers appearing on the input:

11.2 Keeping Structures Ordered

265

public static void main(String[] args)
{

Scanner s = new Scanner(System.in);
OrderedStructure<Integer> o = new OrderedVector<Integer>();
// read in integers
while (s.hasNextInt())
{

Sort

o.add(s.nextInt());

}
// and print them out, in order
for (Integer i : o)
{

System.out.println(i);

}

}

In this simple program a sequence of numbers is read from the input stream.
Each number is placed within an Integer that is then inserted into the Ordered-
Structure, in this case an OrderedVector. The insertion of this value into the
Vector may involve moving, on average, n
2 elements out of the way. As the n
values are added to the Vector, a total of O(n2) values have to be moved. The
overall effect of this loop is to perform insertion sort! Once the values have been
inserted in the ordered structure, we can use an iterator to traverse the Vector
in order and print out the values in order. If the OrderedVector is substituted
with any structure that meets the OrderedStructure interface, similar results
are generated, but the performance of the sorting algorithm is determined by
the complexity of insertion.

Now, what should happen if we don’t have a Comparable data type?

11.2.4 A Comparator-based Approach

Sometimes it is not immediately obvious how we should generally order a spe-
ciﬁc data type, or we are hard-pressed to commit to one particular ordering for
our data. In these cases we ﬁnd it useful to allow ordered structures to be or-
dered in alternative ways. One approach is to have the ordered structure keep
track of a Comparator that can be used when the compareTo method does not
seem appropriate. For example, when constructing a list of Integer values, it
may be useful to have them sorted in descending order.

The approach seems workable, but somewhat difﬁcult when a comparison
needs to actually be made. We must, at that time, check to see if a Comparator
has somehow been associated with the structure and make either a Comparator-
based compare or a class-based compareTo method call. We can greatly simplify
the code if we assume that a Comparator method can always be used: we con-
struct a Comparator, the structure package’s NaturalComparator, that calls
the compareTo method of the particular elements and returns that value for
compare:

266

Ordered Structures

import java.util.Comparator;

public class NaturalComparator<E extends Comparable<E>>

implements Comparator<E>

{

public int compare(E a, E b)
// pre: a, b non-null, and b is same type as a
// post: returns value <, ==, > 0 if a <, ==, > b
{

return a.compareTo(b);

}

Natural-
Comparator

public boolean equals(Object b)
// post: returns true if b is a NaturalComparator
{

return (b != null) && (b instanceof NaturalComparator);

}

}

The NaturalComparator, can then serve as a default comparison method in
classes that wish to make exclusive use of the Comparator-based approach.

To demonstrate the power of the Comparator-based approach we can de-
velop a notion of Comparator composition: one Comparator can be used to
modify the effects of a base Comparator. Besides the NaturalComparator,
the structure package also provides a ReverseComparator class. This class
keeps track of its base Comparator in a protected variable, base. When a
ReverseComparator is constructed, another Comparator can be passed to it
to reverse. Frequently we expect to use this class to reverse the natural or-
der of values, so we provide a parameterless constructor that forces the base
Comparator to be NaturalComparator:

protected Comparator<E> base; // comparator whose ordering is reversed

Reverse-
Comparator

public ReverseComparator()
// post: constructs a comparator that orders in reverse order
{

base = new NaturalComparator<E>();

}

public ReverseComparator(Comparator<E> base)
// post: constructs a Comparator that orders in reverse order of base
{

this.base = base;

}

We are now ready to implement the comparison method. We simply call the
compare method of the base Comparator and reverse its sign. This effectively
reverses the relation between values.

public int compare(E a, E b)

11.2 Keeping Structures Ordered

267

// pre: a, b non-null, and b is of type of a
// post: returns value <, ==, > 0 if a <, ==, > b
{

return -base.compare(a,b);

}

Note that formerly equal values are still equal under the ReverseComparator
transformation.

We now turn to an implementaton of an OrderedStructure that makes ex-

clusive use of Comparators to keep its elements in order.

11.2.5 The Ordered List

Arbitrarily inserting an element into a list is difﬁcult, since it requires moving to
the middle of the list to perform the addition. The lists we have developed are
biased toward addition and removal of values from their ends. Thus, we choose
to use the underlying structure of a SinglyLinkedList to provide the basis for
our OrderedList class. As promised, we will also support orderings through
the use of Comparators. First, we declare the class as an implementation of the
OrderedStructure interface:

public class OrderedList<E extends Comparable<E>>

OrderedList

extends AbstractStructure<E> implements OrderedStructure<E>

The instance variables describe a singly linked list as well as a Comparator to
determine the ordering. The constructors set up the structure by initializing the
protected variables using the clear method:

protected Node<E> data; // smallest value
protected int count;
protected Comparator<? super E> ordering;

// number of values in list

// the comparison function

public OrderedList()
// post: constructs an empty ordered list
{

this(new NaturalComparator<E>());

}

public OrderedList(Comparator<? super E> ordering)
// post: constructs an empty ordered list ordered by ordering
{

this.ordering = ordering;
clear();

}

public void clear()
// post: the ordered list is empty
{

data = null;

268

Ordered Structures

count = 0;

}

Again, the advantage of this technique is that changes to the initialization of the
underlying data structure can be made in one place within the code.

By default, the OrderedList keeps its elements in the order determined by
the compareTo method. The NaturalOrder comparator does precisely that. If
an alternative ordering is desired, the constructor for the OrderedList can be
given a Comparator that can be used to guide the ordering of the elements in
the list.

To warm up to the methods that we will soon have to write, let’s consider
implementation of the contains method. It uses the ﬁnger technique from our
work with SinglyLinkedLists:

public boolean contains(E value)
// pre: value is a non-null comparable object
// post: returns true iff contains value
{

Node<E> finger = data; // target
// search down list until we fall off or find bigger value
while ((finger != null) &&

ordering.compare(finger.value(),value) < 0)

{

finger = finger.next();

}
return finger != null && value.equals(finger.value());

}

This code is very similar to the linear search contains method of the Singly-
LinkedList class. However, because the list is always kept in order, it can stop
searching if it ﬁnds an element that is larger than the desired element. This
leads to a behavior that is linear in the size of the list, but in the case when a
value is not in the list, it terminates—on average—halfway down the list. For
programs that make heavy use of looking up values in the structure, this can
yield dramatic improvements in speed.

Note the use of the compare method in the ordering Comparator. No matter
what order the elements have been inserted, the ordering is responsible for
keeping them in the order speciﬁed.

Exercise 11.1 What would be necessary to allow the user of an OrderedStruc-
ture to provide an alternative ordering during the lifetime of a class? This method
might be called sortBy and would take a Comparator as its sole parameter.

Now, let us consider the addition of an element to the OrderedList. Since
the elements of the OrderedList are kept in order constantly, we must be care-
ful to preserve that ordering after we have inserted the value. Here is the code:

11.2 Keeping Structures Ordered

269

public void add(E value)
// pre: value is non-null
// post: value is added to the list, leaving it in order
{

Node<E> previous = null; // element to adjust
Node<E> finger = data;
// search for the correct location
while ((finger != null) &&

// target element

ordering.compare(finger.value(),value) < 0)

{

previous = finger;
finger = finger.next();

}
// spot is found, insert
if (previous == null) // check for insert at top
{

data = new Node<E>(value,data);

} else {

previous.setNext(

new Node<E>(value,previous.next()));

}
count++;

}

Here we use the ﬁnger technique with an additional previous reference to help
the insertion of the new element. The ﬁrst loop takes, on average, linear time
to ﬁnd a position where the value can be inserted. After the loop, the previous
reference refers to the element that will refer to the new element, or is null, if
the element should be inserted at the head of the list. Notice that we use the
Node methods to ensure that we reuse code that works and to make sure that
the elements are constructed with reasonable values in their ﬁelds.

One of the most common mistakes made is to forget to do important book-
keeping. Remember to increment count when inserting a value and to decre-
ment count when removing a value. When designing and implementing struc-
tures, it is sometimes useful to look at each method from the point of view of
each of the bookkeeping variables that you maintain.

Principle 18 Consider your code from different points of view.

Removing a value from the OrderedList ﬁrst performs a check to see if the
value is included, and then, if it is, removes it. When removing the value, we
return a reference to the value found in the list.

public E remove(E value)
// pre: value is non-null
// post: an instance of value is removed, if in list
{

Node<E> previous = null; // element to adjust
Node<E> finger = data;

// target element

NNWSWSENEWSE270

Ordered Structures

// search for value or fall off list
while ((finger != null) &&

ordering.compare(finger.value(),value) < 0)

{

previous = finger;
finger = finger.next();

}
// did we find it?
if ((finger != null) && value.equals(finger.value())) {

// yes, remove it
if (previous == null) // at top?
{

data = finger.next();

} else {

previous.setNext(finger.next());

}
count--;
// return value
return finger.value();

}
// return nonvalue
return null;

}

Again, because the SinglyLinkedListIterator accepts a SinglyLinked-
ListElement as its parameter, the implementation of the OrderedList’s iter-
ator method is particularly simple:

public Iterator<E> iterator()
{

return new SinglyLinkedListIterator<E>(data);

}

The remaining size-related procedures follow those found in the implemen-

tation of SinglyLinkedLists.

11.2.6 Example: The Modiﬁed Parking Lot

Renter—
ambiguous
noun:
(1) one who
rents from
others,
(2) one who
rents to others.

In Section 9.2 we implemented a system for maintaining rental contracts for a
small parking lot. With our knowledge of ordered structures, we now return to
that example to incorporate a new feature—an alphabetical listing of contracts.
As customers rent spaces from the parking ofﬁce, contracts are added to
a generic list of associations between renter names and lot assignments. We
now change that structure to reﬂect a better means of keeping track of this
information—an ordered list of comparable associations. This structure is de-
clared as an OrderedStructure but is assigned an instance of an OrderedList:

OrderedStructure<ComparableAssociation<String,Space>> rented =

new OrderedList<ComparableAssociation<String,Space>>(); // rented spaces

11.2 Keeping Structures Ordered

271

ParkingLot2

When a renter ﬁlls out a contract, the name of the renter and the parking space
information are bound together into a single ComparableAssociation:

String renter = r.readString();
// link renter with space description
rented.add(new ComparableAssociation<String,Space>(renter,location));
System.out.println("Space "+location.number+" rented.");

Notice that the renter’s name is placed into a String. Since Strings support
the compareTo method, they implement the Comparable interface. The default
ordering is used because the call to the constructor did not provide a speciﬁc
ordering.

At this point, the rented structure has all contracts sorted by name. To print

these contracts out, we accept a new command, contracts:

if (command.equals("contracts"))
{

// print out contracts in alphabetical order
for (ComparableAssociation<String,Space> contract : rented) {

// extract contract from iterator
// extract person from contract
String person = contract.getKey();
// extract parking slot description from contract
Space slot = contract.getValue();
// print it out
System.out.println(person+" is renting "+slot.number);

}

}

An iterator for the OrderedStructure is used to retrieve each of the Compar-
ableAssociations, from which we extract and print the renters’ names in al-
phabetical order. Here is an example run of the program (the user’s input is
indented):

rent small Alice

Space 0 rented.

rent large Bob

Space 9 rented.

rent small Carol

Space 1 rented.
return Alice

Space 0 is now free.

return David

No space rented to David.

rent small David

Space 2 rented.

rent small Eva

Space 0 rented.

quit

6 slots remain available.

272

Ordered Structures

Note that, for each of the requests for contracts, the contracts are listed in al-
phabetical order. This example is particularly interesting since it demonstrates
that use of an ordered structure eliminates the need to sort the contracts before
they are printed each time and that the interface meshes well with software that
doesn’t use ordered structures. While running an orderly parking lot can be a
tricky business, it is considerably simpliﬁed if you understand the subtleties of
ordered structures.

Exercise 11.2 Implement an alternative Comparator that compares two parking
spaces, based on slot numbers. Demonstrate that a single line will change the order
of the records in the ParkingLot2 program.

11.3 Conclusions

Computers spend a considerable amount of time maintaining ordered data struc-
tures. In Java we described an ordering of data values using the comparison
operator, compareTo, or a Comparator. Objects that fail to have an operator
such as compareTo cannot be totally ordered in a predetermined manner. Still,
a Comparator might be constructed to suggest an ordering between otherwise
incomparable values. Java enforces the development of an ordering using the
Comparable interface—an interface that simply requires the implementation of
the compareTo method.

Once data values may be compared and put in order, it is natural to design a
data structure that keeps its values in order. Disk directories, dictionaries, ﬁling
cabinets, and zip-code ordered mailing lists are all obvious examples of abstract
structures whose utility depends directly on their ability to efﬁciently maintain
a consistently ordered state. Here we extend various unordered structures in a
way that allows them to maintain the natural ordering of the underlying data.

Self Check Problems

Solutions to these problems begin on page 448.

11.1 What is the primary feature of an OrderedStructure?

How does the user communicate the order in which elements are to be

11.2
stored in an OrderedStructure?

11.3 What is the difference between a compareTo method and a comparator
with a compare method?

11.4
method) to be close together in an OrderedStructure?

Are we likely to ﬁnd two objects that are equal (using their equals

11.5

Is an OrderedVector a Vector?

Is it reasonable to have an OrderedStack, a class that implements both

11.6
the Stack and OrderedStructure interfaces?

11.3 Conclusions

273

People queue up to enter a movie theater. They are stored in an OrderedStructure.

11.7
How would you go about comparing two people?
Sally and Harry implement two different compareTo methods for a class
11.8
they are working on together. Sally’s compareTo method returns −1, 0, or +1,
depending on the relationship between two objects. Harry’s compareTo method
returns −6, 0, or +3. Which method is suitable?
Sally and Harry are working on an implementation of Coin. Sally de-
11.9
clares the sole parameter to her compareTo method as type Object. Harry
knows the compareTo method will always be called on objects of type Coin
and declares the parameter to be of that type. Which method is suitable for
storing Coin objects in a OrderedVector?

Problems

Describe the contents of an OrderedList after each of the following

Suppose duplicate values are added to an OrderedList. Where is the

Describe the contents of an OrderedVector after each of the following

Show that the expected insertion time of an element into an Ordered-

Show that the expected insertion time of an element into an Ordered-

Suppose duplicate values are added to an OrderedVector. Where is the

Solutions to the odd-numbered problems begin on page 475.
11.1
values has been added: 1, 9, 0, −1, and 3.
11.2
values has been added: 1, 9, 0, −1, and 3.
11.3
oldest value found (with respect to the newest)?
11.4
oldest value found (with respect to the newest)?
11.5
List is O(n) in the worst case.
11.6
Vector is O(n).
11.7
deredList?
11.8
a non-Comparable value to an OrderedVector?
11.9 Write the compareTo method for the String class.
11.10 Write the compareTo method for a class that is to be ordered by a ﬁeld,
key, which is a double. Be careful: The result of compareTo must be an int.
11.11 Write the compareTo method for a class describing a person whose name
is stored as two Strings: first and last. A person is “less than” another if they
appear before the other in a list alphabetized by last name and then ﬁrst name
(as is typical).
11.12 Previous editions of the structures package opted for the use of a
lessThan method instead of a compareTo method. The lessThan method would
return true exactly when one value was lessThan another. Are these ap-
proaches the same, or is one more versatile?

Under what conditions would you use an OrderedVector over an Or-

At what point does the Java environment complain about your passing

274

Ordered Structures

11.13 Suppose we consider the use of an OrderedStructure get method that
takes an integer i. This method returns the ith element of the OrderedStructure.
What are the best- and worst-case running times for this method on Ordered-
Vector and OrderedList?
11.14 Your department is interested in keeping track of information about
majors. Design a data structure that will maintain useful information for your
department. The roster of majors, of course, should be ordered by last name
(and then by ﬁrst, if there are multiple students with the same last name).

11.4 Laboratory: Computing the “Best Of”

Objective. To efﬁciently select the largest k values of n.

Discussion. One method to select the largest k values in a sequence of n is to
sort the n values and to look only at the ﬁrst k. (In Chapter 13, we will learn of
another technique: insert each of the n values into a max-heap and extract the
ﬁrst k values.) Such techniques have two important drawbacks:

• The data structure that keeps track of the values must be able to hold
n >> k values. This may not be possible if, for example, there are more
data than may be held easily in memory.

• The process requires O(n log n) time. It should be possible to accomplish

this in O(n) time.

One way to reduce these overheads is to keep track of, at all times, the best k
values found. As the n values are passed through the structure, they are only
remembered if they are potentially one of the largest k values.

Procedure. In this lab we will implement a BestOf OrderedStructure. The
constructor for your BestOf structure should take a value k, which is an upper
bound on the number of values that will be remembered. The default construc-
tor should remember the top 10.

An add method takes an Object and adds the element (if reasonable) in
the correct location. The get(i) method should return the ith largest value en-
countered so far. The size method should return the number of values currently
stored in the structure. This value should be between 0 and k, inclusive. The
iterator method should return an Iterator over all the values. The clear
method should remove all values from the structure.

Here are the steps that are necessary to construct and test this data structure:

1. Consider the underlying structure carefully. Because the main consider-
ations of this structure are size and speed, it would be most efﬁcient to
implement this using a ﬁxed-size array. We will assume that here.

2. Implement the add method. This method should take the value and, like a
pass of insertion sort, it should ﬁnd the correct location for the new value.
If the array is full and the value is no greater than any of the values,
nothing changes. Otherwise, the value is inserted in the correct location,
possibly dropping a smaller value.

3. Implement the get(i), size, and clear methods.

4. Implement the iterator method. A special AbstractIterator need not
be constructed; instead, values can be added to a linear structure and the
result of that structure’s iterator method is returned.

276

Ordered Structures

To test your structure, you can generate a sequence of integers between 0 and
n − 1. By the end, only values n − k . . . n − 1 should be remembered.
Thought Questions. Consider the following questions as you complete the lab:

1. What is the (big-O) complexity of one call to the add method? What is the

complexity of making n calls to add?

2. What are the advantages and disadvantages of keeping the array sorted at
all times? Would it be more efﬁcient to, say, only keep the smallest value
in the ﬁrst slot of the array?

3. Suppose that f (n) is deﬁned to be n/2 if n is even, and 3n + 1 if n is
odd. It is known that for small values of n (less than 1040) the sequence
of values generated by repeated application of f starting at n eventually
reaches 1. Consider all the sequences generated from n < 10, 000. What
are the maximum values encountered?

4. The BestOf structure can be made more general by providing a third con-
structor that takes k and a Comparator. The comparisons in the BestOf
class can now be recast as calls to the compare method of a Comparator.
When a Comparator is not provided, an instance of the structure pack-
age’s NaturalComparator is used, instead. Such an implementation al-
lows the BestOf class to order non-Comparable values, and Comparable
values in alternative orders.

Notes:

Chapter 12

Binary Trees

Concepts:
. Binary trees
.
. Recursion

Tree traversals

I think that I shall never see
A poem lovely as a binary tree.
—Bill Amend as Jason Fox

RECURSION IS A BEAUTIFUL APPROACH TO STRUCTURING. We commonly think of
recursion as a form of structuring the control of programs, but self-reference can
be used just as effectively in the structuring of program data. In this chapter, we
investigate the use of recursion in describing branching structures called trees.
Most of the structures we have already investigated are linear—their natural
presentation is in a line. Trees branch. The result is that where there is an
inherent ordering in linear structures, we ﬁnd choices in the way we order the
elements of a tree. These choices are an indication of the reduced “friction” of
the structure and, as a result, trees provide us with the fastest ways to solve
many problems.

Before we investigate the implementation of trees, we must develop a con-

cise terminology.

12.1 Terminology

A tree is a collection of elements, called nodes, and relations between them,
called edges. Usually, data are stored within the nodes of a tree. Two trees are
disjoint if no node or edge is found common to both. A trivial tree has no nodes
and thus no data. An isolated node is also a tree.

From these primitives we may recursively construct more complex trees. Let
r be a new node and let T1, T2, . . . , Tn be a (possibly empty) set—a forest—of
distinct trees. A new tree is constructed by making r the root of the tree, and
establishing an edge between r and the root of each tree, Ti, in the forest. We
refer to the trees, Ti, as subtrees. We draw trees with the root above and the
trees below. Figure 12.1g is an aid to understanding this construction.

The parent of a node is the adjacent node appearing above it (see Fig-
ure 12.2). The root of a tree is the unique node with no parent. The ancestors of
a node n are the roots of trees containing n: n, n’s parent, n’s parent’s parent,

278

Binary Trees

Examples of trees. Trees (a) and (b) are three-node trees. Trees are
Figure 12.1
sometimes symbolized abstractly, as in (c). Tree (b) is full, but (d) is not. Tree (e) is not
full but is complete. Complete trees are symbolized as in (f). Abstract tree (g) has root r
and subtrees (a) and (b).

(a)(b)(c)(d)(e)(f)(g)bra12.1 Terminology

279

Figure 12.2 Anatomy of trees. (a) A full (and complete) tree. (b) A complete tree that
is not full. Here, the unique path from node i to root r is bold: i has depth 2. Also,
indicated in bold is a longest path from s to a leaf l: s has height 2 and depth 1. The
subtree rooted at s has 5 nodes.

and so on. The root is the ancestor shared by every node in the tree. A child
of a node n is any node that has n as its parent. The descendants of a node n
are those nodes that have n as an ancestor. A leaf is a node with no children.
Note that n is its own ancestor and descendant. A node m is the proper ancestor
(proper descendant) of a node n if m is an ancestor (descendant) of n, but not
vice versa. In a tree T , the descendants of n form the subtree of T rooted at n.
Any node of a tree T that is not a leaf is an interior node. Roots can be interior
nodes. Nodes m and n are siblings if they share a parent.

A path is the unique shortest sequence of edges from a node n to an ancestor.
The length of a path is the number of edges it mentions. The height of a node n
in a tree is the length of any longest path between a leaf and n. The height of a
tree is the height of its root. This is the maximum height of any node in the tree.
The depth (or level) of a node n in its tree T is the length of the path from n to
T ’s root. The sum of a node’s depth and height is no greater than the height of
the tree. The degree of a node n is the number of its children. The degree of a
tree (or its arity) is the maximum degree of any of its nodes. A binary tree is a
tree with arity less than or equal to 2. A 1-ary binary tree is termed degenerate.
A node n in a binary tree is full if it has degree 2. In an oriented tree we will call
one child the left child and the other the right child. A full binary tree of height h

Level 2’s parent’s child(a)rils(b)bccaaRootInterior nodeLeafLevel 0Level 1280

Binary Trees

has leaves only on level h, and each of its internal nodes is full. The addition of
a node to a full binary tree causes its height to increase. A complete binary tree
of height h is a full binary tree with 0 or more of the rightmost leaves of level h
removed.

12.2 Example: Pedigree Charts

With the growth of the Internet, many people have been able to make contact
with long-lost ancestors, not through some new technology that allows contact
with spirits, but through genealogical databases. One of the reasons that ge-
nealogy has been so successful on computers is that computers can organize
treelike data more effectively than people.

One such organizational approach is a pedigree chart. This is little more than
a binary tree of the relatives of an individual. The root is an individual, perhaps
yourself, and the two subtrees are the pedigrees of your mother and father.1
They, of course, have two sets of parents, with pedigrees that are rooted at your
grandparents.

To demonstrate how we might make use of a BinaryTree class, we might
imagine the following code that develops the pedigree for someone named
George Bush:2

Steve points
out:
relationships in
these trees is
upside down!

// ancestors of George H. W. Bush
// indentation is provided to aid in understanding relations

Pedigree

BinaryTree<String> JSBush = new BinaryTree<String>("Rev. James");
BinaryTree<String> HEFay = new BinaryTree<String>("Harriet");

BinaryTree<String> SPBush = new BinaryTree<String>("Samuel",JSBush,HEFay);

BinaryTree<String> RESheldon = new BinaryTree<String>("Robert");
BinaryTree<String> MEButler = new BinaryTree<String>("Mary");

BinaryTree<String> FSheldon = new BinaryTree<String>("Flora",RESheldon,MEButler);

BinaryTree<String> PSBush = new BinaryTree<String>("Prescott",SPBush,FSheldon);

BinaryTree<String> DDWalker = new BinaryTree<String>("David");
BinaryTree<String> MABeaky = new BinaryTree<String>("Martha");

BinaryTree<String> GHWalker = new BinaryTree<String>("George",DDWalker,MABeaky);

BinaryTree<String> JHWear = new BinaryTree<String>("James II");
BinaryTree<String> NEHolliday = new BinaryTree<String>("Nancy");

BinaryTree<String> LWear = new BinaryTree<String>("Lucretia",JHWear,NEHolliday);

BinaryTree<String> DWalker = new BinaryTree<String>("Dorothy",GHWalker,LWear);

1 At the time of this writing, modern technology has not advanced to the point of allowing nodes
of degree other than 2.
2 This is the Texan born in Massachusetts; the other Texan was born in Connecticut.

12.3 Example: Expression Trees

281

BinaryTree<String> GHWBush = new BinaryTree<String>("George",PSBush,DWalker);

For each person we develop a node that either has no links (the parents were
not included in the database) or has references to other pedigrees stored as
BinaryTrees. Arbitrarily, we choose to maintain the father’s pedigree on the
left side and the mother’s pedigree along the right. We can then answer simple
questions about ancestry by examining the structure of the tree. For example,
who are the direct female relatives of the President?

following the mother’s side?

// Question: What are George H. W. Bush’s ancestors’ names,
//
BinaryTree<String> person = GHWBush;
while (!person.right().isEmpty())
{

person = person.right();
System.out.println(person.value()); // value is name

// right branch is mother

}

The results are

Dorothy
Lucretia
Nancy

Exercise 12.1 These are, of course, only some of the female relatives of President
Bush. Write a program that prints all the female names found in a BinaryTree
representing a pedigree chart.

One feature that would be useful, would be the ability to add branches to
a tree after the tree was constructed. For example, we might determine that
James Wear had parents named William and Sarah. The database might be
updated as follows:

// add individual directly
JHWear.setLeft(new BinaryTree<String>("William"));
// or keep a reference to the pedigree before the update:
BinaryTree<String> SAYancey = new BinaryTree<String>("Sarah");
JHWear.setRight(SAYancey);

A little thought suggests a number of other features that might be useful in
supporting the pedigree-as-BinaryTree structure.

12.3 Example: Expression Trees

Most programming languages involve mathematical expressions that are com-
posed of binary operations applied to values. An example from Java is the
simple expression R = 1 + (L - 1) * 2. This expression involves four opera-
tors (=, +, -, and *), and 5 values (R, 1, L, 1, and 2). Languages often represent
expressions using binary trees. Each value in the expression appears as a leaf,

282

Binary Trees

while the operators are internal nodes that represent the reduction of two values
to one (for example, L - 1 is reduced to a single value for use on the left side
of the multiplication sign). The expression tree associated with our expression
is shown in Figure 12.3a. We might imagine that the following code constructs
the tree and prints −1:

BinaryTree<term> v1a,v1b,v2,vL,vR,t;

Calc

// set up values 1 and 2, and declare variables
v1a = new BinaryTree<term>(new value(1));
v1b = new BinaryTree<term>(new value(1));
v2 = new BinaryTree<term>(new value(2));
vL = new BinaryTree<term>(new variable("L",0));// L=0
vR = new BinaryTree<term>(new variable("R",0));// R=0

// set up expression
t = new BinaryTree<term>(new operator(’-’),vL,v1a);
t = new BinaryTree<term>(new operator(’*’),t,v2);
t = new BinaryTree<term>(new operator(’+’),v1b,t);
t = new BinaryTree<term>(new operator(’=’),vR,t);

// evaluate and print expression
System.out.println(eval(t));

Once an expression is represented as an expression tree, it may be evaluated
by traversing the tree in an agreed-upon manner. Standard rules of mathemati-
cal precedence suggest that the parenthesized expression (L-1) should be eval-
uated ﬁrst. (The L represents a value previously stored in memory.) Once the
subtraction is accomplished, the result is multiplied by 2. The product is then
added to 1. The result of the addition is assigned to R. The assignment opera-
tor is treated in a manner similar to other common operators; it just has lower
precedence (it is evaluated later) than standard mathematical operators. Thus
an implementation of binary trees would be aided by a traversal mechanism
that allows us to manipulate values as they are encountered.

12.4 Implementation

We now consider the implementation of binary trees. As with List implementa-
tions, we will construct a self-referential BinaryTree class. The recursive design
motivates implementation of many of the BinaryTree operations as recursive
methods. However, because the base case of recursion often involves an empty
tree we will make use of a dedicated node that represents the empty tree. This
simple implementation will be the basis of a large number of more advanced
structures we see throughout the remainder of the text.

12.4 Implementation

283

Figure 12.3
R=1+(L-1)*2. (b) A possible connectivity of an implementation using references.

(a) An abstract expression tree representing

Expression trees.

R=+1*2-L1=+1*1-2LR(a)(b)284

Binary Trees

12.4.1 The BinaryTree Implementation

Our ﬁrst step toward the development of a binary tree implementation is to rep-
resent an entire subtree as a reference to its root node. The node will maintain
a reference to user data and related nodes (the node’s parent and its two chil-
dren) and directly provides methods to maintain a subtree rooted at that node.
All empty trees will be represented by one or more instances of BinaryTrees
called “empty” trees. This approach is not unlike the “dummy nodes” provided
in our study of linked lists. Allowing the empty tree to be an object allows
programs to call methods on trees that are empty. If the empty tree were rep-
resented by a null pointer, it would be impossible to apply methods to the
structure. Here is the interface (again we have omitted right-handed versions
of handed operations):

public class BinaryTree<E>
{

BinaryTree

public BinaryTree()
// post: constructor that generates an empty node

public BinaryTree(E value)
// post: returns a tree referencing value and two empty subtrees

public BinaryTree(E value, BinaryTree<E> left, BinaryTree<E> right)
// post: returns a tree referencing value and two subtrees

public BinaryTree<E> left()
// post: returns reference to (possibly empty) left subtree

public BinaryTree<E> parent()
// post: returns reference to parent node, or null

public void setLeft(BinaryTree<E> newLeft)
// post: sets left subtree to newLeft
//

re-parents newLeft if not null

protected void setParent(BinaryTree<E> newParent)
// post: re-parents this node to parent reference, or null

public Iterator<E> iterator()
// post: returns an in-order iterator of the elements

public boolean isLeftChild()
// post: returns true if this is a left child of parent

public E value()
// post: returns value associated with this node

public void setValue(E value)
// post: sets the value associated with this node

12.4 Implementation

285

Figure 12.4
right child reference in parent node.

The structure of a BinaryTree. The parent reference opposes a left or

}

Figure 12.3b depicts the use of BinaryTrees in the representation of an entire
tree. We visualize the structure of a BinaryTree as in Figure 12.4. To construct
such a node, we require three pieces of information: a reference to the data
that the user wishes to associate with this node, and left and right references
to binary tree nodes that are roots of subtrees of this node. The parent refer-
ence is determined implicitly from opposing references. The various methods
associated with constructing a BinaryTree are as follows:

protected E val; // value associated with node
protected BinaryTree<E> parent; // parent of node
protected BinaryTree<E> left, right; // children of node

public BinaryTree()
// post: constructor that generates an empty node
{

val = null;
parent = null; left = right = this;

}

public BinaryTree(E value)
// post: returns a tree referencing value and two empty subtrees
{

Assert.pre(value != null, "Tree values must be non-null.");
val = value;
right = left = new BinaryTree<E>();
setLeft(left);
setRight(right);

}

public BinaryTree(E value, BinaryTree<E> left, BinaryTree<E> right)
// post: returns a tree referencing value and two subtrees
{

Assert.pre(value != null, "Tree values must be non-null.");

parentvalueleftright286

Binary Trees

val = value;
if (left == null) { left = new BinaryTree<E>(); }
setLeft(left);
if (right == null) { right = new BinaryTree<E>(); }
setRight(right);

}

The ﬁrst constructor is called when an empty BinaryTree is needed. The result
of this constructor are empty nodes that will represent members of the fringe
of empty trees found along the edge of the binary tree. In the three-parameter
variant of the constructor we make two calls to “setting” routines. These rou-
tines allow one to set the references of the left and right subtrees, but also
ensure that the children of this node reference this node as their parent. This is
the direct cost of implementing forward and backward references along every
link. The return, though, is the considerable simpliﬁcation of other code within
the classes that make use of BinaryTree methods.

Principle 19 Don’t let opposing references show through the interface.

When maintenance of opposing references is left to the user, there is an oppor-
tunity for references to become inconsistent. Furthermore, one might imagine
implementations with fewer references (it is common, for example, to avoid the
parent reference); the details of the implementation should be hidden from the
user, in case the implementation needs to be changed.
Here is the code for setLeft (setRight is similar):

public void setLeft(BinaryTree<E> newLeft)
// post: sets left subtree to newLeft
//
{

re-parents newLeft if not null

if (isEmpty()) return;
if (left != null && left.parent() == this) left.setParent(null);
left = newLeft;
left.setParent(this);

}

If the setting of the left child causes a subtree to be disconnected from this node,
and that subtree considers this node to be its parent (it should), we disconnect
the node by setting its parent to null. We then set the left child reference to the
value passed in. Any dereferenced node is explicitly told to set its parent refer-
ence to null. We also take care to set the opposite parent reference by calling
the setParent method of the root of the associated non-trivial tree. Because
we want to maintain consistency between the downward child references and
the upward parent references, we declare setParent to be protected to make
it impossible for the user to refer to directly:

protected void setParent(BinaryTree<E> newParent)
// post: re-parents this node to parent reference, or null

NNWSWSENEWSE12.5 Example: An Expert System

287

{

}

if (!isEmpty()) {

parent = newParent;

}

It is, of course, useful to be able to access the various references once they
have been set. We accomplish this through the accessor functions such as left:

public BinaryTree<E> left()
// post: returns reference to (possibly empty) left subtree
{

return left;

}

Once the node has been constructed, its value can be inspected and modiﬁed
using the value-based functions that parallel those we have seen with other
types:

public E value()
// post: returns value associated with this node
{

return val;

}

public void setValue(E value)
// post: sets the value associated with this node
{

val = value;

}

Once the BinaryTree class is implemented, we may use it as the basis for our
implementation of some fairly complex programs and structures.

12.5 Example: An Expert System

Anyone who has been on a long trip with children has played the game Twenty
Questions. It’s not clear why this game has this name, because the questioning
often continues until the entire knowledge space of the child is exhausted. We
can develop a very similar program, here, called InfiniteQuestions. The cen-
tral component of the program is a database, stored as a BinaryTree. At each
leaf is an object that is a possible guess. The interior nodes are questions that
help distinguish between the guesses.

Figure 12.5 demonstrates one possible state of the database. To simulate a
questioner, one asks the questions encountered on a path from the root to some
leaf. If the response to the question is positive, the questioning continues along
the left branch of the tree; if the response is negative, the questioner considers
the right.

288

Binary Trees

Figure 12.5

The state of the database in the midst of playing InfiniteQuestions.

Exercise 12.2 What questions would the computer ask if you were thinking of a
truck?

Of course, we can build a very simple database with a single value—perhaps

a computer. The game might be set up to play against a human as follows:

public static void main(String args[])
{

Infinite-
Questions

Scanner human = new Scanner(System.in);
// construct a simple database -- knows only about a computer
BinaryTree<String> database = new BinaryTree<String>("a computer");

System.out.println("Do you want to play a game?");
while (human.nextLine().equals("yes"))
{

System.out.println("Think of something...I’ll guess it");
play(human,database);
System.out.println("Do you want to play again?");

}
System.out.println("Have a good day!");

}

When the game is played, we are likely to lose. If we lose, we can still beneﬁt by
incorporating information about the losing situation. If we guessed a computer
and the item was a car, we could incorporate the car and a question “Does it
have wheels?” to distinguish the two objects. As it turns out, the program is not
that difﬁcult.

public static void play(Scanner human, BinaryTree<String> database)
// pre: database is non-null
// post: the game is finished, and if we lost, we expanded the database
{

if (!database.left().isEmpty())
{ // further choices; must ask a question to distinguish them

System.out.println(database.value());
if (human.nextLine().equals("yes"))
{

a computerDoes it have a horn?Is it magical?a unicorna car12.5 Example: An Expert System

289

play(human,database.left());

} else {

play(human,database.right());

}

} else { // must be a statement node

System.out.println("Is it "+database.value()+"?");
if (human.nextLine().equals("yes"))
{

System.out.println("I guessed it!");

} else {

What were you thinking of?");

System.out.println("Darn.
// learn!
BinaryTree<String> newObject = new BinaryTree<String>(human.nextLine());
BinaryTree<String> oldObject = new BinaryTree<String>(database.value());
database.setLeft(newObject);
database.setRight(oldObject);
System.out.println("What question would distinguish "+

newObject.value()+" from "+
oldObject.value()+"?");

database.setValue(human.nextLine());

}

}

}

The program can distinguish questions from guesses by checking to see there is
a left child. This situation would suggest this node was a question since the two
children need to be distinguished.

The program is very careful to expand the database by adding new leaves
at the node that represents a losing guess. If we aren’t careful, we can easily
corrupt the database by growing a tree with the wrong topology.

Here is the output of a run of the game that demonstrates the ability of the

database to incorporate new information—that is to learn:

What were you thinking of?

Do you want to play a game?
Think of something...I’ll guess it
Is it a computer?
Darn.
What question would distinguish a car from a computer?
Do you want to play again?
Think of something...I’ll guess it
Does it have a horn?
Is it a car?
Darn.
What question would distinguish a unicorn from a car?
Do you want to play again?
Think of something...I’ll guess it
Does it have a horn?
Is it magical?
Is it a car?
I guessed it!

What were you thinking of?

290

Binary Trees

Do you want to play again?
Have a good day!

Exercise 12.3 Make a case for or against this program as a (simple) model for
human learning through experience.

We now discuss the implementation of a general-purpose Iterator for the
BinaryTree class. Not surprisingly a structure with branching (and therefore a
choice in traversal order) makes traversal implementation more difﬁcult. Next,
we consider the construction of several Iterators for binary trees.

12.6 Traversals of Binary Trees

We have seen, of course, there is a great industry in selling calculators that
allow users to enter expressions in what appear to be arbitrary ways. For ex-
ample, some calculators allow users to specify expressions in inﬁx form, where
keys associated with operators are pressed between operands. Other brands of
calculators advocate a postﬁx3 form, where the operator is pressed only after the
operands have been entered. Reconsidering our representation of expressions
as trees, we observe that there must be a similar variety in the ways we traverse
a BinaryTree structure. We consider those here.

When designing iterators for linear structures there are usually few useful
choices: start at one end and visit each element until you get to the other end.
Many of the linear structures we have seen provide an elements method that
constructs an iterator for traversing the structure. For binary trees, there is no
obvious order for traversing the structure. Here are four rather obvious but
distinct mechanisms:

Preorder traversal. Each node is visited before any of its children are visited.
Typically, we visit a node, and then each of the nodes in its left subtree,
followed by each of the nodes in the right subtree. A preorder traversal of
the expression tree in the margin visits the nodes in the order: =, R, +, 1,
∗, −, L, 1, and 2.

In-order traversal. Each node is visited after all the nodes of its left subtree
have been visited and before any of the nodes of the right subtree. The in-
order traversal is usually only useful with binary trees, but similar traver-
sal mechanisms can be constructed for trees of arbitrary arity. An in-order
traversal of the expression tree visits the nodes in the order: R, =, 1, +, L,
−, 1, ∗, and 2. Notice that, while this representation is similar to the ex-
pression that actually generated the binary tree, the traversal has removed
the parentheses.

3 Reverse Polish Notation (RPN) was developed by Jan Lukasiewicz, a philosopher and mathemati-
cian of the early twentieth century, and was made popular by Hewlett-Packard in their calculator
wars with Texas Instruments in the early 1970s.

2-*R+=1L112.6 Traversals of Binary Trees

291

Postorder traversal. Each node is visited after its children are visited. We visit
all the nodes of the left subtree, followed by all the nodes of the right sub-
tree, followed by the node itself. A postorder traversal of the expression
tree visits the nodes in the order: R, 1, L, 1, −, 2, ∗, +, and =. This is
precisely the order that the keys would have to be pressed on a “reverse
Polish” calculator to compute the correct result.

Level-order traversal. All nodes of level i are visited before the nodes of level
i + 1. The nodes of the expression tree are visited in the order: =, R, +,
1, ∗, −, 2, L, and 1. (This particular ordering of the nodes is motivation
for another implementation of binary trees we shall consider later and in
Problem 12.12.)

As these are the most common and useful techniques for traversing a binary tree
we will investigate their respective implementations. Traversing BinaryTrees
involves constructing an iterator that traverses the entire set of subtrees. For
this reason, and because the traversal of subtrees proves to be just as easy, we
discuss implementations of iterators for BinaryTrees.

Most implementations of iterators maintain a linear structure that keeps
track of the state of the iterator. In some cases, this auxiliary structure is not
strictly necessary (see Problem 12.22) but may reduce the complexity of the
implementation and improve its performance.

12.6.1 Preorder Traversal

For a preorder traversal, we wish to traverse each node of the tree before any of
its proper descendants (recall the node is a descendant of itself). To accomplish
this, we keep a stack of nodes whose right subtrees have not been investigated.
In particular, the current node is the topmost element of the stack, and elements
stored deeper within the stack are more distant ancestors.

We develop a new implementation of an Iterator that is not declared
public. Since it will be a member of the structure package, it is available
for use by the classes of the structure package, including BinaryTree. The
BinaryTree class will construct and return a reference to the preorder iterator
when the preorderElements method is called:

public AbstractIterator<E> preorderIterator()
// post: the elements of the binary tree rooted at node are
//
{

traversed in preorder

return new BTPreorderIterator<E>(this);

}

Note that the constructor for the iterator accepts a single parameter—the root
of the subtree to be traversed. Because the iterator only gives access to values
stored within nodes, this is not a breach of the privacy of our binary tree imple-
mentation. The actual implementation of the BTPreorderIterator is short:

BinaryTree

BTPreorder-
Iterator

292

Binary Trees

class BTPreorderIterator<E> extends AbstractIterator<E>
{

protected BinaryTree<E> root; // root of tree to be traversed
protected Stack<BinaryTree<E>> todo; // stack of unvisited nodes whose

public BTPreorderIterator(BinaryTree<E> root)
// post: constructs an iterator to traverse in preorder
{

todo = new StackList<BinaryTree<E>>();
this.root = root;
reset();

}

public void reset()
// post: resets the iterator to retraverse
{

todo.clear(); // stack is empty; push on root
if (root != null) todo.push(root);

}

public boolean hasNext()
// post: returns true iff iterator is not finished
{

return !todo.isEmpty();

}

public E get()
// pre: hasNext()
// post: returns reference to current value
{

return todo.get().value();

}

public E next()
// pre: hasNext();
// post: returns current value, increments iterator
{

BinaryTree<E> old = todo.pop();
E result = old.value();

if (!old.right().isEmpty()) todo.push(old.right());
if (!old.left().isEmpty()) todo.push(old.left());
return result;

}

}

As we can see, todo is the private stack used to keep track of references to
unvisited nodes whose nontrivial ancestors have been visited. Another way to
think about it is that it is the frontier of nodes encountered on paths from the
root that have not yet been visited. To construct the iterator we initialize the

12.6 Traversals of Binary Trees

293

Three cases of determining the next current node for preorder traversals.
Figure 12.6
Node A has a left child A0 as the next node; node B has no left, but a right child B0; and
node C is a leaf and ﬁnds its closest, “right cousin,” C 0.

stack. We also keep a reference to the root node; this will help reset the iterator
to the correct node (when the root of the traversal is not the root of the tree,
this information is vital). We then reset the iterator to the beginning of the
traversal.

Resetting the iterator involves clearing off the stack and then pushing the
root on the stack to make it the current node. The hasNext method needs only
to check to see if there is a top node of the stack, and value returns the reference
stored within the topmost BinaryTree of the todo stack.

The only tricky method is next. Recall that this method returns the value
of the current element and then increments the iterator, causing the iterator to
reference the next node in the traversal. Since the current node has just been
visited, we push on any children of the node—ﬁrst any right child, then any left.
If the node has a left child (see node A of Figure 12.6), that node (A0) is the
next node to be visited. If the current node (see node B) has only a right child
(B0), it will be visited next. If the current node has no children (see node C),
the effect is to visit the closest unvisited right cousin or sibling (C 0).

It is clear that over the life of the iterator each of the n values of the tree
is pushed onto and popped off the stack exactly once; thus the total cost of
traversing the tree is O(n). A similar observation is possible for each of the
remaining iteration techniques.

12.6.2 In-order Traversal

The most common traversal of trees is in order. For this reason, the BTInorder-
Iterator is the value returned when the elements method is called on a BinaryTree.
Again, the iterator maintains a stack of references to nodes. Here, the stack con-
tains unvisited ancestors of the current (unvisited) node.

A’ABB’CC’294

Binary Trees

Thus, the implementation of this traversal is similar to the code for other
iterators, except for the way the stack is reset and for the mechanism provided
in the nextElement method:

protected BinaryTree<E> root;

// root of subtree to be traversed

protected Stack<BinaryTree<E>> todo; // stack of unvisited ancestors

BTInorder-
Iterator

public void reset()
// post: resets the iterator to retraverse
{

todo.clear();
// stack is empty. Push on nodes from root to
// leftmost descendant
BinaryTree<E> current = root;
while (!current.isEmpty()) {
todo.push(current);
current = current.left();

}

}

public E next()
// pre: hasNext()
// post: returns current value, increments iterator
{

BinaryTree<E> old = todo.pop();
E result = old.value();
// we know this node has no unconsidered left children;
// if this node has a right child,
//
// else
//
if (!old.right().isEmpty()) {

top element of stack is next node to be visited

we push the right child and its leftmost descendants:

BinaryTree<E> current = old.right();
do {

todo.push(current);
current = current.left();
} while (!current.isEmpty());

}
return result;

}

Since the ﬁrst element considered in an in-order traversal is the leftmost de-
scendant of the root, resetting the iterator involves pushing each of the nodes
from the root down to the leftmost descendant on the auxiliary stack.

When the current node is popped from the stack, the next element of the

traversal must be found. We consider two scenarios:

1. If the current node has a right subtree, the nodes of that tree have not
been visited. At this stage we should push the right child, and all the

12.6 Traversals of Binary Trees

295

nodes down to and including its leftmost descendant, on the stack.

2. If the node has no right child, the subtree rooted at the current node has
been fully investigated, and the next node to be considered is the closest
unvisited ancestor of the former current node—the node just exposed on
the top of the stack.

As we shall see later, it is common to order the nodes of a binary tree so
that left-hand descendants of a node are smaller than the node, which is, in
turn, smaller than any of the rightmost descendants. In such a situation, the
in-order traversal plays a natural role in presenting the data of the tree in order.
For this reason, the elements method returns the iterator constructed by the
inorderElements method.

12.6.3 Postorder Traversal

Traversing a tree in postorder also maintains a stack of uninvestigated nodes.
Each of the elements on the stack is a node whose descendants are currently
being visited. Since the ﬁrst element to be visited is the leftmost descendant of
the root, the reset method must (as with the in-order iterator) push on each
of the nodes from the root to the leftmost descendant. (Note that the leftmost
descendant need not be a leaf—it does not have a left child, but it may have a
right.)

protected BinaryTree<E> root; // root of traversed subtree
protected Stack<BinaryTree<E>> todo; // stack of nodes
// whose descendants are currently being visited

public void reset()
// post: resets the iterator to retraverse
{

todo.clear();
// stack is empty; push on nodes from root to
// leftmost descendant
BinaryTree<E> current = root;
while (!current.isEmpty()) {
todo.push(current);
if (!current.left().isEmpty())

current = current.left();

else

current = current.right();

}

}

public E next()
// pre: hasNext();
// post: returns current value, increments iterator
{

BinaryTree<E> current = todo.pop();

BTPostorder-
Iterator

296

Binary Trees

E result = current.value();
if (!todo.isEmpty())
{

BinaryTree<E> parent = todo.get();
if (current == parent.left()) {

current = parent.right();
while (!current.isEmpty())
{

todo.push(current);
if (!current.left().isEmpty())
current = current.left();

else current = current.right();

}

}

}
return result;

}

Here an interior node on the stack is potentially exposed twice before be-
coming current. The ﬁrst time it may be left on the stack because the element
recently popped off was the left child. The right child should now be pushed
on. Later the exposed node becomes current because the popped element was
its right child.

It is interesting to observe that the stack contains the ancestors of the current
node. This stack describes, essentially, the path to the root of the tree. As a
result, we could represent the state of the stack by a single reference to the
current node.

12.6.4 Level-order Traversal

This is the
family values
traversal.

A level-order traversal visits the root, followed by the nodes of level 1, from
left to right, followed by the nodes of level 2, and so on. This can be easily
accomplished by maintaining a queue of the next few nodes to be visited. More
precisely, the queue contains the current node, followed by a list of all siblings
and cousins to the right of the current node, followed by a list of “nieces and
nephews” to the left of the current node. After we visit a node, we enqueue the
children of the node. With a little work it is easy to see that these are either
nieces and nephews or right cousins of the next node to be visited.

BTLevelorder-
Iterator

class BTLevelorderIterator<E> extends AbstractIterator<E>
{

protected BinaryTree<E> root; // root of traversed subtree
protected Queue<BinaryTree<E>> todo; // queue of unvisited relatives

public BTLevelorderIterator(BinaryTree<E> root)
// post: constructs an iterator to traverse in level order
{

todo = new QueueList<BinaryTree<E>>();

12.6 Traversals of Binary Trees

297

this.root = root;
reset();

}

public void reset()
// post: resets the iterator to root node
{

todo.clear();
// empty queue, add root
if (!root.isEmpty()) todo.enqueue(root);

}

public boolean hasNext()
// post: returns true iff iterator is not finished
{

return !todo.isEmpty();

}

public E get()
// pre: hasNext()
// post: returns reference to current value
{

return todo.get().value();

}

public E next()
// pre: hasNext();
// post: returns current value, increments iterator
{

BinaryTree<E> current = todo.dequeue();
E result = current.value();
if (!current.left().isEmpty())

todo.enqueue(current.left());

if (!current.right().isEmpty())

todo.enqueue(current.right());

return result;

}

}

To reset the iterator, we need only empty the queue and add the root. When
the queue is empty, the traversal is ﬁnished. When the next element is needed,
we need only enqueue references to children (left to right). Notice that, unlike
the other iterators, this method of traversing the tree is meaningful regardless
of the degree of the tree.

12.6.5 Recursion in Iterators

Trees are recursively deﬁned structures, so it would seem reasonable to con-
sider recursive implementations of iterators. The difﬁculty is that iterators must

298

Binary Trees

maintain their state across many calls to nextElement. Any recursive approach
to traversal would encounter nodes while deep in recursion, and the state of the
stack must be preserved.

One way around the difﬁculties of suspending the recursion is to initially
perform the entire traversal, generating a list of values encountered. Since the
entire traversal happens all at once, the list can be constructed using recursion.
As the iterator pushes forward, the elements of the list are consumed.

Using this idea, we rewrite the in-order traversal:

protected BinaryTree<T> root; // root of traversed subtree
protected Queue<BinaryTree<T>> todo; // queue of unvisited elements

Recursive-
Iterators

public BTInorderIteratorR(BinaryTree<T> root)
// post: constructs an iterator to traverse in in-order
{

todo = new QueueList<BinaryTree<T>>();
this.root = root;
reset();

}

public void reset()
// post: resets the iterator to retraverse
{

todo.clear();
enqueueInorder(root);

}

protected void enqueueInorder(BinaryTree<T> current)
// pre: current is non-null
// post: enqueue all values found in tree rooted at current
//
{

in in-order

if (current.isEmpty()) return;
enqueueInorder(current.left());
todo.enqueue(current);
enqueueInorder(current.right());

}
public T next()
// pre: hasNext();
// post: returns current value, increments iterator
{

BinaryTree<T> current = todo.dequeue();
return current.value();

}

12.7 Property-Based Methods

299

The core of this implementation is the protected method enqueueInorder. It
simply traverses the tree rooted at its parameter and enqueues every node en-
countered. Since it recursively enqueues all its left descendants, then itself, and
then its right descendants, it is an in-order traversal. Since the queue is a FIFO,
the order is preserved and the elements may be consumed at the user’s leisure.
For completeness and demonstration of symmetry, here are the pre- and

postorder counterparts:

protected void enqueuePreorder(BinaryTree<T> current)
// pre: current is non-null
// post: enqueue all values found in tree rooted at current
//
{

in preorder

if (current.isEmpty()) return;
todo.enqueue(current);
enqueuePreorder(current.left());
enqueuePreorder(current.right());

}

protected void enqueuePostorder(BinaryTree<T> current)
// pre: current is non-null
// post: enqueue all values found in tree rooted at current
//
{

in postorder

if (current.isEmpty()) return;
enqueuePostorder(current.left());
enqueuePostorder(current.right());
todo.enqueue(current);

}

It is reassuring to see the brevity of these implementations. Unfortunately, while
the recursive implementations are no less efﬁcient, they come at the obvious
cost of a potentially long delay whenever the iterator is reset. Still, for many
applications this may be satisfactory.

12.7 Property-Based Methods

At this point, we consider the implementation of a number of property-based
methods. Properties such as the height and fullness of a tree are important
to guiding updates of a tree structure. Because the binary tree is a recursively
deﬁned data type, the proofs of tree characteristics (and the methods that verify
them) often have a recursive feel. To emphasize the point, in this section we
allow theorems about trees and methods that verify them to intermingle. Again,
the methods described here are written for use on BinaryTrees, but they are
easily adapted for use with more complex structures.

Our ﬁrst method makes use of the fact that the root is a common ancestor of
every node of the tree. Because of this fact, given a BinaryTree, we can identify
the node as the root, or return the root of the tree containing the node’s parent.

300

Binary Trees

public BinaryTree<E> root()
// post: returns the root of the tree node n
{

if (parent() == null) return this;
else return parent().root();

BinaryTree

}

A proof that this method functions correctly could make use of induction, based
on the depth of the node involved.

If we count the number of times the root routine is recursively called, we
compute the number of edges from the node to the root—the depth of the node.
Not surprisingly, the code is very similar:

public int depth()
// post: returns the depth of a node in the tree
{

if (parent() == null) return 0;
return 1 + parent.depth();

}

The time it takes is proportional to the depth of the node. For full trees, we
will see that this is approximately O(log n). Notice that in the empty case we
return a height of −1. This is consistent with our recursive deﬁnition, even if
it does seem a little unusual. We could avoid the strange case by avoiding it
in the precondition. Then, of course, we would only have put off the work to
the calling routine. Often, making tough decisions about base cases can play
an important role in making your interface useful. Generally, a method is more
robust, and therefore more usable, if you handle as many cases as possible.

Principle 20 Write methods to be as general as possible.

Having computed the depth of a node, it follows that we should be able to
determine the height of a tree rooted at a particular BinaryTree. We know that
the height is simply the length of a longest path from the root to a leaf, but we
can adapt a self-referential deﬁnition: the height of a tree is one more than the
height of the tallest subtree. This translates directly into a clean implementation
of the height function:

public int height()
// post: returns the height of a node in its tree
{

if (isEmpty()) return -1;
return 1 + Math.max(left.height(),right.height());

}

This method takes O(n) time to execute on a subtree with n nodes (see Prob-
lem 12.9).

NNWSWSENEWSE12.7 Property-Based Methods

301

Figure 12.7

Several full (and complete) binary trees.

At this point, we consider the problem of identifying a tree that is full (see

Figure 12.7). Our approach uses recursion:

public boolean isFull()
// post: returns true iff the tree rooted at node is full
{

if (isEmpty()) return true;
if (left().height() != right().height()) return false;
return left().isFull() && right().isFull();

}

Again, the method is compact. Unfortunately, detecting this property appears to
be signiﬁcantly more expensive than computing the height. Note, for example,
that in the process of computing this function on a full tree, the height of every
node must be computed from scratch. The result is that the running time of the
algorithm on full trees is O(n log n). Can it be improved upon?

To ﬁnd the answer, we ﬁrst prove a series of theorems about the structure of
trees, with hope that we can develop an inexpensive way to test for a full tree.
Our ﬁrst result determines the number of nodes that are found in full trees:

Observation 12.1 A full binary tree of height h ≥ 0 has 2h+1 − 1 nodes.

Proof: We prove this by induction on the height of the tree. Suppose the tree
has height 0. Then it has exactly one node, which is also a leaf. Since 21 −1 = 1,
the observation holds, trivially.

Our inductive hypothesis is that full trees of height k < h have 2k+1 − 1
nodes. Since h > 0, we can decompose the tree into two full subtrees of height
h − 1, under a common root. Each of the full subtrees has 2(h−1)+1 − 1 = 2h − 1
nodes, so there are 2(2h − 1) + 1 = 2h+1 − 1 nodes. This is the result we sought
to prove, so by induction on tree height we see the observation must hold for
all full binary trees. (cid:5)

This observation suggests that if we can compute the height and size of a
tree, we have a hope of detecting a full tree. First, we compute the size of the
tree using a recursive algorithm:

302

Binary Trees

public int size()
// post: returns the size of the subtree
{

if (isEmpty()) return 0;
return left().size() + right().size() + 1;

}

This algorithm is similar to the height algorithm: each call to size counts one
more node, so the complexity of the routine is O(n). Now we have an alternative
implementation of isFull that compares the height of the tree to the number
of nodes:

public boolean isFull()
// post: returns true iff the tree rooted at n is full
{

int h = height();
int s = size();
return s == (1<<(h+1))-1;

}

Notice the return statement makes use of shifting 1 to the left h + 1 binary
places. This efﬁciently computes 2h+1. The result is that, given a full tree,
the function returns true in O(n) steps. Thus, it is possible to improve on our
previous implementation.

There is one signiﬁcant disadvantage, though. If you are given a tree with
height greater than 100, the result of the return statement cannot be accurately
computed: 2100 is a large enough number to overﬂow Java integers. Even rea-
sonably sized trees can have height greater than 100. The ﬁrst implementation
is accurate, even if it is slow. Problem 12.21 considers an efﬁcient and accurate
solution.

We now prove some useful facts about binary trees that help us evaluate per-
formance of methods that manipulate them. First, we consider a pretty result:
if a tree has lots of leaves, it must branch in lots of places.

Observation 12.2 The number of full nodes in a binary tree is one less than the
number of leaves.

Proof: Left to the reader.(cid:5)

With this result, we can now demonstrate that just over half the nodes of a full
tree are leaves.

Observation 12.3 A full binary tree of height h ≥ 0 has 2h leaves.

Proof:
In a full binary tree, all nodes are either full interior nodes or leaves.
The number of nodes is the sum of full nodes F and the number of leaves L.
Since, by Observation 12.2, F = L − 1, we know that the count of nodes is
F + L = 2L − 1 = 2h+1 − 1. This leads us to conclude that L = 2h and that
F = 2h − 1. (cid:5) This result demonstrates that for many simple tree methods (like

Redwoods and
sequoias come
to mind.

12.8 Example: Huffman Compression

303

Figure 12.8
represent. Paths from root to leaves provide Huffman bit strings.

The woodchuck Huffman tree. Leaves are labeled with the characters they

size) half of the time is spent processing leaves. Because complete trees can be
viewed as full trees with some rightmost leaves removed, similar results hold
for complete trees as well.

12.8 Example: Huffman Compression

Information within machines is stored as a series of bits, or 1’s and 0’s. Because
the distribution of the patterns of 1’s and 0’s is not always uniform, it is possible
to compress the bit patterns that are used and reduce the amount of storage
that is necessary. For example, consider the following 32-character phrase:

If a woodchuck could chuck wood!

If each letter in the string is represented by 8 bits (as they often are), the entire
string takes 256 bits of storage. Clearly this catchy phrase does not use the full
range of characters, and so perhaps 8 bits are not needed. In fact, there are 13
distinct characters so 4 bits would be sufﬁcient (4 bits can represent any of 16
values). This would halve the amount of storage required, to 128 bits.

If each character were represented by a unique variable-length string of bits,
further improvements are possible. Huffman encoding of characters allows us to
reduce the size of this string to only 111 bits by assigning frequently occurring
letters (like “o”) short representations and infrequent letters (like “a”) relatively
long representations.

Huffman encodings can be represented by binary trees whose leaves are the
characters to be represented. In Figure 12.8 left edges are labeled 0, while right
edges are labeled 1. Since there is a unique path from the root to each leaf, there

Huffman

du000000000111111111110!alkwIfhc‘ ’o010304

Binary Trees

Figure 12.9
cies of descendant characters.

The Huffman tree of Figure 12.8, but with nodes labeled by total frequen-

is a unique sequence of 1’s and 0’s encountered as well. We will use the string
of bits encountered along the path to a character as its representation in the
compressed output. Note also that no string is a preﬁx for any other (otherwise
one character would be an ancestor of another in the tree). This means that,
given the Huffman tree, decoding a string of bits involves simply traversing the
tree and writing out the leaves encountered.

The construction of a Huffman tree is an iterative process.

Initially, each
character is placed in a Huffman tree of its own. The weight of the tree is the
frequency of its associated character. We then iteratively merge the two most
lightweight Huffman trees into a single new Huffman tree whose weight is the
sum of weights of the subtrees. This continues until one tree remains. One
possible tree for our example is shown in Figure 12.9.

Our approach is to use BinaryTrees to maintain the structure. This allows
the use of recursion and easy merging of trees. Leaves of the tree carry descrip-
tions of characters and their frequencies:

class node
{

Huffman

int frequency; // frequency of char
char ch;

// the character

public node(int f)
// post: construct an entry with frequency f

public node(char c)
// post: construct character entry with frequency 1

‘ ’:5o:5c:5h:2f:1I:1w:2k:2d:3u:3l:1a:1!:1224367134910193212.8 Example: Huffman Compression

305

public boolean equals(Object other)
// post: return true if leaves represent same character

}

Intermediate nodes carry no data at all. Their relation to their ancestors deter-
mines their portion of the encoding. The entire tree is managed by a wrapper
class, huffmanTree:

class huffmanTree implements Comparable<huffmanTree>
{

BinaryTree<node> empty;
BinaryTree<node> root; // root of tree
// weight of tree
int totalWeight;

public huffmanTree(node e)
// post: construct a node with associated character

public huffmanTree(huffmanTree left, huffmanTree right)
// pre: left and right non-null
// post: merge two trees together and merge their weights

public int compareTo(huffmanTree other)
// pre: other is non-null
// post: return integer reflecting relation between values

public boolean equals(Object that)
// post: return true if this and that are same tree instance

public void print()
// post: print out strings associated with characters in tree

protected void print(BinaryTree r, String representation)
// post: print out strings associated with chars in tree r,
//

prefixed by representation

}

This class is a Comparable because it implements the compareTo method. That
method allows the trees to be ordered by their total weight during the merging
process. The utility method print generates our output recursively, building up
a different encoding along every path.

We now consider the construction of the tree:

public static void main(String args[])
{

// read System.in one character at a time
Scanner s = new Scanner(System.in).useDelimiter("");
List<node> freq = new SinglyLinkedList<node>();

// read data from input

306

Binary Trees

while (s.hasNext())
{

// s.next() returns string; we’re interested in first char
char c = s.next().charAt(0);
// look up character in frequency list
node query = new node(c);
node item = freq.remove(query);
if (item == null)
{

// not found, add new node
freq.addFirst(query);

} else { // found, increment node

item.frequency++;
freq.addFirst(item);

}

}

// insert each character into a Huffman tree
OrderedList<huffmanTree> trees = new OrderedList<huffmanTree>();
for (node n : freq)
{

trees.add(new huffmanTree(n));

}

// merge trees in pairs until one remains
Iterator ti = trees.iterator();
while (trees.size() > 1)
{

// construct a new iterator
ti = trees.iterator();
// grab two smallest values
huffmanTree smallest = (huffmanTree)ti.next();
huffmanTree small = (huffmanTree)ti.next();
// remove them
trees.remove(smallest);
trees.remove(small);
// add bigger tree containing both
trees.add(new huffmanTree(smallest,small));

}
// print only tree in list
ti
= trees.iterator();
Assert.condition(ti.hasNext(),"Huffman tree exists.");
huffmanTree encoding = (huffmanTree)ti.next();
encoding.print();

}

There are three phases in this method: the reading of the data, the construction
of the character-holding leaves of the tree, and the merging of trees into a single
encoding. Several things should be noted:

1. We store characters in a list. Since this list is likely to be small, keeping it

12.9 Example Implementation: Ahnentafel

307

ordered requires more code and is not likely to improve performance.

2. The huffmanTrees are kept in an OrderedList. Every time we remove
values we must construct a fresh iterator and remove the two smallest
trees. When they are merged and reinserted, the wrappers for the two
smaller trees can be garbage-collected. (Even better structures for man-
aging these details in Chapter 13.)

3. The resulting tree is then printed out. In an application, the information
in this tree would have to be included with the compressed text to guide
the decompression.

When the program is run on the input

If a woodchuck could chuck wood!

it generates the output:

Encoding of ! is 0000 (frequency was 1)
Encoding of a is 00010 (frequency was 1)
Encoding of l is 00011 (frequency was 1)
Encoding of u is 001 (frequency was 3)
Encoding of d is 010 (frequency was 3)
Encoding of k is 0110 (frequency was 2)
Encoding of w is 0111 (frequency was 2)
Encoding of I is 10000 (frequency was 1)
Encoding of f is 10001 (frequency was 1)
Encoding of h is 1001 (frequency was 2)
Encoding of c is 101 (frequency was 5)
Encoding of
is 110 (frequency was 5)
Encoding of o is 111 (frequency was 5)

Again, the total number of bits that would be used to represent our com-
pressed phrase is only 111, giving us a compression rate of 56 percent. In these
days of moving bits about, the construction of efﬁcient compression techniques
is an important industry—one industry that depends on the efﬁcient implemen-
tation of data structures.

12.9 Example Implementation: Ahnentafel

Having given, in Section 12.2, time to the Republican genealogists, we might
now investigate the heritage of a Democrat, William Jefferson Clinton. In Fig-
ure 12.10 we see the recent family tree presented as a list. This arrangement is
called an ahnentafel, or ancestor table. The table is generated by performing a
level-order traversal of the pedigree tree, and placing the resulting entries in a
table whose index starts at 1.

This layout has some interesting features. First, if we have an individual
with index i, the parents of the individual are found in table entries 2i and

308

Binary Trees

Lou Birchie Ayers
Eldridge Cassidy
Edith Grisham

1 William Jefferson Clinton
2 William Jefferson Blythe III
3 Virginia Dell Cassidy
4 William Jefferson Blythe II
5
6
7
8 Henry Patton Foote Blythe
9
10
11 Hattie Hayes
12
13
14
15

James M. Cassidy
Sarah Louisa Russell
Lemma Newell Grisham
Edna Earl Adams

Frances Ellen Hines
Simpson Green Ayers

The genealogy of President Clinton, presented as a linear table. Each
Figure 12.10
individual is assigned an index i. The parents of the individual can be found at locations
2i and 2i + 1. Performing an integer divide by 2 generates the index of a child. Note the
table starts at index 1.

2i + 1. Given the index i of a parent, we can ﬁnd the child (there is only one
child for every parent in a pedigree), by dividing the index by 2 and throwing
away the remainder.

We can use this as the basis of an implementation of short binary trees. Of
course, if the tree becomes tall, there is potentially a great amount of data in the
tree. Also, if a tree is not full, there will be empty locations in the table. These
must be carefully managed to keep from interpreting these entries as valid data.
While the math is fairly simple, our Lists are stored with the ﬁrst element at
location 0. The implementor must either choose to keep location 0 blank or to
modify the indexing methods to make use of zero-origin indices.

One possible approach to storing tree information like this is to store entrees
in key-value pairs in the list structure, with the key being the index. In this way,
the tree can be stored compactly and, if the associations are kept in an ordered
structure, they can be referenced with only a logarithmic slowdown.

Exercise 12.4 Describe what would be necessary to allow support for trees with
degrees up to eight (called octtrees). At what cost do we achieve this increased
functionality?

In Chapter 13 we will make use of an especially interesting binary tree called
a heap. We will see the ahnentafel approach to storing heaps in a vector shortly.

12.10 Conclusions

12.10 Conclusions

309

The tree is a nonlinear structure. Because of branching in the tree, we will
ﬁnd it is especially useful in situations where decisions can guide the process of
adding and removing nodes.

Our approach to implementing the binary tree—a tree with degree 2 or
less—is to visualize it as a self-referential structure. This is somewhat at odds
with an object-oriented approach. It is, for example, difﬁcult to represent empty
self-referential structures in a manner that allows us to invoke methods. To re-
lieve the tension between these two approaches, we represent the empty tree
with class instances that represent “empty” trees.

The power of recursion on branching structures is that signiﬁcant work can
be accomplished with very little code. Sometimes, as in our implementation
of the isFull method, we ﬁnd ourselves subtly pushed away from an efﬁcient
solution because of overzealous use of recursion. Usually we can eliminate such
inefﬁciencies, but we must always verify that our methods act reasonably.

Self Check Problems

In a binary tree, which node (or nodes) have greatest height?
Is it possible to have two different paths between a root and a leaf?

Can a tree have no root? Can a tree have no leaves?
Can a binary tree have more leaves than interior nodes? Can it have

Solutions to these problems begin on page 448.
12.1
12.2
more interior nodes than leaves?
12.3
12.4
12.5 Why are arithmetic expressions naturally stored in binary trees?
12.6 Many spindly trees look like lists. Is a BinaryTree a List?
12.7
provide indices to the elements of the tree?
12.8
a stack?
12.9
the tree is usually associated with the base case?
12.10 In code that recursively traverses binary trees, how many recursive calls
are usually found within the code?
12.11 What is the average degree of a node in an n-node binary tree?

Suppose we wanted to make a List from a BinaryTree. How might we

Recursion is used to compute many properties of trees. What portion of

Could the queue in the level-order traversal of a tree be replaced with

Problems

Solutions to the odd-numbered problems begin on page 477.
12.1
characters encountered in pre-, post- and in-order traversals.

In the following binary tree containing character data, describe the

310

Binary Trees

In the following tree what are the ancestors of the leaf D? What are the
12.2
descendants of the node S? The root of the tree is the common ancestor of what
nodes?

12.3

Draw an expression tree for each of the following expressions.

a. 1

b. 1 + 5 ∗ 3 − 4/2

c. 1 + 5 ∗ (3 − 4)/2

d. (1 + 5) ∗ (3 − 4/2)

e. (1 + (5 ∗ (3 − (4/2))))

Circle the nodes that are ancestors of the node containing the value 1.
12.4 What topological characteristics distinguish a tree from a list?
Demonstrate how the expression tree associated with the expression
12.5
R = 1 + (L − 1) ∗ 2 can be simpliﬁed using ﬁrst the distributive property and
then reduction of constant expressions to constants. Use pictures to forward
your argument.

For each of the methods of BinaryTree, indicate which method can
12.6
be implemented in terms of other public methods of that class or give a rea-
soned argument why it is not possible. Explain why it is useful to cast methods
in terms of other public methods and not base them directly on a particular
implementation.

The BinaryTree class is a recursive data structure, unlike the List class.
12.7
Describe how the List class would be different if it were implemented as a
recursive data structure.

The parent reference in a BinaryTree is declared protected and is
12.8
accessed through the accessor methods parent and setParent. Why is this any
different than declaring parent to be public.
12.9
take time proportional to the number of nodes in the tree.

Prove that efﬁcient computation of the height of a BinaryTree must

TIMSADDUTIMSADDU12.10 Conclusions

311

12.10 Write an equals method for the BinaryTree class. This function should
return true if and only if the trees are similarly shaped and refer to equal values
(every Object, including the Objects of the tree, has an equals method).

12.11 Write a static method, copy, that, given a binary tree, returns a copy of
the tree. Because not every object implements the copy method, you should not
copy objects to which the tree refers. This is referred to as a shallow copy.

12.12 Design a nonrecursive implementation of a binary tree that maintains
node data in a Vector, data. In this implementation, element 0 of data refer-
ences the root (if it exists). Every non-null element i of data ﬁnds its left and
right children at locations 2i + 1 and 2(i + 1), respectively. (The inverse of these
index relations suggests the parent of a nonroot node at i is found at location
b(i − 1)/2c.) Any element of data that is not used to represent a node should
maintain a null reference.

12.13 Design an interface for general trees—trees with unbounded degree.
Make this interface as consistent as possible with BinaryTrees when the degree
of a tree is no greater than 2.

12.14 Implement the general tree structure of Problem 12.13 using Binary-
TreeNodes. In this implementation, we interpret the left child of a BinaryTree-
Node to be the leftmost child, and the right child of the BinaryTree to be the
leftmost right sibling of the node.

12.15 Write a preorder iterator for the general tree implementation of Prob-
lem 12.14.

12.16 Implement the general tree structure of Problem 12.13 using a tree
node of your own design. In this implementation, each node maintains (some
sort of) collection of subtrees.

12.17 Write an in-order iterator for the general tree implementation of Prob-
lem 12.16.

12.18 Determine the complexity of each of the methods implemented in Prob-
lem 12.14.

12.19 Write a method, isComplete, that returns true if and only if the subtree
rooted at a BinaryTree on which it acts is complete.

12.20 A tree is said to be an AVL tree or height balanced if, for every node
n, the heights of the subtrees of n differ by no more than 1. Write a static
BinaryTree method that determines if a tree rooted at the referenced node is
height balanced.
12.21 The BinaryTree method isFull takes O(n log n) time to execute on
full trees, which, as we’ve seen, is not optimal. Careful thought shows that calls
to height (an O(n) operation) are made more often than is strictly necessary.
Write a recursive method info that computes two values—the height of the tree
and whether or not the tree is full. (This might be accomplished by having the
sign of the height be negative if it is not full. Make sure you do not call this
method on empty trees.) If info makes no call to height or isFull, its per-
formance is O(n). Verify this on a computer by counting procedure calls. This

312

Binary Trees

process is called strengthening, an optimization technique that often improves
performance of recursive algorithms.
12.22 Demonstrate how, in an in-order traversal, the associated stack can be
removed and replaced with a single reference. (Hint: We only need to know the
top of the stack, and the elements below the stack top are determined by the
stack top.)
12.23 Which other traversals can be rewritten by replacing their Linear struc-
ture with a single reference? How does this change impact the complexity of
each of the iterations?
12.24 Suppose the nodes of a binary tree are unique and that you are given
the order of elements as they are encountered in a preorder traversal and the
order of the elements as they are encountered in a postorder traversal. Under
what conditions can you accurately reconstruct the structure of the tree from
these two traversal orders?
12.25 Suppose you are to store a k-ary tree where each internal node has k
If k = 2, we see that Observa-
children and (obviously) each leaf has none.
tion 12.2 suggests that there is one more leaf than internal node. Prove that a
similar situation holds for k-ary trees with only full nodes and leaves: if there
are n full nodes, there are (k − 1)n + 1 leaves. (Hint: Use induction.)
12.26 Assume that the observation of Problem 12.25 is true and that you are
given a k-ary tree with only full nodes and leaves constructed with references
between nodes. In a k-ary tree with n nodes, how many references are null?
Considerable space might be saved if the k references to the children of an
internal node were stored in a k-element array, instead of k ﬁelds. In leaves,
the array needn’t be allocated. In an 8-ary tree with only full nodes and leaves
(an “octtree”) with one million internal nodes, how many bytes of space can be
saved using this array technique (assume all references consume 4 bytes).

12.11 Laboratory: Playing Gardner’s Hex-a-Pawn

Objective. To use trees to develop a game-playing strategy.

Discussion. In this lab we will write a simulator for the game, Hex-a-Pawn. This
game was developed in the early sixties by Martin Gardner. Three white and
three black pawns are placed on a 3 × 3 chessboard. On alternate moves they
may be either moved forward one square, or they may capture an opponent on
the diagonal. The game ends when a pawn is promoted to the opposite rank, or
if a player loses all his pieces, or if no legal move is possible.

In his article in the March 1962 Scientiﬁc American, Gardner discussed a
method for teaching a computer to play this simple game using a relatively
small number of training matches. The process involved keeping track of the
different states of the board and the potential for success (a win) from each
board state. When a move led directly to a loss, the computer forgot the move,
thereby causing it to avoid that particular loss in the future. This pruning of
moves could, of course, cause an intermediate state to lead indirectly to a loss,
in which case the computer would be forced to prune out an intermediate move.
Gardner’s original “computer” was constructed from matchboxes that con-
tained colored beads. Each bead corresponded to a potential move, and the
pruning involved disposing of the last bead played. In a modern system, we can
use nodes of a tree stored in a computer to maintain the necessary information
about each board state. The degree of each node is determined by the number
of possible moves.

Procedure. During the course of this project you are to

1. Construct a tree of Hex-a-Pawn board positions. Each node of the tree is
called a GameTree. The structure of the class is of your own design, but it
is likely to be similar to the BinaryTree implementation.

2. Construct three classes of Players that play the game of Hex-a-Pawn.

These three classes may interact in pairs to play a series of games.

Available for your use are three Javaﬁles:

HexBoard This class describes the state of a board. The default board is the 3×3
starting position. You can ask a board to print itself out (toString) or to
return the HexMoves (moves) that are possible from this position. You can
also ask a HexBoard if the current position is a win for a particular color—
HexBoard.WHITE or HexBoard.BLACK. A static utility method, opponent,
takes a color and returns the opposite color. The main method of this class
demonstrates how HexBoards are manipulated.

HexBoard

HexMove This class describes a valid move. The components of the Vector re-
turned from the HexBoard.moves contains objects of type HexMove. Given
a HexBoard and a HexMove one can construct the resulting HexBoard using
a HexBoard constructor.

HexMove

314

Binary Trees

Player When one is interested in constructing players that play Hex-a-Pawn,
the Player interface describes the form of the play method that must
be provided. The play method takes a GameTree node and an opposing
Player. It checks for a loss, plays the game according to the GameTree,
and then turns control over to the opposing player.

Read these class ﬁles carefully. You should not expect to modify them.

There are many approaches to experimenting with Hex-a-Pawn. One series

of experiments might be the following:

Player

608? No win
test
370? Wrong
win test
150? early stop

1. Compile HexBoard.java and run it as a program. Play a few games
against the computer. You may wish to modify the size of the board. Very
little is known about the games larger than 3 × 3.

2. Implement a GameTree class. This class should have a constructor that,

given a HexBoard and a color (a char, HexBoard.WHITE or HexBoard.BLACK),
generates the tree of all boards reachable from the speciﬁed board posi-
tion during normal game play. Alternate levels of the tree represent boards
that are considered by alternate players. Leaves are winning positions for
the player at hand. The references to other GameTree nodes are suggested
by the individual moves returned from the moves method. A complete
game tree for 3 × 3 boards has 252 nodes.

3. Implement the ﬁrst of three players. It should be called HumanPlayer. If it
hasn’t already lost (i.e., if the opponent hasn’t won), this player prints the
board, presents the moves, and allows a human (through a ReadStream)
to select a move. The play is then handed off to the opponent.

4. The second player, RandPlayer, should play randomly. Make sure you

check for a loss before attempting a move.

5. The third player, called CompPlayer, should attempt to have the CompPlayer

object modify the game tree to remove losing moves.

Clearly, Players may be made to play against each other in any combination.
Thought Questions. Consider the following questions as you complete the lab:

1. How many board positions are there for the 3 × 4 board? Can you deter-

mine how many moves there are for a 3 × 5 board?

2. If you implement the learning machine, pit two machines against each
other. Gardner called the computer to move ﬁrst H.I.M., and the ma-
chine to move second H.E.R. Will H.I.M. or H.E.R. ultimately win more
frequently? Explain your reasoning in a short write-up. What happens for
larger boards?

3. In Gardner’s original description of the game, each matchbox represented
a board state and its reﬂection. What modiﬁcations to HexBoard and
HexMove would be necessary to support this collapsing of the game tree?

Chapter 13

Priority Queues

Priority queues

Concepts:
.
. Heaps
.
.
.

Skew heaps
Sorting with heaps
Simulation

“Exactly!” said Mr. Wonka.
“I decided to invite ﬁve children
to the factory, and the one I liked best
at the end of the day
would be the winner!”
—Roald Dahl

SOMETIMES A RESTRICTED INTERFACE IS A FEATURE. The priority queue, like an
ordered structure, appears to keep its data in order. Unlike an ordered structure,
however, the priority queue allows the user only to access its smallest element.
The priority queue is also similar to the Linear structure: values are added to
the structure, and they later may be inspected or removed. Unlike their Linear
counterpart, however, once a value is added to the priority queue it may only
be removed if it is the minimum value.1 It is precisely this restricted interface
to the priority queue that allows many of its implementations to run quickly.

Priority queues are used to schedule processes in an operating system, to
schedule future events in a simulation, and to generally rank choices that are
generated out of order.

Think triage.

13.1 The Interface

Because we will see many contrasting implementations of the priority queue
structure, we describe it as abstractly as possible in Java—with an interface:

public interface PriorityQueue<E extends Comparable<E>>
{

public E getFirst();
// pre: !isEmpty()
// post: returns the minimum value in priority queue

public E remove();

PriorityQueue

1 We will consider priority queues whose elements are ranked in ascending order. It is, of course,
possible to maintain these queues in descending order with only a few modiﬁcations.

316

Priority Queues

// pre: !isEmpty()
// post: returns and removes minimum value from queue

public void add(E value);
// pre: value is non-null comparable
// post: value is added to priority queue

public boolean isEmpty();
// post: returns true iff no elements are in queue

public int size();
// post: returns number of elements within queue

public void clear();
// post: removes all elements from queue

}

Because they must be kept in order, the elements of a PriorityQueue are
Comparable. In this interface the smallest values are found near the front of
the queue and will be removed soonest.2 The add operation is used to insert
a new value into the queue. At any time a reference to the minimum value
can be obtained with the getFirst method and is removed with remove. The
remaining methods are similar to those we have seen before.

Notice that the PriorityQueue does not extend any of the interfaces we
have seen previously. First, as a matter of convenience, PriorityQueue methods
consume Comparable parameters and return Comparable values. Most struc-
tures we have encountered manipulate unconstrained generic Objects. Though
similar, the PriorityQueue is not a Queue. There is, for example, no dequeue
method. Though this might be remedied, it is clear that the PriorityQueue
need not act like a ﬁrst-in, ﬁrst-out structure. At any time, the value about to
be removed is the current minimum value. This value might have been the
ﬁrst value inserted, or it might have just recently “cut in line” before larger val-
ues. Still, the priority queue is just as general as the stack and queue since,
with a little work, one can associate with inserted values a priority that forces
any Linear behavior in a PriorityQueue. Finally, since the PriorityQueue
has no elements method, it may not be traversed and, therefore, cannot be a
Collection.

Exercise 13.1 An alternative deﬁnition of a PriorityQueue might not take and
return Comparable values. Instead, the constructor for a PriorityQueue could be
made to take a Comparator. Recall that the compare method of the Comparator
class needn’t take a Comparable value. Consider this alternative deﬁnition. Will
the code be simpler? When would we expect errors to be detected?

2 If explicit priorities are to be associated with values, the user may insert a ComparableAssoci-
ation whose key value is a Comparable such as an Integer. In this case, the associated value—the
data element—need not be Comparable.

13.2 Example: Improving the Huffman Code

317

The simplicity of the abstract priority queue makes its implementation rela-
tively straightforward. In this chapter we will consider three implementations:
one based on use of an OrderedStructure and two based on a novel structure
called a heap. First, we consider an example that emphasizes the simplicity of
our interface.

13.2 Example: Improving the Huffman Code

In the Huffman example from Section 12.8 we kept track of a pool of trees.
At each iteration of the tree-merging phase of the algorithm, the two lightest-
weight trees were removed from the pool and merged. There, we used an
OrderedStructure to maintain the collection of trees:

Huffman

OrderedList<huffmanTree> trees = new OrderedList<huffmanTree>();
// merge trees in pairs until one remains
Iterator ti = trees.iterator();
while (trees.size() > 1)
{

// construct a new iterator
ti = trees.iterator();
// grab two smallest values
huffmanTree smallest = (huffmanTree)ti.next();
huffmanTree small = (huffmanTree)ti.next();
// remove them
trees.remove(smallest);
trees.remove(small);
// add bigger tree containing both
trees.add(new huffmanTree(smallest,small));

}
// print only tree in list
= trees.iterator();
ti
Assert.condition(ti.hasNext(),"Huffman tree exists.");
huffmanTree encoding = (huffmanTree)ti.next();

To remove the two smallest objects from the OrderedStructure, we must con-
struct an Iterator and indirectly remove the ﬁrst two elements we encounter.
This code can be greatly simpliﬁed by storing the trees in a PriorityQueue. We
then remove the two minimum values:

PriorityQueue<huffmanTree> trees = new PriorityVector<huffmanTree>();
// merge trees in pairs until one remains
while (trees.size() > 1)
{

Huffman2

// grab two smallest values
huffmanTree smallest = (huffmanTree)trees.remove();
huffmanTree small = (huffmanTree)trees.remove();
// add bigger tree containing both
trees.add(new huffmanTree(smallest,small));

318

Priority Queues

}
huffmanTree encoding = trees.remove();

After the merging is complete, access to the ﬁnal result is also improved.

A number of interesting algorithms must have access to the minimum of a
collection of values, and yet do not require the collection to be sorted. The
extra energy required by an OrderedVector to keep all the values in order may,
in fact, be excessive for some purposes.

13.3 A Vector-Based Implementation

Perhaps the simplest implementation of a PriorityQueue is to keep all the val-
ues in ascending order in a Vector. Of course, the constructor is responsible for
initialization:

protected Vector<E> data;

Priority-
Vector

public PriorityVector()
// post: constructs a new priority queue
{

data = new Vector<E>();

}

From the standpoint of adding values to the structure, the priority queue
is very similar to the implementation of the OrderedVector structure. In fact,
the implementations of the add method and the “helper” method indexOf are
similar to those described in Section 11.2.2. Still, values of a priority queue are
removed in a manner that differs from that seen in the OrderedVector. They
are not removed by value.
Instead, getFirst and the parameterless remove
operate on the Vector element that is smallest (leftmost). The implementation
of these routines is straightforward:

public E getFirst()
// pre: !isEmpty()
// post: returns the minimum value in the priority queue
{

return data.get(0);

}

public E remove()
// pre: !isEmpty()
// post: removes and returns minimum value in priority queue
{

return data.remove(0);

}

The getFirst operation takes constant time. The remove operation caches and
removes the ﬁrst value of the Vector with a linear-time complexity. This can-

13.4 A Heap Implementation

319

not be easily avoided since the cost is inherent in the way we use the Vector
(though see Problem 13.8).

It is interesting to see the evolution of the various types encountered in
the discussion of the PriorityVector. Although the Vector took an entire
chapter to investigate, the abstract notion of a vector seems to be a relatively
natural structure here. Abstraction has allowed us to avoid considering the
minute details of the implementation. For example, we assume that Vectors
automatically extend themselves. The abstract notion of an OrderedVector, on
the other hand, appears to be insufﬁcient to directly support the speciﬁcation of
the PriorityVector. The reason is that the OrderedVector does not support
Vector operations like the index-based get(i) and remove(i). These could, of
course, be added to the OrderedVector interface, but an appeal for symmetry
might then suggest implementation of the method add(i). This would be a
poor decision since it would then allow the user to insert elements out of order.

Principle 21 Avoid unnaturally extending a natural interface.

Designers of data structures spend considerable time weighing these design
trade-offs. While it is tempting to make the most versatile structures support
a wide variety of extensions, it surrenders the interface distinctions between
structures that often allow for novel, efﬁcient, and safe implementations.

Exercise 13.2 Although the OrderedVector class does not directly support the
PriorityQueue interface, it nonetheless can be used in a protected manner. Im-
plement the PriorityVector using a protected OrderedVector? What are the
advantages and disadvantages?

In Section 13.4 we discuss a rich class of structures that allow us to maintain
It turns out that even a loose ordering is

a loose ordering among elements.
sufﬁcient to implement priority queues.

13.4 A Heap Implementation

In actuality, it is not necessary to develop a complete ranking of the elements
of the priority queue in order to support the necessary operations.
It is only
necessary to be able to quickly identify the minimum value and to maintain a
relatively loose ordering of the remaining values. This realization is the motiva-
tion for a structure called a heap.

Deﬁnition 13.1 A heap is a binary tree whose root references the minimum value
and whose subtrees are, themselves, heaps.

An alternate deﬁnition is also sometimes useful.

Deﬁnition 13.2 A heap is a binary tree whose values are in ascending order on
every path from root to leaf.

NNWSWSENEWSE320

Priority Queues

Figure 13.1
among siblings. Only heap (b) is complete.

Four heaps containing the same values. Note that there is no ordering

We will draw our heaps in the manner shown in Figure 13.1, with the mini-
mum value on the top and the possibly larger values below. Notice that each of
the four heaps contains the same values but has a different structure. Clearly,
there is a great deal of freedom in the way that the heap can be oriented—for
example, exchanging subtrees does not violate the heap property (heaps (c)
and (d) are mirror images of each other). While not every tree with these four
values is a heap, many are (see Problems 13.17 and 13.18). This ﬂexibility
reduces the friction associated with constructing and maintaining a valid heap
and, therefore, a valid priority queue. When friction is reduced, we have the
potential for increasing the speed of some operations.

Principle 22 Seek structures with reduced friction.

This is
completely
obvious.

We will say that a heap is a complete heap if the binary tree holding the values
of the heap is complete. Any set of n values may be stored in a complete heap.
(To see this we need only sort the values into ascending order and place them
in level order in a complete binary tree. Since the values were inserted in as-
cending order, every child is at least as great as its parent.) The abstract notion
of a complete heap forms the basis for the ﬁrst of two heap implementations of
a priority queue.

13.4.1 Vector-Based Heaps

As we saw in Section 12.9 when we considered the implementation of Ah-
nentafel structures, any complete binary tree (and therefore any complete heap)
may be stored compactly in a vector. The method involves traversing the tree in
level order and mapping the values to successive slots of the vector. When we
are ﬁnished with this construction, we observe the following (see Figure 13.2):

1. The root of the tree is stored in location 0.
references the minimum value of the heap.

If non-null, this location

2132213222311223(a)(b)(c)(d)NNWSWSENEWSE13.4 A Heap Implementation

321

Figure 13.2 An abstract heap (top) and its vector representation. Arrows from parent
to child are not physically part of the vector, but are indices computed by the heap’s left
and right methods.

2. The left child of a value stored in location i is found at location 2i + 1.

3. The right child of a value stored in location i may be found at the location

following the left child, location 2(i + 1) = (2i + 1) + 1.

4. The parent of a value found in location i can be found at location b i−1

2 c.
Since division of integers in Java-like languages throws away the remain-
der for positive numbers, this expression is written (i-1)/2.

These relations may, of course, be encoded as functions. In Figure 13.2 we see
the mapping of a heap to a vector, with tree relations indicated by arrows in the
vector. Notice that while the vector is not maintained in ascending order, any
path from the root to a leaf encounters values in ascending order. If the vector is
larger than necessary, slots not associated with tree nodes can maintain a null
reference. With this mapping in mind, we consider the constructor and static
methods:

protected Vector<E> data; // the data, kept in heap order

public VectorHeap()
// post: constructs a new priority queue
{

data = new Vector<E>();

}

VectorHeap

1324436558404201433265584042401234567891011121314-10-133322

Priority Queues

public VectorHeap(Vector<E> v)
// post: constructs a new priority queue from an unordered vector
{

int i;
data = new Vector<E>(v.size()); // we know ultimate size
for (i = 0; i < v.size(); i++)
// add elements to heap
{
add(v.get(i));

}

}

protected static int parent(int i)
// pre: 0 <= i < size
// post: returns parent of node at location i
{

return (i-1)/2;

}

protected static int left(int i)
// pre: 0 <= i < size
// post: returns index of left child of node at location i
{

return 2*i+1;

}

protected static int right(int i)
// pre: 0 <= i < size
// post: returns index of right child of node at location i
{

return 2*(i+1);

}

The functions parent, left, and right are declared static to indicate that
they do not actually have to be called on any instance of a heap. Instead, their
values are functions of their parameters only.

Principle 23 Declare object-independent functions static.

Now consider the addition of a value to a complete heap. We know that
the heap is currently complete. Ideally, after the addition of the value the heap
will remain complete but will contain one extra value. This realization forces
us to commit to inserting a value in a way that ultimately produces a correctly
structured heap. Since the ﬁrst free element of the Vector will hold a value,
we optimistically insert the new value in that location (see Figure 13.3).
If,
considering the path from the leaf to the root, the value is in the wrong location,
then it must be “percolated upward” to the correct entry. We begin by comparing
and, if necessary, exchanging the new value and its parent. If the values along
the path are still incorrectly ordered, it must be because of the new value, and
we continue to percolate the value upward until either the new value is the

NNWSWSENEWSE13.4 A Heap Implementation

323

root or it is greater than or equal to its current parent. The only values possibly
exchanged in this operation are those appearing along the unique path from the
insertion point. Since locations that change only become smaller, the integrity
of other paths in the tree is maintained.

The code associated with percolating a value upward is contained in the
function percolateUp. This function takes an index of a value that is possibly
out of place and pushes the value upward toward the root until it reaches the
correct location. While the routine takes an index as a parameter, the parameter
passed is usually the index of the rightmost leaf of the bottom level.

protected void percolateUp(int leaf)
// pre: 0 <= leaf < size
// post: moves node at index leaf up to appropriate position
{

int parent = parent(leaf);
E value = data.get(leaf);
while (leaf > 0 &&

(value.compareTo(data.get(parent)) < 0))

{

data.set(leaf,data.get(parent));
leaf = parent;
parent = parent(leaf);

}
data.set(leaf,value);

}

Adding a value to the priority queue is then only a matter of appending it to
the end of the vector (the location of the newly added leaf) and percolating the
value upward until it ﬁnds the correct location.

public void add(E value)
// pre: value is non-null comparable
// post: value is added to priority queue
{

data.add(value);
percolateUp(data.size()-1);

}

Let us consider how long it takes to accomplish the addition of a value to the
heap. Remember that the tree that we are working with is an n-node complete
binary tree, so its height is blog2 nc. Each step of the percolateUp routine takes
constant time and pushes the new value up one level. Of course, it may be
positioned correctly the ﬁrst time, but the worst-case behavior of inserting the
new value into the tree consumes O(log n) time. This performance is consid-
erably better than the linear behavior of the PriorityVector implementation
described in Section 13.3. What is the best time? It is constant when the value
added is large compared to the values found on the path from the new leaf to
the root.

What is the
expected time?
Be careful!

324

Priority Queues

The addition of a value (2) to a vector-based heap. (a) The value is
Figure 13.3
inserted into a free location known to be part of the result structure. (b) The value is
percolated up to the correct location on the unique path to the root.

−11024436558404223−10143265584042401234567891011121314232−101433265584042401234567891011121314132443655840422−103333(a) Before(b) After13.4 A Heap Implementation

325

Next, we consider the removal of the minimum value (see Figures 13.4 and
13.5). It is located at the root of the heap, in the ﬁrst slot of the vector. The
removal of this value leaves an empty location at the top of the heap. Ultimately,
when the operation is complete, the freed location will be the rightmost leaf of
the bottom level, the last element of the underlying vector. Again, our approach
is ﬁrst to construct a tree that is the correct shape, but potentially not a heap,
and then perform transformations on the tree that both maintain its shape and We’re “heaping
bring the structure closer to being a heap. Thus, when the minimum value is
removed, the rightmost leaf on the bottom level is removed and re-placed at
the root of the tree (Figure 13.4a and b). At this point, the tree is the correct
shape, but it may not be a heap because the root of the tree is potentially too
large. Since the subtrees remain heaps, we need to ensure the root of the tree
is the minimum value contained in the tree. We ﬁrst ﬁnd the minimum child
If the root value is no
and compare this value with the root (Figure 13.5a).
greater, the minimum value is at the root and the entire structure is a heap.
If the root is larger, then it is exchanged with the true minimum—the smallest
child—pushing the large value downward. At this point, the root of the tree has
the correct value. All but one of the subtrees are unchanged, and the shape of
the tree remains correct. All that has happened is that a large value has been
pushed down to where it may violate the heap property in a subtree. We then
perform any further exchanges recursively, with the value sinking into smaller
subtrees (Figure 13.5b), possibly becoming a leaf. Since any single value is a
heap, the recursion must stop by the time the newly inserted value becomes a
leaf.

in shape.”

Here is the code associated with the pushing down of the root:

protected void pushDownRoot(int root)
// pre: 0 <= root < size
// post: moves node at index root down
//
{

to appropriate position in subtree

int heapSize = data.size();
E value = data.get(root);
while (root < heapSize) {

int childpos = left(root);
if (childpos < heapSize)
{

if ((right(root) < heapSize) &&

((data.get(childpos+1)).compareTo

(data.get(childpos)) < 0))

{

childpos++;

}
// Assert: childpos indexes smaller of two children
if ((data.get(childpos)).compareTo

(value) < 0)

{

data.set(root,data.get(childpos));

326

Priority Queues

root = childpos; // keep moving down

} else { // found right location

data.set(root,value);
return;

}

} else { // at a leaf! insert and halt

data.set(root,value);
return;

}

}

}

The remove method simply involves returning the smallest value of the heap,
but only after the rightmost element of the vector has been pushed downward.

public E remove()
// pre: !isEmpty()
// post: returns and removes minimum value from queue
{

E minVal = getFirst();
data.set(0,data.get(data.size()-1));
data.setSize(data.size()-1);
if (data.size() > 1) pushDownRoot(0);
return minVal;

}

Each iteration in pushDownRoot pushes a large value down into a smaller heap
on a path from the root to a leaf. Therefore, the performance of remove is
O(log n), an improvement over the behavior of the PriorityVector implemen-
tation.

Since we have implemented all the required methods of the PriorityQueue,
the VectorHeap implements the PriorityQueue and may be used wherever a
priority queue is required.

The advantages of the VectorHeap mechanism are that, because of the
unique mapping of complete trees to the Vector, it is unnecessary to explicitly
store the connections between elements. Even though we are able to get im-
proved performance over the PriorityVector, we do not have to pay a space
penalty. The complexity arises, instead, in the code necessary to support the
insertion and removal of values.

13.4.2 Example: Heapsort

Any priority queue, of course, can be used as the underlying data structure
for a sorting mechanism. When the values of a heap are stored in a Vector,
an empty location is potentially made available when they are removed. This
location could be used to store a removed value. As the heap shrinks, the values
are stored in the newly vacated elements of the Vector. As the heap becomes
empty, the Vector is completely ﬁlled with values in descending order.

13.4 A Heap Implementation

327

Figure 13.4 Removing a value from the heap shown in (a) involves moving the right-
most value of the vector to the top of the heap as in (b). Note that this value is likely to
violate the heap property but that the subtrees will remain heaps.

132443655840420-130143265584042401234567891011121314-1330143265584042401234567891011121314331324436558404203(a)(b)328

Priority Queues

Figure 13.5 Removing a value (continued).
In (a) the newly inserted value at the
root is pushed down along a shaded path following the smallest children (lightly shaded
In (b) the root value ﬁnds, over
nodes are also considered in determining the path).
several iterations, the correct location along the path. Smaller values shift upward to
make room for the new value.

13265434204358401324365584042304(b)143655840420123456789101112131430342(a)01432655840420123456789101112131434313.4 A Heap Implementation

329

Unfortunately, we cannot make assumptions about the structure of the val-
ues initially found in the Vector; we are, after all, sorting them. Since this
approach depends on the values being placed in a heap, we must consider
one more operation: a constructor that “heapiﬁes” the data found in a Vector
passed to it:

public VectorHeap(Vector<E> v)
// post: constructs a new priority queue from an unordered vector
{

int i;
data = new Vector<E>(v.size()); // we know ultimate size
for (i = 0; i < v.size(); i++)
// add elements to heap
{
add(v.get(i));

}

}

The process of constructing a heap from an unordered Vector obviously takes
the time of n add operations, each of which is O(log n). The worst-case cost
of “heapifying” is, therefore, O(n log n).
(This can be improved—see Prob-
lem 13.10.)

Now, the remaining part of the heapsort—removing the minimum values
and placing them in the newly freed locations—requires n remove operations.
This phase also has worst-case complexity O(n log n). We have, therefore, an-
other sorting algorithm with O(n log n) behavior and little space overhead.

The feature of a heap that makes the sort so efﬁcient is its short height. The
values are always stored in as full a tree as possible and, as a result, we may
place a logarithmic upper bound on the time it takes to insert and remove val-
ues. In Section 13.4.3 we investigate the use of unrestricted heaps to implement
priority queues. These structures have amortized cost that is equivalent to heaps
built atop vectors.

13.4.3 Skew Heaps

The performance of Vector-based heaps is directly dependent on the fact that
these heaps are complete. Since complete heaps are a minority of all heaps, it
is reasonable to ask if efﬁcient priority queues might be constructed from unre-
stricted heaps. The answer is yes, if we relax the way we measure performance.
We consider, here, the implementation of heaps using dynamically struc-
tured binary trees. A direct cost of this decision is the increase in space. Whereas
a Vector stores a single reference, the binary tree node keeps an additional
three references. These three references allow us to implement noncomplete
heaps, called skew heaps, in a space-efﬁcient manner (see Problem 13.21). Here
are the protected data and the constructor for this structure:

protected BinaryTree<E> root;

SkewHeap

330

Priority Queues

protected final BinaryTree<E> EMPTY = new BinaryTree<E>();
protected int count;

public SkewHeap()
// post: creates an empty priority queue
{

root = EMPTY;
count = 0;

}

Notice that we keep track of the size of the heap locally, rather than asking the
BinaryTree for its size. This is simply a matter of efﬁciency, but it requires us
to maintain the value within the add and remove procedures. Once we commit
to implementing heaps in this manner, we need to consider the implementation
of each of the major operators.

The implementation of getFirst simply references the value stored at the

root. Its implementation is relatively straightforward:

public E getFirst()
// pre: !isEmpty()
// post: returns the minimum value in priority queue
{

return root.value();

}

Before we consider the implementation of the add and remove methods, we
consider a (seemingly unnecessary) operation, merge. This method takes two
heaps and merges them together. This is a destructive operation: the elements
of the participating heaps are consumed in the construction of the result. Our
approach will be to make merge a recursive method that considers several cases.
First, if either of the two heaps participating in the merge is empty, then the
result of the merge is the other heap. Otherwise, both heaps contain at least a
value—assume that the minimum root is found in the left heap (if not, we can
swap them). We know, then, that the result of the merge will be a reference to
the root node of the left heap. To see how the right heap is merged into the left
we consider two cases:

1. If the left heap has no left child, make the right heap the left child of the

left heap (see Figure 13.6b).

2. Otherwise, exchange the left and right children of the left heap. Then
merge (the newly made) left subheap of the left heap with the right heap
(see Figure 13.6d).

Notice that if the left heap has one subheap, the right heap becomes the left
subheap and the merging is ﬁnished. Here is the code for the merge method:

As with all good
things, this will
eventually seem
necessary.

13.4 A Heap Implementation

331

In (a) one of the
Figure 13.6 Different cases of the merge method for SkewHeaps.
heaps is empty. In (b) and (c) the right heap becomes the left child of the left heap. In
(d) the right heap is merged into what was the right subheap.

Left133(a)(b)(c)3111110310(d)+++++11331LeftRightResultLeftRightResultLeftRightResultResultRight332

Priority Queues

protected static <E extends Comparable<E>>

BinaryTree<E> merge(BinaryTree<E> left, BinaryTree<E> right)

{

}

if (left.isEmpty()) return right;
if (right.isEmpty()) return left;
E leftVal = left.value();
E rightVal = right.value();
BinaryTree<E> result;
if (rightVal.compareTo(leftVal) < 0)
{

result = merge(right,left);

} else {

result = left;
// assertion left side is smaller than right
// left is new root
if (result.left().isEmpty())
{

result.setLeft(right);

} else {

BinaryTree<E> temp = result.right();
result.setRight(result.left());
result.setLeft(merge(temp,right));

}

}
return result;

Once the merge method has been deﬁned, we ﬁnd that the process of adding
a value or removing the minimum is relatively straightforward. To add a value,
we construct a new BinaryTree containing the single value that is to be added.
This is, in essence, a one-element heap. We then merge this heap with the
existing heap, and the result is a new heap with the value added:

public void add(E value)
// pre: value is non-null comparable
// post: value is added to priority queue
{

BinaryTree<E> smallTree = new BinaryTree<E>(value,EMPTY,EMPTY);
root = merge(smallTree,root);
count++;

}

To remove the minimum value from the heap we must extract and return
the value at the root. To construct the smaller resulting heap we detach both
subtrees from the root and merge them together. The result is a heap with all
the values of the left and right subtrees, but not the root. This is precisely the
result we require. Here is the code:

public E remove()
// pre: !isEmpty()

13.5 Example: Circuit Simulation

333

Figure 13.7 A circuit for detecting a rising logic level.

// post: returns and removes minimum value from queue
{

E result = root.value();
root = merge(root.left(),root.right());
count--;
return result;

}

The remaining priority queue methods for skew heaps are implemented in a
relatively straightforward manner.

Because a skew heap has unconstrained topology (see Problem 13.16), it
is possible to construct examples of skew heaps with degenerate behavior. For
example, adding a new maximum value can take O(n) time. For this reason
we cannot put very impressive bounds on the performance of any individual
operation. The skew heap, however, is an example of a self-organizing structure:
inefﬁcient operations spend some of their excess time making later operations
run more quickly. If we are careful, time “charged against” early operations can
be amortized or redistributed to later operations, which we hope will run very
efﬁciently. This type of analysis can be used, for example, to demonstrate that
m > n skew heap operations applied to a heap of size n take no more than
O(m log n) time. On average, then, each operation takes O(log n) time. For
applications where it is expected that a signiﬁcant number of requests of a heap
will be made, this performance is appealing.

13.5 Example: Circuit Simulation

Consider the electronic digital circuit depicted in Figure 13.7. The two devices
shown are logic gates. The wires between the gates propagate electrical signals.
Typically a zero voltage is called false or low, while a potential of 3 volts or more
is true or high.

The triangular gate, on the left, is an inverter. On its output (pin 0) it “in-
verts” the logic level found on the input (pin 1): false becomes true and true
becomes false. The gate on the right is an and-gate. It generates a true on pin 0
exactly when both of its inputs (pins 1 and 2) are true.

2101outputinput0334

Priority Queues

The action of these gates is the result of a physical process, so the effect
of the inputs on the output is delayed by a period of time called a gate delay.
Gate delays depend on the complexity of the gate, the manufacturing process,
and environmental factors. For our purposes, we’ll assume the gate delay of the
inverter is 0.2 nanosecond (ns) and the delay of the and-gate is 0.8 ns.

The question is, what output is generated when we toggle the input from
low to high and back to low several times? To determine the answer we can
build the circuit, or simulate it in software. For reasons that will become clear
in a moment, simulation will be more useful.

The setup for our simulation will consist of a number of small classes. First,
there are a number of components, including an Inverter; an And; an input, or
Source; and a voltage sensor, or Probe. When constructed, gates are provided
gate delay values, and Sources and Probes are given names. Each of these
components has one or more pins to which wires can be connected. (As with
real circuits, the outputs should connect only to inputs of other components!)
Finally, the voltage level of a particular pin can be set to a particular level. As
an example of the interface, we list the public methods for the And gate:

class And extends Component
{

Circuit

public And(double delay)
// pre: delay >= 0.0ns
// post: constructs and gate with indicated gate delay

public void set(double time, int pinNum, int level)
// pre: pinNum = 1 or 2, level = 0/3
// post: updates inputs and generates events on
//

devices connected to output

}

Notice that there is a time associated with the set method. This helps us doc-
ument when different events happen in the component. These events are sim-
ulated by a comparable Event class. This class describes a change in logic level
on an input pin for some component. As the simulation progresses, Events are
created and scheduled for simulation in the future. The ordering of Events is
based on an event time. Here are the details:

class Event implements Comparable<Event>
{

protected double time;
protected int level;
protected Connection c;

// time of event
// voltage level
// gate/pin

public Event(Connection c, double t, int l)
// pre: c is a valid pin on a gate
// post: constructs event for time t to set pin to level l
{

this.c = c;

13.5 Example: Circuit Simulation

335

time = t;
level = l;

}

public void go()
// post: informs target component of updated logic on pin
{

c.component().set(time,c.pin(),level);

}

public int compareTo(Event other)
// pre: other is non-null
// post: returns integer representing relation between values
{

Event that = (Event)other;
if (this.time < that.time) return -1;
else if (this.time == that.time) return 0;
else return 1;

}

}

The Connection mentioned here is simply a component’s input pin.

Finally, to orchestrate the simulation, we use a priority queue to correctly
simulate the order of events. The following method simulates a circuit by re-
moving events from the priority queue and setting the logic level on the appro-
priate pins of the components. The method returns the time of the last event to
help monitor the progress of the simulation.

public class Circuit
{

static PriorityQueue<Event> eventQueue; // main event queue

public static double simulate()
// post: run simulation until event queue is empty;
//
{

returns final clock time

// voltage of low logic
// voltage of high logic

double low = 0.0;
double high = 3.0;
double clock = 0.0;
while (!eventQueue.isEmpty())
// remove next event
{
Event e = eventQueue.remove();
// keep track of time
clock = e.time;
// simulate the event
e.go();

}
System.out.println("-- circuit stable after "+clock+" ns --");
return clock;

}

336

Priority Queues

}

As events are processed, the logic level on a component’s pins are updated. If
the inputs to a component change, new Events are scheduled one gate delay
later for each component connected to the output pin. For Sources and Probes,
we write a message to the output indicating the change in logic level. Clearly,
when there are no more events in the priority queue, the simulated circuit is
stable. If the user is interested, he or she can change the logic level of a Source
and resume the simulation by running the simulate method again.

We are now equipped to simulate the circuit of Figure 13.7. The ﬁrst portion
of the following code sets up the circuit, while the second half simulates the
effect of toggling the input several times:

public static void main(String[] args)
{

int low = 0;
int high = 3;
eventQueue = new SkewHeap<Event>();
double time;

// voltage of low logic
// voltage of high logic

// set up circuit
Inverter not = new Inverter(0.2);
And and = new And(0.8);
Probe output = new Probe("output");
Source input = new Source("input",not.pin(1));

input.connectTo(and.pin(2));
not.connectTo(and.pin(1));
and.connectTo(output.pin(1));

// simulate circuit
time = simulate();
input.set(time+1.0,0,high); // first: set input high
time = simulate();
input.set(time+1.0,0,low);
time = simulate();
input.set(time+1.0,0,high); // later: set input high
time = simulate();
input.set(time+1.0,0,low);
simulate();

// later: set input low

// later: set input low

}

When run, the following output is generated:

1.0 ns: output now 0 volts
-- circuit stable after 1.0 ns --
2.0 ns: input set to 3 volts
2.8 ns: output now 3 volts
3.0 ns: output now 0 volts

13.6 Conclusions

337

-- circuit stable after 3.0 ns --
4.0 ns: input set to 0 volts
-- circuit stable after 5.0 ns --
6.0 ns: input set to 3 volts
6.8 ns: output now 3 volts
7.0 ns: output now 0 volts
-- circuit stable after 7.0 ns --
8.0 ns: input set to 0 volts
-- circuit stable after 9.0 ns --

When the input is moved from low to high, a short spike is generated on the
output. Moving the input to low again has no impact. The spike is generated by
the rising edge of a signal, and its width is determined by the gate delay of the
inverter. Because the spike is so short, it would have been difﬁcult to detect it
using real hardware.3 Devices similar to this edge detector are important tools
for detecting changing states in the circuits they monitor.

13.6 Conclusions

We have seen three implementations of priority queues: one based on a Vector
that keeps its entries in order and two others based on heap implementations.
The Vector implementation demonstrates how any ordered structure may be
adapted to support the operations of a priority queue.

Heaps form successful implementations of priority queues because they relax
the conditions on “keeping data in priority order.” Instead of maintaining data
in sorted order, heaps maintain a competition between values that becomes
progressively more intense as the values reach the front of the queue. The cost
of inserting and removing values from a heap can be made to be as low as
O(log n).

If the constraint of keeping values in a Vector is too much (it may be im-
possible, for example, to allocate a single large chunk of memory), or if one
wants to avoid the uneven cost of extending a Vector, a dynamic mechanism is
useful. The SkewHeap is such a mechanism, keeping data in general heap form.
Over a number of operations the skew heap performs as well as the traditional
Vector-based implementation.

Self Check Problems

Solutions to these problems begin on page 449.

13.1

13.2

Is a PriorityQueue a Queue?

Is a PriorityQueue a Linear structure?

3 This is a very short period of time. During the time the output is high, light travels just over
2 inches!

338

Priority Queues

13.3
interpret the depth of a node in the tree?

How do you interpret the weight of a Huffman tree? How do you

13.4 What is a min-heap?

Vector-based heaps have O(log n) behavior for insertion and removal

13.5
of values. What structural feature of the underlying tree guarantees this?

13.6 Why is a PriorityQueue useful for managing simulations base on events?

Problems

Solutions to the odd-numbered problems begin on page 481.

Draw the state of a HeapVector after each of the values 3, 4, 7, 0, 2, 8,

13.1
and 6 are added, in that order.

13.2

Consider the heap 0

2

1

3

7

4

6

8

a. What does this heap look like when drawn as a tree?

b. What does this heap look like (in array form) when a value is removed?

13.3

Below is a SkewHeap. What does it look like after a value is removed?

13.4 How might you use priorities to simulate a LIFO structure with a priority
queue?

13.5

Is a VectorHeap a Queue? Is it an OrderedStructure?

13.6 How might you use priorities to simulate a FIFO structure with a priority
queue?

Suppose a user built an object whose compareTo and equals methods
13.7
were inconsistent. For example, values that were equals might also return a
negative value for compareTo. What happens when these values are added to a
PriorityVector? What happens when these values are added to a SkewHeap?

13.8 We have seen that the cost of removing a value from the Priority-
Vector takes linear time.
If elements were stored in descending order, this
could be reduced to constant time. Compare the ascending and descending
implementations, discussing the circumstances that suggest the use of one im-
plementation over the other.

13.9 What methods would have to be added to the OrderedVector class to
make it possible to implement a PriorityVector using only a private Ordered-
Vector?

301452413.6 Conclusions

339

13.10 Reconsider the “heapifying” constructor discussed in Section 13.4.2. In-
stead of adding n values to an initially empty heap (at a cost of O(n log n)),
suppose we do the following: Consider each interior node of the heap in order
of decreasing array index. Think of this interior node as the root of a poten-
tial subheap. We know that its subtrees are valid heaps. Now, just push this
node down into its (near-)heap. Show that the cost of performing this heapify
operation is linear in the size of the Vector.

13.11 Design a more efﬁcient version of HeapVector that keeps its values
in order only when necessary: When values are added, they are appended to
the end of the existing heap and a nonHeap ﬂag is set to true. When values
are removed, the nonHeap ﬂag is checked and the Vector is heapiﬁed if nec-
essary. What are the worst-case and best-case running times of the add and
remove operations? (You may assume that you have access to the heapify of
Problem 13.10.)

13.12 Consider the unordered data:
3

4

7
What does this Vector look like after it has been heapiﬁed?

2

0

5

6

1

13.13 Consider the in-place Vector-based heapsort.

a. A min-heap is particularly suited to sorting data in place into which order:

ascending or descending?

b. What is the worst-case time complexity of this sort?

c. What is the best-case time complexity of this sort?

13.14 Suppose we are given access to a min-heap, but not the code that sup-
ports it. What changes to the comparable data might we make to force the
min-heap to work like a max-heap?

13.15 Suppose we are to ﬁnd the kth largest element of a heap of n values.
Describe how we might accomplish this efﬁciently. What is the worst-case run-
ning time of your method? Notice that if the problem had said “set of n values,”
we would require a heapify operation like that found in Problem 13.10.

13.16 Demonstrate that any binary tree that has the heap property can be
generated by inserting values into a skew heap in an appropriate order. (This
realization is important to understanding why an amortized accounting scheme
is necessary.)

13.17 Suppose you are given n distinct values to store in a full heap—a heap
that is maintained in a full binary tree. Since there is no ordering between
children in a heap, the left and right subheaps can be exchanged. How many
equivalent heaps can be produced by only swapping children of a node?

13.18 Given n distinct values to be stored in a heap, how many heaps can
store the values? (Difﬁcult.)

13.19 What proportion of the binary trees holding n distinct values are heaps?

340

Priority Queues

13.20 Suppose that n randomly selected (and uniformly distributed) numbers
are inserted into a complete heap. Now, select another number and insert it into
the heap. How many levels is the new number expected to rise?
13.21 The mapping strategy that takes a complete binary tree to a vector can
actually be used to store general trees, albeit in a space-inefﬁcient manner. The
strategy is to allocate enough space to hold the lowest, rightmost leaf, and to
maintain null references in nodes that are not currently being used. What is the
worst-case Vector length needed to store an n-element binary tree?
13.22 Write an equals method for a PriorityVector. It returns true if each
pair of corresponding elements removed from the structures would be equal.
What is the complexity of the equals method? (Hint: You may not need to
remove values.)
13.23 Write an equals method for a HeapVector. It returns true if each pair
of corresponding elements removed from the structures would be equal. What is
the complexity of the equals method? (Hint: You may need to remove values.)
13.24 Write an equals method for a SkewHeap. It returns true if each pair of
corresponding elements removed from the structures would be equal. What is
the complexity of the equals method? (Hint: You may need to remove values.)
13.25 Show that the implementation of the PriorityVector can be improved
by not actually keeping the values in order. Instead, only maintain the mini-
mum value at the left. Demonstrate the implementation of the add and remove
methods.
13.26 Suppose you are a manufacturer of premium-quality videocassette re-
corders. Your XJ-6 recorder allows the “user” to “program” 4096 different future
events to be recorded. Of course, as the time arrives for each event, your ma-
chine is responsible for turning on and recording a program.

a. What information is necessary to correctly record an event?

b. Design the data structure(s) needed to support the XJ-6.

13.7 Laboratory: Simulating Business

Objective. To determine if it is better to have single or multiple service lines.
Discussion. When we are waiting in a fast food line, or we are queued up at a
bank, there are usually two different methods of managing customers:

1. Have a single line for people waiting for service. Every customer waits in
a single line. When a teller becomes free, the customer at the head of the
queue moves to the teller. If there are multiple tellers free, one is picked
randomly.

2. Have multiple lines—one for each teller. When customers come in the
door they attempt to pick the line that has the shortest wait. This usu-
ally involves standing in the line with the fewest customers. If there are
multiple choices, the appropriate line is selected randomly.

It is not clear which of these two methods of queuing customers is most efﬁ-
cient. In the single-queue technique, tellers appear to be constantly busy and
no customer is served before any customer that arrives later. In the multiple-
queue technique, however, customers can take the responsibility of evaluating
the queues themselves.

Note, by the way, that some industries (airlines, for example) have a mixture
of both of these situations. First class customers enter in one line, while coach
customers enter in another.
Procedure. In this lab, you are to construct a simulation of these two service
mechanisms. For each simulation you should generate a sequence of customers
that arrive at random intervals. These customers demand a small range of ser-
vices, determined by a randomly selected service time. The simulation is driven
by an event queue, whose elements are ranked by the event time. The type of
event might be a customer arrival, a teller freeing up, etc.

For the single line simulation, have the customers all line up in a single
queue. When a customer is needed, a single customer (if any) is removed from
the customer queue, and the teller is scheduled to be free at a time that is
determined by the service time. You must ﬁgure out how to deal with tellers
that are idle—how do they wait until a customer arrives?

For the multiple line simulation, the customers line up at their arrival time,
in one of the shortest teller queues. When a teller is free, it selects the next
customer from its dedicated queue (if any). A single event queue is used to
drive the simulation.

To compare the possibilities of these two simulations, it is useful to run the
same random customers through both types of queues. Think carefully about
how this might be accomplished.
Thought Questions. Consider the following questions as you complete the lab:

1. Run several simulations of both types of queues. Which queue strategy

seems to process all the customers fastest?

342

Priority Queues

2. Is their a difference between the average wait time for customers between

the two techniques?

3. Suppose you simulated the ability to jump between lines in a multiple line
simulation. When a line has two or more customers than another line,
customers move from the end one line to another until the lines are fairly
even. You see this behavior frequently at grocery stores. Does this change
the type of underlying structure you use to keep customers in line?

4. Suppose lines were dedicated to serving customers of varying lengths of
service times. Would this improve the average wait time for a customer?

Notes:

Chapter 14

Search Trees

Concepts:
.
BinarySearchTrees
.
Tree Sort
.
Splay Trees
. Red-Black Trees

He looked to the right of him.
No caps.
He looked to the left of him.
No caps.
...
Then he looked up into the tree.
And what do you think he saw?
—Esphyr Slobodkina

STRUCTURES ARE OFTEN THE SUBJECT OF A SEARCH. We have seen, for example,
that binary search is a natural and efﬁcient algorithm for ﬁnding values within
ordered, randomly accessible structures. Recall that at each point the algorithm
compares the value sought with the value in the middle of the structure. If they
are not equal, the algorithm performs a similar, possibly recursive search on one
side or the other. The pivotal feature of the algorithm, of course, was that the
underlying structure was in order. The result was that a value could be efﬁ-
ciently found in approximately logarithmic time. Unfortunately, the modifying
operations—add and remove—had complexities that were determined by the
linear nature of the vector.

Heaps have shown us that by relaxing our notions of order we can improve
on the linear complexities of adding and removing values. These logarithmic
operations, however, do not preserve the order of elements in any obviously
useful manner. Still, if we were somehow able to totally order the elements of
a binary tree, then an algorithm like binary search might naturally be imposed
on this branching structure.

14.1 Binary Search Trees

The binary search tree is a binary tree whose elements are kept in order. This is
easily stated as a recursive deﬁnition.

Deﬁnition 14.1 A binary tree is a binary search tree if it is trivial, or if every
node is simultaneously greater than or equal to each value in its left subtree, and
less than or equal to each value in its right subtree.

To see that this is a signiﬁcant restriction on the structure of a binary tree,
one need only note that if a maximum of n distinct values is found at the root,

344

Search Trees

Figure 14.1 Binary search trees with three nodes.

all other values must be found in the left subtree. Figure 14.1 demonstrates
the many trees that can contain even three distinct values. Thus, if one is not
too picky about the value one wants to have at the root of the tree, there are
still a signiﬁcant number of trees from which to choose. This is important if
we want to have our modifying operations run quickly: ideally we should be as
nonrestrictive about the outcome as possible, in order to reduce the friction of
the operation.

One important thing to watch for is that even though our deﬁnition allows
duplicate values of the root node to fall on either side, our code will prefer to
have them on the left. This preference is arbitrary. If we assume that values
equal to the root will be found in the left subtree, but in actuality some are
located in the right, then we might expect inconsistent behavior from methods
that search for these values. In fact, this is not the case.

To guide access to the elements of a binary search tree, we will consider it an

implementation of an OrderedStructure, supporting the following methods:

public class BinarySearchTree<E extends Comparable<E>>

extends AbstractStructure<E> implements OrderedStructure<E>

{

Binary-
SearchTree

public BinarySearchTree()
// post: constructs an empty binary search tree

public BinarySearchTree(Comparator<E> alternateOrder)
// post: constructs an empty binary search tree

public boolean isEmpty()
// post: returns true iff the binary search tree is empty

public void clear()
// post: removes all elements from binary search tree

public int size()
// post: returns the number of elements in binary search tree

public void add(E value)

(a)(b)(c)(d)(e)23312331211312214.2 Example: Tree Sort

345

Maybe even
with no
thought!

// post: adds a value to binary search tree

public boolean contains(E value)
// post: returns true iff val is a value found within the tree

public E get(E value)
// post: returns object found in tree, or null

public E remove(E value)
// post: removes one instance of val, if found

public Iterator<E> iterator()
// post: returns iterator to traverse BST

}

Unlike the BinaryTree, the BinarySearchTree provides only one iterator
method. This method provides for an in-order traversal of the tree, which, with
some thought, allows access to each of the elements in order.

14.2 Example: Tree Sort

Because the BinarySearchTree is an OrderedStructure it provides the natural
basis for sorting. The algorithm of Section 11.2.3 will work equally well here,
provided the allocation of the OrderedStructure is modiﬁed to construct a
BinarySearchTree. The binary search structure, however, potentially provides
If the tree can be kept reasonably
signiﬁcant improvements in performance.
short, the cost of inserting each element is O(log n). Since n elements are ulti-
mately added to the structure, the total cost is O(n log n).1 As we have seen in
Chapter 12, all the elements of the underlying binary tree can be visited in lin-
ear time. The resulting algorithm has a potential for O(n log n) time complexity,
which rivals the performance of sorting techniques using heaps. The advantage
of binary search trees is that the elements need not be removed to determine
their order. To attain this performance, though, we must keep the tree as short
as possible. This will require considerable attention.

14.3 Example: Associative Structures

Associative structures play an important role in making algorithms efﬁcient. In
these data structures, values are associated with keys. Typically (though not
necessarily), the keys are unique and aid in the retrieval of more complete
information—the value. In a Vector, for example, we use integers as indices
to ﬁnd values. In an AssociativeVector we can use any type of object. The

1 This needs to be proved! See Problem 14.11.

346

Search Trees

SymbolTable associated with the PostScript lab (Section 10.5) is, essentially,
an associative structure.

Associative structures are an important feature of many symbol-based sys-
tems. Here, for example, is a ﬁrst approach to the construction of a general-
purpose symbol table, with potentially logarithmic performance:

import structure5.*;
import java.util.Iterator;
import java.util.Scanner;
public class SymTab<S extends Comparable<S>,T>
{

SymTab

protected BinarySearchTree<ComparableAssociation<S,T>> table;
public SymTab()
// post: constructs empty symbol table
{

table = new BinarySearchTree<ComparableAssociation<S,T>>();

}

public boolean contains(S symbol)
// pre: symbol is non-null string
// post: returns true iff string in table
{

ComparableAssociation<S,T> a =

new ComparableAssociation<S,T>(symbol,null);

return table.contains(a);

}

public void add(S symbol, T value)
// pre: symbol non-null
// post: adds/replaces symbol-value pair in table
{

ComparableAssociation<S,T> a =

new ComparableAssociation<S,T>(symbol,value);

if (table.contains(a)) table.remove(a);
table.add(a);

}

public T get(S symbol)
// pre: symbol non null
// post: returns token associated with symbol
{

ComparableAssociation<S,T> a =

new ComparableAssociation<S,T>(symbol,null);

if (table.contains(a)) {
a = table.get(a);
return a.getValue();

} else {

return null;

}

}

14.3 Example: Associative Structures

347

public T remove(S symbol)
// pre: symbol non null
// post: removes value associated with symbol and returns it
//
{

if error returns null

ComparableAssociation<S,T> a =

new ComparableAssociation<S,T>(symbol,null);

if (table.contains(a)) {
a = table.remove(a);
return a.getValue();

} else {

return null;

}

}

}

Based on such a table, we might have a program that reads in a number of alias-
name pairs terminated by the word END. After that point, the program prints out
the fully translated aliases:

public static void main(String args[])
{

SymTab<String,String> table = new SymTab<String,String>();
Scanner s = new Scanner(System.in);
String alias, name;
// read in the alias-name database
do
{

alias = s.next();
if (!alias.equals("END"))
{

name = s.next();
table.add(alias,name);

}

} while (!alias.equals("END"));
// enter the alias translation stage
do
{

name = s.next();
while (table.contains(name))
{

// translate alias
name = table.get(name);

}
System.out.println(name);

} while (s.hasNext());

}

Given the input:

348

Search Trees

three 3
one unity
unity 1
pi three
END

one
two
three
pi

the program generates the following output:

1
two
3
3

We will consider general associative structures in Chapter 15, when we dis-
cuss Dictionaries. We now consider the details of actually supporting the
BinarySearchTree structure.

14.4 Implementation

In considering the implementation of a BinarySearchTree, it is important to
remember that we are implementing an OrderedStructure. The methods of
the OrderedStructure accept and return values that are to be compared with
one another. By default, we assume that the data are Comparable and that the
natural order suggested by the NaturalComparator is sufﬁcient. If alternative
orders are necessary, or an ordering is to be enforced on elements that do not
directly implement a compareTo method, alternative Comparators may be used.
Essentially the only methods that we depend upon are the compatibility of the
Comparator and the elements of the tree.

We begin by noticing that a BinarySearchTree is little more than a binary
tree with an imposed order. We maintain a reference to a BinaryTree and
explicitly keep track of its size. The constructor need only initialize these two
ﬁelds and suggest an ordering of the elements to implement a state consistent
with an empty binary search tree:

protected BinaryTree<E> root;

protected final BinaryTree<E> EMPTY = new BinaryTree<E>();

BinarySearch-
Tree

protected int count;
protected Comparator<E> ordering;

public BinarySearchTree()

14.4 Implementation

349

// post: constructs an empty binary search tree
{

this(new NaturalComparator<E>());

}

public BinarySearchTree(Comparator<E> alternateOrder)
// post: constructs an empty binary search tree
{

root = EMPTY;
count = 0;
ordering = alternateOrder;

}

As with most implementations of OrderedStructures, we develop a method
to ﬁnd the correct location to insert the value and then use that method as the
basis for implementing the public methods—add, contains, and remove. Our
approach to the method locate is to have it return a reference to the location
that identiﬁes the correct point of insertion for the new value. This method, of
course, makes heavy use of the ordering. Here is the Java code for the method:

protected BinaryTree<E> locate(BinaryTree<E> root, E value)
// pre: root and value are non-null
// post: returned: 1 - existing tree node with the desired value, or
//
{

2 - the node to which value should be added

E rootValue = root.value();
BinaryTree<E> child;

// found at root: done
if (rootValue.equals(value)) return root;
// look left if less-than, right if greater-than
if (ordering.compare(rootValue,value) < 0)
{

child = root.right();

} else {

child = root.left();

}
// no child there: not in tree, return this node,
// else keep searching
if (child.isEmpty()) {

return root;

} else {

return locate(child, value);

}

}

The approach of the locate method parallels binary search. Comparisons are
made with the root, which serves as a median value.
If the value does not
match, then the search is refocused on either the left side of the tree (among
In either
smaller values) or the right side of the tree (among larger values).

350

Search Trees

case, if the search is about to step off the tree, the current node is returned: if
the value were added, it would be a child of the current node.

Once the locate method is written, the contains method must check to see

if the node returned by locate actually equals the desired value:2

public boolean contains(E value)
// post: returns true iff val is a value found within the tree
{

if (root.isEmpty()) return false;

BinaryTree<E> possibleLocation = locate(root,value);
return value.equals(possibleLocation.value());

}

It now becomes a fairly straightforward task to add a value. We simply
locate the value in the tree using the locate function.
If the value was not
found, locate returned a node off of which a leaf with the desired value may
be added. If, however, locate has found an equivalent value, we must insert
the new value as the right child of the predecessor of the node returned by
locate.3

public void add(E value)
// post: adds a value to binary search tree
{

BinaryTree<E> newNode = new BinaryTree<E>(value,EMPTY,EMPTY);

// add value to binary search tree
// if there’s no root, create value at root
if (root.isEmpty())
{

root = newNode;

} else {

BinaryTree<E> insertLocation = locate(root,value);
E nodeValue = insertLocation.value();
// The location returned is the successor or predecessor
// of the to-be-inserted value
if (ordering.compare(nodeValue,value) < 0) {

insertLocation.setRight(newNode);

} else {

if (!insertLocation.left().isEmpty()) {

// if value is in tree, we insert just before
predecessor(insertLocation).setRight(newNode);

} else {

2 We reemphasize at this point the importance of making sure that the equals method for an object
is consistent with the ordering suggested by the compare method of the particular Comparator.
3 With a little thought, it is clear to see that this is a correct location. If there are two copies of a
value in a tree, the second value added is a descendant and predecessor (in an in-order traversal)
of the located value. It is also easy to see that a predecessor has no right child, and that if one is
added, it becomes the predecessor.

14.4 Implementation

351

insertLocation.setLeft(newNode);

}

}

}
count++;

}

Our add code makes use of the protected “helper” function, predecessor, which
returns a pointer to the node that immediately precedes the indicated root:

protected BinaryTree<E> predecessor(BinaryTree<E> root)
{

Assert.pre(!root.isEmpty(), "No predecessor to middle value.");
Assert.pre(!root.left().isEmpty(), "Root has left child.");
BinaryTree<E> result = root.left();
while (!result.right().isEmpty()) {

result = result.right();

}
return result;

}

A similar routine can be written for successor, and would be used if we preferred
to store duplicate values in the right subtree.

We now approach the problem of removing a value from a binary search
tree. Observe that if it is found, it might be an internal node. The worst case
occurs when the root of a tree is involved, so let us consider that problem.

There are several cases. First (Figure 14.2a), if the root of a tree has no
left child, the right subtree can be used as the resulting tree. Likewise (Fig-
ure 14.2b), if there is no right child, we simply return the left. A third case
(Figure 14.2c) occurs when the left subtree has no right child. Then, the right
subtree—a tree with values no smaller than the left root—is made the right sub-
tree of the left. The left root is returned as the result. The opposite circumstance
could also be true.

We are, then, left to consider trees with a left subtree that, in turn, contains
a right subtree (Figure 14.3). Our approach to solving this case is to seek out
the predecessor of the root and make it the new root. Note that even though the
predecessor does not have a right subtree, it may have a left. This subtree can
take the place of the predecessor as the right subtree of a nonroot node. (Note
that this is the result that we would expect if we had recursively performed our
node-removing process on the subtree rooted at the predecessor.)

Finally, here is the Java code that removes the top BinaryTree of a tree and

returns the root of the resulting tree:

protected BinaryTree<E> removeTop(BinaryTree<E> topNode)
// pre: topNode contains the value we want to remove
// post: we return an binary tree rooted with the predecessor of topnode.
{

// remove topmost BinaryTree from a binary search tree

352

Search Trees

Figure 14.2

The three simple cases of removing a root value from a tree.

Figure 14.3 Removing the root of a tree with a rightmost left descendant.

(a)ABAB(c)(b)xxx321213xpredecessor(  )x14.4 Implementation

353

= topNode.left();

easy: right subtree is new tree

BinaryTree<E> left
BinaryTree<E> right = topNode.right();
// disconnect top node
topNode.setLeft(EMPTY);
topNode.setRight(EMPTY);
// Case a, no left BinaryTree
//
if (left.isEmpty()) { return right; }
// Case b, no right BinaryTree
//
if (right.isEmpty()) { return left; }
// Case c, left node has no right subtree
//
easy: make right subtree of left
BinaryTree<E> predecessor = left.right();
if (predecessor.isEmpty())
{

easy: left subtree is new tree

left.setRight(right);
return left;

harder: successor of root becomes new root

parent always points to parent of predecessor

}
// General case, slide down left tree
//
//
BinaryTree<E> parent = left;
while (!predecessor.right().isEmpty())
{

parent = predecessor;
predecessor = predecessor.right();

}
// Assert: predecessor is predecessor of root
parent.setRight(predecessor.left());
predecessor.setLeft(left);
predecessor.setRight(right);
return predecessor;

}

With the combined efforts of the removeTop and locate methods, we can now
simply locate a value in the search tree and, if found, remove it from the tree.
We must be careful to update the appropriate references to rehook the modiﬁed
subtree back into the overall structure.

Notice that inserting and removing elements in this manner ensures that the
in-order traversal of the underlying tree delivers the values stored in the nodes
in a manner that respects the necessary ordering. We use this, then, as our
preferred iteration method.

public Iterator<E> iterator()
// post: returns iterator to traverse BST
{

return root.inorderIterator();

}

354

Search Trees

The remaining methods (size, etc.) are implemented in a now-familiar manner.

Exercise 14.1 One possible approach to keeping duplicate values in a binary search
tree is to keep a list of the values in a single node.
In such an implementation,
each element of the list must appear externally as a separate node. Modify the
BinarySearchTree implementation to make use of these lists of duplicate values.

Each of the time-consuming operations of a BinarySearchTree has a worst-
case time complexity that is proportional to the height of the tree. It is easy to
see that checking for or adding a leaf, or removing a root, involves some of the
most time-consuming operations. Thus, for logarithmic behavior, we must be
sure that the tree remains as short as possible.

Unfortunately, we have no such assurance. In particular, one may observe
what happens when values are inserted in descending order: the tree is heavily
skewed to the left. If the same values are inserted in ascending order, the tree
can be skewed to the right. If these values are distinct, the tree becomes, essen-
tially, a singly linked list. Because of this behavior, we are usually better off if
we shufﬂe the values beforehand. This causes the tree to become, on average,
shorter and more balanced, and causes the expected insertion time to become
O(log n).

Considering that the tree is responsible for maintaining an order among data
values, it seems unreasonable to spend time shufﬂing values before ordering
them. In Section 14.5 we ﬁnd out that the process of adding and removing a
node can be modiﬁed to maintain the tree in a relatively balanced state, with
only a little overhead.

14.5 Splay Trees

Because the process of adding a new value to a binary search tree is determin-
istic—it produces the same result tree each time—and because inspection of the
tree does not modify its structure, one is stuck with the performance of any
degenerate tree constructed. What might work better would be to allow the
tree to reconﬁgure itself when operations appear to be inefﬁcient.

The splay tree quickly overcomes poor performance by rearranging the tree’s
nodes on the ﬂy using a simple operation called a splay. Instead of perform-
ing careful analysis and optimally modifying the structure whenever a node is
added or removed, the splay tree simply moves the referenced node to the top
of the tree. The operation has the interesting characteristic that the average
depth of the ancestors of the node to be splayed is approximately halved. As
with skew heaps, the performance of a splay tree’s operators, when amortized
over many operations, is logarithmic.

The basis for the splay operation is a pair of operations called rotations (see
Figure 14.4). Each of these rotations replaces the root of a subtree with one of
its children. A right rotation takes a left child, x, of a node y and reverses their
relationship. This induces certain obvious changes in connectivity of subtrees,

Splay: to spread
outward.

14.5 Splay Trees

355

Figure 14.4

The relation between rotated subtrees.

Finally, a right
handed
method!

BinaryTree-
Node

In particular, there is no
but in all other ways, the tree remains the same.
structural effect on the tree above the original location of node y. A left rotation
is precisely the opposite of a right rotation; these operations are inverses of each
other.

The code for rotating a binary tree about a node is a method of the Binary-
Tree class. We show, here, rotateRight; a similar method performs a left
rotation.

protected void rotateRight()
// pre: this node has a left subtree
// post: rotates local portion of tree so left child is root
{

BinaryTree<E> parent = parent();
BinaryTree<E> newRoot = left();
boolean wasChild = parent != null;
boolean wasLeftChild = isLeftChild();

// hook in new root (sets newRoot’s parent, as well)
setLeft(newRoot.right());

// puts pivot below it (sets this’s parent, as well)
newRoot.setRight(this);

if (wasChild) {

if (wasLeftChild) parent.setLeft(newRoot);
else

parent.setRight(newRoot);

}

}

For each rotation accomplished, the nonroot node moves upward by one
level. Making use of this fact, we can now develop an operation to splay a tree
at a particular node. It works as follows:

yLeft rotationRight rotationBACABCyxx356

Search Trees

Figure 14.5
are mirror images of those shown here.

Two of the rotation pairs used in the splaying operation. The other cases

• If x is the root, we are done.

• If x is a left (or right) child of the root, rotate the tree to the right (or left)

about the root. x becomes the root and we are done.

• If x is the left child of its parent p, which is, in turn, the left child of its
grandparent g, rotate right about g, followed by a right rotation about p
(Figure 14.5a). A symmetric pair of rotations is possible if x is a left child
of a left child. After double rotation, continue splay of tree at x with this
new tree.

• If x is the right child of p, which is the left child of g, we rotate left about
p, then right about g (Figure 14.5b). The method is similar if x is the left
child of a right child. Again, continue the splay at x in the new tree.

After the splay has been completed, the node x is located at the root of the
If node x were to be immediately accessed again (a strong possibility),
tree.
the tree is clearly optimized to handle this situation. It is not the case that the
tree becomes more balanced (see Figure 14.5a). Clearly, if the tree is splayed at
an extremal value, the tree is likely to be extremely unbalanced. An interesting
feature, however, is that the depth of the nodes on the original path from x to
the root of the tree is, on average, halved. Since the average depth of these

(a)(b)gpxxpxggxpgxpgpxgp14.6 Splay Tree Implementation

357

nodes is halved, they clearly occupy locations closer to the top of the tree where
they may be more efﬁciently accessed.

To guarantee that the splay has an effect on all operations, we simply per-
form each of the binary search tree operations as before, but we splay the tree
at the node accessed or modiﬁed during the operation. In the case of remove,
we splay the tree at the parent of the value removed.

14.6 Splay Tree Implementation

Because the splay tree supports the binary search tree interface, we extend the
BinarySearchTree data structure. Methods written for the SplayTree hide or
override existing code inherited from the BinarySearchTree.

SplayTree

public class SplayTree<E extends Comparable<E>>

extends BinarySearchTree<E> implements OrderedStructure<E>

{

}

public SplayTree()
// post: construct a new splay tree

public SplayTree(Comparator<E> alternateOrder)
// post: construct a new splay tree

public void add(E val)
// post: adds a value to the binary search tree

public boolean contains(E val)
// post: returns true iff val is a value found within the tree

public E get(E val)
// post: returns object found in tree, or null

public E remove(E val)
// post: removes one instance of val, if found

protected void splay(BinaryTree<E> splayedNode)

public Iterator<E> iterator()
// post: returns iterator that traverses tree nodes in order

As an example of how the splay operation is incorporated into the existing
binary tree code, we look at the contains method. Here, the root is reset to the
value of the node to be splayed, and the splay operation is performed on the
tree. The postcondition of the splay operation guarantees that the splayed node
will become the root of the tree, so the entire operation leaves the tree in the
correct state.

358

Search Trees

public boolean contains(E val)
// post: returns true iff val is a value found within the tree
{

if (root.isEmpty()) return false;

BinaryTree<E> possibleLocation = locate(root,val);
if (val.equals(possibleLocation.value())) {

splay(root = possibleLocation);
return true;

} else {

return false;

}

}

One difﬁculty with the splay operation is that it potentially modiﬁes the
structure of the tree. For example, the contains method—a method normally
considered nondestructive—potentially changes the underlying topology of the
tree. This makes it difﬁcult to construct iterators that traverse the SplayTree
since the user may use the value found from the iterator in a read-only opera-
tion that inadvertently modiﬁes the structure of the splay tree. This can have
disastrous effects on the state of the iterator. A way around this difﬁculty is
to have the iterator keep only that state information that is necessary to help
reconstruct—with help from the structure of the tree—the complete state of our
traditional nonsplay iterator. In the case of the SplayTreeIterator, we keep
track of two references: a reference to an “example” node of the tree and a
reference to the current node inspected by the iterator. The example node helps
recompute the root whenever the iterator is reset. To determine what nodes
would have been stored in the stack in the traditional iterator—the stack of un-
visited ancestors of the current node—we consider each node on the (unique)
path from the root to the current node. Any node whose left child is also on
the path is an element of our “virtual stack.” In addition, the top of the stack
maintains the current node (see Figure 14.6).

The constructor sets the appropriate underlying references and resets the it-
erator into its initial state. Because the SplayTree is dynamically restructuring,
the root value passed to the constructor may not always be the root of the tree.
Still, one can easily ﬁnd the root of the current tree, given a node: follow parent
pointers until one is null. Since the ﬁrst value visited in an inorder traversal
is the leftmost descendant, the reset method travels down the leftmost branch
(logically pushing values on the stack) until it ﬁnds a node with no left child.

It can also
wreck your day.

protected BinaryTree<E> tree; // node of splay tree, root computed
protected final BinaryTree<E> LEAF;
protected BinaryTree<E> current; // current node
// In this iterator, the "stack" normally used is implied by
// looking back up the path from the current node.
// for which the path goes left are on the stack

Those nodes

SplayTree-
Iterator

public SplayTreeIterator(BinaryTree<E> root, BinaryTree<E> leaf)
// pre: root is the root of the tree to be traversed

14.6 Splay Tree Implementation

359

Figure 14.6 A splay tree iterator, the tree it references, and the contents of the virtual
stack driving the iterator.

// post: constructs a new iterator to traverse splay tree
{

tree = root;
LEAF = leaf;
reset();

}

public void reset()
// post: resets iterator to smallest node in tree
{

current = tree;
if (!current.isEmpty()) {

current = current.root();
while (!current.left().isEmpty()) current = current.left();

}

}

The current node points to, by deﬁnition, an unvisited node that is, logically,
on the top of the outstanding node stack. Therefore, the hasNext and get
methods may access the current value immediately.

public boolean hasNext()
// post: returns true if there are unvisited nodes
{

return !current.isEmpty();

}

public E get()
// pre: hasNext()
// post: returns current value

Virtual Stackrootcurrent360

Search Trees

{

}

return current.value();

All that remains is to move the iterator from one state to the next. The next
method ﬁrst checks to see if the current (just visited) element has a right child.
If so, current is set to the leftmost descendant of the right child, effectively
popping off the current node and pushing on all the nodes physically linking
the current node and its successor. When no right descendant exists, the sub-
tree rooted at the current node has been completely visited. The next node to
be visited is the node under the top element of the virtual stack—the closest
ancestor whose left child is also an ancestor of the current node. Here is how
we accomplish this in Java:

public E next()
// pre: hasNext()
// post: returns current element and increments iterator
{

E result = current.value();
if (!current.right().isEmpty()) {
current = current.right();
while (!current.left().isEmpty())
{

current = current.left();

}
} else {

// we’re finished with current’s subtree.
// nodes until we come to the parent of a leftchild ancestor
// of current
boolean lefty;
do
{

We now pop off

lefty = current.isLeftChild();
current = current.parent();
} while (current != null && !lefty);
if (current == null) current = new BinaryTree<E>();

}
return result;

}

The iterator is now able to maintain its position through splay operations.

Again, the behavior of the splay tree is logarithmic when amortized over a
number of operations. Any particular operation may take more time to execute,
but the time is usefully spent rearranging nodes in a way that tends to make the
tree shorter.

From a practical standpoint, the overhead of splaying the tree on every oper-
ation may be hard to justify if the operations performed on the tree are relatively
random. On the other hand, if the access patterns tend to generate degenerate
binary search trees, the splay tree can improve performance.

14.7 An Alternative: Red-Black Trees

361

14.7 An Alternative: Red-Black Trees

A potential issue with both traditional binary search trees and splay trees is
the fact that they potentially have bad performance if values are inserted or
accessed in a particular order. Splay trees, of course, work hard to make sure
that repeated accesses (which seem likely) will be efﬁcient. Still, there is no
absolute performance guarantee.

One could, of course, make sure that the values in a tree are stored in as
perfectly balanced a manner as possible. In general, however, such techniques
are both difﬁcult to implement and costly in terms of per-operation execution
time.

Exercise 14.2 Describe a strategy for keeping a binary search tree as short as
possible. One example might be to unload all of the values and to reinsert them in
a particular order. How long does your approach take to add a value?

Because we consider the performance of structures using big-O notation, we
implicitly suggest we might be happy with performance that is within a constant
of optimal. For example, we might be happy if we could keep a tree balanced
within a factor of 2. One approach is to develop a structure called a red-black
tree.

For accounting purposes only, the nodes of a red-black tree are imagined to
be colored red or black. Along with these colors are several simple rules that
are constantly enforced:

1. Every red node has two black children.

2. Every leaf has two black (EMPTY is considered black) children.

3. Every path from a node to a descendent leaf contains the same number of

black nodes.

The result of constructing trees with these rules is that the height of the tree
measured along two different paths cannot differ by more than a factor of 2:
two red nodes may not appear contiguously, and every path must have the
same number of black nodes. This would imply that the height of the tree is
O(log2 n).

Exercise 14.3 Prove that the height of the tree with n nodes is no worse than
O(log2 n).

Of course, the purpose of data abstraction is to be able to maintain the con-
sistency of the structure—in this case, the red-black tree rules—as the structure
is probed and modiﬁed. The methods add and remove are careful to maintain
the red-black structure through at most O(log n) rotations and re-colorings of
nodes. For example, if a node that is colored black is removed from the tree, it
is necessary to perform rotations that either convert a red node on the path to
the root to black, or reduce the black height (the number of black nodes from

362

Search Trees

root to leaf) of the entire tree. Similar problems can occur when we attempt to
add a new node that must be colored black.

The code for red-black trees can be found online as RedBlackTree. While
the code is too tedious to present here, it is quite elegant and leads to binary
search trees with very good performance characteristics.

The implementation of the RedBlackTree structure in the structure pack-
age demonstrates another approach to packaging a binary search tree that is
important to discuss. Like the BinaryTree structure, the RedBlackTree is de-
ﬁned as a recursive structure represented by a single node. The RedBlackTree
also contains a dummy-node representation of the EMPTY tree. This is useful in
reducing the complexity of the tests within the code, and it supports the notion
that leaves have children with color, but most importantly, it allows the user to
call static methods that are deﬁned even for red-black trees with no nodes.
This approach—coding inherently recursive structures as recursive classes—
leads to side-effect free code. Each method has an effect on the tree at hand
but does not modify any global structures. This means that the user must be
very careful to record any side effects that might occur. In particular, it is im-
portant that methods that cause modiﬁcations to the structure return the “new”
value of the tree. If, for example, the root of the tree was the object of a remove,
that reference is no longer useful in maintaining contact with the tree.

To compare the approaches of the BinarySearchTree wrapper and the re-
cursive RedBlackTree, we present here the implementation of the SymTab struc-
ture we investigated at the beginning of the chapter, but cast in terms of Red-
BlackTrees. Comparison of the approaches is instructive (important differences
are highlighted with uppercase comments).

import structure5.*;
import java.util.Iterator;
public class RBSymTab<S extends Comparable<S>,T>
{

protected RedBlackTree<ComparableAssociation<S,T>> table;

RBSymTab

public RBSymTab()
// post: constructs empty symbol table
{

table = new RedBlackTree<ComparableAssociation<S,T>>();

}

public boolean contains(S symbol)
// pre: symbol is non-null string
// post: returns true iff string in table
{

return table.contains(new ComparableAssociation<S,T>(symbol,null));

}

public void add(S symbol, T value)
// pre: symbol non-null
// post: adds/replaces symbol-value pair in table

14.8 Conclusions

363

{

}

ComparableAssociation<S,T> a = new ComparableAssociation<S,T>(symbol,value);
if (table.contains(a)) table = table.remove(a);
table = table.add(a);

public T get(S symbol)
// pre: symbol non-null
// post: returns token associated with symbol
{

ComparableAssociation<S,T> a = new ComparableAssociation<S,T>(symbol,null);
if (table.contains(a)) {
a = table.get(a);
return a.getValue();

} else {

return null;

}

}

public T remove(S symbol)
// pre: symbol non-null
// post: removes value associated with symbol and returns it
//
{

if error returns null

ComparableAssociation<S,T> a = new ComparableAssociation<S,T>(symbol,null);
if (table.contains(a)) {
a = table.get(a);
table = table.remove(a);
return a.getValue();

} else {

return null;

}

}

}

The entire deﬁnition of RedBlackTrees is available in the structure package,
when O(log n) performance is desired. For more details about the structure,
please see the documentation within the code.

14.8 Conclusions

A binary search tree is the product of imposing an order on the nodes of a binary
tree. Each node encountered in the search for a value represents a point where
a decision can be accurately made to go left or right. If the tree is short and
fairly balanced, these decisions have the effect of eliminating a large portion of
the remaining candidate values.

The binary search tree is, however, a product of the history of the insertion
of values. Since every new value is placed at a leaf, the internal nodes are left

364

Search Trees

untouched and make the structure of the tree fairly static. The result is that
poor distributions of data can cause degenerate tree structures that adversely
impact the performance of the various search tree methods.

To combat the problem of unbalanced trees, various rotation-based opti-
In splay trees, rotations are used to force a recently
mizations are possible.
accessed value and its ancestors closer to the root of the tree. The effect is often
to shorten degenerate trees, resulting in an amortized logarithmic behavior. A
remarkable feature of this implementation is that there is no space penalty: no
accounting information needs to be maintained in the nodes.

Self Check Problems

Suppose values have only been added into a BinarySearchTree. Where

Solutions to these problems begin on page 449.
14.1 What motivates the use of binary search trees?
14.2
is the ﬁrst node added to the tree? Where is the last node added to the tree?
14.3 What is an associative structure?
14.4 Which node becomes the root after a tree is rotated left?
14.5
14.6
in a BinarySearchTree?
14.7 Why is it so difﬁcult to construct an Iterator for a SplayTree?
14.8 What is the primary advantage of a red-black tree over a splay tree?

Is the right rotation the reverse of the left rotation?
If two values are equal (using equals) are they found near each other

14.8 Conclusions

Problems

365

Solutions to the odd-numbered problems begin on page 483.

14.1 What distinguishes a binary search tree from a binary tree?

Draw all three-node integer-valued trees whose nodes are visited in the

14.2
order 1-2-3 in an in-order traversal. Which trees are binary search trees?

Draw all three-node integer-valued trees whose nodes are visited in the

14.3
order 1-2-3 in a preorder traversal. Which trees are binary search trees?

Draw all three-node integer-valued trees whose nodes are visited in the

14.4
order 1-2-3 in a postorder traversal. Which trees are binary search trees?

14.5
moved.

14.6
moved.

Redraw the following binary search tree after the root has been re-

Redraw the tree shown in Problem 14.5 after the leaf labeled 3 is re-

14.7
labeled 3.

Redraw the tree shown in Problem 14.5 after it is splayed at the leaf

The locate methods from OrderedVectors and BinarySearchTrees
14.8
are very similar. They have, for example, similar best-case behaviors. Explain
why their behaviors differ in the worst case.

14.9
structed by appropriately ordering insertion operations.

Prove that, if values are distinct, any binary search tree can be con-

14.10 In splay trees rotations are performed, possibly reversing the parent-
child relationship between two equal values. It is now possible to have a root
node with a right child that is equal. Explain why this will not cause problems
with each of the current methods locate, add, and remove.
14.11 Describe the topology of a binary search tree after the values 1 through
n have been inserted in order. How long does the search tree take to construct?
14.12 Describe the topology of a splay tree after the values 1 through n have
been inserted in order. How long does the splay tree take to construct?
14.13 Because the remove method of binary search trees prefers to replace a
node with its predecessor, one expects that a large number of removes will cause
the tree to lean toward the right. Describe a scheme to avoid this problem.
14.14 Suppose n distinct values are stored in a binary tree. It is noted that the
tree is a min-heap and a binary search tree. What does the tree look like?

14.15 As we have seen, the splay tree requires the construction of an iterator
that stores a single reference to the tree, rather than an unlimited number of

325463035366

Search Trees

references to ancestors. How does this reduction in space utilization impact the
running time of the iterator?
14.16 Write an equals method for binary search trees. It should return true
if both trees contain equal values.
14.17 Having answered Problem 14.16, is it possible to accurately use the
same method for splay trees?
14.18 Write a copy method for binary search trees. The result of the copy
should be equal to the original. Carefully argue the utility of your approach.
14.19 Prove that the expected time to perform the next method of the splay
tree iterator is constant time.

14.9 Laboratory: Improving the BinarySearchTree

Objective. To understand it is possible to improve an implementation.

Discussion. As we have seen in the implementation of the BinarySearchTree
class, the insertion of values is relative to the root of the tree. One of the
situations that must be handled carefully is the case where more than one node
can have the same key. If equal keys are allowed in the binary search tree, then
we must be careful to have them inserted on one side of the root. This behavior
increases the complexity of the code, and when there are many duplicate keys,
it is possible that the tree’s depth can be increased considerably.

Procedure. An alternative approach is to have all the nodes with similar keys
stored in the same location. When the tree is constructed in this manner, then
there is no need to worry about keeping similar keys together—they’re always
together.

In this lab, we will implement a BinaryMultiTree—a BinarySearchTree-
like structure that stores a multiset (a set of values with potential duplicates).
We are not so concerned with the set features, but we are demanding that dif-
ferent values are kept in sorted order in the structure. In particular, the traversal
of the BinaryMultiTree should return the values in order.

In this implementation, a BinaryTree is used to keep track of a List of val-
ues that are equal when compared with the compare method of the ordering
Comparator. From the perspective of the structure, there is no distinguishing
the members of the list. Externally, the interface to the BinaryMultiTree is
exactly the same as the BinarySearchTree, but the various methods work with
values stored in Lists, as opposed to working with the values directly. For ex-
ample, when we look at a value stored in a node, we ﬁnd a List. A getFirst
of this List class picks out an example that is suitable, for example, for com-
parison purposes.

Here are some things to think about during your implementation:

1. The size method does not return the number of nodes; it returns the
number of values stored in all the nodes. The bookkeeping is much the
same as it was before, but size is an upper bound on the actual size of
the search tree.

2. The add method compares values to the heads of lists found at each node
along the way. A new node is created if the value is not found in the
tree; the value is inserted in a newly created List in the BinaryTreeNode.
When an equal key is found, the search for a location stops, and the value
is added to the List. A carefully considered locate method will help
considerably here.

3. The contains method is quite simple: it returns true if the getFirst of

any of the Lists produces a similar value.

368

Search Trees

4. The get method returns one of the matching values, if found. It should
probably be the same value that would be returned if a remove were exe-
cuted in the same situation.

5. The iterator method returns an Iterator that traverses all the values of
the BinarySearchTree. When a list of equal values is encountered, they
are all considered before a larger value is returned.

When you are ﬁnished, test your code by storing a large list of names of peo-
ple, ordered only by last name (you will note that this is a common tech-
nique used by stores that keep accounts: “Smith?” “Yes!” “Are you Paul or
John?”). You should be able to roughly sort the names by inserting them into a
BinaryMultiTree and then iterating across its elements.
Thought Questions. Consider the following questions as you complete the lab:

1. Recall: What is the problem with having equal keys stored on either side

of an equal-valued root?

2. Does it matter what type of List is used? What kinds of operations are to

be efﬁcient in this List?

3. What is the essential difference between implementing the tree as de-
scribed and, say, just directly storing linked lists of equivalent nodes in the
BinarySearchTree?

4. An improved version of this structure might use a Comparator for primary
and secondary keys. The primary comparison is used to identify the cor-
rect location for the value in the BinaryMultiTree, and the secondary
key could be used to order the keys that appear equal using the primary
key. Those values that are equal using the primary key are kept within an
OrderedStructure that keeps track of its elements using the secondary
key Comparator.

Notes:

Chapter 15

Maps

Concepts:
. Maps
. Hash tables
.
Tables

X is very useful
if your name is
Nixie Knox.
It also
comes in handy
spelling ax
and extra fox.
—Theodor Seuss Geisel

WE HAVE SEEN THAT AN ASSOCIATION ESTABLISHES A LINK between a key and
a value. An associative array or map is a structure that allows a disjoint set of
keys to become associated with an arbitrary set of values. The convenience of
an associative array is that the values used to index the elements need not be
comparable and their range need not be known ahead of time. Furthermore,
there is no upper bound on the size of the structure. It is able to maintain an
arbitrary number of different pieces of information simultaneously. The analogy
with a mathematical map or function stems from the notion that every key has
at most associated value. Maps are sometimes called dictionaries because of the
uniqueness of the association of words and deﬁnitions in a household dictionary.
Needless to say, a map structure would nicely support the storage of dictionary
deﬁnitions.

15.1 Example Revisited: The Symbol Table

In Chapter 14 we stored the words and their translations (name-alias pairs) in a
structure called a SymTab. This structure forms a good basis for a more general-
purpose approach. Here, we suggest a slightly modiﬁed program to accomplish
exactly the same task. The names of the methods, however, have been changed
to suggest slight improvements in the semantics of structure:

public static void main(String args[])
{

Map<String,String> table = new MapList<String,String>();
Scanner s = new Scanner(System.in);
String alias, name;

SymMap

// read in the alias-name database

370

Maps

do
{

alias = s.next();
if (!alias.equals("END"))
{

name = s.next();
table.put(alias,name); // was called add, but may modify

}

} while (!alias.equals("END"));

// enter the alias translation stage
do
{

name = s.next();
while (table.containsKey(name)) // was contains; more explicit
{

name = table.get(name); // translate alias

}
System.out.println(name);

} while (s.hasNext());

}

The differences between this implementation and that of Section 14.3 involve
improvements in clarity. The method add was changed to put. The difference
is that put suggests that the key-value pair is replaced if it is already in the Map.
We also check for a value in the domain of the Map with containsKey. There
might be a similar need to check the range; that would be accomplished with
containsValue. Finally, we make use of a method, keySet, that returns a Set of
values that are possible keys. This suggests aliases that might be typed in during
the translation phase. Other methods might return a collection of values.

Thus we see that the notion of a Map formalizes a structure we have found
useful in the past. We now consider a more complete description of the inter-
face.

15.2 The Interface

In Java, a Map can be found within the java.util package. Each Map structure
must have the following interface:

public interface Map<K,V>
{

Map

public int size();
// post: returns the number of entries in the map

public boolean isEmpty();
// post: returns true iff this map does not contain any entries

public boolean containsKey(K k);

15.2 The Interface

371

// pre: k is non-null
// post: returns true iff k is in the domain of the map

public boolean containsValue(V v);
// pre: v is non-null
// post: returns true iff v is the target of at least one map entry;
// that is, v is in the range of the map

public V get(K k);
// pre: k is a key, possibly in the map
// post: returns the value mapped to from k, or null

public V put(K k, V v);
// pre: k and v are non-null
// post: inserts a mapping from k to v in the map

public V remove(K k);
// pre: k is non-null
// post: removes any mapping from k to a value, from the mapping

public void putAll(Map<K,V> other);
// pre: other is non-null
// post: all the mappings of other are installed in this map,
// overriding any conflicting maps

public void clear();
// post: removes all map entries associated with this map

public Set<K> keySet();
// post: returns a set of all keys associated with this map

public Structure<V> values();
// post: returns a structure that contains the range of the map

public Set<Association<K,V>> entrySet();
// post: returns a set of (key-value) pairs, generated from this map

public boolean equals(Object other);
// pre: other is non-null
// post: returns true iff maps this and other are entry-wise equal

public int hashCode();
// post: returns a hash code associated with this structure

}

372

Maps

The put method places a new key-value pair within the Map. If the key was
already used to index a value, that association is replaced with a new association
between the key and value.
In any case, the put method returns the value
replaced or null. The get method allows the user to retrieve, using a key, the
value from the Map. If the key is not used to index an element of the Map, a
null value is returned. Because this null value is not distinguished from a
stored value that is null, it is common to predicate the call to get with a call
to the containsKey method. This method returns true if a key matching the
parameter can be found within the Map. Sometimes, like human associative
memory, it is useful to check to see if a value is found in the array. This can be
accomplished with the containsValue method.

Aside from the fact that the keys of the values stored within the Map should
be distinct, there are no other constraints on their type. In particular, the keys
of a Map need only be accurately compared using the equals method. For this
reason, it is important that a reasonable key equality test be provided.

There are no iterators provided with maps. Instead, we have a Map return a
Set of keys (a keySet as previously seen), a Set of key-value pairs (entrySet),
or any Structure of values (values). (The latter must not be a Set because
values may be duplicated.) Each of these, in turn, can generate an Iterator
with the iterator method. Because keys might not implement the Comparable
class, there is no obvious ordering of the entries in a Map. This means that the
keys generated from the keySet and the values encountered during an iteration
over the values structure may appear in different orders. To guarantee the
correct association, use the Iterator associated with the entrySet method.

15.3 Simple Implementation: MapList

One approach to this problem, of course, is to store the values in a List. Each
mapping from a key to a value is kept in an Association which, in turn, is
stored in a List. The result is what we call a MapList; we saw this in Sec-
tion 15.2, though we referred to it as a generic Map structure. The approach is
fairly straightforward. Here is the protected data declaration and constructors:

public MapList()
// post: constructs an empty map, based on a list
{

data = new SinglyLinkedList<Association<K,V>>();

MapList

}

public MapList(Map<K,V> source)
// post: constructs a map with values found in source
{

this();
putAll(source);

}

15.3 Simple Implementation: MapList

373

It is conventional for complex structures to have a copy constructor that gen-
erates a new structure using the entries found in another Map. Notice that we
don’t make any assumptions about the particular implementation of the Map we
copy from; it may be a completely different implementation.

Most of the other methods are fairly straightforward. For example, the put
method is accomplished by ﬁnding a (possible) previous Association and re-
placing it with a fresh construction. The previous value (if any) is returned.

public V put(K k, V v)
// pre: k and v are non-null
// post: inserts a mapping from k to v in the map
{

Association<K,V> temp = new Association<K,V>(k,v);
Association<K,V> result = data.remove(temp);
data.add(temp);
if (result == null) return null;
else return result.getValue();

}

The Set constructions make use of the Set implementations we have discussed
in passing in our discussion of Lists:

public Set<K> keySet()
// post: returns a set of all keys associated with this map
{

Set<K> result = new SetList<K>();
Iterator<Association<K,V>> i = data.iterator();
while (i.hasNext())
{

Association<K,V> a = i.next();
result.add(a.getKey());

}
return result;

}

public Set<Association<K,V>> entrySet()
// post: returns a set of (key-value) pairs, generated from this map
{

Set<Association<K,V>> result = new SetList<Association<K,V>>();
Iterator<Association<K,V>> i = data.iterator();
while (i.hasNext())
{

Association<K,V> a = i.next();
result.add(a);

}
return result;

}

(We will discuss the implementation of various Iterators in Section 15.4; they
are ﬁltering iterators that modify Associations returned from subordinate it-
erators.) Notice that the uniqueness of keys in a Map suggests they form a Set,

374

Maps

yet this is checked by the Set implementation in any case. The values found
in a Map are, of course, not necessarily unique, so they are stored in a general
Structure. Any would do; we make use of a List for its simplicity:

public Structure<V> values()
// post: returns a structure that contains the range of the map
{

Structure<V> result = new SinglyLinkedList<V>();
Iterator<V> i = new ValueIterator<K,V>(data.iterator());
while (i.hasNext())
{

result.add(i.next());

}
return result;

}

Exercise 15.1 What would be the cost of performing a containsKey check on a
MapList? How about a call to containsValue?

Without giving much away, it is fairly clear the answers to the above exercise
are not constant time. It would seem quite difﬁcult to get a O(1) performance
from operators like contains and remove. We discuss the possibilities in the
next section.

15.4 Constant Time Maps: Hash Tables

Clearly a collection of associations is a useful approach to ﬁlling the needs of
the map. The costs associated with the various structures vary considerably. For
Vectors, the cost of looking up data has, on average, O(n) time complexity. Be-
cause of limits associated with being linear, all the O(n) structures have similar
performance. When data can be ordered, sorting the elements of the Linear
structure improves the performance in the case of Vectors: this makes sense
because Vectors are random access structures whose intermediate values can
be accessed given an index.

When we considered binary search trees—a structure that also stores Com-
parable values—we determined the values could be found in logarithmic time.
At each stage, the search space can be reduced by a factor of 2. The difference
between logarithmic and linear algorithms is very dramatic. For example, a
balanced BinarySearchTree or an ordered Vector might ﬁnd one number in a
million in 20 or fewer compares. In an unordered Vector the expected number
of compares increases to 500,000.

Is it possible to improve on this behavior? With hash tables, the answer
is, amazingly, yes. With appropriate care, the hash table can provide access
to an arbitrary element in roughly constant time. By “roughly,” we mean that
as long as sufﬁcient space is provided, each potential key can be reserved an
undisturbed location with probability approaching 1.

15.4 Constant Time Maps: Hash Tables

How is this possible? The technique is, actually, rather straightforward. Here

is an example of how hashing occurs in real life:

We head to a local appliance store to pick up a new freezer. When we
arrive, the clerk asks us for the last two digits of our home telephone
number! Only then does the clerk ask for our last name. Armed with
that information, the clerk walks directly to a bin in a warehouse of
hundreds of appliances and comes back with the freezer in tow.

The technique used by the appliance store was hashing. The “bin” or bucket that
contains the object is identiﬁed by the last two digits of the phone number of
the future owner. If two or more items were located in the bin, the name could
be used to further distinguish the order.

An alternative approach to the “addressing” of the bins might be to identify
each bin with the ﬁrst letter of the name of the customer. This, however, has a
serious ﬂaw, in that it is likely that there will be far more names that begin with
S than with, say, K. Even when the entire name is used, the names of customers
are unlikely to be evenly distributed. These techniques for addressing bins are
less likely to uniquely identify the desired parcel.

The success of the phone number technique stems from generating an identi-
ﬁer associated with each customer that is both random and evenly distributed.1

15.4.1 Open Addressing

We now implement a hash table, modeled after the Hashtable of Java’s java.-
util package. All elements in the table are stored in a ﬁxed-length array whose
length is, ideally, prime. Initialization ensures that each slot within the array
is set to null. Eventually, slots will contain references to associations between
keys and values. We use an array for speed, but a Vector would be a logical
alternative.

protected static final String RESERVED = "RESERVED";
protected Vector<HashAssociation<K,V>> data;
protected int count;

protected final double maximumLoadFactor = 0.6;

public Hashtable(int initialCapacity)
// pre: initialCapacity > 0
// post: constructs a new Hashtable
//
{

holding initialCapacity elements

375

I was just going
to say that.

That would be a
large number of
bins!

Hashtable

Assert.pre(initialCapacity > 0, "Hashtable capacity must be positive.");

1 Using the last two digits of the telephone number makes for an evenly distributed set of values. It
is not the case that the ﬁrst two digits of the exchange would be useful, as that is not always random.
In our town, where the exchange begins with 45, no listed phones have extensions beginning with
45.

376

Maps

Figure 15.1 Hashing color names of antique glass. (a) Values are hashed into the ﬁrst
available slot, possibly after rehashing. (b) The lookup process uses a similar approach
to possibly ﬁnd values.

data = new Vector<HashAssociation<K,V>>();
data.setSize(initialCapacity);
count = 0;

}

public Hashtable()
// post: constructs a new Hashtable
{

this(997);

}

The key and value management methods depend on a function, locate,
that ﬁnds a good location for a value in the structure. First, we use an index-
producing function that “hashes” a value to a slot or bucket (see Figure 15.1).
In Java, every Object has a function, called hashCode, that returns an integer
to be used for precisely this purpose. For the moment, we’ll assume the hash
code is the alphabet code (a = 0, b = 1, etc.) of the ﬁrst letter of the word. The

(a)(b)21alexandritecrystalemeraldtangerineflamingoalexandriteemeraldmoongleamflamingohawthornedawncrystalmarigolddawnmarigoldmarigold?cobalt?moongleamvaselinetangerinevaselinevaseline?hawthorne031245678910121113141516171819202122324567891012111314151617181920102215.4 Constant Time Maps: Hash Tables

377

Figure 15.2
A reserved cell is considered empty during insertion and full during lookup.

(a) Deletion of a value leaves a shaded reserved cell as a place holder. (b)

hash code for a particular key (2 for the word “crystal”) is used as an index to
the ﬁrst slot to be considered for storing or locating the value in the table. If the
slot is empty, the value can be stored there. If the slot is full, it is possible that
another value already occupies that space (consider the insertion of “marigold”
in Figure 15.1). When the keys of the two objects do not match, we have a
collision. A perfect hash function guarantees that (given prior knowledge of the
set of potential keys) no collisions will occur. When collisions do occur, they can
be circumvented in several ways. With open addressing, a collision is resolved
by generating a new hash value, or rehashing, and reattempting the operation
at a new location.

Slots in the hash table logically have two states—empty (null) or full (a
reference to an object)—but there is also a third possibility. When values are
removed, we replace the value with a reserved value that indicates that the
location potentially impacts the lookup process for other cells during insertions.
That association is represented by the empty shaded cell in Figure 15.2a. Each
time we come across the reserved value in the search for a particular value in
the array (see Figure 15.2b), we continue the search as though there had been
a collision. We keep the ﬁrst reserved location in mind as a possible location for
an insertion, if necessary. In the ﬁgure, this slot is used by the inserted value
“custard.”

When large numbers of different-valued keys hash or rehash to the same
locations, the effect is called clustering (see Figure 15.3). Primary clustering is
when several keys hash to the same initial location and rehash to slots with
potential collisions with the same set of keys. Secondary clustering occurs when

(b)(a)101113141003124567891012111314alexandritecrystalflamingohawthornedawnmarigoldalexandritecrystalflamingohawthornedawnmarigoldcustardmarigold?delete emeralddelete moongleam3245678912378

Maps

(a) Primary clustering occurs when two values that hash to the same slot
Figure 15.3
continue to compete during rehashing. (b) Rehashing causes keys that initially hash to
different slots to compete.

keys that initially hash to different locations eventually rehash to the same se-
quence of slots.

In this simple implementation we use linear probing (demonstrated in Fig-
ures 15.1 to 15.3). Any rehashing of values occurs a constant distance from
the last hash location. The linear-probing approach causes us to wrap around
the array and ﬁnd the next available slot. It does not solve either primary or
secondary clustering, but it is easy to implement and quick to compute. To
avoid secondary clustering we use a related technique, called double hashing,
that uses a second hash function to determine the magnitude of the constant
offset (see Figure 15.4). This is not easily accomplished on arbitrary keys since
we are provided only one hashCode function. In addition, multiples and factors
of the hash table size (including 0) must also be avoided to keep the locate
function from going into an inﬁnite loop. Still, when implemented correctly,
the performance of double hashing can provide signiﬁcant improvements over
linear-probing.

We now discuss our implementation of hash tables. First, we consider the
locate function. Its performance is important to the efﬁciency of each of the
public methods.

protected int locate(K key)
{

// compute an initial hash code
int hash = Math.abs(key.hashCode() % data.size());
// keep track of first unused slot, in case we need it
int reservedSlot = -1;
boolean foundReserved = false;
while (data.get(hash) != null)
{

if (data.get(hash).reserved()) {

(b)(a)duncancanarycustardalexandriteflamingohawthornedawncustardalexandritecrystalflamingohawthornedawn245678103crystalcobaltcobalt15.4 Constant Time Maps: Hash Tables

379

// remember reserved slot if we fail to locate value
if (!foundReserved) {

reservedSlot = hash;
foundReserved = true;

}

} else

{

// value located? return the index in table
if (key.equals(data.get(hash).getKey())) return hash;

}
// linear probing; other methods would change this line:
hash = (1+hash)%data.size();

}
// return first empty slot we encountered
if (!foundReserved) return hash;
else return reservedSlot;

}

To measure the difﬁculty of ﬁnding an empty slot by hashing, we use the
load factor, α, computed as the ratio of the number of values stored within the
table to the number of slots used. For open addressing, the load factor cannot
exceed 1. As we shall see, to maintain good performance we should keep the
load factor small as possible. Our maximum allowable load factor is a constant
maximumLoadFactor. Exceeding this value causes the array to be reallocated
and copied over (using the method extend).

When a value is added, we simply locate the appropriate slot and insert a
new association. If the ideal slot already has a value (it must have an equal key),
we return the replaced association. If we replace the reference to an empty cell
with the reserved association, we return null instead.

public V put(K key, V value)
// pre: key is non-null object
// post: key-value pair is added to hash table
{

if (maximumLoadFactor*data.size() <= (1+count)) {

extend();

}
int hash = locate(key);
if (data.get(hash) == null || data.get(hash).reserved())
{

// logically empty slot; just add association
data.set(hash,new HashAssociation<K,V>(key,value));
count++;
return null;

} else {

// full slot; add new and return old value
HashAssociation<K,V> a = data.get(hash);
V oldValue = a.getValue();
a.setValue(value);
return oldValue;

}

}

380

Maps

The keys of Figure 15.3 are rehashed by an offset determined by the
Figure 15.4
alphabet code (a = 1, b = 2, etc.) of the second letter. No clustering occurs, but strings
must have two letters!

marigoldtangerinevaseline324567891012111314151617181920251024232221cobaltcobaltduncancanarycustardalexandritecrystalflamingohawthornedawnmarigoldcustardalexandritecrystalvaselineflamingohawthornedawntangerine15.4 Constant Time Maps: Hash Tables

381

The get function works similarly—we simply return the value from within

the key-located association or null, if no association could be found.

public V get(K key)
// pre: key is non-null Object
// post: returns value associated with key, or null
{

int hash = locate(key);
if (data.get(hash) == null ||

data.get(hash).reserved()) return null;

return data.get(hash).getValue();

}

The containsKey method is similar. To verify that a value is within the table

we build contains from the elements iterator:

public boolean containsValue(V value)
// pre: value is non-null Object
// post: returns true iff hash table contains value
{

for (V tableValue : this) {

if (tableValue.equals(value)) return true;

}
// no value found
return false;

}

public boolean containsKey(K key)
// pre: key is a non-null Object
// post: returns true if key appears in hash table
{

int hash = locate(key);
return data.get(hash) != null && !data.get(hash).reserved();

}

The containsValue method is difﬁcult to implement efﬁciently. This is one of
the trade-offs of having a structure that is fast by most other measures.

To remove a value from the Hashtable, we locate the correct slot for the
In its place, we leave a reserved mark to

value and remove the association.
maintain consistency in locate.

public V remove(K key)
// pre: key is non-null object
// post: removes key-value pair associated with key
{

int hash = locate(key);
if (data.get(hash) == null || data.get(hash).reserved()) {

return null;

}
count--;

382

Maps

V oldValue = data.get(hash).getValue();
data.get(hash).reserve(); // in case anyone depends on us
return oldValue;

}

Hash tables are not made to be frequently traversed. Our approach is to
construct sets of keys, values, and Associations that can be, themselves, tra-
versed. Still, to support the Set construction, we build a single iterator (a
HashtableIterator) that traverses the Hashtable and returns the Associa-
tions. Once constructed, the association-based iterator can be used to generate
the key- and value-based iterators.

The protected iterator is similar to the Vector iterator. A current index
points to the cell of the current non-null (and nonreserved) association. When
the iterator is incremented, the underlying array is searched from the current
point forward to ﬁnd the next non-null entry. The iterator must eventually
inspect every element of the structure, even if very few of the elements are
currently used.2

Given an iterator that returns Associations, we can construct two different
public ﬁltering iterators, a ValueIterator and a KeyIterator. Each of these
maintains a protected internal “slave” iterator and returns, as the iterator is in-
cremented, values or keys associated with the respective elements. This design
is much like the design of the UniqueFilter of Section 8.5. The following code,
for example, implements the ValueIterator:

class ValueIterator<K,V> extends AbstractIterator<V>
{

protected AbstractIterator<Association<K,V>> slave;

ValueIterator

public <T extends Association<K,V>> ValueIterator(Iterator<T> slave)
// pre: slave is an iterator returning Association elements
// post: creates a new iterator returning associated values
{

this.slave = (AbstractIterator<Association<K,V>>)slave;

}

public boolean hasNext()
// post: returns true if current element is valid
{

return slave.hasNext();

}

public V next()
// pre: hasNext()
// post: returns current value and increments iterator
{

2 The performance of this method could be improved by linking the contained associations together.
This would, however, incur an overhead on the add and remove methods that may not be desirable.

15.4 Constant Time Maps: Hash Tables

383

Association<K,V> pair = ((AbstractIterator<Association<K,V>>)slave).next();
return pair.getValue();

}

}

Once these iterators are deﬁned, the Set and Structure returning methods
are relatively easy to express. For example, to return a Structure that contains
the values of the table, we simply construct a new ValueIterator that uses the
HashtableIterator as a source for Associations:

public Structure<V> values()
// post: returns a Structure that contains the (possibly repeating)
// values of the range of this map.
{

List<V> result = new SinglyLinkedList<V>();
Iterator<V> i = new ValueIterator<K,V>(new HashtableIterator<K,V>(data));
while (i.hasNext())
{

Hashtable

result.add(i.next());

}
return result;

}

It might be useful to have direct access to iterators that return keys and values.
If that choice is made, the keys method is similar but constructs a KeyIterator
instead. While the ValueIterator and KeyIterator are protected, they may
be accessed publicly when their identity has been removed by the elements and This is a form of
keys methods, respectively.

identity
laundering.

15.4.2 External Chaining

Open addressing is a satisfactory method for handling hashing of data, if one
can be assured that the hash table will not get too full. When open addressing
is used on nearly full tables, it becomes increasingly difﬁcult to ﬁnd an empty
slot to store a new value.

One approach to avoiding the complexities of open addressing—reserved
associations and table extension—is to handle collisions in a fundamentally dif-
ferent manner. External chaining solves the collision problem by inserting all
elements that hash to the same bucket into a single collection of values. Typi-
cally, this collection is a singly linked list. The success of the hash table depends
heavily on the fact that the average length of the linked lists (the load factor of
the table) is small and the inserted objects are uniformly distributed. When the
objects are uniformly distributed, the deviation in list size is kept small and no
list is much longer than any other.

The process of locating the correct slot in an externally chained table in-
volves simply computing the initial hashCode for the key and “modding” by the
table size. Once the appropriate bucket is located, we verify that the collection
is constructed and the value in the collection is updated. Because our List

384

Maps

classes do not allow the retrieval of internal elements, we may have to remove
and reinsert the appropriate association.

Chained-
HashTable

public V put(K key, V value)
// pre: key is non-null object
// post: key-value pair is added to hash table
{

List<Association<K,V>> l = locate(key);
Association<K,V> newa = new Association<K,V>(key,value);
Association<K,V> olda = l.remove(newa);
l.addFirst(newa);
if (olda != null)
{

return olda.getValue();

}
else
{

}

}

count++;
return null;

Most of the other methods are implemented in a similar manner: they locate
the appropriate bucket to get a List, they search for the association within the
List to get the association, and then they manipulate the key or value of the
appropriate association.

One method, containsValue, essentially requires the iteration over two di-
mensions of the hash table. One loop searches for non-null buckets in the hash
table—buckets that contain associations in collections—and an internal loop
that explicitly iterates across the List (the containsKey method can directly
use the containsValue method provided with the collection). This is part of
the price we must pay for being able to store arbitrarily large numbers of keys
in each bucket of the hash table.

public boolean containsValue(V value)
// pre: value is non-null Object
// post: returns true iff hash table contains value
{

for (V v : this) {

if (value.equals(v)) return true;

}
return false;

}

At times the implementations appear unnecessarily burdened by the inter-
faces of the underlying data structure. For example, once we have found an
appropriate Association to manipulate, it is difﬁcult to modify the key. This
is reasonable, though, since the value of the key is what helped us locate the

15.4 Constant Time Maps: Hash Tables

385

Figure 15.5
ues.

The time required to construct large ordered structures from random val-

bucket containing the association. If the key could be modiﬁed, we could insert
a key that was inconsistent with its bucket’s location.

Another subtle issue is the selection of the collection class associated with
the bucket. Since linked lists have poor linear behavior for most operations,
it might seem reasonable to use more efﬁcient collection classes—for example,
tree-based structures—for storing data with common hash codes. The graph of
Figure 15.5 demonstrates the performance of various ordered structures when
asked to construct collections of various sizes. It is clear that while SplayTrees
provide better ultimate performance, the simple linear structures are more ef-
ﬁcient when the structure size is in the range of expected use in chained hash
tables (see Figure 15.6). When the average collection size gets much larger than
this, it is better to increase the size of the hash table and re-insert each of the
elements (this is accomplished with the Hashtable method, extend).

15.4.3 Generation of Hash Codes

Because any object might eventually be stored within a hash table, and because
data abstraction hides the details of implementation, it is important for imple-
mentors to provide a hashCode method for their classes whenever possible.

Principle 24 Provide a method for hashing the objects you implement.

05e+061e+071.5e+072e+072.5e+073e+07010002000300040005000600070008000Time (in units of integer compares)Size of StructurePerformance of Ordered StructuresSplayTreeOrderedListOrderedVectorNNWSWSENEWSE386

Maps

Figure 15.6
values.

The time required to construct small ordered structures from random

When a hashCode method is provided, it is vital that the method return the
same hashCode for any pair of objects that are identiﬁed as the same under the
equals method. If this is not the case, then values indexed by equivalent keys
can be stored in distinct locations within the hash table. This can be confusing
for the user and often incorrect.

Principle 25 Equivalent objects should return equal hash codes.

The generation of successful hash codes can be tricky. Consider, for example,
the generation of hash codes for Strings. Recall that the purpose of the hash
code generation function is to distribute String values uniformly across the
hash table.

Most of the approaches for hashing strings involve manipulations of the
characters that make up the string. Fortunately, when a character is cast as
an integer, the internal representation (often the ASCII encoding) is returned,
usually an integer between 0 and 255. Our ﬁrst approach, then, might be to
use the ﬁrst character of the string. This has rather obvious disadvantages: the
ﬁrst letters of strings are not uniformly distributed, and there isn’t any way of
generating hash codes greater than 255.

Our next approach would be to sum all the letters of the string. This is
a simple method that generates large-magnitude hash codes if the strings are
long. The main disadvantage of this technique is that if letters are transposed,

01002003004005006000246810121416Time (in units of integer compares)Size of StructurePerformance of Ordered Structures (Detail)SplayTreeOrderedListOrderedVectorNNWSWSENEWSE15.4 Constant Time Maps: Hash Tables

387

Figure 15.7 Numbers of words from the UNIX spelling dictionary hashing to each of
the 997 buckets of a default hash table, if sum of characters is used to generate hash
code.

then the strings generate the same hash values. For example, the string "dab"
has 100 + 97 + 98 = 295 as its sum of ASCII values, as does the string "bad".
The string "bad" and "bbc" are also equivalent under this hashing scheme. Fig-
ure 15.7 is a histogram of the number of words that hash, using this method, to
each slot of a 997 element hash table. The periodic peaks demonstrate the fact
that some slots of the table are heavily preferred over others. The performance
of looking up and modifying values in the hash table will vary considerably, de-
pending on the slot that is targeted by the hash function. Clearly, it would be
useful to continue our search for a good mechanism.

Another approach might be to weight each character of the string by its
position. To ensure that even very short strings have the potential to generate
large hash values, we can provide exponential weights: the hash code for an l
character string, s, is

l−1
X

i=0

s[i]ci

where c is usually a small integer value. When c is 2, each character is weighted
by a power of 2, and we get a distribution similar to that of Figure 15.8. While
this is closer to being uniform, it is clear that even with exponential behavior,
the value of c = 2 is too small: not many words hash to table elements with

01020304050607080900100200300400500600700800900FrequencyBucket388

Maps

Figure 15.8
are weighted by powers of 2 to generate hash code.

Frequency of dictionary words hashing to each of 997 buckets if characters

Figure 15.9
hash code is generated by weighting characters by powers of 256.

Frequency of words from dictionary hashing to each of 997 buckets if

01020304050607080900100200300400500600700800900FrequencyBucket01020304050607080900100200300400500600700800900FrequencyBucket15.4 Constant Time Maps: Hash Tables

389

Figure 15.10
using the Java String hash code generation.

Frequency of words from dictionary hashing to each of 997 buckets,

large indices. When c = 256, the hash code represents the ﬁrst few characters
of the string exactly (see Figure 15.9). Java currently hashes with c = 31.

The hashing mechanism used by Java Strings in an early version of Java’s
development environment (see Figure 15.10) used a combination of weight-
ings that provided a wide range of values for short strings and was efﬁcient
to compute for longer strings. Unfortunately, the constant-time algorithm was
not suitable for distinguishing between long and nearly identical strings often
found, say, in URLs.

Method

Linear probes

Double hashing

External chaining

Successful
(cid:16)
1 + 1

(1−α)

1
2

(cid:17)

1
α ln
1 + 1

1
(1−α)
2 α

Unsuccessful
(cid:17)
1 + 1
1
2

(1−α)2

(cid:16)

1
1−α
α + e−α

Expected theoretical performance of hashing methods, as a function of
Figure 15.11
α, the current load factor. Formulas are for the number of association compares needed
to locate the correct value or to demonstrate that the value cannot be found.

01020304050607080900100200300400500600700800900FrequencyBucket390

Maps

The shape of the theoretical performance curves for various hashing
Figure 15.12
techniques. (These graphs demonstrate theoretical predictions and not experimental re-
sults which are, of course, dependant on particular data and hashing functions.) Our
hash table implementation uses linear probing.

01234500.20.40.60.811.2Number of ProbesLoad Factor (Stored Values/Table Size)linear, not foundlinear, founddouble, not founddouble, foundchaining, not foundchaining, found15.4 Constant Time Maps: Hash Tables

391

Many of the data structures we have investigated are classes that contain
multiple objects of unspeciﬁed type. When hashing entire container classes, it
can be useful to compose the codes of the contained elements.

15.4.4 Hash Codes for Collection Classes

Each of the techniques used to generate hash codes from a composition of char-
acters of Strings can be used to compose hash codes of objects in collection
classes. The features of primary importance for the construction of hash codes
are:

1. Whenever two structures are equal, using the equals methods, the hash-

Code method should return the same value.

2. For container structures—structures whose only purpose is to hold values—
the state of the structure itself should be transparent; the state of the
structure should not be included in the hashCode.

The ﬁrst item was discussed before, but it is the most common error leading to
difﬁculties with the use of Hashtables. When the hashCodes do not match for
objects that are logically equal, the objects have a high probability of entering
into different locations of the table. Accesses that should interact do not.

The second consideration is understood if we consider self-modifying struc-
tures like SplayTrees or Hashtables whose external state may be modeled by
several distinct internal states. The construction of the hash code should con-
sider those bits of information that enter into identifying equal structures. In
the case of the SplayTree, for example, we might use the sum of the hash codes
of the values that appear within the tree.

In general, the following ﬁrst attempt at a hashCode method, taken from the

AbstractStructure type, is reasonable:

public int hashCode()
// post: generate a hashcode for the structure: sum of
// all the hash codes of elements
{

Iterator<E> i = iterator();
int result = 0;
while (i.hasNext())
{

E o = i.next();
result = result * 31;
if (o != null) result += o.hashCode();

}
return result;

}

Abstract-
Structure

As we can see here, we must constantly be on the watch for values that may po-
tentially be null references. For some structures, of course, such an approach

392

Maps

may lead to intolerable amounts of time computing hashCodes for large struc-
tures.

One last point concerns the hashCodes associated with recursively deﬁned
structures. If the recursive structure is visible externally, that is, the structure
could be referenced at several points, it may be suitable to deﬁne the hashCode
to be the value contained within a single node of the structure. This certainly
fulﬁlls the primary obligation of a hashing function, but it also serves to separate
the structure from the hash code. In our case, we choose to make a recursive
deﬁnition, similar to the following deﬁnition found in BinaryTree:

BinaryTree

public int hashCode()
// post: return sum of hashcodes of the contained values
{

if (isEmpty()) return 0;
int result = left().hashCode() + right().hashCode();
if (value() != null) result += value().hashCode();
return result;

}

15.4.5 Performance Analysis

For open addressing, the load factor α obviously cannot exceed 1. As the load
factor approaches 1, the performance of the table decreases dramatically. By
counting the number of probes or association compares needed to ﬁnd a value
(a successful search) or to determine that the value is not among the elements
of the map (an unsuccessful search), we can observe the relative performance
of the various hashing techniques (see Figures 15.11 and 15.12). Notice that
the number of probes necessary to ﬁnd an appropriate key is a function of the
load factor, and not directly of the number of keys found in the table.

When a hash table exceeds the maximum allowable load factor, the entire
table is forced to expand, and each of the keys is rehashed. Looking at the graph
in Figure 15.12, we select our threshold load factor to be 60 percent, the point
at which the performance of linear probing begins to degrade. When we expand
the hash table, we make sure to at least double its size. For the same reasons
that doubling is good when a Vector is extended, doubling the size of the hash
table improves the performance of the hash table without signiﬁcant overhead.

15.5 Ordered Maps and Tables

In fact, better
hash functions
probably avoid
order!

A signiﬁcant disadvantage of the Map interface is the fact that the values stored
within the structure are not kept in any particular order. Often we wish to
efﬁciently maintain an ordering among key-value pairs. The obvious solution
is to construct a new OrderedMap that builds on the interface of the Map, but
where methods may be allowed to assume parameters that are Comparable:

15.5 Ordered Maps and Tables

393

OrderedMap

public interface OrderedMap<K extends Comparable<K>,V> extends Map<K,V>
{
}

When we do this, the methods of the Map are inherited. As a result, the types
of the key-based methods manipulate Objects and not Comparables. Because
we desire to maintain order among comparable keys, we have a general pre-
condition associated with the use of the data structure—that keys provided and
returned must be objects supporting the Comparable interface.

Even with comparable keys, it is not easy to construct a Hashtable whose
keys iterator returns the keys in order. The hash codes provided for Comparable
objects are not required (and unlikely) to be ordered in a way consistent with
the compareTo function. We therefore consider other OrderedStructures to
maintain the order among ComparableAssociations.

We will call our implementation of the OrderedMap a Table. As a basis for
the implementation, we depend on the SplayTree class. OrderedLists and
OrderedVectors could also provide suitable implementations for small applica-
tions. The Table maintains a single protected data item—the SplayTree. The
constructor is responsible for allocating the SplayTree, leaving it initialized in
its empty state:

protected OrderedStructure<ComparableAssociation<K,V>> data;

public Table()
// post: constructs a new table
{

data = new SplayTree<ComparableAssociation<K,V>>();

}

Table

public Table(Table<K,V> other)
{

data = new SplayTree<ComparableAssociation<K,V>>();
Iterator<Association<K,V>> i = other.entrySet().iterator();
while (i.hasNext())
{

Association<K,V> o = i.next();
put(o.getKey(),o.getValue());

}

}

When a key-value pair is to be put into the Table, a ComparableAssociation
is constructed with the key-value pair, and it is used to look up any previous
association using the same key. If the association is present, it is removed. In
either case, the new association is inserted into the tree. While it seems indi-
rect to remove the pair from the table to update it, it maintains the integrity of
the ComparableAssociation and therefore the SplayTree. In addition, even
though two keys may be logically equal, it is possible that they may be distin-
guishable. We insert the actual key-value pair demanded by the user, rather

394

Maps

than perform a partial modiﬁcation. Theoretically, removing and inserting a
value into the SplayTree costs the same as ﬁnding and manipulating the value
in place. Next, we see the method for put (get is similar):

public V put(K key, V value)
// pre: key is non-null object
// post: key-value pair is added to table
{

ComparableAssociation<K,V> ca =

new ComparableAssociation<K,V>(key,value);

// fetch old key-value pair
ComparableAssociation<K,V> old = data.remove(ca);
// insert new key-value pair
data.add(ca);
// return old value
if (old == null) return null;
else return old.getValue();

}

While most of the other methods follow directly from considering Hash-
tables and SplayTrees, the contains method—the method that returns true
exactly when a particular value is indexed by a key in the table—potentially re-
quires a full traversal of the SplayTree. To accomplish this, we use an Iterator
returned by the SplayTree’s elements methods. We then consider each associ-
ation in turn, returning as soon as an appropriate value is found:

public boolean containsValue(V value)
// pre: value is non-null object
// post: returns true iff value in table
{

Iterator<V> i = iterator();
while (i.hasNext())
{

V nextValue = i.next();
if (nextValue != null &&

nextValue.equals(value)) return true;

}
return false;

}

Next, our Table must provide an Iterator to be used in the construction
of the keySet and entrySet. The approach is similar to the Hashtable—
we construct a private Association-returning Iterator and then return its
KeyIterator or ValueIterator. Because every value returned from the Splay-
Tree’s iterator is useful,3 we need not implement a special-purpose iterator for

3 Compare this with, perhaps, a Vector iterator that might be used to traverse a Vector-based
Hashtable.

15.6 Example: Document Indexing

395

Tables; instead, we use the SplayTree’s iterator directly. Since Comparable-
Associations extend Associations, the KeyIterator generates an Iterator
that returns the comparable keys as Objects to be cast later.

public Set<K> keySet()
// post: returns a set containing the keys referenced
// by this data structure.
{

Set<K> result = new SetList<K>();
Iterator<K> i = new KeyIterator<K,V>(data.iterator());
while (i.hasNext())
{

result.add(i.next());

}
return result;

}

public Set<Association<K,V>> entrySet()
// post: returns a structure containing all the entries in
// this Table
{

Set<Association<K,V>> result = new SetList<Association<K,V>>();
Iterator<ComparableAssociation<K,V>> i = data.iterator();
while (i.hasNext())
{

result.add(i.next());

}
return result;

}

Previous hard work greatly simpliﬁes this implementation! Since no hashing
occurs, it is not necessary for any of the keys of a Table to implement the
hashCode method. They must, though, implement the compareTo method since
they are Comparable. Thus, each of the methods runs in amortized logarithmic
time, instead of the near-constant time we get from hashing.

Exercise 15.2 Modify the Table structure to make use of RedBlackTrees, instead
of SplayTrees.

Exercise 15.3 It is common to allow ordered structures, like OrderedMap, to use a
Comparator to provide an alternative ordering. Describe how this approach might
be implemented.

15.6 Example: Document Indexing

Indexing is an important task, especially for search engines that automatically
index keywords from documents retrieved from the Web. Here we present the

396

Maps

skeleton of a document indexing scheme that makes use of a Map to keep track
of the vocabulary.

Given a document, we would like to generate a list of words, each followed
by a list of lines on which the words appear. For example, when provided
Gandhi’s seven social sins:

politics without principle
pleasure without conscience

wealth without work

knowledge without character
business without morality
science without humanity
and

worship without sacrifice

(It is interesting to note that programming without comments is not among
these!) The indexing program should generate the following output:

and: 7
business: 5
character: 4
conscience: 2
humanity: 6
knowledge: 4
morality: 5
pleasure: 2
politics: 1
principle: 1
sacrifice: 8
science: 6
wealth: 3
without: 1 2 3 4 5 6 8
work: 3
worship: 8

In this program we make use of Java’s StreamTokenizer class. This class
takes a stream of data and converts it into a stream of tokens, some of which
are identiﬁed as words. The process for constructing this stream is a bit difﬁcult,
so we highlight it here.

public static void main(String args[])
{

try {

Index

InputStreamReader isr = new InputStreamReader(System.in);
java.io.Reader r = new BufferedReader(isr);
StreamTokenizer s = new StreamTokenizer(r);

...

} catch (java.io.IOException e) {

Assert.fail("Got an I/O exception.");

}

}

15.6 Example: Document Indexing

397

Each of the objects constructed here provides an additional layer of ﬁltering on
the base stream, System.in. The body of the main method is encompassed by
the try statement in this code. The try statement catches errors generated by
the StreamTokenizer and rewraps the exception as an assertion failure.

We begin by associating with each word of the input an initially empty list
of line numbers.
It seems reasonable, then, to use the vocabulary word as a
key and the list of lines as the value. Our Map provides an ideal mechanism to
maintain the data. The core of the program consists of reading word tokens
from the stream and entering them into the Map:

// allocate the symbol table (uses comparable keys)
Map<String,List<Integer>> t = new Table<String,List<Integer>>();
int token;
// we’ll not consider period as part of identifier
s.ordinaryChar(’.’);
// read in all the tokens from file
for (token = s.nextToken();

token != StreamTokenizer.TT_EOF;
token = s.nextToken())

{

// only tokens we care about are whole words
if (token == StreamTokenizer.TT_WORD)
{

// each set of lines is maintained in a List
List<Integer> l;

// look up symbol
if (t.containsKey(s.sval))
{

// symbol is there, get line # list
l = t.get(s.sval);
l.addLast(s.lineno());

} else {

// not found, create new list
l = new DoublyLinkedList<Integer>();
l.addLast(s.lineno());
t.put(s.sval,l);

}

}

}

Here, we use a Table as our Map because it is important that the entries be
sorted alphabetically. As the tokens are read from the input stream, they are
looked up in the Map. Since the Map accepts comparable keys, it is important
to use a (comparable) String to allow the words to index the structure. If the
key is within the Map, the value associated with the key (a list) is updated by
appending the current line number (provided by the stream’s lineno method)
to the end of the list. If the word is not found, a new list is allocated with the
current line appended, and the fresh word–list pair is inserted into the table.
The next section of the program is responsible for generating the output:

398

Maps

// printing table involves tandem key-value iterators
Iterator<List<Integer>> ki = t.values().iterator();
for (String sym : t.keySet())
{

// print symbol
System.out.print(sym+": ");
// print out (and consume) each line number
for (Integer lineno : ki.next())
{

System.out.print(lineno+" ");

}
System.out.println();
// increment iterators

}

Here, two iterators—one for keys and one for values—are constructed for the
Map and are incremented in parallel. As each word is encountered, it is printed
out along with the list of line numbers, generated by traversing the list with an
iterator.

Because we used a Table as the underlying structure, the words are kept
and printed in sorted order. If we had elected to use a Hashtable instead, the
output would appear shufﬂed. The order is neither alphabetical nor the order in
which the words are encountered. It is the result of the particular hash function
we chose to locate the data.

15.7 Conclusions

In this chapter we have investigated two structures that allow us to access values
using a key or index from an arbitrary domain. When the keys can be uniformly
distributed across a wide range of values, hashing is an excellent technique for
providing constant-time access to values within the structure. The cost is extra
time necessary to hash the value, as well as the extra space needed to keep the
load factor small enough to provide the expected performance.

When the keys are comparable, and order is to be preserved, we must de-
pend on logarithmic behavior from ordered structures we have seen before. In
our implementation of Tables, the SplayTree was used, although any other
OrderedStructure could be used instead.

Because of the nonintuitive nature of hashing and hash tables, one of the
more difﬁcult tasks for the programmer is to generate useful, effective hash
code values. Hash functions should be designed speciﬁcally for each new class.
They should be fast and deterministic and have wide ranges of values. While
all Objects inherit a hashCode function, it is important to update the hashCode
method whenever the equals method is changed; failure to do so leads to subtle
problems with these useful structures.

15.7 Conclusions

Self Check Problems

399

Solutions to these problems begin on page 450.

15.1 What access feature distingishes Map structures from other structrues
we have seen?

15.2 What is the load factor of a hash table?

15.3

15.4

In a hash table is it possible to have a load factor of 2?

Is a constant-time performance guaranteed for hash tables?

15.5 What is a hash collision?

15.6 What are the qualities we seek in a hash function?

15.7

Under what condition is a MapList preferable to a Hashtable?

15.8 Many of our more complex data structures have provided the under-
pinnings for efﬁcient sorts. Is that the case for the Hashtable? Does the Table
facilitate sorting?

Problems

Solutions to the odd-numbered problems begin on page 484.

15.1

15.2

Is it possible for a hash table to have two entries with equal keys?

Is it possible for a hash table to have two entries with equal values?

Suppose you have a hash table with seven entries (indexed 0 through
15.3
6). This table uses open addressing with the hash function that maps each letter
to its alphabet code (a = A = 0, etc.) modulo 7. Rehashing is accomplished
using linear-probing with a jump of 1. Describe the state of the table after each
of the letters D, a, d, H, a, and h are added to the table.

Suppose you have a hash table with eight entries (indexed 0 through 7).
15.4
The hash mechanism is the same as for Problem 15.3 (alphabet code modulo 8),
but with a linear probe jump of 2. Describe what happens when one attempts to
add each of the letters A, g, g, a, and g, in that order. How might you improve
the hashing mechanism?

15.5 When using linear probing with a rehashing jump size of greater than
1, why is it necessary to have the hash table size and jump size be relatively
prime?

15.6
ber.

15.7

Design a hashCode method for a class that represents a telephone num-

Design a hashCode method for a class that represents a real number.

Suppose two identiﬁers—Strings composed of letters—were consid-
15.8
ered equal even if their cases were different. For example, AGEdwards would be
equal to AgedWards. How would you construct a hash function for strings that
was “case insensitive”?

400

Maps

15.9 When 23 randomly selected people are brought together, chances are
greater than 50 percent that two have the same birthday. What does this tell us
about uniformly distributed hash codes for keys in a hash table?
15.10 Write a hashCode method for an Association.
15.11 Write a hashCode method for a Vector. It should only depend on hash
codes of the Vector’s elements.
15.12 Write a hashCode method for a BinaryTree. Use recursion.
15.13 Write a hashCode method for a Hashtable. (For some reason, you’ll be
hashing hash tables into other hash tables!) Must the hashing mechanism look
at the value of every element?
15.14 The Java hash function for Strings computes a hash code based on a
ﬁxed maximum number of characters of the string. Given that Strings have
no meaningful upper bound in length, describe how an effective, constant-time
hashing algorithm can be constructed. (Hint: If you were to pick, say, eight
characters to represent a string of length l, which would you choose?)
15.15 Since URLs differ mostly toward their end (at high indices), write code
that efﬁciently computes a hash code based on characters l − xi where xi =
2i and i = 0, 1, 2, . . . How fast does this algorithm run? Is it better able to
distinguish different URLs?
15.16 A hash table with ordered linear probing maintains an order among
keys considered during the rehashing process. When the keys are encountered,
say, in increasing order, the performance of a failed lookup approaches that of
a successful search. Describe how a key might be inserted into the ordered
sequence of values that compete for the same initial table entry.
15.17 Isn’t the hash table resulting from Problem 15.16 just an ordered Vector?
(Hint: No.) Why?
15.18 If we were to improve the iterators for Maps, we might add an iterator
that returned key-value pairs. Is this an improvement in the interface?
15.19 Design a hash function for representing the state of a checkerboard.
15.20 Design a hash function for representing the state of a tic-tac-toe board.
(It would—for strategy reasons—be useful to have mirror images of a board be
considered equal.)
15.21 One means of potentially reducing the complexity of computing the
hash code for Strings is to compute it once—when the String is constructed.
Future calls to hashCode would return the precomputed value. Since the value
of a String never changes, this has potential promise. How would you evaluate
the success of such a method?
15.22 Explain how a Map might be useful in designing a spelling checker.
(Would it be useful to have the words bible and babble stored near each
other?)

15.8 Laboratory: The Soundex Name Lookup Sys-

tem

Objective. To use a Map structure to keep track of similar sounding names.

Discussion. The United States National Archives is responsible for keeping track
of the census records that, according to the Constitution, must be gathered every
10 years. After a signiﬁcant amount of time has passed (70 or more years), the
census records are made public. Such records are of considerable historical
interest and a great many researchers spend time looking for lost ancestors
among these records.

To help researchers ﬁnd individuals, the censuses are often indexed using a
phonetic system called Soundex. This system takes a name and produces a short
string called the Soundex key. The rules for producing the Soundex key of a
name are precisely:

1. The entire name is translated into a series of digit characters:

Letter of name
b, p, f, v
c, s, k, g, j, q, x, z
d, t
l

Character
‘1’
‘2’
‘3’
‘4’
’5’ m, n
’6’
’7’

r
all other letters

For example, Briggs would be translated into the string 167222.

2. All double digits are reduced to single digits. Thus, 167222 would become

1672.

3. The ﬁrst digit is replaced with the ﬁrst letter of the original name, in

uppercase. Thus, 1672 would become B672.

4. All 7’s are removed. Thus, B672 becomes B62.

5. The string is truncated to four characters. If the resulting string is shorter
than four characters, it is packed with enough ’0’ characters to bring the
length to four. The result for Briggs would be B620. Notice that, for the
most part, the nonzero characters represent the signiﬁcant sounded letters
of the beginning of the name.

Other names translate to Soundex keys as follows:

402

Maps

Bailey
Ballie
Knuth
Scharstein
Lee

becomes B400
becomes B400
becomes K530
S623
becomes
L000
becomes

Procedure. You are to write a system that takes a list of names (the UNIX
spelling dictionary is a good place to ﬁnd names) and generates an ordered
map whose entries are indexed by the Soundex key. The values are the actual
names that generated the key. The input to the program is a series of names,
and the output is the Soundex key associated with the name, along with all the
names that have that same Soundex key, in alphabetical order.

Pick a data structure that provides performance: the response to the query

should be nearly instantaneous, even with a map of several thousand names.
Thought Questions. Consider the following questions as you complete the lab:

1. What is the Soundex system attempting to do when it encodes many let-
ters into one digit? For example, why are ‘d’ and ‘t’ both encoded as ’3’?

2. Why does the Soundex system ignore any more than the ﬁrst four sounds

in a name?

Notes:

Chapter 16

Graphs

Concepts:
. Graphs
. Adjacency Lists
. Adjacency Matrices
. Graph Algorithms

. . . 314159 . . .
—π (digits 176452–176457)

RELATIONS ARE OFTEN AS USEFUL AS DATA. The process of building and ac-
cessing a data structure can be thought of as a means of effectively focusing
the computation. Linear structures record the history of their accesses, ordered
structures perform incremental sorting, and binary trees encode decisions about
the partitioning of collections of data.

The most general mechanism for encoding relations between data is the
graph. Simple structures, like arrays, provide implicit connections, such as ad-
jacency, between stored values. Graphs are more demanding to construct but,
as a result, they can encode more detailed information. Indeed, the versatility of
graphs allows them to represent many of the most difﬁcult theoretical problems
of computer science.

This chapter investigates two traditional implementations of graphs, as well
as several standard algorithms for analyzing their structure. We ﬁrst agree on
some basic terminology.

16.1 Terminology
A graph G consists of a collection of vertices v ∈ VG and relations or edges
(u, v) ∈ EG between them (see Figure 16.1). An edge is incident to (or mentions)
each of its two component vertices. A graph is undirected if each of its edges
is considered a set of two unordered vertices, and directed if the mentioned
vertices are ordered (e.g., referred to as the source and destination). A graph S
is a subgraph of G if and only if VS ⊆ VG and ES ⊆ EG. Simple examples of
graphs include the list and the tree.

In an undirected graph, the number of edges (u, v) incident to a vertex u
is its degree. In a directed graph, the outgoing edges determine its out-degree
(or just degree) and incoming edges its in-degree. A source is a vertex with no
incoming edges, while a sink is a vertex with no outgoing edges.

404

Graphs

Some graphs. Each node a is adjacent to node b, but never to d. Graph
Figure 16.1
G has two components, one of which is S. The directed tree-shaped graph is a directed,
acyclic graph. Only the top left graph is complete.

Components are
always
connected.

Two edges (u, v) and (v, w) are said to be adjacent. A path is a sequence
of n distinct, adjacent edges (v0, v1), (v1, v2), . . . , (vn−1, vn).
In a simple path
the vertices are distinct, except for, perhaps, the end points v0 and vn. When
v0 = vn, the simple path is a cycle.

Two vertices u and v are connected (written u ; v) if and only if a simple
path of the graph mentions u and v as its end points. A subgraph S is a connected
component (or, often, just a component) if and only if S is a largest subgraph of
G such that for every pair of vertices u, v ∈ VS either u ; v or v ; u. A
connected component of a directed graph G is strongly connected if u ; v and
v ; u for all pairs of vertices u, v ∈ VS.

A graph containing no cycles is acyclic. A directed, acyclic graph (DAG) plays
an important role in solving many problems. A complete graph G contains an
edge (u, v) for all vertices u, v ∈ VG.

16.2 The Graph Interface

Vertices of a graph are usually labeled with application-speciﬁc information. As
a result, our implementations of a graph structure depend on the user specifying
unique labels for vertices. In addition, edges may be labeled, but not necessarily
uniquely. It is common, for example, to specify weights or lengths for edges. All
the graph implementations allow addition and removal of vertices and edges:

Graph

SabefdabcbacbadcedcG16.2 The Graph Interface

405

public interface Graph<V,E> extends Structure<V>
{

public void add(V label);
// pre: label is a non-null label for vertex
// post: a vertex with label is added to graph
//

if vertex with label is already in graph, no action

public void addEdge(V vtx1, V vtx2, E label);
// pre: vtx1 and vtx2 are labels of existing vertices
// post: an edge (possibly directed) is inserted between
//

vtx1 and vtx2.

public V remove(V label);
// pre: label is non-null vertex label
// post: vertex with "equals" label is removed, if found

public E removeEdge(V vLabel1, V vLabel2);
// pre: vLabel1 and vLabel2 are labels of existing vertices
// post: edge is removed, its label is returned

public V get(V label);
// post: returns actual label of indicated vertex

public Edge<V,E> getEdge(V label1, V label2);
// post: returns actual edge between vertices

public boolean contains(V label);
// post: returns true iff vertex with "equals" label exists

public boolean containsEdge(V vLabel1, V vLabel2);
// post: returns true iff edge with "equals" label exists

public boolean visit(V label);
// post: sets visited flag on vertex, returns previous value

public boolean visitEdge(Edge<V,E> e);
// pre: sets visited flag on edge; returns previous value

public boolean isVisited(V label);
// post: returns visited flag on labeled vertex

public boolean isVisitedEdge(Edge<V,E> e);
// post: returns visited flag on edge between vertices

public void reset();
// post: resets visited flags to false

public int size();
// post: returns the number of vertices in graph

406

Graphs

public int degree(V label);
// pre: label labels an existing vertex
// post: returns the number of vertices adjacent to vertex

public int edgeCount();
// post: returns the number of edges in graph

public Iterator<V> iterator();
// post: returns iterator across all vertices of graph

public Iterator<V> neighbors(V label);
// pre: label is label of vertex in graph
// post: returns iterator over vertices adj. to vertex
//

each edge beginning at label visited exactly once

public Iterator<Edge<V,E>> edges();
// post: returns iterator across edges of graph
//

iterator returns edges; each edge visited once

public void clear();
// post: removes all vertices from graph

public boolean isEmpty();
// post: returns true if graph contains no vertices

public boolean isDirected();
// post: returns true if edges of graph are directed

}

Because edges can be fully identiﬁed by their constituent vertices, edge opera-
tions sometimes require pairs of vertex labels. Since it is useful to implement
both directed and undirected graphs, we can determine the type of a speciﬁc
In undirected graphs, the addition of
graph using the isDirected method.
an edge effectively adds a directed edge in both directions. Many algorithms
keep track of their progress by visiting vertices and edges. This is so com-
mon that it seems useful to provide direct support for adding (visit), checking
(isVisited), and removing (reset) marks on vertices and edges.

Two iterators—generated by iterator and edges—traverse the vertices and
edges of a graph, respectively. A special iterator—generated by neighbors—
traverses the vertices adjacent to a given vertex. From this information, out-
bound edges can be determined.

Before we discuss particular implementations of graphs, we consider the
abstraction of vertices and edges. From the user’s point of view a vertex is a
label. Abstractly, an edge is an association of two vertices and an edge label. In
addition, we must keep track of objects that have been visited. These features
of vertices and edges are independent of the implementation of graphs; thus we
commit to an interface for these objects early. Let’s consider the Vertex class.

16.2 The Graph Interface

407

class Vertex<E>
{

public Vertex(E label)
// post: constructs unvisited vertex with label

public E label()
// post: returns user label associated w/vertex

Vertex

public boolean visit()
// post: returns, then marks vertex as being visited

public boolean isVisited()
// post: returns true iff vertex has been visited

public void reset()
// post: marks vertex unvisited

public boolean equals(Object o)
// post: returns true iff vertex labels are equal

}

This class is similar to an Association: the label portion of the Vertex cannot
be modiﬁed, but the visited ﬂag can be freely set and reset. Two Vertex objects
are considered equal if their labels are equal. It is a bare-bones interface. It
should also be noted that the Vertex is a nonpublic class. Since a Vertex is
not visible through the Graph interface, there is no reason for the user to have
access to the Vertex class.

Because the Edge class is visible “through” the Graph interface (you might

ask why—see Problem 16.8), the Edge class is declared public:

public class Edge<V,E>
{

public Edge(V vtx1, V vtx2, E label,

boolean directed)

// post: edge associates vtx1 and vtx2; labeled with label
//

directed if "directed" set true

Edge

public V here()
// post: returns first node in edge

public V there()
// post: returns second node in edge

public void setLabel(E label)
// post: sets label of this edge to label

public E label()
// post: returns label associated with this edge

public boolean visit()

408

Graphs

// post: visits edge, returns whether previously visited

public boolean isVisited()
// post: returns true iff edge has been visited

public boolean isDirected()
// post: returns true iff edge is directed

public void reset()
// post: resets edge’s visited flag to initial state

public boolean equals(Object o)
// post: returns true iff edges connect same vertices

}

As with the Vertex class, the Edge can be constructed, visited, and reset. Unlike
its Vertex counterparts, an Edge’s label may be changed. The methods here
and there provide access to labels of the vertices mentioned by the edge. These
method names are sufﬁciently ambiguous to be easily used with undirected
edges and convey a slight impression of direction for directed edges. Naming
of these methods is important because they are used by those who wish to get
vertex information while traversing a (potentially directed) graph.

16.3 Implementations

As “traditional”
as this science
gets, anyway!

Now that we have a good feeling for the graph interface, we consider traditional
implementations. Nearly every implementation of a graph has characteristics
of one of these two approaches. Our approach to specifying these implemen-
tations, however, will be dramatically impacted by the availability of object-
oriented features. We ﬁrst discuss the concept of a partially speciﬁed abstract
class in Java.

16.3.1 Abstract Classes Reemphasized

Normally, when a class is declared, code for each of the methods must be pro-
vided. Then, when an instance of the class is constructed, each of the methods
can be applied to the resulting object. As is common with our design approach,
however, it is useful to partially implement a class and later ﬁnish the imple-
mentation by extending the class in a particular direction. The partial base class
is abstract; it cannot be constructed because some of the methods are not com-
pletely deﬁned. The extension to the class inherits the methods that have been
deﬁned and speciﬁes any incomplete code to make the class concrete.

Again, we use abstract classes in our design of various graph implemen-
tations. Each implementation will be declared abstract, with the abstract
keyword:

GraphMatrix

16.3 Implementations

409

abstract public class GraphMatrix<V,E>

extends AbstractStructure<V> implements Graph<V,E>

Our approach will be to provide all the code that can be written without con-
sidering whether the graph is undirected or directed. When we must write code
that is dependent on the “directedness” of the graph, we delay it by writing
just an abstract header for the particular method. For example, we will need
to add edges to our graph, but the implementation depends on whether or not
the graph is directed. Looking ahead, here is what the declaration for addEdge
looks like in the abstract class GraphMatrix:

abstract public void addEdge(V v1, V v2, E label);
// pre: vtx1 and vtx2 are labels of existing vertices
// post: an edge (possibly directed) is inserted between
//

vtx1 and vtx2.

That’s it! It is simply a promise that code will eventually be written.

Once the abstract class is described as fully as possible, we extend it, commit-
ting the graph to being undirected or directed. The directed version of the Graph
implementation, called GraphMatrixDirected, speciﬁes the addEdge method as
follows:

public class GraphMatrixDirected<V,E> extends GraphMatrix<V,E>
{

public void addEdge(V vLabel1, V vLabel2, E label)
// pre: vLabel1 and vLabel2 are labels of existing vertices
// post: an edge is inserted between vLabel1 and vLabel2;
//
{

if edge is new, it is labeled with label (can be null)

GraphMatrixVertex<V> vtx1,vtx2;

GraphMatrix-
Directed

}

}

Because we declare the class GraphMatrixDirected to be an extension of the
GraphMatrix class, all the code written for the GraphMatrix class is inherited;
it is as though it had been written for the GraphMatrixDirected class. By
providing the missing pieces of code (tailored for directed graphs), the ex-
tension class becomes concrete. We can actually construct instances of the
GraphMatrixDirected class.

A related concept, subtyping, allows us to use any extension of a class wher-
ever the extended class could be used. We call the class that was extended
the base type or superclass, and the extension the subtype or subclass. Use of
subtyping allows us to write code like

GraphMatrix<String,String> g = new GraphMatrixDirected<String,String>();

g.add("Alice");
g.add("Bob");
g.addEdge("Alice","Bob","helps"); // "Alice helps Bob!"

410

Graphs

(a) An undirected graph and (b) its adjacency matrix representation. Each
Figure 16.2
nontrivial edge is represented twice across the diagonal—once in the gray and once in
the white—making the matrix symmetric.

Because GraphMatrixDirected is an extension of GraphMatrix, it is a Graph-
Matrix. Even though we cannot construct a GraphMatrix, we can correctly
manipulate concrete subtypes using the methods described in the abstract class.
In particular, a call to the method addEdge calls the method of GraphMatrixDi-
rected.

We now return to our normally scheduled implementations!

16.3.2 Adjacency Matrices

An n × n matrix of booleans is sufﬁcient to represent an arbitrary graph of rela-
tions among n vertices. We simply store true in the boolean at matrix location
[u][v] to represent the fact that there is an edge between u and v (see Fig-
ure 16.2), and false otherwise. Since entries [u][v] and [v][u] are independent,
the representation is sufﬁcient to describe directed graphs as well. Our conven-
tion is that the ﬁrst index (the row) speciﬁes the source and the second index
(the column) indicates the destination. To represent undirected graphs, we sim-
ply duplicate the entry [u][v] at entry [v][u]. This is called an adjacency matrix
representation of a graph. The abstract graphs of Figures 16.2a and 16.3a are
represented, respectively, by the matrices of Figures 16.2b and 16.3b.

One difﬁcult feature of our implementation is the arbitrary labeling of ver-
tices and edges. To facilitate this, we maintain a Dictionary that translates a
vertex label to a Vertex object. To help each vertex keep track of its associated
index we extend the Vertex class to include methods that manipulate an index
ﬁeld. Each index is a small integer that identiﬁes the dedicated row and column
that maintain adjacency information about each vertex. To help allocate the
indices, we keep a free list (see Section 9.2) of available indices.

One feature of our implementation has the potential to catch the unwary
programmer by surprise. Because we keep a Map of vertex labels, it is important
that the vertex label class implement the hashCode function in such a way as to

Beware: Edges
on the diagonal
appear exactly
once.

(a)(b)FTFFTFFFFFFFFFFFFTTTTTTTT01234012343124016.3 Implementations

411

Figure 16.3
edge appears exactly once in the matrix.

(a) A directed graph and (b) its adjacency matrix representation. Each

guarantee that if two labels are equal (using the equals method), they have the
same hashCode.

We can now consider the protected data and constructors for the GraphMat-

rix class:

protected int size;
protected Object data[][];
protected Map<V,GraphMatrixVertex<V>> dict;
protected List<Integer> freeList;
protected boolean directed; // graph is directed

// allocation size for graph

// matrix - array of arrays

// labels -> vertices
// available indices in matrix

GraphMatrix

protected GraphMatrix(int size, boolean dir)
{

// fix direction of edges

this.size = size; // set maximum size
directed = dir;
// the following constructs a size x size matrix
data = new Object[size][size];
// label to index translation table
dict = new Hashtable<V,GraphMatrixVertex<V>>(size);
// put all indices in the free list
freeList = new SinglyLinkedList<Integer>();
for (int row = size-1; row >= 0; row--)
freeList.add(new Integer(row));

}

To construct the graph, the user speciﬁes an upper bound on the number of
vertices. We allocate size arrays of length size—a two-dimensional array. By
default, the array elements are null, so initially there are no edges. We then
put each of the indices into the list of available vertex indices.

This constructor is declared protected.

It takes a second parameter, di-
rected, that identiﬁes whether or not the graph constructed is to act like a

(a)(b)03214FFFFFFFFFFFT0123401234TTFTFTFFFFFTDestinationSourceT412

Graphs

GraphMatrix-
Directed

directed graph. When we extend the graph to implement either directed or
undirected graphs, we write a public constructor to call the abstract protected
class’s constructor with an appropriate boolean value:

public GraphMatrixDirected(int size)
// pre: size > 0
// post: constructs an empty graph that may be expanded to
//
//
{

at most size vertices.
and undirected otherwise

Graph is directed if dir true

super(size,true);

}

As we discussed before, this technique allows the implementor to selectively
inherit the code that is common between directed and undirected graphs. Since
we hide the implementation, we are free to reimplement either type of graph
without telling our users, perhaps allowing us to optimize our code.

Returning to the GraphMatrix class, the add method adds a labeled vertex.
If the vertex already exists, the operation does nothing. If it is new to the graph,
an index is allocated from the free list, a new Vertex object is constructed, and
the label-vertex association is recorded in the Map. The newly added vertex
mentions no edges, initially.

GraphMatrix

public void add(V label)
// pre: label is a non-null label for vertex
// post: a vertex with label is added to graph;
//
{

if vertex with label is already in graph, no action

// if there already, do nothing
if (dict.containsKey(label)) return;

Assert.pre(!freeList.isEmpty(), "Matrix not full");
// allocate a free row and column
int row = freeList.removeFirst().intValue();
// add vertex to dictionary
dict.put(label, new GraphMatrixVertex<V>(label, row));

}

Removing a vertex reverses the add process. We must, however, be sure to
set each element of the vertex’s matrix row and column to null, removing any
mentioned edges (we may wish to add a new, isolated vertex with this index in
the future). When we remove the vertex from the Map, we “recycle” its index
by adding it to the list of free indices. As with all of our remove methods, we
return the previous value of the label. (Even though the labels match using
equals, they may not be precisely the same; once returned the user can extract
any unknown information from the previous label before the value is collected
as garbage.)

public V remove(V label)

16.3 Implementations

413

// pre: label is non-null vertex label
// post: vertex with "equals" label is removed, if found
{

// find and extract vertex
GraphMatrixVertex<V> vert;
vert = dict.remove(label);
if (vert == null) return null;
// remove vertex from matrix
int index = vert.index();
// clear row and column entries
for (int row=0; row<size; row++) {

data[row][index] = null;
data[index][row] = null;

}
// add node index to free list
freeList.add(new Integer(index));
return vert.label();

}

Within the graph we store references to Edge objects. Each Edge records all
of the information necessary to position it within the graph, including whether it
is directed or not. This allows the equals method to work on undirected edges,
even if the vertices were provided in the opposite order (see Problem 16.12).
To add an edge to the graph, we require two vertex labels and an edge label.
The vertex labels uniquely identify the vertices within the graph, and the edge
label is used to form the value inserted within the matrix at the appropriate row
and column. To add the edge, we construct a new Edge with the appropriate
information. This object is written to appropriate matrix entries: undirected
graphs update one or two locations; directed graphs update just one. Here is
the addEdge method for undirected graphs:

public void addEdge(V vLabel1, V vLabel2, E label)
// pre: vLabel1 and vLabel2 are labels of existing vertices, v1 & v2
// post: an edge (undirected) is inserted between v1 and v2;
//
{

if edge is new, it is labeled with label (can be null)

GraphMatrixVertex<V> vtx1,vtx2;
// get vertices
vtx1 = dict.get(vLabel1);
vtx2 = dict.get(vLabel2);
// update matrix with new edge
Edge<V,E> e = new Edge<V,E>(vtx1.label(),vtx2.label(),label,false);
data[vtx1.index()][vtx2.index()] = e;
data[vtx2.index()][vtx1.index()] = e;

}

GraphMatrix-
Undirected

Here is a similar method for directed graphs:

public void addEdge(V vLabel1, V vLabel2, E label)

GraphMatrix-
Directed

414

Graphs

// pre: vLabel1 and vLabel2 are labels of existing vertices
// post: an edge is inserted between vLabel1 and vLabel2;
//
{

if edge is new, it is labeled with label (can be null)

GraphMatrixVertex<V> vtx1,vtx2;
// get vertices
vtx1 = dict.get(vLabel1);
vtx2 = dict.get(vLabel2);
// update matrix with new edge
Edge<V,E> e = new Edge<V,E>(vtx1.label(),vtx2.label(),label,true);
data[vtx1.index()][vtx2.index()] = e;

}

The differences are quite minor, but the two different subtypes allow us to write
specialized code without performing explicit run-time tests.1

The removeEdge method removes and returns the label associated with the
Edge found between two vertices. Here is the undirected version (the directed
version is similar):

public E removeEdge(V vLabel1, V vLabel2)
// pre: vLabel1 and vLabel2 are labels of existing vertices
// post: edge is removed, its label is returned
{

// get indices
int row = dict.get(vLabel1).index();
int col = dict.get(vLabel2).index();
// cache old value
Edge<V,E> e = (Edge<V,E>)data[row][col];
// update matrix
data[row][col] = null;
data[col][row] = null;
if (e == null) return null;
else return e.label();

}

GraphMatrix-
Undirected

The get, getEdge, contains, and containsEdge methods return informa-
tion about the graph in an obvious way. Modifying the objects returned by these
methods can be dangerous: they have the potential of invalidating the state of
the underlying graph implementation.

Each of the visit-type methods passes on requests to the underlying ob-
ject. For example, the visit method simply refers the request to the associated
Vertex:

public boolean visit(V label)
// post: sets visited flag on vertex, returns previous value

GraphMatrix

1 This is somewhat misleading, as the obvious run-time tests are replaced by less obvious decreases
in performance due to subtyping. Still, the logical complexity of the code can be dramatically
reduced using these techniques.

16.3 Implementations

415

{

}

Vertex<V> vert = dict.get(label);
return vert.visit();

The process of resetting the visitation marks on a graph traverses each of the
vertices and edges, resetting them along the way.

We now consider the implementation of each of the three iterators. The
ﬁrst, generated by iterator, traverses the vertices. The values returned by the
Iterator are vertex labels. This Iterator is easily constructed by returning
the value of the Map’s keys function!

But I reiterate
myself.

public Iterator<V> iterator()
// post: returns traversal across all vertices of graph
{

return dict.keySet().iterator();

}

The neighbors iterator, which traverses the edges adjacent to a single ver-
tex, considers only the outgoing edges. We simply look up the index associated
with the vertex label and scan across the row, building up a list of vertex labels
that are adjacent using each of the edges. By putting these values in a list, we
can return a ListIterator that will give us iterative access to each of the adja-
cent vertex labels. With this information we may retrieve the respective edges
with getEdge if necessary.

public Iterator<V> neighbors(V label)
// pre: label is label of vertex in graph
// post: returns traversal over vertices adj. to vertex
//
{

each edge beginning at label visited exactly once

GraphMatrixVertex<V> vert;
vert = dict.get(label);
List<V> list = new SinglyLinkedList<V>();
for (int row=size-1; row>=0; row--)
{

Edge<V,E> e = (Edge<V,E>)data[vert.index()][row];
if (e != null) {

if (e.here().equals(vert.label()))

list.add(e.there());

else list.add(e.here());

}

}
return list.iterator();

}

All that remains is to construct an iterator over the edges of the graph. Again,
we construct a list of the edges and return the result of the iterator method
invoked on the list. For directed edges, we include every edge; for undirected

416

Graphs

edges we include only the edges found in, say, the lower half of the array (in-
cluding the diagonal). Here is the version for the undirected graph:

public Iterator<Edge<V,E>> edges()
// post: returns traversal across all edges of graph (returns Edges)
{

GraphMatrix-
Undirected

List<Edge<V,E>> list = new SinglyLinkedList<Edge<V,E>>();
for (int row=size-1; row>=0; row--)

for (int col=size-1; col >= row; col--) {

Edge<V,E> e = (Edge<V,E>)data[row][col];
if (e != null) list.add(e);

}

return list.iterator();

}

The great advantage of the adjacency matrix representation is its simplicity.
The access to a particular edge in a graph of size n can be accomplished in con-
stant time. Other operations, like remove, appear to be more complex, taking
O(n) time. The disadvantage is that the implementation may vastly overesti-
mate the storage required for edges. While we have room for storing O(n2)
directed edges, some graphs may only need to make use of O(n) edges. Graphs
with superlinear numbers of edges are called dense; all other graphs are sparse.
When graphs are sparse, most of the elements of the adjacency matrix are not
used, leading to a signiﬁcant waste of space. Our next implementation is par-
ticularly suited for representing sparse graphs.

16.3.3 Adjacency Lists

Recalling the many positive features of a linked list over a ﬁxed-size array, we
now consider the use of an adjacency list. As with the adjacency matrix repre-
sentation, we maintain a Map for identifying the relationship between a vertex
label and the associated Vertex object. Within the vertex, however, we store a
collection (usually a linked list) of edges that mention this vertex. Figures 16.4
and 16.5 demonstrate the adjacency list representations of undirected and di-
rected graphs. The great advantage of using a collection is that it stores only
edges that appear as part of the graph.

As with the adjacency matrix implementation, we construct a privately used
extension to the Vertex class.
In this extension we reference a collection of
edges that are incident to this vertex. In directed graphs, we collect edges that
mention the associated vertex as the source. In undirected graphs any edge inci-
dent to the vertex is collected. Because the edges are stored within the vertices,
most of the actual implementation of graphs appears within the implementation
of the extended vertex class. We see most of the implementation here:

class GraphListVertex<V,E> extends Vertex<V>
{

GraphListVertex

protected Structure<Edge<V,E>> adjacencies; // adjacent edges
public GraphListVertex(V key)

16.3 Implementations

417

// post: constructs a new vertex, not incident to any edge
{

super(key); // init Vertex fields
adjacencies = new SinglyLinkedList<Edge<V,E>>(); // new list

}

public void addEdge(Edge<V,E> e)
// pre: e is an edge that mentions this vertex
// post: adds edge to this vertex’s adjacency list
{

if (!containsEdge(e)) adjacencies.add(e);

}

public boolean containsEdge(Edge<V,E> e)
// post: returns true if e appears on adjacency list
{

return adjacencies.contains(e);

}

public Edge<V,E> removeEdge(Edge<V,E> e)
// post: removes and returns adjacent edge "equal" to e
{

return adjacencies.remove(e);

}

public Edge<V,E> getEdge(Edge<V,E> e)
// post: returns the edge that "equals" e, or null
{

Iterator<Edge<V,E>> edges = adjacencies.iterator();
while (edges.hasNext())
{

Edge<V,E> adjE = edges.next();
if (e.equals(adjE)) return adjE;

}
return null;

}

public int degree()
// post: returns the degree of this node
{

return adjacencies.size();

}

public Iterator<V> adjacentVertices()
// post: returns iterator over adj. vertices
{

return new GraphListAIterator<V,E>(adjacentEdges(), label());

}

public Iterator<Edge<V,E>> adjacentEdges()

418

Graphs

Figure 16.4
edge is represented twice in the structure. (Compare with Figure 16.2.)

(a) An undirected graph and (b) its adjacency list representation. Each

Figure 16.5
appears once in the source list. (Compare with Figure 16.3.)

(a) A directed graph and (b) its adjacency list representation. Each edge

// post: returns iterator over adj. edges
{

return adjacencies.iterator();

}

}

The constructor initializes its Vertex ﬁelds, and then constructs an empty adja-
cency list. Elements of this list will be Edge objects. Most of the other methods
have obvious behavior.

The only difﬁcult method is getEdge. This method returns an edge from
the adjacency list that logically equals (i.e., is determined to be equal through
a call to Edge’s equals method) the edge provided. In an undirected graph the
order of the vertex labels may not correspond to the order found in edges in the
edge list. As a result, getEdge returns a canonical edge that represents the edge
speciﬁed as the parameter. This ensures that there are not multiple instances of
edges that keep track of shared information.

We are now ready to implement most of the methods required by the Graph

(a)(b)30124012344342423020(a)(b)03214023014301234Source16.3 Implementations

419

interface. First, we consider the protected GraphList constructor:

protected Map<V,GraphListVertex<V,E>> dict; // label -> vertex
protected boolean directed; // is graph directed?

protected GraphList(boolean dir)
{

dict = new Hashtable<V,GraphListVertex<V,E>>();
directed = dir;

}

Our approach to extending the abstract GraphList type to support directed and
undirected graphs is similar to that described in the adjacency matrix imple-
mentation. With the list-based implementation, though, we need not provide
an upper bound on the number of vertices that will appear in the graph. This is
because the underlying structures automatically extend themselves, if necessary.
The process of adding and removing a vertex involves simple manipulations
of the Map. Here, for example, is the code for adding a new vertex to the graph:

public void add(V label)
// pre: label is a non-null label for vertex
// post: a vertex with label is added to graph;
//
{

if vertex with label is already in graph, no action

if (dict.containsKey(label)) return; // vertex exists
GraphListVertex<V,E> v = new GraphListVertex<V,E>(label);
dict.put(label,v);

}

To add an edge to the graph we insert a reference to the Edge object in
the appropriate adjacency lists. For a directed graph, we insert the edge in
the list associated with the source vertex. For an undirected graph, a reference
to the edge must be inserted into both lists. It is important, of course, that a
reference to a single edge be inserted in both lists so that changes to the edge
are maintained consistently. Here, we show the undirected version:

public void addEdge(V vLabel1, V vLabel2, E label)
// pre: vLabel1 and vLabel2 are labels of existing vertices, v1 & v2
// post: an edge (undirected) is inserted between v1 and v2;
//
{

if edge is new, it is labeled with label (can be null)

GraphListVertex<V,E> v1 = dict.get(vLabel1);
GraphListVertex<V,E> v2 = dict.get(vLabel2);
Edge<V,E> e = new Edge<V,E>(v1.label(), v2.label(), label, false);
v1.addEdge(e);
v2.addEdge(e);

}

GraphList-
Undirected

Removing an edge simply reverses this process:

420

Graphs

public E removeEdge(V vLabel1, V vLabel2)
// pre: vLabel1 and vLabel2 are labels of existing vertices
// post: edge is removed, its label is returned
{

GraphListVertex<V,E> v1 = dict.get(vLabel1);
GraphListVertex<V,E> v2 = dict.get(vLabel2);
Edge<V,E> e = new Edge<V,E>(v1.label(), v2.label(), null, false);
v2.removeEdge(e);
e = v1.removeEdge(e);
if (e == null) return null;
else return e.label();

}

Steve doesn’t
like this
I agree.

Notice that to remove an edge a “pattern” edge must be constructed to identify
(through equals) the target of the remove.

Now that we can remove edges, we can remove a vertex. Since the removal
of a vertex should remove incident edges, it is important that each of the ad-
jacency lists be checked. Our approach is to iterate across each of the vertices
and remove any edge that mentions that vertex. This requires some care. Here
is the directed version:

public V remove(V label)
// pre: label is non-null vertex label
// post: vertex with "equals" label is removed, if found
{

GraphListVertex<V,E> v = dict.get(label);

GraphList-
Directed

Iterator<V> vi = iterator();
while (vi.hasNext())
{

V v2 = vi.next();
if (!label.equals(v2)) removeEdge(v2,label);

}
dict.remove(label);
return v.label();

}

The complexity of this method counterbalances the simplicity of adding a vertex
to the graph.

Many of the remaining edge and vertex methods have been greatly simpli-
ﬁed by our having extended the Vertex class. Here, for example, is the degree
method:

public int degree(V label)
// pre: label labels an existing vertex
// post: returns the number of vertices adjacent to vertex
{

Assert.condition(dict.containsKey(label), "Vertex exists.");
return dict.get(label).degree();

}

16.3 Implementations

421

This code calls the GraphListVertex degree method. That, in turn, calls the
size method of the underlying collection, a SinglyLinkedList. Most of the
remaining methods are simply implemented.

At this point, it is useful to discuss the implementation of iterators for the
adjacency list representation. Like the adjacency matrix implementation, the
iterator method simply returns the result of the keys iterator on the underly-
ing Map. Each of the values returned by the iterator is a vertex label, which is
exactly what we desire.

The neighbors iterator should return an iterator over the neighbors of the
provided vertex. Since each vertex maintains a Collection of edges, the iterator
method of the collection returns Edge values. Our approach is similar to the ap-
proach we used in constructing the iterators for Maps: we construct a private,
special-purpose iterator that drives the Collection iterator as a slave. The pro-
cess of extracting the “other” vertex from each edge encountered is made com-
plex by the fact that “this” vertex can appear as either the source or destination
vertex when the graph is undirected.

The Edge’s iterator has similar complexities. The easiest approach is to con-
struct a list of edges by traversing each of the edge lists found in each of the
vertices. The result is an iterator over the resulting list. Here is the code for the
constructor of our private GraphListEIterator class:

protected AbstractIterator<Edge<V,E>> edges;

public GraphListEIterator(Map<V,GraphListVertex<V,E>> dict)
// post: constructs a new iterator across edges of
//
{

vertices within dictionary

List<Edge<V,E>> l = new DoublyLinkedList<Edge<V,E>>();
Iterator<GraphListVertex<V,E>> dictIterator = dict.values().iterator();
while (dictIterator.hasNext())
{

GraphListVertex<V,E> vtx =

(GraphListVertex<V,E>)dictIterator.next();

Iterator<Edge<V,E>> vtxIterator = vtx.adjacentEdges();
while (vtxIterator.hasNext())
{

Edge<V,E> e = vtxIterator.next();
if (vtx.label().equals(e.here())) l.addLast(e);

}

}
edges = (AbstractIterator<Edge<V,E>>)l.iterator();

}

Each of the edges is traversed in the construction of the iterator, so there is
considerable overhead just during initialization. Once constructed, however, the
traversal is quick. An alternative implementation would distribute the cost over
each step of the traversal. Construction of the iterator would be less expensive,
but each step of the traversal would be slightly slower. In the end, both methods

422

Graphs

consume similar amounts of time. If, however, partial traversals of the edge lists
are expected, the alternative implementation has its merits.

With two implementations of graphs in mind, we now focus on a number of

examples of their use.

16.4 Examples: Common Graph Algorithms

Because the graph structure is so ﬂexible there are many good examples of
graph applications. In this section, we investigate a number of beautiful algo-
rithms involving graphs. These algorithms provide a cursory overview of the
problems that may be cast as graph problems, as well as techniques that are
commonly used to solve them.

16.4.1 Reachability

Once data are stored within a graph, it is often desirable to identify vertices
that are reachable from a common source (see Figure 16.6). One approach is
to treat the graph as you would a maze and, using search techniques, ﬁnd the
reachable vertices. For example, we may use depth-ﬁrst search: each time we
visit an unvisited vertex we seek to further deepen the traversal.

The following code demonstrates how we might use recursion to search for

unvisited vertices:

Reachability

static void reachableFrom(Graph<V,E> g, V vertexLabel)
// pre: g is a non-null graph, vertexLabel labels a vertex of g
// post: unvisited vertices reachable from vertex are visited
{

g.visit(vertexLabel);

// visit this vertex

// recursively visit unvisited neighbor vertices
Iterator<V> ni = g.neighbors(vertexLabel);
while (ni.hasNext())
{

V neighbor = ni.next(); // adjacent node label
if (!g.isVisited(neighbor))
{

reachableFrom(g,neighbor); // depth-first search

}

}

}

We clear each Vertex’s visited ﬂag with a call to reset, and then call reach-
ableFrom with the graph and the source vertex for the reachability test. Before
the call to reachableFrom, the vertex labeled with the vertexLabel has not
been visited. After the call, every vertex reachable from the vertex has been
visited. Some vertices may be left unvisited and are not reachable from the

16.4 Examples: Common Graph Algorithms

423

Figure 16.6 Courses you might be expected to have taken if you’re in a compiler de-
sign class. (a) A typical prerequisite graph (classes point to prerequisites). Note the
central nature of data structures! (b) Bold courses can be reached as requisite courses
for compiler design.

(a)TheoryCompiler designData structuresOperating systemsLinear algebraParallel systemsDiscrete math(b)GraphicsNetworksA.I.SurfingModelingVisionLanguagesJavaOrganizationTheoryAlgorithmsOrganizationJavaGraphicsLanguagesAlgorithmsNetworksA.I.SurfingModelingVisionLinear algebraDiscrete mathParallel systemsOperating systemsCompiler designData structures424

Graphs

source. So, to determine whether you may reach one vertex from another, the
following code can be used:

g.reset();
reachableFrom(g,sourceLabel);
canGetThere = g.isVisited(destinationLabel);

In Section 10.3 we discussed the use of a Linear structure to maintain the state
of a search of a maze. The use of a Stack led to a depth-ﬁrst search. Here,
however, no Stack appears! The reason is that the act of calling the procedure
recursively maintains an implied stack of local variables.

How long does it take to execute the procedure? Suppose that, ultimately,
we visit the reachable vertices Vr. Let Er be the edges of the graph found among
the vertices of Vr. Clearly, each vertex of Vr is visited, so there is one call to
reachableFrom from each vertex v ∈ Vr. For each call, we ask each destination
vertex if it has been visited or not. There is one such test for every edge within
Er. Thus, the total time is O(|Vr| + |Er|). Since |Er| ≥ |Vr − 1| (every new
vertex is visited by traversing a new edge), the algorithm is dominated by the
number of edges actually investigated. Of course, if the graph is dense, this is
bounded above by the square of the number of vertices.

In an undirected graph the reachable vertices form a component of the
graph. To count the components of a graph (the undirected version of the
graph of Figure 16.6 has three components), we iterate across the vertices of
the graph, calling the reachableFrom procedure on any vertex that has not yet
been visited. Since each unvisited vertex is not reachable from those that have
been encountered before, the number of searches determines the number of
components.

16.4.2 Topological Sorting

Occasionally it is useful to list the vertices of a graph in such a way as to make
the edges point in one direction, for example, toward the front of the list. Such
graphs have to be directed and acyclic (see Problem 16.13). A listing of vertices
with this property is called a topological sort.

One technique for developing a topological sort involves keeping track of a
counter or virtual timer. The timer is incremented every time it is read. We now
visit each of the nodes using a depth-ﬁrst search, labeling each node with two
time stamps. These time stamps determine the span of time that the algorithm
spends processing the descendants of a node. When a node is ﬁrst encountered
during the search, we record the start time. When the recursive depth-ﬁrst
search returns from processing a node, the timer is again read and the ﬁnish
time is recorded. Figure 16.7 depicts the intervals associated with each vertex
of the graph of Figure 16.6. (As arbitrary convention, we assume that a vertex
iterator would encounter nodes in the diagram in “reading” order.)

One need only observe that the ﬁnish time of a node is greater than the
ﬁnish time of any node it can reach. (This depth-ﬁrst search may have to be

16.4 Examples: Common Graph Algorithms

425

The progress of a topological sort of the course graph. The time inter-
Figure 16.7
val following a node label indicates the time interval spent processing that node or its
descendants. Dark nodes reachable from compiler design are all processed during the
interval [1–16]—the interval associated with compiler design.

Parallel systems: 33−34Operating systems: 17−18Data structures: 3−6Linear algebra: 26−27Discrete math: 10−11A.I.: 21−22Modeling: 23−24Vision: 19−20Organization: 14−15Java: 4−5Algorithms: 9−12Surfing: 31−32Networks: 29−30Theory: 8−13Graphics: 25−28Languages: 2−7Compiler design: 1−16426

Graphs

started at several nodes if there are several independent components, or if the
graph is not strongly connected.) The algorithm, then, simply lists the vertices
in the order in which they are ﬁnished. For our course graph we generate one
of many course schedules that allow students to take courses without violating
course requirements:

Vertices Ordered by Finish Time

Java

5.
6. Data structures
Languages
7.

11. Discrete math
12. Algorithms
Theory
13.

15. Organization
16. Compiler design
18. Operating systems
20. Vision
22. A.I.
24. Modeling

Linear algebra

27.
28. Graphics
30. Networks
32.
34.

Surﬁng
Parallel systems

TopoSort

Actually, the time stamps are useful only for purposes of illustration. In fact,
we can simply append vertices to the end of a list at the time that they would
normally be ﬁnished. Here is a sample code:

public static List<V> topoSort(Graph<V,E> g)
// pre: g is non-null
// post: returns list of all vertices of g, topologically ordered
{

// construct result list
List<V> l = new DoublyLinkedList<V>();
Iterator<V> vi = g.elements();
while (vi.hasNext())
{

V v = vi.next();
// perform depth-first search on unvisited vertices
if (!g.isVisited(v))
{

DFS(g,v,l);

}

}
// result is queue of vertex labels
return l;

}

static protected void DFS(Graph<V,E> g, V n, List<V> l)
// post: performs depth-first search enqueuing
//
{

unvisited descendants of node n into l

g.visit(n); // mark node visited
Iterator<V> ei = g.neighbors(n); // get neighbors
while (ei.hasNext())
{

V neighbor = ei.next();
// potentially deepen search if neighbor not visited
if (!g.isVisited(neighbor)) {

16.4 Examples: Common Graph Algorithms

427

DFS(g,neighbor,l);

}

}
l.addLast(n); // add this value once decendants added

}

These functions are declared as static procedures of a program that might
make use of a topological sort. Alternatively, they could be written as methods
of a graph, reducing the complexity of method calls.

16.4.3 Transitive Closure

Previously we discussed a reachability algorithm that determines if it is possi-
ble to reach any particular vertex from a particular source. It is also useful to
compute the transitive closure of a graph: for each pair of vertices u, v ∈ V , is
v reachable from u? These questions can be answered by O(|V |) calls to the
depth-ﬁrst search algorithm (leading to an algorithm that is O(|V |(|V | + |E|))),
or we can look for a more direct algorithm that has similar behavior.

One algorithm, Warshall’s algorithm, computes reachability for each pair of
vertices by modifying the graph. When the algorithm is applied to a graph,
edges are added until there is an edge for every pair of connected vertices (u, v).
The concept behind Warshall’s algorithm is relatively simple. Two connected
vertices u and v are either directly connected, or the path from u to v passes
through an intermediate node w. The algorithm simply considers each node
and connects all pairs of nodes u and v that can be shown to use w as an
intermediate node. Here is a Java implementation:

static void warshall(Graph<V,E> g)
// pre: g is non-null
// post: g contains edge (a,b) if there is a path from a to b
{

Iterator<V> witer = g.iterator();

Warshall

while (witer.hasNext())
{

Iterator<V> uiter = g.iterator();
V w = witer.next();
while (uiter.hasNext())
{

Iterator<V> viter = g.iterator();
V u = uiter.next();
while (viter.hasNext())
{

V v = viter.next();
// check for edge from u to v via w
if (g.containsEdge(u, w) &&

g.containsEdge(w, v))

{

428

Graphs

g.addEdge(u, v, null);

}

}

}

}

}

This algorithm is clearly O(|V |3): each iterator visits |V | vertices and (for adja-
cency matrices) the check for existence of an edge can be performed in constant
time.

To see how the algorithm works, we number the vertices in the order they
are encountered by the vertex iterator. After k iterations of the outer loop,
all “reachability edges” of the subgraph containing just the ﬁrst k vertices are
completely determined. The next iteration extends this result to a subgraph of
k + 1 vertices. An inductive approach to proving this algorithm correct (which
we avoid) certainly has merit.

16.4.4 All Pairs Minimum Distance

A slight modiﬁcation of Warshall’s algorithm gives us a method for computing
the minimum distance between all pairs of points. The method is due to Floyd.
Again, we use three loops to compute the new edges representing reachability,
but these edges are now labeled, or weighted, with integer distances that indi-
cate the current minimum distance between each pair of nodes. As we consider
intermediate nodes, we merge minimum distance approximations by computing
and updating the distance if the sum of path lengths through an intermediate
node w is less than our previous approximation. Object orientation makes this
code somewhat cumbersome:

static void floyd(Graph<V,E> g)
// post: g contains edge (a,b) if there is a path from a to b
{

Floyd

Iterator<V> witer = g.iterator();

while (witer.hasNext())
{

Iterator<V> uiter = g.iterator();
V w = witer.next();
while (uiter.hasNext())
{

Iterator<V> viter = g.iterator();
V u = uiter.next();
while (viter.hasNext())
{

V v = viter.next();
if (g.containsEdge(u,w) && g.containsEdge(w,v))
{

Edge<V,E> leg1 = g.getEdge(u,w);

16.4 Examples: Common Graph Algorithms

429

Edge<V,E> leg2 = g.getEdge(w,v);
int leg1Dist = leg1.label();
int leg2Dist = leg2.label();
int newDist = leg1Dist+leg2Dist;

if (g.containsEdge(u,v))
{

Edge<V,E> across = g.getEdge(u,v);
int acrossDist = across.label();
if (newDist < acrossDist)

across.setLabel(newDist);

} else {

g.addEdge(u,v,newDist);

}

}

}

}

}

}

Clearly, edge labels could contain more information than just the path length.
For example, the path itself could be constructed, stored, and produced on re-
quest, if necessary. Again, the complexity of the algorithm is O(|V |3). This is
satisfactory for dense graphs, especially if they’re stored in adjacency matrices,
but for sparse graphs the checking of all possible edges seems excessive. Indeed,
other approaches can improve these bounds. We leave some of these for your
next course in algorithms!

16.4.5 Greedy Algorithms

We now consider two examples of greedy algorithms—algorithms that compute
optimal solutions to problems by acting in the optimal or “most greedy” manner
at each stage in the algorithm. Because both algorithms seek to ﬁnd the best
choice for the next step in the solution process, both make use of a priority
queue.

Minimum Spanning Tree

The solution to many network problems involves identifying a minimum span-
ning tree of a graph. A minimum spanning tree of an edge-weighted graph is
a tree that connects every vertex of a component whose edges have minimum
total edge weight. Such a tree might represent the most inexpensive way to con-
nect several cities with telephone trunk lines. For this reason, we will interpret
the weights as edge lengths. For the purposes of our discussion, we will assume
that the graph under consideration is composed of a single component. (If the
graph contains multiple components, we can compute a minimum spanning
forest with multiple applications of the minimum spanning tree algorithm.)

430

Graphs

It is useful to note that if the vertices of a connected graph are partitioned
into any two sets, any minimum spanning tree contains a shortest edge that
connects nodes between the two sets. We will make use of this fact by segre-
gating visited nodes from unvisited nodes. The tree (which spans the visited
nodes and does so minimally) is grown by iteratively incorporating a shortest
edge that incorporates an unvisited node to the tree. The process stops when
|V | − 1 edges have been added to the tree.

MCST

static public void mcst(Graph<String,Integer> g)
// pre: g is a graph
// post: edges of minimum spanning tree of a component are visited
{

// keep edges ranked by length
PriorityQueue<ComparableEdge<String,Integer>> q =

new SkewHeap<ComparableEdge<String,Integer>>();

// looking for a nearby vertex
// clear visited flags

// current vertex

String v = null;
Edge<String,Integer> e; // current edge
boolean searching;
g.reset();
// select a node from the graph, if any
Iterator<String> vi = g.iterator();
if (!vi.hasNext()) return;
v = vi.next();
do
{

// visit the vertex and add all outgoing edges
g.visit(v);
Iterator<String> ai = g.neighbors(v);
while (ai.hasNext()) {

// turn it into outgoing edge
e = g.getEdge(v,ai.next());
// add the edge to the queue
q.add(new ComparableEdge<String,Integer>(e));

}
searching = true;
while (searching && !q.isEmpty())
{

// grab next shortest edge on tree fringe
e = q.remove();
// does this edge take us somewhere new?
v = e.there();
if (g.isVisited(v)) v = e.here();
if (!g.isVisited(v)) {
searching = false;
g.visitEdge(g.getEdge(e.here(),e.there()));

}

}

} while (!searching);

}

16.4 Examples: Common Graph Algorithms

431

The progress of a minimum spanning tree computation. Bold vertices
Figure 16.8
and edges are part of the tree. Harrisburg and Trenton, made adjacent by the graph’s
shortest edge, are visited ﬁrst. At each stage, a shortest external edge adjacent to the
tree is incorporated.

First, we use a priority queue to rank edges based on length. As we re-
move the edges from the queue, the smallest edges are considered ﬁrst (see
Figure 16.8). When an edge is considered that includes an unvisited vertex,
we visit it, logically adding it to the minimum spanning tree. We then add any
edges that are outward-bound from the newly visited node. At any time, the
priority queue contains only edges that mention at least one node of the tree.
If, of course, an edge is considered that mentions two previously visited nodes,
the edge is superﬂuous, as the nodes are already connected by a path in the tree
(albeit a potentially long one). When the priority queue “runs dry,” the tree is
fully computed. The result of the algorithm will be visited marks on all nodes
and edges that participate in the tree. (If the graph has multiple components,
some vertices will not have been visited.)

Our implementation begins by ﬁnding a source vertex (v) to “prime” the
greedy algorithm. The main loop of the algorithm then runs until no new ver-
tices can be added to the tree. Each new vertex is marked as visited and its
outbound edges2 are then added to the priority queue (q) of those to be con-
sidered. Short edges are removed from the queue until an unvisited vertex is
mentioned by a new tree edge (e), or the queue is emptied.

Over the course of the algorithm, consideration of each edge and vertex
results in a priority queue operation. The running time, then is O((|V | +

2 We use ComparableEdges here, an extension to an edge that assumes that the labels implement
Comparable.

AlbanyTrentonMontpelierDoverBoston1501304200120250100Athens47001500Salt Lake CityHarrisburgPhoenixSacramentoBangkok450550650Kuala LumpurConnecting the worldAll distances are approximate nautical miles.432

Graphs

|E|) log (|V |)).

Notice that the ﬁrst edge added may not be the graph’s shortest.

Single-Source Shortest Paths

The minimum spanning tree algorithm is related to a fast, single-source, shortest-
path algorithm attributed to Dijkstra. In this algorithm, we desire the minimum-
length paths from a single source to all other nodes. We expect the algorithm,
of course, to run considerably faster than the all-pairs version. This algorithm
also runs in time proportional to O((|V | + |E|) log (|V |)) due to the fact that it
uses much the same control as the minimum spanning tree. Here is the code:

public static

Map<String,ComparableAssociation<Integer,Edge<String,Integer>>>
dijkstra(Graph<String,Integer> g, String start)

Dijkstra

// pre: g is a graph; start is source vertex
// post: returns a dictionary of vertex-based results
//
{

value is association (total-distance,prior-edge)

// keep a priority queue of distances from source
PriorityQueue<ComparableAssociation<Integer,Edge<String,Integer>>>

q = new SkewHeap<ComparableAssociation<Integer,

Edge<String,Integer>>>();

// results, sorted by vertex
Map<String,ComparableAssociation<Integer,Edge<String,Integer>>>

result = new Table<String,

ComparableAssociation<Integer,

Edge<String,Integer>>>();

String v = start;
// result is a (total-distance,previous-edge) pair
ComparableAssociation<Integer,Edge<String,Integer>> possible =

// last vertex added

new ComparableAssociation<Integer,Edge<String,Integer>>(0,null);

// as long as we add a new vertex...
while (v != null)
{

if (!result.containsKey(v))
{

// visit node v - record incoming edge
result.put(v,possible);
// vDist is shortest distance to v
int vDist = possible.getKey();

// compute and consider distance to each neighbor
Iterator<String> ai = g.neighbors(v);
while (ai.hasNext())
{

// get edge to neighbor
Edge<String,Integer> e = g.getEdge(v,ai.next());
// construct (distance,edge) pair for possible result

16.4 Examples: Common Graph Algorithms

433

possible = new ComparableAssociation<Integer,

Edge<String,Integer>>(vDist+e.label(), e);

q.add(possible);

// add to priority queue

}

}
// now, get closest (possibly unvisited) vertex
if (!q.isEmpty())
{

possible = q.remove();
// get destination vertex (take care w/undirected graphs)
v = possible.getValue().there();
if (result.containsKey(v))

v = possible.getValue().here();

} else {

// no new vertex (algorithm stops)
v = null;

}

}
return result;

}

Unlike the minimum cost spanning tree algorithm, we return a Table of
results. Each entry in the Table has a vertex label as a key. The value is an
association between the total distance from the source to the vertex, and (in the
nontrivial case) a reference to the last edge that supports the minimum-length
path.

We initially record trivial results for the source vertex (setting its distance
to zero) and place every outgoing edge in the priority queue (see Figure 16.9).
Unlike the minimum spanning tree algorithm, we rank the edges based on total
distance from the source. These edges describe how to extend, in a nearest-
ﬁrst, greedy manner, the paths that pass from the source through visited nodes.
If, of course, an edge is dequeued that takes us to a vertex with previously
recorded results, it may be ignored: some other path from the source to the
If the vertex has not been visited, it is placed in the Table
vertex is shorter.
with the distance from the source (as associated with the removed edge). New
outbound edges are then enqueued.

The tricky part is to rank the edges by the distance of the destination vertex
from the source. We can think of the algorithm as considering edges that fall
within a neighborhood of increasing radius from the source vertex. When the
boundary of the neighborhood includes a new vertex, its minimum distance
from the source has been determined.

Since every vertex is considered once and each edge possibly twice, the
worst-case performance is O(|V | + |E|), an improvement over the O(|V |3) per-
formance for sparse graphs.

434

Graphs

The progress of a single-source, shortest-path computation from source a.
Figure 16.9
As nodes are incorporated, a minimum distance is associated with the vertex. Compare
with Figure 16.8.

16.5 Conclusions

In this chapter we have investigated two traditional implementations of graphs.
The adjacency matrix stores information about each edge in a square matrix
while the adjacency list implementation keeps track of edges that leave each
vertex. The matrix implementation is ideal for dense graphs, where the number
of actual edges is high, while the list implementation is best for representing
sparse graphs.

Our approach to implementing graph structures is to use partial implemen-
tations, called abstract classes, and extend them until they are concrete, or com-
plete. Other methods are commonly used, but this has the merit that common
code can be shared among similar classes. Indeed, this inheritance is one of the
features commonly found in object-oriented languages.

This last section is, in effect, a stepping stone to an investigation of algo-
rithms. There are many approaches to answering graph-related questions, and
because of the dramatic differences in complexities in different implementa-
tions, the solutions are often affected by the underlying graph structure.

Finally, we note that many of the seemingly simple graph-related problems
cannot be efﬁciently solved with any reasonable representation of graphs. Those
problems are, themselves, a suitable topic for many future courses of study.

345511108bcdef51110c1111103443a:034(1)(3)(4)(2)5c88810da:0a:0a:0f:3f:3f:3e:7e:7e:7b:8b:8b:8c:185555d:12d:1216.5 Conclusions

Self Check Problems

435

Under what conditions would you use an adjacency matrix over an

Solutions to these problems begin on page 450.
16.1 What is the difference between a graph and a tree?
16.2 What is the difference between an undirected graph and a directed
graph?
16.3
adjacency list implementation of a graph?
16.4 What do we know if the adjacency matrix is symmetric?
16.5 What is the time potentially required to add an edge to a graph repre-
sented as an adjacency list? What if the graph is represented using an adjacency
matrix?
16.6 What is a spanning tree of a graph?
16.7 What is a minimum spanning tree of a weighted graph?
16.8 What is the transitive closure of a graph?
16.9 What is the topological ordering of vertices of a graph?
16.10 Under what conditions is a topological sort of the vertices of a graph
possible?

Problems

Solutions to the odd-numbered problems begin on page 486.
16.1
(undirected and complete) graph:

Draw the adjacency matrix and list representations of the following

16.2
(directed) graph:

Draw the adjacency matrix and list representations of the following

Draw the adjacency matrix and list representations of a complete tree

16.3
with seven nodes and undirected edges.
16.4 What are the transitive closures of each of the following graphs?

abcdabcd436

Graphs

Suppose that we use an n × n boolean matrix to represent the edges
16.5
of a directed graph. Assume, as well, that the diagonal elements are all true.
How should we interpret the nth power of this adjacency matrix?

16.6 What topological characteristics distinguish a general graph from a gen-
eral tree?

16.7

Consider the following (simpliﬁed) map of the world:

a. Compute the shortest air distance to each of the cities from scenic Mont-
pelier, Vermont. Redraw the map with distances to cities and include only
the air routes that support the most direct travel to Vermont.

b. Suppose you’re interested in setting up a network among the capitals.

Redraw the map to depict the minimum spanning network.

c. Suppose you’re interested in setting up a Camel Express system. Redraw
the map to depict the minimum spanning road systems that don’t cross
bodies of water (indicated by crossed edges).

Explain why it is necessary that the Edge class “show through” the Graph
16.8
interface. (Hint: Consider implementations of the Iterator constructed by the
edges method.)

Compare and contrast the performance of the adjacency list and adja-

16.9
cency matrix implementations of graphs.

16.10 For both implementations of graphs, write a method, isSink, that re-
turns true if and only if the vertex indicated is a sink. (A sink has out-degree 0.)

abdeabd(a)(b)eccAlbanyTrentonMontpelierDoverBoston750015013043001202504200100200Athens47001500Salt Lake CityHarrisburg80007500PhoenixSacramentoBangkok450550650Kuala LumpurDrive camels only on noncrossed edges.All distances are approximate nautical miles.Part of the world16.5 Conclusions

437

Topological
sorting solves
this, given no
cycles.

16.11 For both implementations of graphs, write a method, isSource, that
returns true if and only if the vertex indicated is a source. (A source has in-
degree 0.)
16.12 In an undirected graph, it is possible for a single edge to be represented
by Edge objects whose vertices appear in opposite orders. Describe how a gen-
eral equals method for Edges might be written.
16.13 Explain why graphs with topologically sortable vertices must be (1)
directed and (2) acyclic.
16.14 Suppose we have a cycle-free graph that describes the dependencies
between Java modules. How would you compute the order of compilations that
had to occur?
16.15 It is a fairly common practice to traverse the vertices and edges of a
graph. Consider a new implementation of graphs that keeps a Map of vertices
as well as an unordered List of edges. This makes traversal of edges simple.
What is the complexity of each of the other Graph operations?
16.16 Extend the all-pairs minimum distance algorithm to keep track of the
shortest path between the nodes.
16.17 Explain why it is sometimes more efﬁcient to compute the distance
from a single source to all other nodes, even though a particular query may be
answered with a partial solution.
16.18 Under what conditions can a graph component have nonunique mini-
mum spanning trees?
16.19 Prove that a minimum spanning tree of a graph component must in-
clude a shortest edge.
16.20 It is possible, in Dijkstra’s algorithm, that an edge removed from the
priority queue is not useful: it takes us to a previously visited node. Some of
these extraneous edges can be avoided by not placing an edge in the priority
queue if the destination has already been visited. Is it still possible to encounter
an edge to a previously visited node?

16.6 Laboratory: Converting Between Units

Objective. To perform the transitive closure of a graph.
Discussion. An interesting utility available on UNIX systems is the units pro-
gram. With this program you can convert between one unit and another. For
example, if you were converting between feet and yards, you might have the
following interaction with the units program:

You have: yard
You want: inch
Multiply by 36.0

The program performs its calculations based on a database of values which has
entries that appear as follows:

1 yard 3 foot
1 foot 12 inch
1 meter 39.37 inch

Notice that there is no direct conversion between yards and inches.

In this lab you are to write a program that computes the relations between
units. When the program starts, it reads in a database of values that describe the
ratios between units. Each unit becomes a node of a graph, and the conversions
are directed edges between related units. Note that the edges of the graph must
be directed because the factor that converts inches to yards is the reciprocal of
the factor that converts yards to inches.

In order to deduce conversions between distantly related units, it will be
necessary for you to construct the closure of the graph. The labels associated
with adjacent edges are multiplied together to label the direct edge.

Once your program reads the unit conversion database, the program should
prompt for the source units and the destination units. It should then print out
the conversion factor used to multiply the source units by to get the destination
units. If the units do not appear to be related, you should print out a message
that indicates that fact. Prompting continues until the user enters a blank line
for the source units.
Thought Questions. Consider the following questions as you complete the lab:

1. There are two approaches to this problem: (1) construct the closure of the
graph, or (2) perform a search from the source unit node to the destina-
tion unit node. What are the trade-offs to the two approaches?

2. What does it mean if the graph is composed of several disconnected com-

ponents?

Notes:

Appendix A

Answers

Concepts:
.
.

Self check problem solutions
Solutions to selected problems

From north, south, east, and west
every man who had a shade of
red in his hair had tramped into the city
to answer the advertisement.
—Sir Arthur Conan Doyle

This section contains answers to many problems presented to the reader in
the text. In the ﬁrst section, we provide answers to self check problems. In the
second section, we provide answers to odd-numbered problems found in the
text.

A.1 Solutions to Self Check Problems

Chapter 0

Problems begin on page 3.
Answers to all self check problems are found, here, in Appendix A.
0.1
Large programs take time to write and involve complex structures. Large
0.2
programs may take a long time to run, however many programs are large be-
cause they more effectively manipulate the data and, in the end, they may run
faster than an equivalent short program.
The author of this text made it briefer than most texts to make it pos-
0.3
sible for students to spend less time reading and more time practicing writing
programs.
No. They are, essentially, rules of thumb. Because programmers spend
0.4
considerable time attacking new problems, principles are useful for guiding the
design process, but not determining the design.

Chapter 1

Abstraction, in computer science, is the process of removing or hiding

Problems begin on page 26.
1.1
unnecessary details of a deﬁnition.
1.2
tion by hiding them within the bodies of procedures or methods.

Procedural abstraction removes unnecessary details of an implementa-

442

Answers

Data abstraction removes the unnecessary details of an implementation

A method is identiﬁed by its name and the types of each of its parame-

1.3
by hiding them within the protected portion of a data structure.
In Java, a message is sent to an object by calling a method.
1.4
A class provides the details of an implementation of a data structure.
1.5
An object is a particular instance (or, sometimes, an instantiation) of a class.
Objects take up memory; classes are simply models for objects. A class need not
be instantiated. java.lang.Math is an example of a instance-free class.
1.6
ters. The return type, in Java, is not part of its signature.
An interface describes those aspects of a data structure or class that are
1.7
publicly visible; it forms a contract between the implementor and the user of a
class. The implementation describes the particular details necessary to make a
workable object. Those portions of the implementation that are not part of the
interface are usually hidden from the user.
An accessor provides read-only access to an implementation’s data, di-
1.8
rectly or indirectly. It does not modify the structure. A mutator allows the user
to modify the state of an object. It may also return a structure’s data.
The Object class is the most general class in Java. Every object is of type
1.9
Object. A general purpose class manipulates objects of the most general type
possible.
1.10 An object is an instance of a class. A reference provides a means of
refering to a particular object. If a reference refers to no particular object, this
is indicated by a null value.
1.11
programs. For some classes you implement you may be the only user.

Programmers make use of a class when they import them into their

Chapter 2

Problems begin on page 37.
The pre- and postconditions provide a slight increase in formality that
2.1
allows the user of the data structure to identify exactly what will happen when
a method is run.
2.2
2.3
a method does not have a predictable result, it is not generally useful.
2.4
essentially, like users of your hidden code.

If a method has no precondition it may be called at any time.
A method with no postcondition does not have any predictable result. If

Because hidden code is called by other methods. These methods act,

Chapter 3

Problems begin on page 64.
Arrays and Vectors are both randomly accessed, container objects. Both
3.1
can have their sizes determined at run-time. Since both are objects, either may
be null, a condition that is often to be avoided.

A.1 Solutions to Self Check Problems

443

Arrays are a feature of Java and enjoy access using square bracket ([]) no-
tation, while Vectors are accessed using their get and set methods. Arrays
are declared to hold a speciﬁc type of object, determined by the user, and when
accessed, they return objects of that type. Vectors always hold objects of type
Object, the most general type possible. A value accessed from a vector is al-
ways of type Object; the result must be cast to suggest and verify the desired
type.

The remove method either removes the ﬁrst occurrence of a value v, or

The most important feature of the Vector class is the notion of extensibility.
Vectors may be lengthened by the add method. Arrays must be explictly reallo-
cated and copied. The vector data abstraction supports this automatically, and
efﬁciently.
The add(v) method appends a value to the end of the Vector. The
3.2
add(i,v) method inserts a value into a Vector so that v will be found at loca-
tion i. Any values found at i or higher are shifted to higher slots in the process.
The add methods insert new values into the Vector, logically increasing
3.3
its size. No existing values are changed. The set(i,v) method replaces the
value at location i.
3.4
a value found at a location i. Both decrease the logical size of the Vector.
The size is the number of storage locations logically available in the
3.5
Vector. Usually, the size corresponds to the number of items stored within the
Vector. The capacity is the number of memory references currently allocated
to the Vector. The capacity indicates how large the vector’s size can get before
the underlying array must be reallocated and copied over. Ideally, the capacity
provides enough room so that the size can increase without a signiﬁcant risk of
reallocation.
The use of a Vector allows us to concentrate on the manipulation of the
3.6
words that remain in the list, not the allocation and reallocation of the word
list, itself.
3.7
each moved value to destroy an unmoved value.
The doubling of the size effectively delays the next reallocation of the
3.8
Vector for a long time. Since reallocation involves moving every element of
the Vector, reallocation is to be avoided, if possible. The doubling reduces the
average number of times a single value is copied from approximately n
The code allocates an array of initialCapacity Objects and assigns
3.9
a reference to the array to elementData. The array previously referenced by
elementData is potentially lost. Each element of the array is null; no element
of the array refers to any particular Object.
3.10 Constructors, like other methods, are identiﬁed by the types of their pa-
rameters. A constructor is called when the new keyword is encountered. Follow-
ing the new keyword is the type of object to be constructed, and the parameters
used to select and execute the appropriate constructor. The variable v is con-
structed with the parameterless constructor and has size 0 and capacity 10. The
variable w is constructed with the single-parameter (an int) constructor and
has size 0 and capacity 1000. w will not have to be reallocated as soon as v, but

This avoids destroying data. Starting at the low end of the Vector causes

2 to 1.

444

Answers

it initially consumes more memory.
The row is bounded by the height and the column by the width. The
3.11
row, in Java and most modern languages, is the ﬁrst index provided. This is
motivated by the notation used in mathematics.

Chapter 5

Problems begin on page 111.
5.1
5.2
5.3
5.4
5.5
largest is 30x.
log2 4 = 2,
largest is 30x.

The function f (x) = x is O(x). Let c = 1, and n0 = 1.
The function f (x) = 3x is O(x). Let c = 3 and n0 = 1.
If x + 900 is O(x), set c = 901 and n0 = 1.
It grows with rate O(x). Set c = 450 and n0 = 1.
log2 2 = 1,
√

√

2 ≈ 1.414, x = 2, x! = 2, x2 = 4, 2x = 4, and 30x = 60. The

4 = 2, x = 4, x2 = 16, 2x = 16, x! = 24, and 30x = 120. The

log2 16 = 4,

16 = 4, x = 16, x2 = 256, 30x = 480, 2x = 65, 536, and

x! = 20, 922, 789, 888, 000. The largest is x!.

64 = 8, x = 64, 30x = 1920, x2 = 4096,

2x = 18, 446, 744, 073, 709, 551, 616

√

√

log2 64 = 6,

and

x! = 126, 886, 932, 185, 884, 164, 103, 433, 389, 335,
161, 480, 802, 865, 516, 174, 545, 192, 198, 801,
894, 375, 214, 704, 230, 400, 000, 000, 000, 000.

A base case, self-reference, and, in the self-reference, progress toward

x! is the largest, by far. This ordering of these functions is maintained for all
values of x > 30.
5.6
the base case.
Assume there are 400 million people in the U.S. Reading children (say
5.7
ages 10 to 18) represent, say 20 percent of the population, or 80 million people.
The book is read by, say, 60 million U.S. children. Suppose the book has 750
pages, read at 1 minute per page. Each child takes approximately 10 hours to
read the text (they get faster near the end), yielding a hit of 600 million child-
hours of extra-curricular reading. A child-year we approximate at 2000 hours.
The result is about 300 thousand child-years of reading. (If parents read this
book aloud to their 2 children, this might cut the time in half, but the children
would likely read the text themselves, again.)
It is easy to see that that postmaster would never return more than
5.8
21 penny stamps when a single 21 cent stamp would do. Similary, the post-
master would never return 37 or more 21 cent stamps in change, since 21 letter

A.1 Solutions to Self Check Problems

445

stamps would sufﬁce. Both c1(x) and c21(x) must, then be O(1). The remain-
der of at least x − (0.20 + 36 × 0.21) dollars change must be accommodated by
37 cent stamps. The function c37(x) is clearly linear, O(x).

Chapter 6

You are using insertion sort. As you encounter the values, you insert

It is useful to have a temporary object reference so that the two values

You are using selection sort. You extract minimum values one at a time

Problems begin on page 144.
6.1
can be exchanged without forgetting one of the two references.
6.2
them into the list represented on your piece of paper.
6.3
until all values have been listed in increasing order.
It is valid. This is a ﬁrst stage of a sort involving merging. Merge months
6.4
1 and 2. Then merge months 3 and 4. Then merge each of these two piles. Note
that after the month sorting pass the checks are nearly in order.
This is a form of radix sort. It works because she expects that the mail is
6.5
fairly evenly distributed. Notice that the street number has lower priority than
the street name. If she walks down the even side of each street and up the odd
side, how would you suggest she change this algorithm?
The compareTo method returns a value that is less than, equal to, or
6.6
greater than 0 if this object is less than, equal to, or greater than the parameter.
Essentially, the compareTo method determines if a swap is necessary in most
sorting algorithms.

Chapter 8

Problems begin on page 173.
8.1

Typical code is as follows:

Enumeration e = x.elements();

while (e.hasMoreElements())
{

System.out.println(e.nextElement());

}

8.2

Typical code is as follows:

Iterator i = x.iterator();

while (i.hasNext())
{

System.out.println(i.next());

}

It is important, in this case, to make sure that you assign the value of the

8.3
Iterator variable to an Integer variable using a cast:

446

Answers

// w is a Vector of integer values:
Iterator iter = w.iterator();

while (iter.hasNext())
{

Integer I = (Integer)iter.next();
int n = I.intValue();
if ((n % 2) == 0) System.out.println(n);

}

8.4

The code makes use of a boolean value and the Math.sqrt function.

boolean perfectSoFar = true;

// attempt to read a first value
if (g.hasNext())
{

Integer I = (Integer)g.next();
int i = I.intValue();
// attempt to read second, and on, while perfect
while (perfectSoFar && g.hasNext())
{

Integer J = (Integer)g.next();
int j = J.intValue();
// compute integer square root
int k = (int)Math.sqrt(j+i);
// it’s working if i+j is a perfect square
perfectSoFar = perfectSoFar && ((i+j) == (k*k));
// save current value
i = j;

}

}
if (perfectSoFar) System.out.println("Perfect!");

Chapter 9

Problems begin on page 212.
The List allocates and frees memory in small units, while the Vector
9.1
actually seeks to avoid changes in allocation. This is because the Vector class
must copy any data in the Vector at the time of reallocation, while a List
does not. The List type allows for the efﬁcient insertion and deletion of values
within the structure, while the Vector class requires linear time, on average.
The use of references allows the List structures to rearrange their values
9.2
more easily. For example, in most List structures, it is fairly simple to add or
remove a value at the front of the list
The structure is a List if (1) it implements each of the methods re-
9.3
quired by the List interface, and (2) it implements the List interface directly
or indirectly.

A.1 Solutions to Self Check Problems

447

If C extends SinglyLinkedList, it is a (speciﬁc kind of) SinglyLinked-
9.4
List. Since all SinglyLinkedLists are extensions of AbstractLists, C is an
AbstractList. Since at least one (in actuality, all) of these classes implements
List, C is a List as well. C is not a DoublyLinkedList because C does not
extend that class.
No. Both the DoublyLinkedList and SinglyLinkedList classes are
9.5
direct extensions of AbstractList; they are both AbstractLists, but one is
not an extension of the other.
The tail reference facilitates getting to the far end of the list. It makes it
9.6
possible, for example, to add an element to the end of the list in constant time.
If tail was eliminated, it would be necessary to traverse the entire list to ﬁnd
the far end; this operation is O(n).
It is not directly useful. We might arrange to have the tail reference
9.7
point to the last element, but keeping that reference up to date is relatively
difﬁcult. It is possible, though, and the result is, essentially, a circular list.
Because it uses less space. A ListVector needs room to store one ref-
9.8
erence to each element; other List implementations also need to store one or
more references.
9.9
that are generic, or shared among all implementations of the List class.
9.10 It gets added at the tail of the list. The remove method will remove and
return this value.
9.11 It is the second element. The ﬁrst element occurs in position 0.

The abstract class only attempts to provide implementations of methods

Chapter 10

Problems begin on page 245.
10.1 A Stack is an interface that extends a Linear. It is therefore a Linear.
It is not a List because it does not directly or indirectly extend that inter-
face.
It is not an AbstractLinear; a Stack is an interface and does extend
the AbstractLinear class. (A particular implementation of a Stack is likely to
extend an AbstractStack, which extends an AbstractLinear.) Stacks and
Queues are both Linear objects, but otherwise unrelated.
10.2 The Stack is an interface that does not extend List; they are unrelated.
A StackList is an implementation of the Stack interface that uses a List in its
implementation. It is not, however, a List itself. (You can see this because the
StackList class does not have some List methods: for example, get(i).) The
StackList is an implementation of a Stack.
Queue is more general than QueueList. It makes it impossible for us to
10.3
depend on features of a QueueList implementation. If we wish to change our
actual types from QueueList to QueueVector, the change is relatively simple
and painless.
10.4 No. The Queue is an interface and, for practical purposes cannot be
constructed. If you want to construct a value for a Queue variable, you must
pick an implementation and construct an instance of that type.

448

Answers

10.5 Your car is in a queue. The ﬁrst car in is the ﬁrst car to leave.
10.6 It is a stack. The last page you visited is the ﬁrst page you recall.
10.7 The ﬁrst solution you ﬁnd is the shortest possible.
10.8 The stack-based mechanism is more similar to human maze-solving strat-
egy. The breadth-ﬁrst approach involves keeping track of and reconsidering an
increasing number of options. If you are walking, the stack-based approach is
simpler.
10.9 The reason is that it is possible for the Queue to “wrap around” the array.

Chapter 11

Problems begin on page 272.
11.1 The values within an OrderedStructure are maintained in some agreed-
upon order.
11.2 The user speciﬁes a compareTo method for each class potentially stored
in an ordered structure. This method is at least useful for determining the
relative ordering of two objects of that class.
11.3 The compareTo method is a feature of a class, and is used to order the
instances of the class. The compareTo method is typically ﬁxed, and cannot be
changed to suit the particular need. A Comparator, on the other hand, is an ob-
ject itself, with a compare method that compares two objects. Each Comparator
instance may choose to order the objects in a different manner. Thus, the
Comparator approach provides greater ﬂexibility, while the compareTo approach
leaves the ordering of objects up to the designer of the original class.
11.4 Yes. It is best if the ordering mechanism keeps equals objects near each
other. Otherwise, inconsistencies can result.
11.5 No. A Vector allows the insertion of values in the center of the structure.
An OrderedVector must disallow this, in order to maintain the internal order
of the structure.
11.6 No. The values in a Stack are ordered by their entry time, not necessarily
the object’s value.
11.7 The people are ordered according to their arrival time.
11.8 They’re both equally suitable. As long as the value returned is negative,
0, or positive, that is all that is needed to determine the order of two values.
Harry, though, might do well to see why Sally’s method seems more symmetric.
Sally’s method will work, and Harry’s will not. The OrderedVector
11.9
does not know ahead of time what type will be stored within the structure.
It must call, then, the compareTo method that takes a general Object. Harry
will not have written this method, and the program will complain when the
OrderedVector attempts to compare two of Harry’s Coins.

Chapter 12

Problems begin on page 309.
12.1 Yes, in both cases, but only if the tree is empty.

A.1 Solutions to Self Check Problems

449

Yes, but barely. There is always just one fewer full nodes than leaves.
12.2
A spindly tree—a tree with no nodes of degree 2—can have as many interior
nodes as it wants, but it always has exactly one leaf.
12.3 The root has the greatest height, if there is a root.
12.4 No. If there were two paths, then some node would have two parents.
Nodes have at most one parent.
12.5 Because the operations in arithmetic expressions usually involve at most
two operands. Each operand is represented by an independent subtree of an
operator node.
12.6 A BinaryTree is not a List because, among other things, it has no notion
of indexing.
12.7 Any of the traversals encounter each element at some point. The index
associated with an element in a BinaryTree might be the relative position of
the node in the traversal of the BinaryTree.
12.8 Not easily. The queue makes sure that the nodes on a level are all visited
around the same time. Use of a stack would make it possible to visit nodes
at a deeper level before all the nodes at a relatively shallow level had been
encountered.
12.9 Usually the base case is found near the leaves. The recursion starts at
the root, and follows the branches outward. It stops when it gets to the leaves.
12.10 Usually 2. Each call investigates one subtree.
12.11 It is n−1
n , very close to 1 for large n. The sum of the degrees is the
number of edges in the tree. Since every edge attaches one node directly or
indirectly to the root, there are n − 1 edges. The average, then, is as indicated.

Chapter 13

Problems begin on page 337.
13.1 No. A PriorityQueue cannot guarantee that the ﬁrst value entering the
structure is the ﬁrst out.
13.2 No. A PriorityQueue must accept Comparable values. A Linear struc-
ture has no such constraint.
13.3 The weight corresponds to the total number of occurrences of the letters
found within the tree. The depth of a node is the number of bits associated with
the encoding of the information at the node.
13.4 A min-heap is a tree of values whose minimum is found at the root, and
whose subtrees are either empty or min-heaps themselves.
13.5
(O(log n)) height.
Events have associated times. The time of the event determines when
13.6
it should be simulated. The timing of events is neither LIFO nor FIFO, so a
time-as-priority scheme is useful.

Because the tree is complete, we can be sure that it has minimum

Chapter 14

Problems begin on page 364.

450

Answers

The right child. Everything to the left of the right child in the original

14.1 The compareTo operator identiﬁes one of two orderings for unequal val-
ues. Thus at each stage it is natural to have degree 2.
14.2 The ﬁrst node is at the root, the last node is at a leaf.
14.3 Any structure where a value is selected by providing a key.
14.4
tree moves to the left side of the ﬁnal tree.
14.5 It is. Look at Figure 14.4.
14.6 Not necessarily. One may be at the root of a large tree, and the other,
having just been added, may be found at a leaf quite far away.
14.7 The SplayTree potentially changes topology whenever it is referenced.
The Iterator must be capable of maintaining its state while the tree is splayed.
14.8 The red-black tree is always very close to balanced. It has height that is
always O(log2 n). The splay tree has no such guarantee.

Chapter 15

Yes. If the hash table uses external chaining, the number of entries in

Problems begin on page 399.
The Map structure is associative. It allows the user to specify separate
15.1
keys and values. Previous structures have depended heavily on compareTo
methods to order values.
15.2 It is the percentage of the table entries that contain key-value pairs.
15.3
the table can be smaller than the number of key-value pairs.
15.4 It is not guaranteed. In practice, if the load factor is kept low, the number
of comparisons can be expected to be very low.
15.5 A collision occurs when two different keys attempt to hash to the same
location. This can occur when the two keys directly map to the same bucket, or
as the result of rehashing due to previous collisions for one or both of the keys.
It should be fast, reproducible, and it should spread keys as uniformly
15.6
as possible across the hash table. This is usually accomplished by attempting to
ﬁlter out any nonuniform features of the key distribution.
15.7 When space is tight, the extra space required by a Hashtable’s low load
factor makes a MapList somewhat preferable. Since a MapList is slower than a
Hashtable in the expected case, and since a MapList has the overhead of links
between elements, this preference is relatively infrequent.
15.8 The Hashtable does not help with sorting, precisely because the hashing
mechanism is unlikely to be correlated with any useful ordering of the keys. On
the other hand, the Table structure does provide, essentially, a sorting mecha-
nism based on binary search trees. Key-value pairs are placed in the table, and
they are encountered in sorted order when iterating across the keys.

Chapter 16

Problems begin on page 435.

A.2 Solutions to Odd-Numbered Problems

451

The graph is undirected. Each edge is essentially represented by two

16.1 A tree does not have cycles; a graph can. All trees are graphs, but not all
graphs are trees.
16.2 An undirected graph has oriented edges–edges that establish a poten-
tially one-way relation between vertices. The edges of an undirected graph
establish a symmetric relationship.
16.3 You use an adjacency matrix when more edges appear in the graph than
not. If the percentage of potential edges is low, an adjacency list is more efﬁ-
cient.
16.4
directed edges.
For the adjacency list, it takes O(|V |) potentially, since you must check
16.5
to see if the edge is possibly in the graph already.
In the adjacency matrix
implementation, it is constant time because the edge is represented by one or
two spots in a matrix.
16.6 A tree of edges that includes every vertex. Spanning trees are not possi-
ble for disconnected graphs; in that case, a spanning forest is required.
16.7 Of all of the spanning trees of a graph, a minimum spanning tree is one
whose total edge weight is least.
16.8 The transitive closure of a graph includes edge (u, v) if and only if one
can reach vertex v from vertex u.
16.9
all point in one direction.
16.10 If the graph is directed and acyclic.

It is an ordering of vertices such that the (directed) edges of the graph

A.2 Solutions to Odd-Numbered Problems

Chapter 0

Problems begin on page 3.
0.1
0.3
0.5
0.7
0.9

Right here, in the back of the book.
There are many resources available at http://www.cs.williams.edu/JavaStructures.
There are many resources available from http://www.javasoft.com.
There are a number of great texts associated with Java.
These pages are mirrored at http://www.mhhe.com/javastructures.

Chapter 1

int, double, char, boolean.
(a) java.lang.Double extends java.lang.Number, (b) java.lang.Integer

Problems begin on page 26.
1.1
1.3
extends java.lang.Number, (c) java.lang.Number extends java.lang.Object,
(d) java.util.Stack extends java.util.Vector, (e) java.util.Hashtable
extends java.util.Dictionary.
1.5

The program

452

Answers

import structure.Association;
public class ModAssoc
{

public static void main(String args[])
{

Association a = new Association("key","value");
a.theKey = null; // illegal access; generates compile error

}

}

generates the compile-time error

ModAssoc.java:9: theKey has protected access in structure.Association

a.theKey = null; // illegal access; generates compile error

^

1 error

It is a compile-time error, because ﬁeld access is determined by the compiler.
1.7
version of gcd is particularly fast.

The following code approximates π using the technique stated. This

import java.util.Random;
public class Pi
{

static int gcd(int a, int b)
// post: iteratively compute the gcd of two values
{

while (a*b!=0) // a and b not zero
// reduce larger by smaller
{
if (a < b) b %= a;
a %= b;
else

}
return a+b; // a or b is zero; return other

}

public static void main(String args[])
{

int least = 0;
int trials = 1000000;
Random g = new Random();
// perform the trials
for (int i = 0; i < trials; i++)
{

// pick a random fraction, and test for a gcd of 1;
// the fraction is in lowest terms
if (gcd(Math.abs(g.nextInt())%10000,

Math.abs(g.nextInt())%10000) == 1) least++;

}
System.out.println("pi is approximately:");
System.out.println("# "+Math.sqrt(6.0*trials/least));

}

}

A.2 Solutions to Odd-Numbered Problems

453

The code for StopWatch is precisely the code of the structure package

1.9
Clock object:

package structure;

public class Clock
{

// we use a native-code library for structures
protected boolean running; // is the clock on?
protected long strt;
protected long accum;

// starting millisecond count
// total milliseconds

public Clock()
// post: returns a stopped clock
{

running = false;
strt = 0;
accum = 0;

}

public void start()
// post: clock is stopped
// pre: starts clock, begins measuring possibly accumulated time
{

running = true;
strt = System.currentTimeMillis();

}

public void stop()
// pre: clock is running
// post: stops clock, and accumulates time
{

running = false;
accum += (System.currentTimeMillis()-strt);

}

public double read()
// pre: clock is stopped
// post: returns the accumulated time on the clock
{

return (double)accum/(double)1000.0;

}

public void reset()
// post: stops running clock and clears the accumulated time
{

running = false;
accum = 0;

}

454

Answers

public String toString()
// post: returns a string representation of the clock
{

return "<Clock: "+read()+" seconds>";

}

}

1.11

246.76047636306453 Hz

1.13

import structure.Assert;
public class Radio {

protected boolean isOn;
protected boolean isFM;
protected double tunedTo;
protected double tunedToAM;
protected double tunedToFM;
protected double FMFreq[];
protected double AMFreq[];
protected double AMPreset = 1030.0; // factory setting for AM
// factory setting for FM
protected double FMPreset = 88.5;

// is the radio on
// is the radio tuned to FM
// current radio freq
// current AM radio freq
// current FM radio freq
// presets for FM
// presets for AM

public Radio()
// post: constructs a radio, initially tuned to FM.
// FM presets are 88.5, and AM presets are 1030.
{

FMFreq = new double[12];
AMFreq = new double[12];
for (int i = 0; i < 12; i++)
{

FMFreq[i] = FMPreset;
AMFreq[i] = AMPreset;

}
on();
AM();
press(1);
FM();
press(1);
off();

}

public boolean isOn()
// post: returns true iff the radio is on
{

return isOn;

}

public boolean isFM()

A.2 Solutions to Odd-Numbered Problems

455

// post: returns true iff the radio is set to FM
{

return isFM;

}

protected boolean validAM(double f)
// post: returns true iff f is a valid AM frequency
{

return (530 <= f && f <= 1610) && (0.0 == f % 10.0);

}

protected boolean validFM(double f)
// post: returns true iff f is a valid FM frequency
{

return (87.9 <= f) && (f <= 107.9) && ((((int)(f*10+0.5))-879)%2) == 0;

}

public void set(int button, double frequency)
// pre: radio is on and
//
//
// post: sets the indicated preset if the frequency is valid
{

button is between 1 and 12, and
frequency is a multiple of 10 if set to AM, or .2 if set to FM

if (!isOn()) return;
Assert.pre(button >= 1 && button <= 12, "Button value between 1 and 12.");
if (isFM() && validFM(frequency)) FMFreq[button-1] = frequency;
if ((!isFM()) && validAM(frequency)) AMFreq[button-1] = frequency;

}

public void press(int button)
// pre: button is between 1 and 12
// post: tunes radio to preset indicated by button
{

if (isFM()) tune(FMFreq[button-1]);
else tune(AMFreq[button-1]);

}

public void tune(double freq)
// pre: radio is on and frequency is valid for current AM/FM setting
// post: tunes the indicated radio frequency
{

if (!isOn()) return;
if (isFM() && validFM(freq)) tunedTo = tunedToFM = freq;
else if ((!isFM()) && validAM(freq)) tunedTo = tunedToAM = freq;

}

public double frequency()
// post: returns the frequency that the radio is tuned to.
{

return tunedTo;

456

Answers

}

public void AM()
// post: sets frequency range to AM; tunes radio to last AM freq
{

isFM = false;
tunedTo = tunedToAM;

}

public void FM()
// post: sets frequency range to FM; tunes radio to last FM freq
{

isFM = true;
tunedTo = tunedToFM;

}

public void on()
// post: turns radio on
{

isOn = true;

}

public void off()
// post: turns radio off
{

isOn = false;

}

public String toString()
// post: generates a string representing the radio.
{

String result = "<Radio: ";
if (isOn()) result += "turned on, ";
else result += "turned off, ";
if (isFM()) {

result += "tuned to FM frequency "+tunedTo+"\n";
result += "current AM frequency is "+tunedToAM+"\n";

} else {

result += "tuned to AM frequency "+tunedTo+"\n";
result += "current FM frequency is "+tunedToFM+"\n";;

}
int i;
result += "FM presets: ";
for (i = 1; i <= 12; i++)
{

result += " "+FMFreq[i-1];

}
result += "\n";
result += "AM presets: ";
for (i = 1; i <= 12; i++)

A.2 Solutions to Odd-Numbered Problems

457

{

result += " "+AMFreq[i-1];

}
result += "\n";
result += ">";
return result;

}

}

Chapter 2

Problems begin on page 37.
Precondition: none. Postcondition: returns the length of the string.
2.1
Precondition: the string provided as the parameter is non-null. Post-
2.3
condition: returns a new string consisting of chars of this string followed by
characters of parameter string.
This is a good exercise with most programs.
2.5
Precondition: the value must have magnitude no greater than 1. The
2.7
postcondition can be gleaned from the Sun API web pages: “Returns the arc
sine of an angle, in the range of −π/2 through π/2. Special cases:

• If the argument is NaN or its absolute value is greater than 1, then the

result is NaN.

• If the argument is positive zero, then the result is positive zero; if the

argument is negative zero, then the result is negative zero.”

Programmers should remember that the angle cannot generally be retrieved
from the inverse trigonometric functions unless both the numerator and de-
nominator are remembered.

Chapter 3

The size is the actual number of elements in use; the capacity is the

Problems begin on page 65.
3.1
number of cells allocated in the underlying array.
The trimToSize method allows the programmer to reduce the memory
3.3
used by a Vector to the absolute minimum. In cases where doubling is used, or
where there is shrinkage in the size of the Vector, there may be considerable
savings. It is possible to trim the size of a Vector by copying it to a new Vector
that is constructed using a capacity that is the exact size needed. The old Vector
is reclaimed. In fact, this is often how the process is accomplished internally.
Here is the code, from the structure package implementation.
3.5

public int indexOf(Object elem, int index)
// post: returns index of element equal to object, or -1; starts at index
{

int i;

458

Answers

for (i = index; i < elementCount; i++)
{

if (elem.equals(elementData[i])) return i;

}
return -1;

}

Clearly, the function returns −1 when the value cannot be found. Returning 0
would be incorrect, since 0 is a possible index for the Vector. On average, it
takes n
2 to ﬁnd the value.
This approach will not work. Removing values from the front and insert-
3.7
ing them at the rear of the Vector keeps the elements in the original order. A
better approach is to iteratively remove the ﬁrst element from the Vector, and
add it to a new result Vector at the zero location.
Using the default settings of Sun’s Java 1.1.2 compiler under Solaris(TM),
3.9
an array of approximately 1 million strings can be allocated. Can you write a
program to determine the upper bound?
3.11 Here is the modiﬁed rewrite method:

public static Vector rewrite(Vector s)
// pre: s is a string of letters and strings
// post: returns a string rewritten by productions
{

Vector result = new Vector();
for (int pos = 0; pos < s.size(); pos++)
{

// rewrite according to two different rules
if (s.get(pos) instanceof Vector)
{

result.add(rewrite((Vector)s.get(pos)));

} else if (S == s.get(pos)) {

result.add(T);

} else if (T == s.get(pos)) {

result.add(U);

} else if (U == s.get(pos)) {

result.add(V);

} else if (V == s.get(pos)) {

result.add(W);

} else if (W == s.get(pos)) {

Vector temp = new Vector();
temp.add(S);
result.add(temp);
result.add(U);

}

}
return result;

}

These methods must both assume, as preconditions, that the values
3.13
stored within the Matrix are of type Double. The add method must require

A.2 Solutions to Odd-Numbered Problems

459

that the matrices are the same shape. The multiply method must verify that
the number of columns of the left Matrix is the same as the number of rows of
the right Matrix. Each of these conditions should be checked by formal asser-
tions.
3.15 This is a partial implementation of an alternative Matrix class:

import structure.Vector;
import structure.Assert;

public class VecMat {

protected int width;
protected int height;
protected Vector data;

// width of the matrix
// height of the matrix

public VecMat(int rows, int cols)
{

data = new Vector(rows*cols);
width = cols;
height = rows;

}
public Object get(int row, int col)
// pre: 0 <= row < height(), 0 <= col < width()
// post: returns object at (row, col)
{

Assert.pre(0 <= row && row < height, "Row in bounds.");
Assert.pre(0 <= col && col < width, "Col in bounds.");
return data.get(row*width+col);

}

public void set(int row, int col, Object value)
// pre: 0 <= row < height(), 0 <= col < width()
// post: changes location (row,col) to value
{

Assert.pre(0 <= row && row < height, "Row in bounds.");
Assert.pre(0 <= col && col < width, "Col in bounds.");
data.set(row*width+col,value);

}

public String toString()
{

int i, j;
String result = "<VecMat:\n";
for (i = 0; i < height; i++)
{

for (j = 0; j < width; j++)
{

result += "\t"+get(i,j);

}
result += "\n";

}

460

Answers

return result + ">";

}

public static void main(String[] args)
{

int w = 3;
int h = 10;
VecMat m = new VecMat(h,w);
int i, j;
for (i = 0; i < h; i++)
{

for (j = 0; j < w; j++)
{

m.set(i,j,new Double(((double)i+1)/((double)j+1)));

}

}
System.out.println(m);

}

}

The advantages of this implementation are increased speed (only one Vector
lookup is necessary), and size (only one Vector is necessary; any overhead to
a Vector occurs once). Most matrix-like structures are implemented in this
manner in C-like languages. The disadvantage is that Matrix structures must
be rectangular: every row must be the same length. This can be avoided by
changing the computation of the index in the alternate implementation, but
that requires recompilation of the code.
3.17

import structure.Assert;
import structure.Vector;
public class SymmetricMatrix
{

protected int size;
protected Vector rows;

// size of matrix

// vector of row vectors

public SymmetricMatrix()
// post: constructs empty matrix
{

this(0);

}

public SymmetricMatrix(int h)
// pre: h >= 0
// post: constructs an h row by w column matrix
{

// initialize size

size = h;
// allocate a vector of rows
rows = new Vector(size);
for (int r = 0; r < size; r++)

A.2 Solutions to Odd-Numbered Problems

461

// each row is allocated and filled with nulls
Vector theRow = new Vector(r+1);
rows.add(theRow);
for (int c = 0; c < r+1; c++)
{

theRow.add(null);

}

{

}

}

public Object get(int row, int col)
// pre: 0 <= row < size(), 0 <= col < size()
// post: returns object at (row, col)
{

Assert.pre(0 <= row && row < size, "Row in bounds.");
Assert.pre(0 <= col && col < size, "Col in bounds.");
if (row < col) return get(col,row);
else {

Vector theRow = (Vector)rows.get(row);
return theRow.get(col);

}

}

public void set(int row, int col, Object value)
// pre: 0 <= row < size(), 0 <= col < size
// post: changes location (row,col) to value
{

Assert.pre(0 <= row && row < size, "Row in bounds.");
Assert.pre(0 <= col && col < size, "Col in bounds.");
if (row < col) set(col,row,value);
else
{

Vector theRow = (Vector)rows.get(row);
theRow.set(col,value);

}

}

public int size()
// post: returns number of rows in matrix
{

return size;

}

public String toString()
// post: returns string description of matrix
{

StringBuffer s = new StringBuffer();
s.append("<SymmetricMatrix:\n");
for (int r = 0; r < size(); r++)
{

462

Answers

for (int c = 0; c < size(); c++)
{

s.append("

"+get(r,c));

}
s.append("\n");

}
s.append(">");
return s.toString();

}

}

Chapter 5

The running time is O(log n). The function loops approximately log2 n

Problems begin on page 112.
Accessing a value in an array is O(1) time. The same is true with the
5.1
get method. The difference is absorbed in the constant associated with big-O
notation.
5.3
times.
This “standard” approach to matrix multiplication is O(n3).
5.5
The lower bound on adding a value to the end of the Vector is constant
5.7
time as well: Ω(1). The doubling of the size of the array may affect the upper
bound, but not the lower bound.
The complexity is O(n): we must construct a new string with space that
5.9
has a.size()+1 characters, and we must copy them all over. Notice that, from
a theoretical standpoint, it is no more complex than adding a character to the
end of a zero-terminated mutable string. The time is still determined, in that
case, by a search for the end of the string: O(n) time.
5.11
strips the logarithmic term log n. Select c = 2 and n0 = 1. (Proof omitted.)
5.13 The rate is O(1). Select c = 1.5 and n0 = 1. (Proof omitted.)
5.15 O(tan n). If no “traditional” bound can be found, we can make use of
the fact that every function f is O(f ).
5.17 Here is a recursive implementation:

The function grows as a linear function, O(n). The linear term n out-

public static void syr(int s0)
{

System.out.print(s0+" ");
if (s0 == 1) System.out.println();
else if ((s0 % 2) == 0) syr(s0/2);
else syr(s0*3+1);

}

5.19 This answer is discussed in detail in Java Elements by Bailey and Bailey:

public static void drawLine(int x0, int y0, int x1, int y1)
// pre: drawing window d is non-null
// post: a line is drawn on d from (x0,y0) to (x1,y1)
{

A.2 Solutions to Odd-Numbered Problems

463

int mx = (x0+x1)/2;
int my = (y0+y1)/2;
int dx = Math.abs(x1-x0);
int dy = Math.abs(y1-y0);
if (dx <= 1 && dy <= 1)
{

// midpoint

// span in x and y direction

// very close end points

d.draw(new Pt(x1,y1));

// draw destination

} else if (dx > 1 || dy > 1)
{

drawLine(x0,y0,mx,my);
drawLine(mx,my,x1,y1);

// draw first half
//

and second half

}

}

5.21 The complexity of this function is O(n) where n is the smaller of the two
values.

public static int mult(int a, int b)
{

if (a < 0) return -mult(-a,b);
if (a == 0) return 0;
if (a == 1) return b;
if (a < b) return mult(a-1,b)+b;
else return mult(b,a);

}

5.23 Let us prove this by induction. First, note that if n = 0, then

5n − 4n − 1 = 1 + 0 − 1 = 0

Since 16 divides 0 evenly, the observation holds for n = 0. Now, assume that
the observation holds for all values less than n, and speciﬁcally in the case of
n − 1. We have, then, that

5n−1 − 4(n − 1) − 1

is divisible by 16. Multiplying by 5 we get

5n − 20(n − 1) − 5

which is also divisible by 16. That is not quite the desired expression. At this
point we add 16(n − 1), a multiple of 16:

5n − 20(n − 1) + 16(n − 1) − 5

= 5n − 4(n − 1) − 5
= 5n − 4n + 4 − 5
= 5n − 4n − 1

Clearly, this expression is divisible by 16, so the observation holds for n. By
induction on n, we see the result holds for all n ≥ 0.

464

Answers

It is clear to see that nd can be rewritten as nc+(d−c) = nc · nd−c. Since
5.25
d ≥ c, then d − c ≥ 0 and, for n ≥ 1, nd−c ≥ 1. We have, therefore, that nd ≥ nc
for all n ≥ 1. This is sufﬁcient to demonstrate that nc is O(nd) for any d ≥ c.

5.27 Clearly, the sum of the ﬁrst n odd integers is

  n
X

!
i

2

i=0

− n = n(n + 1) − n = n2

5.29 Since log n is monotonic and increasing, the log n bounds log i above for
i ≤ n. The result is, then, trivial.

5.31 This is a good problem to perform in a later lab. Keeping a log of errors
is instructive and points to weaknesses in our understanding of the systems we
use.

The best strategy is to decrease the size of the Vector by a factor of 2
5.33
when the number of elements within the Vector falls below 1
3 . An interesting
experiment would be to simulate the expansion and contraction of Vectors
over time with random inserts and deletes. What are the characteristics of the
various expansion and reduction policies?

Chapter 6

Problems begin on page 145.

6.1

The two values a and b can be exchanged with the following code:

a=b-a;
b=b-a;
a=a+b;

Bubble sort, as presented in the text, performs comparisons in a data-

6.3
independent manner. Its behavior, then, is always O(n2).

All cases take O(n2) time. Selection sort does not improve the ordering
6.5
of the data by selecting a value. It is often worse than the modiﬁed bubble sort.

The author ran the following test sorts on arrays of randomly selected

6.7
integers. Winner for each test is indicated in italics.

A.2 Solutions to Odd-Numbered Problems

465

n
2
4
8
16
32
64
128
256
512
1,024
2,048
4,096
8,192
16,384

Insertion Sort

Quicksort

Compares
1
4
17
51
213
1,008
3,923
17,465
64,279
255,250
1,064,222
4,245,882
16,816,801
66,961,345

Time (ms) Compares
1
0.0
2
0.0
6
0.0
42
0.0
125
0.0
217
0.0010
595
0.0020
1,234
0.0010
3,492
0.0040
6,494
0.016
18,523
0.067
31,930
0.265
76,093
1.092
179,017
4.424

Time (ms)
0.0
0.0
0.0
0.0010
0.0
0.0010
0.0040
0.0020
0.0020
0.0010
0.0020
0.0050
0.012
0.027

6.9

It is relatively simple to construct a counterexample array. Here is one:

19

28

29

46

55

An immediate pass through the 10’s digits does not change the array. If this is
followed by a pass on the units bits, you get

55

46

28

19

29

which is clearly not sorted. If this was, however, the ﬁrst pass, a second pass
then generates

19

28

29

46

55

which is sorted.
6.11 In late 2001 the author informally measured the speed of an assignment
using the Java 2 HotSpot(TM) virtual machine under Mac OS/X running on a
Macintosh Powerbook G3 at approximately 2 nanoseconds. The speed of other
operations measured in units of a basic assignment were approximately:

Integer assignment
Add to integer
Multiply by integer
Integer array entry assignment
Iteration of empty loop
Vector entry assignment

1
1
1
3
9
70

6.13 As presented, insertion and mergesort are stable. Selection and quicksort
are not stable. Selection sort can easily be made stable. For quicksort to be made
stable, the partition function needs to be reconsidered.
6.15
gram:

The following modiﬁcations are made to the standard quickSort pro-

466

Answers

protected static Random gen;
public static void quickSort(int data[], int n)
// post: the values in data[0..n-1] are in ascending order
{

gen = new Random();
quickSortRecursive(data,0,n-1);

}

private static int partition(int data[], int left, int right)
// pre: left <= right
// post: data[left] placed in the correct (returned) location
{

if (left < right) {

int x = left+(Math.abs(gen.nextInt()) % (right-left+1));
swap(data,left,x);

}
while (true)

...

}

It is important to construct a new random number generator at most once per
sort or it will lose its random behavior. Here, we have used an instance variable,
but it could be passed along to the partition method.
6.17 This solution was derived from the iterative version. Since the iteration
has a loop that runs the index toward 0, the recursive call happens before the
work loop:

public static void insertionSort(int data[], int n)
// pre: 0 <= n <= data.length
// post: values in data[0..n-1] are in ascending order
{

if (n > 1)
{

// sort the first n-1 values
insertionSort(data,n-1);
// take the last unsorted value
int temp = data[n-1];
int index; // a general index
// ...and insert it among the sorted:
for (index = n-1; 0 < index && temp < data[index-1]; index--)
{

data[index] = data[index-1];

}
// reinsert value
data[index] = temp;

}

}

6.19 The worst-case running time cannot be determined since given any par-
ticular upper bound, there is a nonzero probability that a run of shufﬂe sort will

A.2 Solutions to Odd-Numbered Problems

467

exceed that bound. The revised assumption allows a worst-case running time
to be determined. Here is the questionable code to perform this sort:

public static void shuffleSort(int data[], Random gen)
// pre: gen is initialized, data contain only valid values
// post: data are sorted
{

int n = data.length;
while (true)
{

// check to see if data is in order
boolean inOrder = true;
for (int i = 1; i < n && inOrder; i++)
{

inOrder = inOrder && data[i-1] <= data[i];

}
// if it is sorted, return!
if (inOrder) return;
// nope: shuffle the values again
shuffle(data,gen);

}

}

Notice we check to see if the data are in order before shufﬂing. This is advanta-
geous.

Chapter 8

Problems begin on page 173.
8.1
implement the methods of the Iterator.
8.3
values, rather than ints.

AbstractIterator is an implementation of the Iterator class, it must

The main difference is that a prime number iterator returns Integer

import structure.*;
import java.util.Iterator;

public class PrimeIterator extends AbstractIterator
{

protected Vector primes;
protected int current;

public PrimeIterator()
// post: construct a generator that delivers primes starting at 2
{

reset();

}

public void reset()
// post: reset the generator to return primes starting at 2

468

Answers

{

}

primes = new Vector();
primes.add(new Integer(2));
current = 0;

public boolean hasNext()
// post: returns true - an indication that there are always more primes
{

return true;

}

public Object get()
{

return primes.get(current);

}

public Object next()
// post: geneate the next prime
{

Integer N = (Integer)get();
current++;
if (current >= primes.size())
{

int n = N.intValue();
int i, f;
Integer F;
do
{

if (n == 2) n = 3;
else n = n + 2;
for (i = 0; i < primes.size(); i++)
{

F = (Integer)primes.get(i);
f = F.intValue();
if ((n%f) == 0) break;

}

} while (i < primes.size());
primes.add(new Integer(n));

}
return N;

}

}

This solution uses OrderedLists, which appear in Chapter 11. Another
8.5
approach is to store values in a Vector using insertion sort, or to store all the
values in a Vector, and perform quicksort.

import structure.*;
import java.util.Iterator;

A.2 Solutions to Odd-Numbered Problems

469

public class OrderedIterator extends AbstractIterator
{

protected OrderedList data;
protected AbstractIterator li;
protected Iterator iter;

public OrderedIterator(Iterator subIterator)
// pre: subIterator valid and generates a finite number of elements
// post: returns elements returned by subIterator, in reverse order
{

data = new OrderedList();
iter = subIterator;
while (subIterator.hasNext())
{

data.add(subIterator.next());

}
li = (AbstractIterator)data.iterator();

}

public void reset()
// post: resets the iterator to next return the last element of the
// sub-iterator
{

li.reset();

}

public boolean hasNext()
// post: returns true if there are more values to be returned from iterator.
{

return li.hasNext();

}

public Object next()
// post: returns the previous element of the sub-iterator; advances iter
{

return li.next();

}

public Object get()
// post: returns next element of iterator, does not advance
{

return li.get();

}

}

This involves constructing two classes and an interface. The Predicate

8.7
interface has the following form:

public interface Predicate
{

public boolean select(Object item);

470

Answers

// pre: item is a valid object
// post: returns true or false, depending on the predicate tested

}

A particular Predicate returns true if and only if the String object handed to
it is 10 characters or longer.

public class LongString implements Predicate
{

public boolean select(Object o)
// pre: o is a String
// post: returns true if string is longer than 10 characters
{

String s = (String)o;
return s.length() > 10;

}

}

We now have the speciﬁcation of a PredicateIterator:

import structure.*;
import java.util.Iterator;

public class PredicateIterator extends AbstractIterator
{

protected Iterator iter;
protected Predicate p;
protected Object current;
public PredicateIterator(Predicate p, Iterator subIterator)
{

iter = subIterator;
this.p = p;
reset();

}

public void reset()
{

current = null;
while (iter.hasNext())
{

Object temp = iter.next();
if (p.select(temp)) {

current = temp;
break;

}

}

}

public boolean hasNext()
{

return (current != null);

A.2 Solutions to Odd-Numbered Problems

471

}

public Object get()
{

return current;

}

public Object next()
{

Object result = current;
current = null;
while (iter.hasNext())
{

Object temp = iter.next();
if (p.select(temp)) {

current = temp;
break;

}

}
return result;

}

}

Chapter 9

Problems begin on page 212.
The methods size, isEmpty, clear, addFirst, addLast; remove() and
9.1
remove(Object); all add methods; contains, indexOf, and lastIndexOf meth-
ods; and all methods otherwise inherited from the Structure object.
9.3

It is a suitable structure:

1. It is a naturally dynamic structure: there is no bound on the number of

elements that may be saved in a List.

2. Each element of the List can be accessed randomly by specifying its loca-

tion with respect to the head of the list—its index.

3. Although the access time is likely to be proportional to the index provided,
it is unlikely that users would notice the slowdown associated with the
access.

9.5

public void reverse()
// post: this list is in reverse order.
// post: the list is reversed, destructively
{

Node previous = null;
Node finger = head;

472

Answers

while (finger != null)
{

Node next = finger.next();
finger.setNext(previous);
previous = finger;
finger = next;

}
head = previous;

}

public int size()
// post: returns the number of elements in the list
{

return size(head);

}

protected int size(Node e)
// pre: e is an element of the list, or null
// post: returns the number of elements in the sublist headed by e
{

if (e == null) return 0;
else return 1 + size(e.next());

}

public boolean contains(Object o)
// pre: o is a valid object
// post: returns true if a value in this list is "equals" to o
{

return contains(o,head);

}

protected boolean contains(Object o,Node e)
// pre: o is a valid object, e is the head of a sublist of this
// post: returns true if a value in this sublist is equals to o
{

if (e == null) return false;
if (e.value().equals(o)) return true;
return contains(o,e.next());

}

This is a hard problem.

9.7
linear, and then work out the boundary conditions.

It is made easier if we imagine the list to be

public void reverse()
// post: this list is in reverse order.
{

if (tail == null) return;
Node previous = tail;
Node finger = tail.next();
tail.setNext(null); // now, it’s a singly linked list
tail = finger;

A.2 Solutions to Odd-Numbered Problems

473

while (finger != null)
{

Node next = finger.next();
finger.setNext(previous);
previous = finger;
finger = next;

}

}

9.9

Here’s the add method from the CircularList implementation:

public void add(int i, E o)
// pre: 0 <= i <= size()
// post: adds ith entry of list to value o
{

Assert.pre((0 <= i) && (i <= size()),"Index in range.");
if (i == 0) addFirst(o);
else if (i == size()) addLast(o);
else {

Node<E> previous = tail;
Node<E> next = tail.next();
while (i > 0)
{

previous = next;
next = next.next();
i--;

}
Node<E> current = new Node<E>(o,next);
count++;
previous.setNext(current);

}

}

9.11 Because we have backed up the reference, it is farther away from the ref-
erences that need to be manipulated. In general, the construction of the head
and tail references will take longer and operations will execute more slowly.
Clearly, though, the method removeLast has the potential for great improve-
ment; it becomes an O(1) operation, instead of O(n).
9.13 Theoretically, we expect the following outcome:

1. For addFirst, all linked list implementations perform better than Vector.
Singly linked lists are only slightly better than circular and doubly linked
lists.

2. For remove(Object) a Vector takes O(n) time to ﬁnd and O(n) time to
remove the value. The removal is constant time for linked implementa-
tions.

3. For removeLast, Vector, doubly and circularly linked list classes are all

potentially much faster than singly linked lists.

474

9.15

Answers

public int size()
// post: returns the number of elements in the list
{

return size(head);

}

protected int size(Node e)
// pre: e is an element of the list, or null
// post: returns the number of elements in the sublist headed by e
{

if (e == null) return 0;
else return 1 + size(e.next());

}

9.17 If we add new words to the end of the list, the ﬁnal vocabulary list is in
an order that is well correlated to decreasing frequency. If words are used infre-
quently, they will likely appear later in the input and thus in the list. Frequently
used words are likely to be encountered quickly. This is a help: we want high-
frequency words to appear near the front of the vocabulary list to terminate the
search quickly.

Chapter 10

Problems begin on page 245.
10.1 The stack contains values n − m from 1 (at the bottom) to n − m (on the
top).
10.3 One additional stack is necessary.

public static void copy(Stack s, Stack t)
// pre: s and t are valid stacks; t is empty
// post: the elements of s are copied to t, in order
{

Stack temp = new StackList();
while (!s.isEmpty()) temp.push(s.pop());
// Assertion: the elements of the stack are reversed, in temp
while (!temp.isEmpty())
{

Object item = temp.pop();
s.push(item); t.push(item);

}

}

10.5 Determine the size of the queue, say, n. Perform n dequeue operations,
enqueuing the result both at the end of the original as well as the copy:

public static void copy(Queue q, Queue r)
// pre: q and r are valid queues; r is empty
// post: the elements of q are copied to r, in order

A.2 Solutions to Odd-Numbered Problems

475

{

}

int n = q.size();
for (int i = 0; i < n; i++)
{

Object item = q.dequeue();
q.enqueue(item); r.enqueue(item);

}

10.7 The following recursive program should do the trick. Note that the ele-
ments of the queue are stored on the stack that manages recursion!

public static void reverse(Queue q)
// pre: q is a valid queue
// post: the elements of q are reversed
{

if (!q.isEmpty())
{

Object o = q.dequeue();
reverse(q);
q.enqueue(o);

}

}

10.9 At any time, the elements of the stack are in decreasing order from the
top to the bottom. Any values that are missing from top to bottom must have
been pushed and popped off already. Thus, at any point in the sequence if you
pop off two values l and ultimately k (k < l), then all values between k and l
must already have been popped off before l.

Thus, for n = 4, there are several impossible sequences: those that mention

3, 1, 2 in order, or 4, 1, 2, or 4, 1, 3, or 4, 2, 3.
10.11 It is only possible to remove elements from a queue in the order they
are inserted. Thus, it is only possible to remove the elements in the order
1, 2, 3, . . . , n.
10.13 Suppose that we have two stacks, head and tail. The bottom of head
contains the head of the queue, while the bottom of tail contains the tail of the
queue. Enqueuing takes time proportional to the size of tail, while dequeuing
takes time proportional to head.
10.15 This is a queue structure, where the elements contain lists of food items.

Chapter 11

Problems begin on page 273.
11.1

1

1

1

0

9

9

476

Answers

-1

0

1

9

-1

0

1

3

9

11.3 The older value is found in a location with a lower index.
11.5
procedure:

To insert a value into an OrderedList we must execute the following

public void add(E value)
// pre: value is non-null
// post: value is added to the list, leaving it in order
{

Node<E> previous = null; // element to adjust
Node<E> finger = data;
// search for the correct location
while ((finger != null) &&

// target element

ordering.compare(finger.value(),value) < 0)

{

previous = finger;
finger = finger.next();

}
// spot is found, insert
if (previous == null) // check for insert at top
{

data = new Node<E>(value,data);

} else {

previous.setNext(

new Node<E>(value,previous.next()));

}
count++;

}

The initialization statements do not take any time. The while loop executes
once for each smaller value. In the worst case, the value inserted is larger than
all values within the list; the loop must be executed n times. Actual insertion
involves the construction of a new Node—constant time. The overal time is,
then, O(n).
11.7 An OrderedVector is faster at ﬁnding the correct location for a value,
but an OrderedList is faster at performing the insertion. Both, then, have O(n)
insertion time. The main distinction between the two is in get-type accesses.
The OrderedVector can make use of the binary search approach, while the
OrderedList cannot. When you expect many accesses, the OrderedVector is a
better structure.
11.9 Here is a straightforward implementation:

public int compareTo(Object other)
// pre: other is a valid Strg
// post: returns value <, ==, or > to 0 if this <, ==, or > that
{

Strg that = (Strg)other;
int thisLen = this.length();

A.2 Solutions to Odd-Numbered Problems

477

int thatLen = that.length();
char thisc, thatc;
int i;
// as long as one string has a character
for (i = 0; i < thisLen || i < thatLen; i++)
{

thisc = (i < thisLen) ? this.charAt(i) : 0; // char or 0
thatc = (i < thatLen) ? that.charAt(i) : 0; // char or 0
if (thisc != thatc)
{

return (int)(thisc - thatc);

}

}
// strings are the same
return 0;

}

The proprietary code for the method is also available from Sun at www.sun.com.
11.11 This performs a comparison between objects of type Person, containing
ﬁrst and last names stored as String types:

public int compareTo(Object other)
// pre: this and other are people with first and last names
// post: returns relationship between people based on last name, or
// first name if last names are the same.
{

Person that = (Person)other;
// compare last names
int lastCompare = this.last.compareTo(that.last);
if (lastCompare != 0) return lastCompare;
// compare first names if necessary
return this.first.compareTo(that.first);

}

11.13 For the OrderedVector, the get method consistently involves a binary
search method that demands O(log n) time; best-, expected-, and worst-case
behaviors are logarithmic.

For the OrderedList the get method is linear: searching for larger values
takes a proportionately longer time. The best case is constant time; the average
and worst case are O(n).

Chapter 12

Problems begin on page 309.
12.1 In a preorder traversal: U, T, I, M, S, D, A, D.
In a postorder traversal: M, I, T, A, D, D, S, U.
In an in-order traversal: M, I, T, U, S, A, D, D.
12.3

478

Answers

a. (1)

b.

(-)

(+)

(1)

*
5 3

/
4 2

c.

(+)

(1)

5

*

3

/

-

2

4

d.

(*)
(+) -
(1) 5 3 /

4 2

e. (+)
(1)

*

5

-

3

/
4 2

12.5 We see that

R = 1 + (L − 1) ∗ 2
R = 1 + 2L − 2
R = 2L − 1

In terms of expression trees this becomes:

R

=

1

+

*
- 2

L 1

=

R

+

=>

1

-

=>

*
L 2

*
1 2

=

R

-

*
L 2

-
2 1

12.7 The List classes, if deﬁned recursively, would, probably, have more re-
cursively deﬁned functions. For example, many of the index-based methods
(e.g., get(i) and set(i,v), would be deﬁned recursively: they would work on
the current element if i = 0, and otherwise recursively act on the next element.
12.9 Here is an outline of the proof: If the time is less than linear in the
number of nodes, then some node is not considered during the computation.
Suppose that, in the computation of the height of a tree, node x does not need
to be considered. We can, then, construct a tree of similar structure, except
that a very tall subtree—a tree tall enough to change the height of the overall
tree—can be hung from a descendant of x. Since x is not considered in the
height computation of the original tree, the height of the revised tree will not be
computed correctly, because the height-determining subtree hangs from below
x.
12.11 This is very similar to the implementation of the clone method.

public BT copy()
// post: contructs a structural copy of this binary tree
{

if (isEmpty()) return this;
return new BT(value(),left().copy(),right().copy());

// don’t copy the empty tree

}

12.13 Our implementation is the same as that of Problem 12.14. Another
approach keeps track of a Vector of children.
12.15

import structure.*;

A.2 Solutions to Odd-Numbered Problems

479

public class TreeIterator extends AbstractIterator
{

protected Tree root;
protected Stack todo;
public TreeIterator(Tree root)
{

todo = new StackList();
this.root = root;
reset();

}

public void reset()
// post: resets the iterator to retraverse
{

todo.clear();
// stack is empty. Push on the current node.
if (root != null) todo.push(root);

}

public boolean hasNext()
// post: returns true iff iterator is not finished
{

return !todo.isEmpty();

}

public Object get()
// pre: hasNext()
// post: returns reference to current value
{

return ((Tree)todo.getFirst()).value();

}

public Object next()
// pre: hasNext();
// post: returns current value, increments iterator
{

Tree old = (Tree)todo.pop();
Object result = old.value();
Stack s = new StackList();
Tree child = old.oldest();
while (!child.isEmpty())
{

s.push(child);
child = child.siblings();

}
while (!s.isEmpty())
{

todo.push(s.pop());

}
return result;

480

Answers

}

}

12.17 Left to the reader.
12.19 This version is part of the BinaryTree implementation.

public boolean isComplete()
// post: returns true iff the tree rooted at node is complete
{

int leftHeight, rightHeight;
boolean leftIsFull, rightIsFull;
boolean leftIsComplete, rightIsComplete;
if (isEmpty()) return true;
leftHeight = left().height();
rightHeight = right().height();
leftIsFull = left().isFull();
rightIsFull = right().isFull();
leftIsComplete = left().isComplete();
rightIsComplete = right().isComplete();

// case 1: left is full, right is complete, heights same
if (leftIsFull && rightIsComplete &&

(leftHeight == rightHeight)) return true;

// case 2: left is complete, right is full, heights differ
if (leftIsComplete && rightIsFull &&

(leftHeight == (rightHeight + 1))) return true;

return false;

}

12.21 Here we implement the method as described. Note the important changes
to isFull and height.

protected int info()
// pre: this tree is not empty
// post: height of tree if it is full, or negative of that value
//
{

if the tree is not full

if (left().isEmpty() && right().isEmpty()) return 0;
int leftInfo = left.info();
int rightInfo = right.info();
int result = 1+Math.max(Math.abs(leftInfo),Math.abs(rightInfo));
if (leftInfo<0 || rightInfo < 0 || leftInfo != rightInfo)

result = -result;

return result;

}

public int height()
// post: returns the height of a node in its tree
{

if (isEmpty()) return -1;
return Math.abs(info());

A.2 Solutions to Odd-Numbered Problems

481

}

public boolean isFull()
// post: returns true iff the tree rooted at node is full
{

if (isEmpty()) return true;
return info() >= 0;

}

12.23 The simplicity of the in-order iterator ﬁx is that, given a node that you
are currently processing, ﬁnding the next node is fairly simple. In the in-order
case, for example, ﬁnding the next node to consider is no worse than the height
of the tree.

Preorder traversals simply need to move left if there is a left child, or move
right if there is a right child, or move up until we leave a left subtree and there
is a right subtree: we then go to the right. Complexity is no worse than height
of the tree between iterations.

In the case of postorder, we must always move upward. If we arise from a
left subtree, head back down the right. If we arise from a right subtree, stop at
the parent. Complexities are similar.

For level-order traversal, we must increase the complexity considerably be-
cause of the difﬁculty of moving between sibling nodes. To move right, we go
up until we arise from a left subtree and head down a similar number of levels
to the leftmost child (if any). If there is no such sibling, we must head to the
root and down to the leftmost descendant at the next level down. This code is
quite complex, but can be accomplished. A queue is a better solution!
12.25 If there are n = 0 full nodes, there is one node, a leaf, so the observation
holds. Now, suppose the condition holds for all values less than n. The root of
the tree has degree k. The k subtrees each have subtrees with n0, n1, . . . , nk−1
full nodes. By the inductive hypothesis, these subtrees have (k − 1)ni + 1 leaves.
The total for all subtrees is

k
X

i=0

(k − 1)ni + 1 = (k − 1)(n − 1) + k = (k − 1)n + 1 − k + k = (k − 1)n + 1

and the theorem holds. By mathematical induction on n, the theorem holds for
all g ≥ 0.

Chapter 13

Problems begin on page 338.
13.1

3

4

4

7

3

3

3

0

7

4

482

Answers

0

2

2

2

6

7

7

4

4

4

3

3

3

8

8

7

0

0

13.3

1

3

4

5

2

4

13.5 A VectorHeap is not a Queue because it does not support the FIFO or-
dering. It is not an OrderedStructure.
13.7 For a PriorityVector, only the compareTo method is used to locate the
correct position. Equal priorities may appear separated, but they will appear
to be consistent from the point of view of compareTo. The same is true with
SkewHeaps.
13.9 In fact, no additional methods are necessary.
13.11 Notice ﬁrst that heapify of Problem 13.10 is linear even if the data
are already a heap. For add, the best- and worst-case running times can be
made constant: simply add the value to the end of the Vector. For the remove
method, the best-case time is O(log n) when no heapify is necessary.
In the
worst case, the time is O(n), if a heapify is necessary.
13.13

a. Descending. The minimum value is removed and swapped with the last

element of the Vector.

b. The worst-case time-complexity is O(n log n), when the data are all unique.

c. The best-case time-complexity is O(n), when all the data are the same.

Removing a value from a heap in this case requires constant time.

13.15 First, we suppose that k is a constant determined at compile time. We
can, then, keep an OrderedVector of k of the largest values, and a heap of the
remaining values. When we add a value, we ﬁrst add it into the OrderedVector
(in constant time), and then remove the last element and insert it into a heap
of the remaining values. As we remove values from the Vector, we can absorb
values from the heap as necessary. The worst-case running time is O(log n).
13.17 Every full node has two states: swapped and unswapped. Since no full
node can undo the swapping of another node, every swap has a unique effect.
Since there are n
2 different
heaps related by swapping children.
13.19 This percentage is quite small. Every node (save the root) has a par-
ent. The chance that the child is less than the parent is 50 percent. We have,
then, that 1 in 2n−1 trees is a heap. (This is easy to see if we just consider the
degenerate trees.)
13.21 The Vector could be as large as 2n − 1 elements. Consider a rightmost
degenerate tree.

2 interior nodes that may be swapped, there are 2 n

A.2 Solutions to Odd-Numbered Problems

483

13.23 Certainly this can be accomplished in O(n log n) time.
13.25 We simply add new values to the end of the Vector, checking to see if
the new value should be swapped for the old minimum. When we are asked
to remove a value, we hand over the leftmost value and heapify the rest of the
Vector. Adding is constant time; removing is linear time.

Chapter 14

Problems begin on page 365.
14.1 The nodes of a binary search tree are kept in an order that allows them
to be inspected in increasing order with an in-order traversal.
14.3

1

1

1

2

3

2

3

2

3

1

2

3

1

2

3

Only the last is a binary search tree.

484

Answers

14.5

14.7

3

2

0

3

5

6

4

5

3

2

0

3

3

5

6

4

5

14.9 One need only observe that values are added as a leaf in a unique loca-
tion. A level-order traversal of the binary search tree desired gives an appropri-
ate order for insertion.
14.11 This is, effectively, the construction of a Linear structure, a degener-
ate rightmost search tree. Behavior is O(n2). (The best-case running time for
arbitrary order is O(n log n).)
14.13 Have it ﬁrst use the successor, then predecessor, then successor, and so
on. Another possibility is to pick randomly, with equal probability.
14.15 At each stage of the iteration, there may be a signiﬁcant search—perhaps
as great as O(n)—for the next value. Over the entire iteration, however, the
total expense is only linear.
14.17 Yes. However, one must be careful to avoid causing splays of the trees
by directly accessing the tree itself. Such behavior could increase the cost of the
iteration.
14.19 Provided that no splays occur during the iteration, it is fairly simple to
see that each edge of the splay tree is traversed twice during the iteration. Over
n next calls, this averages out to be two iterations of a loop in the next method
per call.

Chapter 15

Problems begin on page 399.
15.1 No. If the keys were equal, the second key inserted would have found
the ﬁrst.
15.3 The state of the table changes as follows:

4

0

2

1

Index
Insert D
Insert a
Insert d
Insert H a H
Insert a
Insert h

3
D
D
D d
D d
a H a D d
a H a D d

a
a

5

6

h

A.2 Solutions to Odd-Numbered Problems

485

15.5 As was the case with Problem 15.4, it is possible to have a hash table
appear full when, in fact, it is only partially full.
15.7 Get the integer representation of the real number. Simply call Double.-
doubleToLongBits(d). The result is a suitable hash code when reduced to an
int. The Sun implementation performs the exclusive-or of the two ints that
make up the longword.
It is best to design a hash table to be several times larger than the data
15.9
to be stored. If this is not done, the collisions will decrease the performance
considerably.
15.11 This technique has all the disadvantages of exclusive-or, but with little
knowledge of the contained data, it is suitable. The Sun implementation over-
ﬂows after the ﬁrst seven or eight entries.

public int hashCode()
// post: returns the hashcode associated with non-null values of this
// vector
{

int result = 0;
for (int i = 0; i < size(); i++)
{

Object temp = get(i);
if (temp != null) result = result ^ temp.hashCode();

}
return result;

}

15.13 This technique works for all collections:

public int hashCode()
{

int result = 0;
Iterator it = iterator();
while (it.hasNext())
{

Object temp = it.next();
if (temp != null) result = result ^ temp.hashCode();

}
return result;

}

15.15 The value of i can get to be as large as log2 l, so the algorithm is poten-
tially logarithmic in the length of the string.
15.17 No. It’s only ordered from the perspective of rehashing. The smallest
value, for example, may appear in any location to which it hashes—anywhere
in the table.
15.19 One approach would be to hash together all the coordinates for each of
the checker pieces. Another approach is to consider each of the rows to be a
12-bit integer, where each useful square is represented by 3 bits that encode
empty, red, red-king, black, or black-king. The hash code for the entire board is
then the exclusive-or of all rows, where each is shifted by three bits.

486

Answers

15.21 One would have to count the number of times that the hashCode method
for the String class was called and compute the expected frequency per String.
(It is likely that this value is larger than 1; compiled Strings are manipulated
fairly frequently, for example, in the lifetime of a compiled program.) If this
value is larger than 1, the hash code could be computed once, at only the cost
of space. If the average length of a String was short—say 4 or less, the storage
of the hash code might not be space efﬁcient.

Chapter 16

Problems begin on page 435.
16.1

16.3

16.5 The power matrix describes which nodes are connected with paths of n
or fewer edges: those with zero entries cannot be reached in n steps.
16.7

a.

TFFFTTTabcdabcdTTTTFTTTTabcdaababcbcdcdd012333b120340561125FTFFFFFFFFFTTFFFFTFFFFFTFFTFF01234560123456FFTFTFTTTFFFTFFFFFFFDover (380)Montpelier (0)Boston (130)Athens (4600)Kuala Lumpur (9300)Bangkok (9950)Sacramento (2400)Phoenix(2950)Salt Lake City(1950)Harrisburg(450)Trenton(400)Albany (250)750015013012025042001002004700150080007500450550650Distances to Montpelier, VermontAll distances are approximate nautical miles.4300A.2 Solutions to Odd-Numbered Problems

487

b.

c.

This is discussed fully in the text. The adjacency list is used when the
16.9
graph is relatively sparse, and is likely to use less space than the adjacency
matrix implementation. The adjacency matrix is used when graphs with more
than O(n) edges connect n nodes. It allows for fast access to speciﬁc edges—all
in constant time, but at the cost of O(n2) space.
16.11 For the adjacency list, we must check each node’s list to see if the poten-
tial source node is mentioned. In the adjacency matrix we check for non-null
entries in the column associated with the row (don’t check the diagonal entry).
16.13 An undirected edge may be considered a 2-cycle. Topological sorting
depends on the transitivity of the ordering suggested by the edges. If there is a
cycle, the edges do not suggest a consistent order.
16.15 Adding a vertex to the graph can be accomplished in near constant time.
Adding an edge involves a O(e) check for the presence of the edge already.
Removing a vertex involves a O(e) scan through the edges to remove those
connected to the vertex. The edge iterator is ideally constructed from the list
iterator. The vertex iterator is constructed from an iterator over the Map.
16.17 This can happen if multiple queries are made regarding a single source.
One calculation of all solutions is less expensive than computing several partial
solutions; some aspects of different solutions are shared.
16.19 Suppose it did not. This assumption leads to a contradiction.

AlbanyTrentonMontpelierDoverBoston1501304200120250100Athens47001500Salt Lake CityHarrisburgPhoenixSacramentoBangkok450550650Kuala LumpurConnecting the worldAll distances are approximate nautical miles.AlbanyTrentonMontpelierDoverBoston750015013043001202504200100200Athens47001500Salt Lake CityHarrisburg80007500PhoenixSacramentoBangkok450550650Kuala LumpurDrive camels only on noncrossed edges.All distances are approximate nautical miles.Camel Networks488

Answers

Appendix B

Beginning with Java

Concepts:
. A sip of Java

I love the java jive and it loves me.
—Milton Drake and Ben Oakland

THE JAVA PROGRAMMING LANGUAGE WAS DESIGNED at Sun Microsystems as a
simple, object-oriented, portable programming language for supporting Internet-
based products. With large numbers of students, educators, and developers
writing applications once written in languages such as C, C++, Fortran, and
Pascal, the language is developing rapidly. As it gains wider acceptance, it will
be increasingly important for programmers of these other languages to become
familiar with the important features and paradigms introduced by Java. This
Appendix serves as a quick introduction to the features of Java. Other, more
in-depth treatments are available over the Internet or through your bookstore
or local library.

B.1 A First Program

The main focus of a Java programmer is to write classes. These classes are tem-
plates for structures called objects. In most environments, the code supporting
each class is placed in a dedicated ﬁle by the same name (class Sort is found in
Sort.java). Writing a Java program involves writing a class deﬁnition. Here is
a good ﬁrst program to consider:

import structure.*;

public class MyFirstProgram
{

public static void main(String[] arguments)
{

// print a message to the standard output stream
System.out.println("Look Mom: know Java!");

}

}

At the top of the program, the import statement searches the programming

environment for a package or library of classes called structure:

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import structure.*;

and makes all of them (.*) available for use within the program. The structure
and structure5 packages are the subject of this book; the applications of this
text will include this import statement. Other packages available with Java in-
clude java.lang—the package of items automatically imported (this includes,
for example, the deﬁnition of System, String, and Comparable); java.io—the
package that provides access to special-purpose I/O facilities; and java.util—
a package that contains utility classes, including random number generators,
date objects, and simple data structures and interfaces. Documentation for
these packages is freely available online from developer.java.sun.com and
in a wide variety of trade books.
The class is marked public:

public class MyFirstProgram

which means that it is available for access by anyone who wants to use the
class—in particular anyone who wants to run the class as a program.

Classes that contain a method called main may be run as applications:

public static void main(String[] arguments)

When the application is run, the main method is executed. The sole param-
eter (here, arguments) is an array of Strings that are passed to it from the
programming environment. One common mistake is to incorrectly declare the
main method. The result is that a method main is declared, but the method
required to run the application is not present. In such cases, the error

Exception in thread "main" java.lang.NoSuchMethodError: main

is common.

Any text on a line following a double-slash (//) is considered a comment:

// print a message to the standard output stream

and is ignored by the compiler. You can make multiple line comments by en-
closing the text between /* and */.

The computer prints a single string to the standard output:

System.out.println("Look Mom: know Java!");

The print method is part of the out stream from the System object. This dotted
notation is used to provide names of data and methods within an object. For
example, another name for the main procedure is MyFirstProgram.main.

Java is case sensitive; it is important to make sure identiﬁers are all in the
same case. By convention, packages and methods begin with lowercase letters,
while classes begin with uppercase.

Occasionally, two packages contain classes with the same name.

If both
classes are imported simultaneously, they are only accessible if preﬁxed by their

B.2 Declarations

491

respective package names. While the package system is a convenient means
of compartmentalizing name spaces, programmers are encouraged to consider
potential naming conﬂicts with common packages before designing their own
classes. Still, over time with a moving target like the Java run-time environ-
ment, naming conﬂicts are sometimes difﬁcult to avoid.

B.2 Declarations

The scope of an identiﬁer—the name of a variable, method, or class—is the
range of code that may legally access the identiﬁer. Variables in Java may be
declared in a number of different places:

• Variables must be declared within a class or method. Variables declared
within a method are local to the method and may not be accessed outside
the method. Variables declared within the class are instance variables
whose access is determined by access keywords.

• Variables and values declared within a class are visible to any method
declared as part of the class. Depending on the protection of the variable,
it may be seen in subclasses, within the class’s package, or globally.

• Variables and values declared within a method can be seen within code
following the declaration that is delimited by the tightest enclosing braces.
Ideally variables are declared at the top of a method, but they may be
declared at the top of the block where they are used.

Methods may only be declared within a class. Unlike Pascal, they may not be
nested. An understanding of the access control keywords for Java is important
to understanding their scope. Section B.8 discusses this topic in considerable
detail.

B.2.1 Primitive Types

Two types of data are available through Java–primitive types and reference types.
Primitive types include integers, ﬂoating point numbers, booleans, and charac-
ters:

Name

boolean
char
int
long
float
double

Type
Boolean
Character data
Integer
Long integer
Real
Double precision real −1.7977E308 . . . 1.7977E308

Range
true or false
Any character
−231 . . . 231 − 1
−263 . . . 263 − 1
−3.4028E38 . . . 3.4028E38

Default Value

false
’\0’ (null)
0
0
0.0
0.0

Java variables are automatically initialized to their default values, but program-
mers are best advised to initialize variables before they are used.

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Java supports a large variety of operators for primitive data types. Here are

the most common, in decreasing order of precedence:

Operator Meaning

++
--
!
* / %
+ -

Unary increment
Unary decrement
Logical not
Multiplication, division, remainder
Addition, subtraction
< <= => > Numeric comparison

== !=
&&
||
=

Primitive and reference equality test
Logical and
Logical or
Assignment

Notice that assignment (=) is an operator and can be used to generate complex
(and unreadable) expressions:

int a, b, c = 0;
a = 1 + b = 1 + c; // a = (1 + (b = (1 + c)))

leaves c at 0 and sets b to 1 and a to 2. Pascal programmers should note that
the assignment (=) and equality (==) operators are easily confused (although
the compiler is often capable of detecting such mistakes). The assignment and
equality tests work on references as well, but not the objects to which they refer.
Programmers should be aware that the division operator (/) returns only the
integer portion of the quotient when both operands are integers. The remainder
operator (%) recovers the remainder after division. As usual, one should perform
integer division and remainder on negative numbers with care.

Characters in Java are encoded using a very general mechanism called uni-
code, which is a superset of ASCII characters. Characters are surrounded by
apostrophes and have the following shorthands familiar to C and C++ pro-
grammers:

Escaped character Meaning

Escaped character Meaning

’\b’
’\r’
’\t’
’\"’

Backspace
Carriage return
Tab
Quotation mark

’\f’
’\n’
’\’’
’\\’

Form feed
New line
Apostrophe
Backslash

Primitive types may be explicitly converted to other types through casting.
When information would not be lost, Java may perform the type conversion
automatically. The following removes the fractional component of a double
value:

double d = -3.141;
int i = (int)d;

B.2 Declarations

493

Associated with each primitive type is a corresponding class. For example,
the class java.lang.Integer can hold int types, java.lang.Character holds
char values, and so forth. As of Java 5 the conversion between the primitive
and object versions can be handled automatically in a process called autoboxing.
So, for example, the following is possible:

Integer u = 3;
Integer v = 2;
Double result = Math.sqrt((double)(u+v));
System.out.println(result);

which prints

2.23606797749979
*/

Unlike C-style languages, Java does not automatically convert non-boolean ex-
pressions to true or false values.

B.2.2 Reference Types

Java does not have explicit pointers. Instead, more tightly controlled references
are provided. Reference types include arrays and objects. Arrays in Java must
be explicitly allocated and, as a result, may have run-time-determined bounds.
The following code demonstrates several means of initializing arrays; all are
identical.

int primes1[] = {2,3,5,7,11};
int primes2[] = new int[5];
int primes3[] = primes2;
primes[0] = 2;
primes[1] = 3;
primes[2] = 5;
primes[3] = 7;
primes[4] = 11;

The arrays primes2 and primes3 refer to the same dynamically allocated mem-
ory; primes1 refers to another chunk of memory with similar values. An array
or object may be referred to by more than one reference variable at the same
time. Thus, the result of comparing primes1==primes2 is false (the references
refer to different memory), but primes2==primes3 is true. The length of an
array may be determined by inspecting its length property: primes2.length
is 5.

Every reference either refers to no object (called a null reference) or an
instance of an object of the appropriate type. Arrays may be null, in which
case they refer to no memory.

References to objects perform a level of indirection. Thus

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But remember:
Java has no
pointers!

String s;
s = new String("Measure for measure must be answered.");
System.out.println(s.length());

calls the length method referred to by the String s. If s were null, the indirec-
tion would be impossible, leading to a null pointer exception. The following
code, then, generates a null pointer exception:

String s; // a null reference since no string allocated
System.out.println(s.length()); // illegal!

Java works hard to ensure that references do not point to invalid objects. In
traditional languages, signiﬁcant errors can be incorporated into programs by
incorrectly allocating or freeing dynamic memory. Java’s approach is to force
the user to explicitly allocate each nonprimitive object and rely on a garbage
collector to free allocated memory, but only after it is known to be unreferenced.
This approach has some cost associated with it (the garbage collector has no
hints as to where to ﬁnd likely garbage), but a reference is guaranteed to never
point to invalid memory.

B.3 Important Classes

Several classes are used in nearly every Java program. As with all classes, they
are also fully documented online.

B.3.1 The structure.ReadStream Class

Before Java 5Java made little effort to support input of primitive types from
streams. As a result, the structure package provides a simple class for perform-
ing Pascal-like reading of primitive values from the keyboard. A ReadStream can
be attached (much like a ﬁlter) to an input stream and can then be made to read
primitive values. It is, perhaps, best to learn by observing an example:

ReadStream r = new ReadStream();
int i;

ReadStream

for (r.skipWhite(); !r.eof(); r.skipWhite())
{

i = r.readInt();
if (isPrime(i)) System.out.println(i);

}
r.close();

Here, a new ReadStream is created. By default, it is attached to the standard
input stream, System.in. The method skipWhite scans and disposes of char-
acters on the input, as long as they are white space (including tabs and end-of-
line marks). The eof method returns true if no more data are available on the

B.3 Important Classes

495

ReadStream. Finally, readInt reads in an integer from the stream. The value
is stored in i. ReadStreams are capable of reading in all the primitive types,
including doubles and booleans:

Method
r.readChar()
r.readBoolean()
r.readInt()
r.readLong()
r.readFloat()
r.readDouble()
r.readString()
r.readLine()
r.readln()
r.skipWhite()
r.eof()
r.eoln()

Reads
next char from stream r
next boolean from stream r
next int from stream r
next long from stream r
next float from stream r
next double from stream r
next word, returned as String, from stream r
next line, returned as String, from stream r
next line, returning no value, from stream r
until next nonwhitespace from stream r
returns true iff no more data in stream r
returns true iff no more data on line from stream r

The ReadStream class is also available in structure5, but programmers are
encouraged to use the newer Scanner class in the java.util package.

B.3.2 The java.util.Scanner Class

In Java 5 Sun introduced a new class, Scanner. This class provides greatly
improved, portable, and natively supported reading of primitive types from ﬁles.
Here’s a simple example, reading ints from the standard input:

import java.util.Scanner;
...
Scanner s = new Scanner(System.in);
int i;

while (s.hasNextInt())
{

i = s.nextInt();
if (isPrime(i)) System.out.println(i);

}
s.close();

In addition, the Scanner class implements Iterator<String>. The hasNext
and next methods check for and return tokens separated by white space.

Here are the primary methods of Scanner:

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Method
s.hasNext()
s.next()
s.hasNextChar()
s.nextChar()
s.hasNextBoolean()
s.nextBoolean()
s.hasNextInt()
s.nextInt()
s.hasNextLong()
s.nextLong()
s.hasNextFloat()
s.nextFloat()
s.hasNextDouble()
s.nextDouble()
s.hasNextLine()
s.nextLine()

Reads
returns true iff scanner has an unread token
next String (a token) from scanner s
returns true iff scanner has a char next
next char from scanner s
returns true iff scanner has an boolean next
next boolean from scanner s
returns true iff scanner has an integer next
next int from scanner s
returns true iff scanner has an long next
next long from scanner s
returns true iff scanner has a ﬂoat next
next float from scanner s
returns true iff scanner has a double next
next double from scanner s
returns true iff scanner has an unread line
next line (as String), from scanners

If one s.has method fails, the stream is left in the original state so that succes-
sive checks may be accurately performed.

B.3.3 The PrintStream Class

Simple output can be accomplished by writing to a PrintStream, usually System.out.
The interface for the standard output has been greatly improved in Java 5. Use-
ful methods include:

Writes
Write primitive to standard output

Method
print(<primitive>)
println(<primitive>) Write primitive, followed by new line
print(o)
println(o)
printf(format,args) Write arguments using String format

Write toString representation of o
Write object o followed by new line

An important feature of this interface is that any subclass of Object—really,
any class—has a toString method. This method is invoked by the print and
println methods to generate a readable form of the Object.

The printf form is new to Java 5. It requires a format string that determines

how the arguments that follow are printed. For example,

System.out.printf("%d (decimal) is %x (hexadecimal)\n",392,392);

prints the number 392 as decimal and hexadecimal values. The arguments are
consumed by corresponding format descriptors that begin with %. Here are the
common formats:

B.3 Important Classes

497

Descriptor
%b
%c
%d
%o
%x
%f
%g
%%

Format
Write a Boolean value.
Write a Character value.
Write an Integer in base 10 (decimal).
Write an Integer in base 8 (octal).
Write an Integer in base 16 (hexadecimal).
Write a ﬂoating point value.
Write a ﬂoating point value in scientiﬁc notation.
Write a percent sign.

Note that there is no -ln version of printf; newlines must be explicitly pro-
vided in the format string.

B.3.4 Strings

Strings, in Java, are implemented as immutable objects. These objects are spe-
cial in a number of ways. First, strings may be initialized to string constants,
which are delimited by double quotation marks ("). Secondly, the addition
operator (+) is overloaded to mean string concatenation. When strings are con-
catenated with other primitive types, the primitive type is converted to a string
before concatenation. Thus, the following program constructs and prints one
string of the ﬁrst n integers:

String s = "";
int i;
for (i = 1; i <= 10; i++)
{

s = s + " " + i;

}
System.out.println(s);

Each String value is allocated dynamically from the heap, and may not be
modiﬁed. Thus, this program actually constructs 11 different Strings, and we
are printing only the last.

Other String methods include:

Method
s.length()
s.charAt(i)
s.compareTo(t)
s.equals(t)
s.indexOf(c)
s.indexOf(t)
s.substring(start,end)

Computes
s’s length
the ith character (starts at 0)
integer relating s and t
true if string s has same value as t
index of c in s
index of beginning of match of t in s
substring of s between start and end

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B.4 Control Constructs

Java provides many of the forms of control found in other languages, includ-
ing conditional and multiple-choice statements, and various forms of looping
constructs. Also included are a number of convenience operations that allow
unusual forms of control.

B.4.1 Conditional Statements

Java provides several conditional statements. The if statement allows code to
be conditionally executed, based on a boolean expression. Its general form is

if (<condition>) <statement>

If the boolean <condition> is true, the <statement> is executed. If the <condition>
is false, <statement> is ignored. Unlike Pascal-style languages, there is no
then keyword; instead, the parentheses around the <condition> are required.
An alternative form is the if-then-else statement:

if (<condition>)

<then-statement>

else

<else-statement>

In this form the <then-statement> is executed if the <condition> is true;
otherwise the <else-statement> is executed. Since the semicolon is part of
many statements, it is often the case that the then part of the if terminates with
a semicolon just before the keyword else. When if statements are nested, the
else matches the closest if statement that doesn’t yet contain an else. Thus,
the following statement checks for various values of integer i:

if (i == 0)

System.out.println("zero");

else if (i == 1)

System.out.println("one");

else if (i == 2)

System.out.println("two");

Such instances of cascading if statements are so common it is useful to express
them as multiple-choice switch statements:

switch (<expression>)
{

case <constant1>: <statement1>; break;
case <constant2>: <statement2>; break;

...

default: <default-statement>; break;

}

B.4 Control Constructs

499

When the <expression> takes on any of the values <constant1>, <constant2>,
and so forth, the respective <statement> is executed. The break statement is
not strictly part of the switch, but if it is missing, the control falls through the
various cases until a break is encountered. It is a good habit to introduce break
statements regularly in the switch. The default case is executed if none of
the cases match. It is a good idea to include a default statement in switch
statements and perform an appropriate action. It is interesting to note that the
structure package does not make use of the switch statement. The cascading
if statements, given previously, might be recoded as

switch (i)
{

// note: order doesn’t matter if breaks are used:
case 1: System.out.println("one"); break;
case 0: System.out.println("zero"); break;
case 2: System.out.println("two"); break;
default: // do nothing!

}

B.4.2 Loops

Loops are an important part of most programs. Casting your code using the
appropriate loop is important to making your programs as understandable as
possible. Java, like C-based languages, provides three types of looping con-
structs. In order of popularity, they are the for loop, the while loop, and the
do-while loop. Unlike Fortran and Pascal, each of the loops can be used—with
some modiﬁcation—as a substitute for any of the others. As a result, most pro-
grammers use rules of thumb. Some prefer to always cast loops as for loops,
while others attempt to avoid the for loop altogether. The best advice is to
use the style of loop that seems most natural—but be prepared to defend your
choice.

The most general of the three looping constructs is the for loop. It has the

form

for (<initialization>; <continue-test>; <iteration>)

<statement>

When the loop is encountered, the <initialization> is executed. At that
point, the loop is started and the <continue-test> (a boolean) is evaluated.
If false, the loop
If true, the body of the loop (<statement>) is executed.
is ﬁnished, and the execution continues at the statement that follows. After
each execution of the loop body, the <iteration> statement is executed before
reevaluation of the <continue-test>.

Here are some idiomatic for loops. First, to loop n times, the statement

for (i = 0; i < n; i++) ...

is used. Programmers unfamiliar with C-style languages might prefer

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for (i = 1; i <= n; i++) ...

but this often detracts from the readability of the code. Besides, many opera-
tions on Strings and arrays require a range of values from 0 to n-1. Stick with
the idiom, whenever possible.

The statement

for (i = s.nextInt(); i != 0; i = s.nextInt());

reads in integers from a stream until one is zero. Notice that the empty state-
ment (a lone semicolon) provides the body of the loop.

To loop forever, one need only use:

for (;;) ...

Here, the empty <continue-test> is interpreted as true. The equivalent

for (;true;) ...

is less idiomatic and not frequently used but is just as effective.

Often—as in the last example—it is useful to be able to terminate or control
the loop from within its body. If, for example, we wish to consume a stream
positive integers in a stream, we write:

for ( ; !s.hasNextInt(); )
{

int i = s.nextInt();
if (i <= 0) break;

}

This loop terminates when the end-of-ﬁle mark is read on the Scanner or when
a negative integer is encountered; break statements terminate the tightest en-
closing loop or switch statement.

Sometimes it is useful to jump to the next iteration of the loop from within

the body. Here’s an example of the use of continue:

for (i = 0; i < n; i++)
{

if (i == 2) System.out.println("two is prime");
if ((i % 0) == 0) continue;
// only odd numbers remain
... test for prime ...

}

This loop does not look for primes among even numbers (other than 2). When
an even number is encountered, it jumps to the i++ statement to begin the next
iteration of the loop.

New in Java 5 is the foreach form of the for loop. It may be used to traverse
the elements of any array, or any class that has an iterator method. Here, for
example, we print the values of an array, and their squares:

B.4 Control Constructs

501

int a[] = new int[10];
...
for (int i : a) {

System.out.printf("%d squared is %d\n",i,i*i);

}

Alternatively, we may print the values of a SkewHeap of BusStop types as fol-
lows:

PriorityQueue<BusStop> s = new SkewHeap<BusStop>();
...
for (BusStop b : s) {

System.out.println(b);

}

This loop performs exactly as the following:

PriorityQueue<BusStop> s = new SkewHeap<BusStop>();
...
for (Iterator<BusStop> si = s.iterator(); si.hasNext(); ) {

BusStop b = si.next();
System.out.println(b);

}

The two remaining loops are straightforward in comparison. The while loop

has the form

while (<continue-condition>) <statement>

It is equivalent to the while loop in many languages.
equivalent for loop:

It could be cast as an

for (; <continue-condition>; ) <statement>

Another loop, the do-while loop, is a while loop with its test at the bottom:

do {

<statement>

} while (<continue-condition>)

The <statement> is executed at least once, and the loop continues as long as
the <continue-condition> is true. (Compare with the repeat-until loop of
Pascal, whose condition is an exit condition.)

In the structure package, there are approximately 70 for loops, 135 while

loops, and exactly one do-while loop.

I’m rewriting
the do-while!

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B.5 Methods

Methods are declared within classes. Unless declared otherwise, methods may
only act on instances of an object. If there are no instances of an object, these
methods may not be called. If, on the other hand, a method is declared static,
it exists (and may be called) no matter the number of instances. This explains
the need to bootstrap the Java application with a static method, main.

Methods are declared much as they are in C. Here, for example, is a (very

inefﬁcient) method that checks to see if n is prime:

static boolean isPrime(int n)
{

int factor;
for (factor = 2; factor < n; factor++)
{

if ((n % factor) == 0) return false;

}
return true;

}

Somewhere later we ﬁnd:

if (isPrime(2)) System.out.println("Two is prime.");

n is passed by value—changing the value of n will not change the value of the
actual parameter passed to it. Unlike Pascal and C++ there are no means of
passing the parameter by reference or as a variable parameter. And, unlike C,
we have no means of generating a pointer to simulate passing by reference.
All primitive values are passed by value. Being references, all variables referring
to objects pass the object by reference. The reference, of course, cannot be
changed. All this makes it difﬁcult, for example, to exchange the value of two
parameters as you can easily do in most other languages.

B.6 Inheritance and Subtyping

Many object-oriented languages provide support for two useful concepts: inher-
itance and subtyping. Java is no exception.

B.6.1 Inheritance

Suppose, for example, we develop a class to perform the functions of a stop-
watch. It might be declared as follows:

public class stopWatch
{

protected double currentTime;

B.6 Inheritance and Subtyping

503

protected boolean running;
public stopWatch() {...} // code omitted for brevity
public void start() {...} // start the watch
public void stop() {...} // stop the watch
public double read() {...} // read the time

}

This watch provides three methods once the stopWatch is constructed. The
state of the watch is maintained in the protected running and currentTime
variables. A watch with more features could be declared as

public class lapWatch extends stopWatch
{

protected double memory[50];
protected int lapNumber;
public lapWatch() {...}
public void startLap() {...}
public double recallLap(int number) {...}

}

This class inherits the deﬁnitions in the stopWatch deﬁnition. It is as though
all the deﬁnitions found in the stopWatch class were textually included in the
deﬁnition of the lapWatch. In this way we think of the deﬁnitions added by the
lapWatch class as extending the deﬁnition of the stopWatch.

Every class declared in Java is an extension of some other class. Even the
stopWatch deﬁnition is a type extension: it extends the deﬁnition of Object
(short for java.lang.Object). The Object class provides several interesting
methods, including toString—which generates a String representation of the
watch—and hashCode—which generates an integer that ideally represents the
state of the watch (see Section 15.4.1). Since lapWatch extends stopWatch, it
indirectly extends Object, inheriting methods like toString.

Type extension is useful in reducing the amount of code that must be rewrit-
ten for each class. Whenever a new class must be written, a similar type of
object may be extended to inherit the hard work of previous programmers. In
Java, however, the extension of types actually serves to promote a stronger re-
lationship between types of objects—that one type of object may be substituted
for another wherever it appears. This is called subtyping.

B.6.2 Subtyping

It is often the case that programmers do not realize the full potential of the code
they write. Thus, in Pascal-like languages, it is difﬁcult to design data structures
that are capable of holding generic information. For example, it is not obvi-
ous how to write a general-purpose deﬁnition of “a list of things” because the
deﬁnition of “thing” must be completely determined when the code is written.
Another result is that a “list of integers” and a “list of reals” must be distinct
deﬁnitions since the lists contain different types.

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Java increases the utility of types by relaxing the type-checking system so
that wherever a class stopWatch is referred to, any type extension of a stopWatch
(including our lapWatch) works just as well. We say that stopWatch is a super-
type or base class and lapWatch is a subtype or subclass. The rule is that

Wherever a supertype can be used, a subtype works just as well.

The following code makes use of subtyping:

stopWatch minny = new lapWatch();
minny.start();

...

minny.stop();
System.out.println("Elapsed time is "+minny.read());

Even though minny refers to a lapWatch, we are informing the compiler that
we want to use only the stopWatch features of minny. Since minny’s actual type
(lapWatch) is an extension of the stopWatch, certainly the stopWatch methods
are all available (though they may have lapWatch-like behavior if the lapWatch
overrides their behavior).

In C++ the compiler and programmer are responsible for ensuring that op-
erations work on compatible types. In Java, the object to which minny refers
keeps track of its actual type. On occasion it is useful to verify that the type
information known to the compiler is consistent with that known to the ob-
ject. This is accomplished by a form of casting. We can improve the program’s
knowledge of minny’s type in the following manner:

( (lapWatch)minny ).startLap()

By placing the lapWatch type in parentheses, the run-time system veriﬁes that
labWatch is a supertype of minny’s actual type (it is). The object is not modiﬁed
in any way. After casting, the value of the expression is of type lapWatch and
the startLap method can be correctly invoked.

In Java, the way that subtypes are determined is by inheritance. Any type
that extends another is automatically a subtype. Language designers are not
completely convinced that subtyping and inheritance go hand in hand, but in
Java they do.

B.6.3 Interfaces and Abstract Classes

The Java interface allows the programmer to specify a template for verifying
the methods provided by a class deﬁnition. An interface is speciﬁed much like
a class, but methods may not have associated bodies. We might, for example,
have an interface for things that may be started and stopped:

public interface timer
{

B.6 Inheritance and Subtyping

505

public void start();
public void stop();

// this can be started
// and stopped

}

As new classes are developed, their public methods may support one or more
interfaces. Since the stopWatch supports both the start and stop methods,
the following deﬁnition of stopWatch requests that the compiler verify that the
stopWatch supports the timer interface. The result is a stopWatch that is also
a timer:

public class stopWatch implements timer
{

protected double currentTime;
protected boolean running;
public stopWatch() {...}
public void start() {...} // start the watch
public void stop() {...} // stop the watch
public double read() {...} // read the time

}

Since interfaces have no actual code associated with them, it is impossible to
It is possible, however, to use
explicitly create objects of the interface type.
subtyping to generate references to objects that implement timer features:

timer minny = new lapWatch();

minny, in this case, is a timer that (via the stopWatch class) is really a reference
to a lapWatch. The previous discussion of casting would apply to this example
as well.

Sometimes it is useful to develop partial or abstract classes. These classes are
partially implemented and demand extension before the class becomes concrete.
If, for example, we had several choices for actually constructing a stopWatch,
we might use the following deﬁnition:

abstract public class stopWatch
{

protected double currentTime;
protected boolean running;
public stopWatch() {...}
abstract public void start(); // start the watch
abstract public void stop(); // stop the watch
public double read() {...} // read the time

}

Here, we indicate to the compiler that the start and stop methods are part of
the stopWatch deﬁnition, but we’re not willing to commit to an actual imple-
mentation at this time (perhaps there are choices, or we just don’t know how).

506

Beginning with Java

Notice, by the way, that we have committed to the protected data and the deﬁ-
nition of the read method, so this is not like an interface. Still, because there
is no code associated with the start and stop methods, no instances of the
class can actually be constructed. We must depend on references to subtypes—
instances of extensions to this class—if we want to use a stopWatch in our
code. We warn the compiler that we are working with an incompletely speciﬁed
class by attaching the abstract keyword to the deﬁnition of the class and any
method headers that are left unspeciﬁed.

Notice, by the way, any concrete extension of the latest stopWatch class
must specify code for the start and stop methods. Thus, the last deﬁnition of
lapWatch must either specify those methods completely or be declared abstract
itself.

Java suffers somewhat from a desire to avoid mistakes of the past. As a result
it is impossible to have a single class extend multiple superclasses at the same
time. This multiple inheritance introduces some sticky problems into language
design. Still, it is possible for a single class to implement multiple interfaces
at the same time. The careful and conservative programmer is often rewarded
with easily understood code.

B.7 Use of the Assert Command

In Java 1.4, Sun introduced the assert keyword. This control construct al-
lows the programmer to write assertions directly in the code at a native level,
without the need for classes like structure.Assert. Careful use also allows
the programmer to remove all traces of the assertion easily, and for improved
performance.

Programmers with access to Java 1.4 should investigate the assert control

construct. It has the following two forms:

assert <condition>;

assert <condition>:<message>;

The ﬁrst of the two forms evaluates the <condition> and if it is true, continues
exectution with the next statement. If the <condition> is false, an assertion
error is thrown.

The second form of the statement allows the programmer to pass informa-

tion (<message>) about the failed statement to the program.

Because assertions are a relatively new feature to Java, they must be ex-
plicitly enabled. You should read your compiler’s documentation to determine
how to enable this feature. Compilers that do not have the feature enabled will
report a warning or error if the assert keyword is encounterd.

Until the assert statement is supported by a large number of environments,
we still recommend the use of a system like structure.Assert, which allows
the consistent checking of assertions in all current Java programming environ-
ments.

B.8 Use of the Keyword Protected

507

B.8 Use of the Keyword Protected

Before I built a wall I’d ask to know
What I was walling in or walling out,
And to whom I was like to give offense.
—Robert Frost

AT EVERY TURN, this text has advocated the use of the access-control keyword
protected. Because of the complexity of access control in the Java language, it
is important that the case for the use of protected be adequately argued. With
some reservation, we do that here.

To make the best use of data abstraction, we have argued that each data
structure be considered from two vantage points: the interface and the imple-
mentation. Again, the interface describes a public contract between the imple-
mentor and the user. The private implementation informs the machine how the
demands of the contract may be met.

For both parties—the user and implementor—to agree on a structure’s be-
havior, its interface must be visible wherever it is used. The “user,” of course,
may be the structure itself, another structure within the same package, or some
entity external to the package. We suggest the use of the word public to de-
scribe classes, methods, and ﬁelds that are to be visible to the user. (It is almost
never necessary to make data ﬁelds public since they are only accessed through
public methods.)

The implementation, on the other hand, should be hidden from the user
as much as possible. Only the implementor should be able to see the internal
workings of the structure. When implementations are protected in this way,
then it is possible for the implementor to make changes to the implementation
without affecting the user’s applications—as long as contracts are met. Indeed,
languages with dynamic loading of classes, like Java, make it possible for im-
plementors of data structures to update applications that have already been
compiled. For maintenance and reliability reasons, this is a feature.

Unfortunately, the means of protecting the implementation is not very clear
in Java. To understand the problem we review access control brieﬂy, and suggest
further investigation by the reader.

Of the widely used languages, Java is the ﬁrst to provide two levels of en-
capsulation: class and package. Clearly, the class provides a means of grouping
together related ﬁelds and methods. It is important for an implementor to be
able to see all the ﬁelds and methods within a class, but only selected elements
should be visible to the user.

The package provides an (increasingly common) method for grouping re-
lated classes together. It also serves to partition the name space (we can refer
to two types of Vectors: structure.Vector and java.util.Vector). Since
access protections are sensitive to whether or not access to a ﬁeld or method oc-
curs from within the same package, the language designers clearly had a feeling
that there should be a difference in visibility between inter- and intrapackage
access.

508

Beginning with Java

There are two types of access for classes: public and default. Default classes
are private and not visible outside the containing package. For methods (and
data) there are four types of access. In increasing constraint, they are public,
protected, default, and private:

• public—methods declared public are always visible. This is an ideal

keyword for use in describing interfaces.

• protected—class methods declared as protected are visible to any class
declared in the same package, as well as any extension to the class whether
or not it is in the same package.

• default or friendly—class methods declared without any keyword are vis-
ible to any class declared in the same package, as well as any extension to
the class within the same package.

• private—class methods are not visible outside of the class.

For our purposes it seems best to commit to one level of protection for the
implementation: protected. What follows is an informal defense of that deci-
sion.

First, the use of public provides no control over the access of methods and
ﬁelds.
It would be possible, then, for users to access the implementation di-
rectly, undermining the control provided by the interface. For even the simplest
data structures, access to the implementation undermines any effort to keep the
structure in a consistent state. The following principle has been demonstrated
regularly in major software systems:

Principle 26 Make it public and they will use it.

Even if you don’t want them to.

The use of the private access control is far too restrictive. Suppose we are
interested in constructing two types of lists—a SimpleList and an extension, a
ComplexList. The SimpleList has a ﬁeld, head, that references the ﬁrst ele-
ment of the list. Clearly, this ﬁeld should be declared protected; manipulating
the head of the list without going through a method is likely to put the list
into an inconsistent state. Now, suppose that the ComplexList manipulates the
list in a way that requires manipulating the head of the list in a manner not
previously provided. Declaring head to be private makes it impossible for the
ComplexList to access the ﬁeld directly. We might be tempted to provide access
through a method of SimpleList, but we are then forced to restate our argument
as it applies to methods. If, then, we are to have an effective extension of types,
the private keyword cannot be the preferred method of access control.

If one uses the default access control, it is only possible for a class inside the
package to access the associated ﬁeld. While it seems unconstrained to allow
all classes within the same package to access the ﬁeld (after all, we may be in-
terested in protecting ﬁelds from a package co-resident), the protection against
extension outside the package is too constraining. If the concern is access to

NNWSWSENEWSEB.8 Use of the Keyword Protected

509

implementation information outside the package, the class should be declared
final to indicate that extension is not allowed.1 If extension is allowed, all
extensions should have equal access.

What remains, then, is the use of the protected access control. Fields and
methods of a class are available to classes within the same package, as well
as extensions to the class that reside outside the package. Use of this keyword
protects ﬁelds and methods from unrelated agents outside the package. Since all
extensions are provided equal access, extensions to the package are welcomed
and are likely to occur.

A special warning is necessary for the “everyday programmer.” Since the
motivation for use of packages is subtle and their use introduces considerable
bureaucratic overhead in current programming environments, the tendency is
to develop one’s software in the default or user package. This is particularly
dangerous since, within that package, there is no distinction between access
controls that are not private. Therefore absolute hiding of the implementation
is particularly dangerous since, within that package, there is no distinction be-
tween public, protected, and default access. It follows, then, that absolute
hiding of the implementation is only possible with the private keyword.

For most purposes, the protected keyword is suitable. The reader should
be aware, however, that its use here is not an adoption of an ideal, but an
acceptance of what’s available. In short, while Java is not perfect, it is a work
in progress and we may reasonably expect improvements in its access control.

Principle 27 Fight imperfection.

1 It should be pointed out that Sun’s widespread use of final in their own classes serves to destroy
one of the features most sought in this type of language—type extension and code reuse. As it
stands now, it is impossible to implement many reasonable class extensions without rewriting the
base class, and even then the subtype relation is lost.

NNWSWSENEWSE510

Beginning with Java

Appendix C

Collections

PERHAPS ONE OF THE MOST IMPORTANT changes to the Java environment over
the past few years has been the introduction of a set of classes informally called
the collection classes. This group of classes ﬁlls out java.util, bringing many
commonly used data structures under one roof. For example, there is a Stack
class. Unfortunately, there is no Queue.

C.1 Collection Class Features

Whenever possible, this text borrows interfaces from the Collection classes so
that students may better understand what is in store for them when they enter
full time into the wild. The notable cases of this borrowing include:

java.util.Enumeration The old iteration mechanism.

java.util.Iterator The new iteration mechanism, still not outﬁtted with some

methods introduced in prior editions of this text.

java.util.Comparator An interface for specifying wrappers for comparison meth-

ods.

java.lang.Comparable An interface we introduced in a different form in the
ﬁrst edition, now brought into line with the new Comparable interface to
avoid a forced name clash.

java.util.Map.Entry The interface describing Association-like classes.

The user should, as always, be aware that those features that are borrowed from
outside the java.lang pacakge must be explicitly imported. For example, all
programs that use Iterator classes would do well to import java.util.Iterator.
Programmers should avoid importing all features from the java.util package,
since there are likely to be signiﬁcant clashes with Sun’s class deﬁnitions.

C.2 Parallel Features

Whenever possible, the design approach outlined in this text is followed. This
mirrors, in many ways, the approach of the the design of the Collection

512

Collections

interfaces are developed, followed by abstract base classes that im-
classes:
plement the shared features of all implementations of the interface, followed by
solid implementations that extend the abstract base classes.

The following features parallel the Collection classes, but with reduced
functionality. For pedegogical reasons many occasionally useful features have
been dropped to highlight the important features of the parallel class or inter-
face. Included are:

structure.Structure The common interface for all Collection-like structures

in the structure package.

structure.Map The interface for associative structures.

structure.List The interface for random-access, indexable, and extensible struc-

tures.

Each interface is implemented in several ways with, for example, StackList
interpreted as a Stack constructed using an underlying List type.
In many
cases, these structures correspond to similar structures found in java.util.
The programmer should be aware that moving to the use of java.util may
introduce new methods or slight differences in the interface. For the most part,
the structure library is a superset of those classes found in java.util.

C.3 Conversion

Most structures in the structure package extend the AbstractStructure class.
One feature of this base class is the method values. The values method returns
an object that implements Collection. This allows the programmer to convert
structure package objects for use in the Collection system. It avoids, for the
most part, the introduction of the complexities of supporting the needs of the
masses, and focuses on the task at hand, learning about classic data structures.
At the time of this writing, alternative copy constructors are being incor-
porated into most structures to allow the conversion from Collection objects.
This will allow the mixed use of structures.

Appendix D

Documentation

Concepts:
. Class hierarchy
.
Principles

‘My name is Ozymandius, King of Kings,
Look on my Works, ye Mighty, and despair!’
Nothing beside remains. Round the decay
Of the colossal Wreck, boundless and bare
The lone and level sands stretch far away.
—Percy Bysshe Shelley

D.1 Structure Package Hierarchy

You can ﬁnd the release version for structure package software by typing:

java structure5.Version

The structure package contains a large number of interfaces and implemen-
tations of common data structures. The relationship between these is indicated
below. Private structures are, of course, not available for direct use by users.
Indentation indicates extension or implementation: GraphList is an implemen-
tation of the Graph and Structure interfaces.

Assert
BinaryTree
Clock
Collection (java.util)
Comparable (java.lang)

ComparableAssociation

Comparator (java.util)

NaturalComparator
ReverseComparator

DoublyLinkedNode
Edge

ComparableEdge

Error (java.lang)

FailedAssertion

FailedInvariant
FailedPostcondition
FailedPrecondition

section 2.2
section 12.4
A.2
Appendix C
section 11.1
subsection 11.1.2
subsection 11.2.4
subsection 11.2.4
subsection 11.2.4
section 9.5
section 16.2
16.4.5

section 2.2
section 2.2
section 2.2
section 2.2

514

Iterator (java.util)

AbstractIterator

AbstractListIterator
BTInorderIterator
BTLevelorderIterator
BTPostorderIterator
BTPreorderIterator
ChainedHashtableIterator
GraphListAIterator
GraphListEIterator
HashtableIterator
KeyIterator
SinglyLinkedListIterator
SplayTreeIterator
ValueIterator
VectorIterator

Map

Hashtable
ChainedHashtable
MapList
OrderedMap

Table

Map.Entry (java.util)
Association

ComparableAssociation

Entry

Matrix
PriorityQueue

PriorityVector
SkewHeap
VectorHeap

ReadStream
Node
Structure

AbstractStructure
Graph

GraphList

GraphListDirected
GraphListUndirected

GraphMatrix

GraphMatrixDirected
GraphMatrixUndirected

Set

AbstractSet
SetList
SetVector

Documentation

section 8.2
section 8.2

subsection 12.6.2
subsection 12.6.4
subsection 12.6.3
subsection 12.6.1
subsection 15.4.2
subsection 16.3.3
subsection 16.3.3
subsection 15.4.1
subsection 15.4.1
section 9.8
section 14.6
subsection 15.4.1
section 8.3
section 15.2
subsection 15.4.1
subsection 15.4.2
section 15.3
section 15.5
section 15.5

section 1.5
subsection 11.1.2

section 3.8
section 13.1
section 13.3
subsection 13.4.3
subsection 13.4.1
subsection B.3.1
section 9.4
section 1.8

section 16.2
subsection 16.3.3
subsection 16.3.3
subsection 16.3.3
subsection 16.3.2
subsection 16.3.2
subsection 16.3.2
section 1.8

D.2 Principles

515

Linear

AbstractLinear

Queue

AbstractQueue

QueueArray
QueueList
QueueVector

Stack

AbstractStack

StackArray
StackList
StackVector

List

AbstractList
Vector
CircularList
DoublyLinkedList
SinglyLinkedList

OrderedStructure

BinarySearchTree

SplayTree

OrderedList
OrderedVector
RedBlackTree

Version
Vertex

GraphListVertex
GraphMatrixVertex

D.2 Principles

chapter 10
chapter 10
section 10.2

subsection 10.2.4
subsection 10.2.2
subsection 10.2.3
section 10.1
A.1

subsection 10.1.3
subsection 10.1.2
chapter 9
section 9.3
section 3.1
section 9.6
section 9.5
section 9.4
subsection 11.2.1
section 14.1
section 14.6
subsection 11.2.5
subsection 11.2.2
14.7
section D.1
section 16.2
subsection 16.3.3
subsection 16.3.2

1. The principled programmer understands a principle well enough to form

an opinion about it. (Page 3)

2. Free the future: reuse code. (Page 17)

3. Design and abide by interfaces as though you were the user. (Page 25)

4. Declare data ﬁelds protected. (Page 25)

5. Test assertions in your code. (Page 35)

6. Maintaining a consistent interface makes a structure useful. (Page 51)

7. Recursive structures must make “progress” toward a “base case.” (Page 96)

516

Documentation

8. Never modify a data structure while an associated Enumeration is live.

(Page 162)

9. When manipulating references, draw pictures. (Page 189)

10. Every public method of an object should leave the object in a consistent

state. (Page 191)

11. Symmetry is good. (Page 194)

12. Test the boundaries of your structures and methods. (Page 197)

13. Question asymmetry. (Page 201)

14. Assume that values returned by iterators are read-only. (Page 211)

15. Understand the complexity of the structures you use. (Page 235)

16. Declare parameters of overriding methods with the most general types

possible. (Page 256)

17. Avoid multiple casts of the same object by assigning the value to a tempo-

rary variable. (Page ??)

18. Consider your code from different points of view. (Page 269)

19. Don’t let opposing references show through the interface. (Page 286)

20. Write methods to be as general as possible. (Page 300)

21. Avoid unnaturally extending a natural interface. (Page 319)

22. Seek structures with reduced friction. (Page 320)

23. Declare object-independent functions static. (Page 322)

24. Provide a method for hashing the objects you implement. (Page 385)

25. Equivalent objects should return equal hash codes. (Page 386)

26. Make it public and they will use it. (Page 508)

27. Fight imperfection. (Page 509)

Index

α, 379
π, 27, 93, 403
-unboxing, 74

abstract base class, 149, 151, 160, 408,

505

AbstractGenerator example, 152
AbstractCard example, 156
abstraction, 5, 6, 151

data, 5
procedural, 5

access, random, 179
accessors, 13
adjacency list, 416
adjacency matrix, 410
adjacent, 404
algorithm, greedy, 429
allocation, dynamic, 179
Amend, Bill, 277
analysis

amortized, 229, 329, 333
asymptotic, 81–92
average-case, 90
best-case, 90
big-O, 81, 82
worst-case, 90
Analysis example, 85
ancestor, 277

proper, 279

and

logical, 262
short-circuiting, 262

and-gate, 333
API, 45
Application Program Interface, 45
application, scalable, 43
array, 43

associative, 369
two-dimensional, 411

assertion, 33–37
association, 14, 369
associative array, 369
associative memory, 66
associative structure, 345–348
asymptotic, 92
asymptotic analysis, 81–92

atinlay example, 15
autoboxed, 74
autoboxing, 74, 493

balloon, helium-ﬁlled, 189
bank account, 11
BankAccount example, 11
base case, recursive, 95
base type, 409
baseclass, 504
big-Ω, 112
big-O notation, 81–92
binary search, 259–264
binary search tree, 343
binary tree, 279

complete, 280
full, 279
bound, upper, 44
boundary cases, 212
breadth-ﬁrst, 251
bubble sort, 120
BubbleSort example, 119

C++, xi, 253
cache, 99
Calc example, 282
call

frame, 222
stack, 222
Card example, 155
Carroll, Lewis, 219
cast, 143, 256, 492
child, 279

left, 279
right, 279

Circuit example, 334
class, 5, 8

abstract, 149, 191, 408–410, 505
concrete, 408, 505
dynamic loading of, 507
encapsulation, 507
extension, 408
hierarchy, 513
importing, 11
instance of, 5, 8
program is a, 489

518

Index

class cast exception, 143
clone, of object, 211
clustering, 377

primary, 377, 378
secondary, 377

code reuse, 235
CoinPuzzle example, 231
collection classes, 511
collections, 511–512
comment, 33–37

political, 2
comparable, 253

association, 256
ratio, 254
comparator, 140–142
compare, 253
comparing values, 197
compile-time, 69
CompInsSort example, 142
complexity, 1, 81

of method, 235
space, 82–92
time, 82–92
component, 404

connected, 404
strongly connected, 404

concrete class, 505
connected component, 404
ConstantGenerator example, 153
constructive toys, 1
constructor, 10

copy, 59, 373
container types, 72
contract, 6

interface viewed as, 6, 25, 181

control

ﬂow of, 222
structures, 8, 161
correctness, proof of, 102
craftsmen, 81
cycle, 404

DAG, 404
Dahl, Roald, 315
data

comparing, 253
declaring protected, 25, 515

data abstraction, 6–7
data structure, 1, 8
data types, generic, 72

data types, parameterized, 72
dead code, 91
degree, 279
deprecated, xiii
descendant, 279
proper, 279

design

method, 149–160
properties of, 108–110

deterministic, 134
dictionary, 369
Dijkstra example, 432
disjoint, 277
documentation, 513–516
Dodgeson, Charles Lutwidge, 219
doubly linked list, 201
Doyle, Arthur Conan, Sir, 441
Drake, Milton, 489
dummy nodes, 215
dynamic programming, 99

edge

canonical, 418
of tree, 277
weighted, 428

element

current, 164

encapsulation, 6–8
class, 507
package, 507
encoding, 241, 303

unicode, 492
energy, 195, 201, 258
enumeration, 161
live, 162
equals, default, 13
example

AbstractGenerator, 152
AbstractCard, 156
Analysis, 85
atinlay, 15
BankAccount, 11
BubbleSort, 119
Calc, 282
Card, 155
Circuit, 334
CoinPuzzle, 231
CompInsSort, 142
ConstantGenerator, 153
Dijkstra, 432

Index

519

Example, 2
Fibo, 106
Floyd, 428
FullPostage, 99
Generator, 152
HelloWorld, 162, 164
HexBoard, 313
HexMove, 313
HTML, 251
Huffman, 303, 304, 317
Huffman2, 317
Index, 396
InfiniteQuestions, 288
InsertionSort, 125
LinkedList, 216
LongWords, 70
LongWords2, 71
LongWords3, 78
LSystem, 56
MCST, 430
MergeSort, 127
nim, xii
ParkingLot, 183
ParkingLot2, 271
Pedigree, 280
PFGenerator, 168
PhoneBook, 138, 143
PinochleCard, 159
Player, 314
PokerCard, 158
PrimeGenerator, 154
QuickSort, 131, 222
RadixSort, 135
Ratio, 8, 254
RBSymTab, 362
Reachability, 422
Reader, 248
Rect, 20
Recursion, 94, 102
RecursiveIterators, 298
RecursivePostage, 98
SelectionSort, 122
Sort, 265
sqrt, 34
StringReader, 43
StringVector, 71
SymbolTable, 249
SymMap, 369
SymTab, 346
Token, 248

TopoSort, 426
Unique, 182
UniqueFilter, 170
Warshall, 427
WordFreq, 48
WordList, 19, 23, 47

Example example, 2
example, icon for, xii
extensible, 70
extension of class, 408

Fibo example, 106
Fibonacci numbers, 106
ﬁelds, 8
FIFO, see also queue
ﬁlter, 170
ﬁnger method, 192
Floyd example, 428
foreach, 500
forest, 277
format, 496
Fox, Jason, 277
free list, 183–186, 410
friction, 110

deﬁnition, 108, 110
in heaps, 320
Frost, Robert, 507
FullPostage example, 99
function, see also method
functions, static, 322

garbage collection, 199
garbage collector, 189, 494
gate delay, 334
Geisel, Theodor Seuss, 5, 369
generator, 152–155

as iterator, 167–170
Generator example, 152
generic, xv
generic class, 69, 72
Gillespie, Haven, 179
grammars, 57
graph, 403–434
acyclic, 404
adjacency list, 416
arc, see also graph, edge
complete, 404
dense, 416
directed, 403
directed acyclic, 404

520

Index

edge, 403
node, see also graph, vertex
sparse, 416
sub-, 403
transitive closure, 427
undirected, 403
vertex, 403
Guthrie, Arlo, 69

Hangman, 18
hash code, 385–392
hash function

perfect, 377
rehashing, 377
hash table, 374–392

bucket, 375, 376
collision, 377
double hashing, 378
external chaining, 383–385, 392
linear probing, 378
load factor, 379
open addressing, 375–383, 392
ordered linear probing, 400
primary clustering, 378
probe count, 392
rehashing, 377
reserved value, 377

hashing, 375
heap, 308, 319

complete, 320
skew, 329

height, 279

black, 361
of node, 279
of tree, 279

HelloWorld example, 162, 164
HexBoard example, 313
HexMove example, 313
hierarchy, 513–515
HTML example, 251
Huffman encoding, 303
Huffman example, 303, 304, 317
Huffman2 example, 317

icons, marginal, xii
implementation, 6, 19
point-of-view, 507
incident, edge to vertex, 403
Index example, 396
induction, mathematical, 81, 101–106

InfiniteQuestions example, 288
inﬁx, 290
information theoretic limit, 91
inheritance, 408

multiple, 506
insertion sort, 125–127
InsertionSort example, 125
instantiation, 442
interface, 5, 6, 22–24, 149, 504
abiding by, 25, 515
design of, 149
implementation, 190
implementing, 23
list, 180
point-of-view, 507
program, 19
Internet, see also Web
intractable, 85
inverter, 333
iterator, 161–173

ﬁltering, 170–172
is read-only, 211
potato, 161

Java, xi, 1, 5, 489–506
jit compiler, 117
Johnson, Crockett, 149
just-in-time compiler, 117

key

in association, 369

L-systems, 56
language, purpose of, 69
Lao Tzu, 81
larger, 123
length of path, 279
level, 279
LIFO, see also stack
linear structures, 219–246
LinkedList example, 216
list, 179–211

abstract, 186
add, 179
adjacency, 416–422
circular, 206–209
concept, 179
doubly linked, 201–206
head, 179
interface, 180

521

Index

is a graph, 403
iterator, 209–211
remove, 179
singly linked, 188–201
tail, 179

load factor, 383
logic gates, 333
LongWords example, 70
LongWords2 example, 71
LongWords3 example, 78
LSystem example, 56

map, 369–398

ordered, 392–398

Mars, life on, 191
matrix, 60–63

adjacency, 410–416
symmetric, 66, 410
triangular, 66

maze

ﬁnish cell, 242
start cell, 242

mazes

solving, 242–244

McCloskey, Robert, 119
MCST example, 430
mechanics of data structures, 6
median, 259
memory leak, 194
mention, 403
merge, 127
mergesort, 127–131
MergeSort example, 127
message, 26, see also method
method, 8, 490
calling, 5
calls, 8
destructive, 213, 330
helper, 318
nondestructive, 194
postcondition, xii
precondition, xii
this, 17
methods, static, 322
minimum spanning tree, 429
Modula-2, xi
mutator, 13

nanosecond, 93, 115
Napster, 15

new operator, 12, 188
nim example, xii
node

degree, 279
depth, 279
interior, 279
level, 279
parent, in tree, 277
sibling, 279
of tree, 277
nondeterministic, 134
null, 189

Oakland, Ben, 489
object, 5, 8, 489

general, 345
model of, 7–8
object orientation, 5–25
in languages, 5
terminology of, 8–11

object-oriented programming, 5–25
objects, comparable, 253–258
octtrees, 308
optimization, 91
ordered

list, 267
map, 392–398
vector, 259
ordered structure, 144
ordered structures, 253–272
overriding methods, 357
overriding operators, 253

package, 11, 489
default, 509
encapsulation, 507
user, 509

ParkingLot example, 183
ParkingLot2 example, 271
partially ordered, 140
Pascal, xi
path, 279, 404

end points, 404
simple, 404
Pedigree example, 280
performance

exponential, 85
linear, 83
polynomial, 83
superlinear, 83

522

Index

PFGenerator example, 168
philosophy, 1–3
PhoneBook example, 138, 143
Pig Latin, 14
PinochleCard example, 159
pivot, 131, 133
Player example, 314
pointer, see also reference
PokerCard example, 158
postcondition, xii, 34
postﬁx, 290
precedence, operator, 282
precondition, xii, 34
PrimeGenerator example, 154
principle, xii, 3

assertions, 35, 515
code reuse, 17
code review, 269
consistent interfaces, 51
consistent state, 191
drawing pictures, 189
equal hash codes, 386
ﬁghting imperfection, 509
hiding opposing references, 286
icon for, xii
interface design, 25
live enumerations, 162
most general method, 300
natural interface, 319
overriding signatures, 256
principled programmer, 3
progress in recursion, 96
protected data, 25
providing hash function, 385
public protection, 508
questioning asymmetry, 201
read-only iterator values, 211
reducing friction, 320
static functions, 322
symmetry, 194
testing boundaries, 197
understanding complexity, 235

principles, 1–3, 515–516
principles, abstract, xi
priority queue, 315
problems, importance of, 2
procedure, 249, see also method
productions, 56
program counter, 222
program, state, 34

programming, craft of, xi
programs, writing large, 1
progress in recursion, 95
proof by induction, 94
protected, use of keyword, 507–509
protection of ﬁelds, 8, 10

queue, 220, 229–242

dequeue, 229
enqueue, 229
head, 229
priority, 315
tail, 229
quicksort, 131–134
QuickSort example, 131, 222

RadixSort example, 135
random access property, 45
Ratio example, 8, 254
RBSymTab example, 362
Reachability example, 422
reachability, computing, 427
Reader example, 248
record, traditional, 8
Rect example, 20
recursion, 81, 94–101
Recursion example, 94, 102
recursive, 94
recursive data structure, 310
RecursiveIterators example, 298
RecursivePostage example, 98
reference, 19, 188, 189, 493

null, 189
rewriting, 56
Rhoda, 203
Rhonda, 203
rising edge, 337
rotations, 354
runtime, 44, 69

scope, of identiﬁer, 491
search

binary, 260
depth-ﬁrst, 422
linear, 268

selection sort, 122–123
SelectionSort example, 122
self-reference, 81, 94–106
semantics, 69, 369

of data structure, 6

Index

523

implementation independent, 181
implementation speciﬁc, 182

Seuss, Dr., 14
shallow copy, 311
Shelley, Percy Bysshe, 513
signature, 19, 149
Slobodkina, Esphyr, 343
solution

self-referential, 94

sort

bucket, 134
radix, 134–138

Sort example, 265
sorting, 119–144, 264, 345, 424

objects, 138–140
vectors, 143–144

Soundex, 401
splay operation, 354
splay tree, 354
sqrt example, 34
stability

of sort, 146

stack, 220–229

trace of, 36

start string, 56
state, 222

program, 34
structure, 150

string

counted, 7
end mark, 7
terminated, 7

StringReader example, 43
StringVector example, 71
structure

control, 8, 161
data, 8
deterministic, 354
dynamic, 179
linear, 219, 277
nonlinear, 277
symmetric, 201
structure package, 11

AbstractIterator, 163
AbstractLinear, 220
AbstractList, 186
AbstractQueue, 230
AbstractStructure, 391
Assert, 35
Association, 15, 72

Association, 15–17
BinarySearchTree, 348
BinarySearchTree, 344
BinaryTree, 284, 291, 300, 392
BinaryTreeNode, 355
BTInorderIterator, 294
BTLevelorderIterator, 296
BTPostorderIterator, 295
BTPreorderIterator, 291
CaselessComparator, 141
ChainedHashTable, 384
CircularList, 207
Comparable, 254
ComparableAssociation, 256
Comparator, 141
documentation of, 2
DoublyLinkedList, 203
DoublyLinkedNode, 203
downloading, xii, 1
Edge, 407
Enumeration, 161
Graph, 404
GraphListDirected, 420
GraphListUndirected, 419
GraphListVertex, 416
GraphMatrix, 408, 411, 412, 414
GraphMatrixDirected, 409, 412,

413

GraphMatrixUndirected,

413,

414, 416
Hashtable, 375, 383
icon for, xii
Iterator, 163
Linear, 219
List, xii, 180
Map, 370
MapList, 372
Matrix, 60
MazeRunner, 242
NaturalComparator, 266
Node, 189
OrderedMap, 393
OrderedStructure, 259
OrderedVector, 259
OrderedList, 267
PriorityVector, 318
PriorityQueue, 315
Queue, 229
QueueArray, 239
QueueList, 234

524

Index

QueueVector, 237
ReadStream, 494
ReverseComparator, 266
Set, 23
SetVector, 57
SinglyLinkedList, 191
SinglyLinkedListIterator, 210
SkewHeap, 329
SplayTree, 357
SplayTreeIterator, 358
Stack, 221
StackList, 227
StackVector, 225
Structure, 2, 22
Table, 393
ValueIterator, 382
Vector, 45, 50, 74, 96, 151, 165
VectorIterator, 165
VectorHeap, 321
Vertex, 407

structures

associative, 345

subclass, 409, 504
subroutine, see also method
subtype, 409, 503–504
superclass, 409
supertype, 504
SymbolTable, 346
SymbolTable example, 249
SymMap example, 369
symmetry, 108–110, 201
in interface, 108
is predictability, 108

SymTab example, 346
syntax, 69
Syracuse sequence, 113

table, 392–398
tail recursion, 98
the entire argument array, 70
The Phone Company, 1
this, 17
time stamp, 424
time–space trade-off, 193
token, 247
Token example, 248
topological sort, 424
TopoSort example, 426
totally ordered, 140
tree, 277–309

arity, 279
AVL, 311
binary, 279
binary search, 343–364
complete, 278
degenerate, 279
degree, 279
expression, 282
full, 278, 279
height balanced, 311
is a graph, 403
leaf, 279
minimum spanning, 429
oriented, 279
pruning, 313
red-black, 361
root, 277
rotation, 354
splay, 354–360
sub-, 277, 279
traversing expression, 282
trivial, 277

type

conversion, 492
primitive, 491
reference, 491

type bound, 79
type parameter
actual, 76
formal, 75
type parameters, 73

Unique example, 182
UniqueFilter example, 170
user, 24

value

in association, 369

value, comparison of, 197
vector, 43–64

capacity, 46, 50, 65
cost of heapifying, 339
end, 47
extending, 53
extent of, 50
iterator, 165–167
size, 65
size of, 50
tail, 65
two-dimensional, 60

525

Index

vertex

degree, 403
destination, 403
in-degree, 403
out-degree, 403
sink, 403
source, 403

Warshall example, 427
Warshall’s algorithm, 427
Web

programming, xi
resources available on, 2

White, Elwyn Brooks, 253
WordFreq example, 48
WordList example, 19, 23, 47

Colophon

This book was written LATEX 2e by Leslie Lamport,
based on TEX 3.141592 by Donald E. Knuth.

The technical fonts used in this book are primarily from
the Computer Modern family of typefaces, designed by Donald Knuth.

Figures were drawn using xfig by Supoj Stanthavibul,
a ﬂexible multi-target drawing program running under
the X Window System, version 11, release 6, from the X Consortium, Inc.

Typesetting was performed on
an Apple 2.0 Gigahertz (black) PowerBook Intel Core-Duo laptop
running Macintosh OS/X 10.4.9, based on Darwin 8.10.1.

The processor can typeset this book
in approximately 8.5 seconds and costs about $1100.

All code examples are extracted from the original Java sources
using software designed by the author.
The system allows veriﬁcation of the correctness of all examples in less than one minute.

Much of the book was designed on yellow college-ruled pads.