Transcript
Preface
A journey of a thousand miles begins with a single step.
  A Chinese proverb
eople often ask: What is discrete mathematics? It's the mathematics of discrete (distinct and disconnected) objects. In other words, it is the study of discrete objects and relationships that bind them. The geometric representations of discrete objects have gaps in them. For example, integers are discrete objects, therefore (elementary) number theory, for instance, is part of discrete mathematics; so are linear algebra and abstract algebra. On the other hand, calculus deals with sets of connected (without any gaps) objects. The set of real numbers and the set of points on a plane are two such sets; they have continuous pictorial representations. Therefore, calculus does not belong to discrete mathematics, but to continuous mathematics. However, calculus is relevant in the study of discrete mathematics. The sets in discrete mathematics are often finite or countable, whereas those in continuous mathematics are often uncountable. Interestingly, an analogous situation exists in the field of computers. Just as mathematics can be divided into discrete and continuous mathematics, computers can be divided into digital and analog. Digital computers process the discrete objects 0 and 1, whereas analog computers process continuous d a t a ~ t h a t is, data obtained through measurement. Thus the terms discrete and continuous are analogous to the terms digital and analog, respectively. The advent of modern digital computers has increased the need for understanding discrete mathematics. The tools and techniques of discrete mathematics enable us to appreciate the power and beauty of mathematics in designing problemsolving strategies in everyday life, especially in computer science, and to communicate with ease in the language of discrete mathematics.
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The Realization of a Dream
This book is the fruit of many years of many dreams; it is the endproduct of my fascination for the myriad applications of discrete mathematics to a variety of courses, such as Data Structures, Analysis of Algorithms, Programming Languages, Theory of Compilers, and Databases. Data structures and Discrete Mathematics compliment each other. The information in this book is applicable to quite a few areas in mathematics; discrete
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mathematics is also an excellent preparation for number theory and abstract
algebra.
A logically conceived, selfcontained, wellorganized, and a userfriendly book, it is suitable for students and amateurs as well; so the language employed is, hopefully, fairly simple and accessible. Although the book features a wellbalanced mix of conversational and formal writing style, mathematical rigor has not been sacrificed. Also great care has been taken to be attentive to even minute details.
Audience
The book has been designed for students in computer science, electrical engineering, and mathematics as a one or twosemester course in discrete mathematics at the sophomore/junior level. Several earlier versions of the text were classtested at two different institutions, with positive responses from students.
Prerequisites
No formal prerequisites are needed to enjoy the material or to employ its power, except a very strong background in college algebra. A good background in precalculus mathematics is desirable, but not essential. Perhaps the most important requirement is a bit of sophisticated mathematical maturity: a combination of patience, logical and analytical thinking, motivation, systematism, decisionmaking, and the willingness to persevere through failure until success is achieved. Although no programming background is required to enjoy the discrete mathematics, knowledge of a structured programming language, such as Java or C + +, can make the study of discrete mathematics more rewarding.
Coverage
The text contains indepth coverage of all major topics proposed by professional associations for a discrete mathematics course. It emphasizes problemsolving techniques, pattern recognition, conjecturing, induction, applications of varying nature, proof techniques, algorithm development, algorithm correctness, and numeric computations. Recursion, a powerful problemsolving strategy, is used heavily in both mathematics and computer science. Initially, for some students, it can be a bittersweet and demanding experience, so the strategy is presented with great care to help amateurs feel at home with this fascinating and frequently used technique for program development. This book also includes discussions on Fibonacci and Lucas numbers, Fermat numbers, and figurate numbers and their geometric representations, all excellent tools for exploring and understanding recursion.
Preface
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A sufficient amount of theory is included for those who enjoy the beauty in the development of the subject, and a wealth of applications as well for those who enjoy the power of problemsolving techniques. Hopefully, the student will benefit from the nice balance between theory and applications. Optional sections in the book are identified with an asterisk (.) in the left margin. Most of these sections deal with interesting applications or discussions. They can be omitted without negatively affecting the logical development of the topic. However, students are strongly encouraged to pursue the optional sections to maximize their learning.
Historical Anecdotes and Biographies
Biographical sketches of about 60 mathematicians and computer scientists who have played a significant role in the development of the field are threaded into the text. Hopefully, they provide a h u m a n dimension and attach a h u m a n face to major discoveries. A biographical index, keyed to page, appears on the inside of the back cover for easy access.
Examples and Exercises
Each section in the book contains a generous selection of carefully tailored examples to clarify and illuminate various concepts and facts. The backbone of the book is the 560 examples worked out in detail for easy understanding. Every section ends with a large collection of carefully prepared and wellgraded exercises (more than 3700 in total), including thoughtprovoking truefalse questions. Some exercises enhance routine computational skills; some reinforce facts, formulas, and techniques; and some require mastery of various proof techniques coupled with algebraic manipulation. Often exercises of the latter category require a mathematically sophisticated mind and hence are meant to challenge the mathematically curious. Most of the exercise sets contain optional exercises, identified by the letter o in the left margin. These are intended for more mathematically sophisticated students. Exercises marked with one asterisk (.) are slightly more advanced than the ones that precede them. Doublestarred (**) exercises are more challenging than the singlestarred; they require a higher level of mathematical maturity. Exercises identified with the letter c in the left margin require a calculus background; they can be omitted by those with no or minimal calculus. Answers or partial solutions to all oddnumbered exercises are given at the end of the book.
Foundation
Theorems are the backbones of mathematics. Consequently, this book contains the various proof techniques, explained and illustrated in detail.
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They provide a strong foundation in problemsolving techniques, algorithmic approach, verification and analysis of algorithms, as well as in every discrete mathematics topic needed to pursue computer science courses such as Data Structures, Analysis of Algorithms, Programming Languages, Theory of Compilers, Databases, and Theory of Computation.
Proofs
Most of the concepts, definitions, and theorems in the book are illustrated with appropriate examples. Proofs shed additional light on the topic and enable students to sharpen their problemsolving skills. The various proof techniques appear throughout the text.
Applications
Numerous current and relevant applications are woven into the text, taken from computer science, chemistry, genetics, sports, coding theory, banking, casino games, electronics, decisionmaking, and gambling. They enhance understanding and show the relevance of discrete mathematics to everyday life. A detailed index of applications, keyed to pages, is given at the end of the book. Algorithms Clearly written algorithms are presented throughout the text as problemsolving tools. Some standard algorithms used in computer science are developed in a straightforward fashion; they are analyzed and proved to enhance problemsolving techniques. The computational complexities of a number of standard algorithms are investigated for comparison. Algorithms are written in a simpletounderstand pseudocode that can easily be translated into any programming language. In this pseudocode: 9 Explanatory comments are enclosed within the delimeters (* and *). 9 The body of the algorithm begins with a B e g i n and ends in an E n d ; they serve as the outermost parentheses. 9 Every compound statement begins with a b e g i n and ends in an end; again, they serve as parentheses. In particular, for easy readability, a while (for) loop with a compound statement ends in e n d w h i l e (endfor).
Chapter Summaries
Each chapter ends with a summary of important vocabulary, formulas, and properties developed in the chapter. All the terms are keyed to the text pages for easy reference and a quick review.
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Each chapter summary is followed by an extensive set of wellconstructed review exercises. Used along with the summary, these provide a comprehensive review of the chapter. Chapterend supplementary exercises provide additional challenging opportunities for the mathematically sophisticated and curiousminded for further experimentation and exploration. The book contains about 950 review and supplementary exercises.
Computer Assignments
Over 150 relevant computer assignments are given at the end of chapters. They provide handson experience with concepts and an opportunity to enhance programming skills. A computer algebra system, such as Maple, Derive, or Mathematica, or a programming language of choice can be used.
Exploratory Writing Projects
Each chapter contains a set of wellconceived writing projects, for a total of about 600. These expository projects allow students to explore areas not pursued in the book, as well as to enhance research techniques and to practice writing skills. They can lead to original research, and can be assigned as group projects in a real world environment. For convenience, a comprehensive list of references for the writing projects, compiled from various sources, is provided in the S t u d e n t ' s
Solutions Manual.
Enrichment Readings
Each chapter ends with a list of selected references for further exploration and enrichment. Most expand the themes studied in this book.
Numbering System
A concise numbering system is used to label each item, where an item can be an algorithm, figure, example, exercises, section, table, or theorem. Item m . n refers to item n in Chapter "m". For example, Section 3.4 is Section 4 in Chapter 3.
Special Symbols
Colored boxes are used to highlight items that may need special attention. The letter o in the left margin of an exercise indicates that it is optional, whereas a c indicates that it requires the knowledge of calculus. Besides, every theorem is easily identifiable, and the end of every proof and example
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is marked with a solid square (l l). An asterisk (.) next to an exercise indicates that it is challenging, whereas a doublestar (**) indicates that it is even more challenging. While "  " stands for equality, the closely related symbol "~" means is approximately equal to:
0 C
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optional exercises requires a knowledge of calculus end of a proof or a solution a challenging exercise a more challenging exercise is equal to is approximately equal to
Abbreviations
For the sake of brevity, four useful abbreviations are used throughout the text: LHS, RHS, PMI, and IH: LHS RHS PMI IH LeftHand Side RightHand Side Principle of Mathematical Induction Inductive Hypothesis
Symbols Index
An index of symbols used in the text and the page numbers where they occur can be found inside the front and back covers.
Web Links
The World Wide Web can be a useful resource for collecting information about the various topics and algorithms. Web links also provide biographies and discuss the discoveries of major mathematical contributors. Some Web sites for specific topics are listed in the Appendix.
Student's Solutions Manual The Student's Solutions Manual contains detailed solutions of all oddnumbered exercises. It also includes suggestions for studying mathematics, and for preparing to take an math exam. The Manual also contains a comprehensive list of references for the various writing projects and assignments.
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The Instructor's Manual contains detailed solutions to all evennumbered exercises, two sample tests and their keys for each chapter, and two sample final examinations and their keys.
Acknowledgments
A number of people, including many students, have played a major role in substantially improving the quality of the manuscript through its development. I am truly grateful to every one of them for their unfailing encouragement, cooperation, and support. To begin with, I am sincerely indebted to the following reviewers for their unblemished enthusiasm and constructive suggestions: Gerald Alexanderson Stephen Brick Neil Calkin Andre Chapuis Luis E. Cuellar H. K. Dai Michael Daven Henry Etlinger Jerrold R. Griggs John Harding Nan Jiang Warren McGovern Tim O'Neil Michael O'Sullivan Stanley Selkow Santa Clara University University of South Alabama Clemson University Indiana University McNeese State University Oklahoma State University Mt. St. Mary College Rochester Institute of Technology University of South Carolina New Mexico State University University of South Dakota Bowling Green State University University of Notre Dame San Diego State University Worcester Polytechnic Institute
Thanks also go to Henry Etlinger of Rochester Institute of Technology and Jerrold R. Griggs of the University of South Carolina for reading the entire manuscript for accuracy; to Michael Dillencourt of the University of California at Irvine, and Thomas E. Moore of Bridgewater State College for preparing the solutions to the exercises; and to Margarite Roumas for her excellent editorial assistance. My sincere thanks also go to Senior Editor, Barbara Holland, Production Editor, Marcy BarnesHenrie, Copy Editor, Kristin Landon, and Associate Editor, Thomas Singer for their devotion, cooperation, promptness, and patience, and for their unwavering support for the project. Finally, I must accept responsibility for any errors that may still remain. I would certainly appreciate receiving comments about any unwelcome surprises, alternate or better solutions, and exercises, puzzles, and applications you have enjoyed.
Framingham, Massachusetts September 19, 2003
Thomas Koshy tkoshy@frc.mass.edu
A W o r d to t h e S t u d e n t
Tell me a n d I will forget. S h o w me a n d I will remember. Involve me a n d I will u n d e r s t a n d .
n Confucius
The SALT of Life
Mathematics is a science; it is an art; it is a precise and concise language; and it is a great problemsolving tool. Thus mathematics is the SALT of life. To learn a language, such as Greek or Russian, first you have to learn its alphabet, grammar, and syntax; you also have to build up a decent vocabulary to speak, read, or write. Each takes a lot of time and practice.
The Language of Mathematics
Because mathematics is a concise language with its own symbolism, vocabulary, and properties (or rules), to be successful in mathematics, you must know them well and be able to apply them. For example, it is important to know that there is a difference between perimeter and area, area and volume, factor and multiple, divisor and dividend, hypothesis and hypotenuse, algorithm and logarithm, reminder and remainder, computing and solving, disjunction and destruction, conjunction and construction, and negation and negative. So you must be fluent in the language of mathematics, just like you need to be fluent in any foreign language. So keep speaking the language of mathematics. Although mathematics is itself an unambiguous language, algebra is the language of mathematics. Studying algebra develops confidence, improves logical and critical thinking, and enhances what is called mathematical maturity, all needed for developing and establishing mathematical facts, and for solving problems. This book is written in a clear and concise language that is easy to understand and easy to build on. It presents the essential (discrete) mathematical tools needed to succeed in all undergraduate computer science courses.
Theory and Applications
This book features a perfect blend of both theory and applications. Mathematics does not exist without its logically developed theory; in fact, theorems are like the steel beams of mathematics. So study the various xxi
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proof techniques, follow the various proofs presented, and try to reproduce them in your own words. Whenever possible, create your own proofs. Try to feel at home with the various methods and proofs. Besides developing a working vocabulary, pay close attention to facts, properties, and formulas, and enjoy the beautiful development of each topic. This book also draws on a vast array of interesting and practical applications to several disciplines, especially to computer science. These applications are spread throughout the book. Enjoy them, and appreciate the power of mathematics that can be applied to a variety of situations, many of which are found in business, industry, and scientific discovery in today's workplace.
ProblemSolving Strategies
To master mathematics, you must practice it; that is, you must apply and do mathematics. You must be able to apply previously developed facts to solve problems. For this reason, this book emphasizes problemsolving techniques. You will encounter two types of exercises in the exercise sets: The first type is computational, and the second type is algebraic and theoretical. Being able to do computational exercises does not automatically imply that you are able to do algebraic and theoretical exercises. So do not get discouraged, but keep trying until you succeed. Of course, before you attempt the exercises in any section, you will need to first master the section; know the definitions, symbols, and facts, and redo the examples using your own steps. Since the exercises are graded in ascending order of difficulty, always do them in order; save the solutions and refine them as you become mathematically more sophisticated. The chapterend review exercises give you a chance to revisit the chapter. They can be used as a quick review of important concepts.
Recursion
Recursion is an extremely powerful problemsolving strategy, used often in mathematics and computer science. Although some students may need a lot of practice to get used to it, once you know how to approach problems recursively, you will certainly appreciate its great power.
Stay Actively Involved
Professional basketball players Magic Johnson, Larry Bird, and Michael Jordan didn't become superstars overnight by reading about basketball or by watching others play on television. Besides knowing the rules and the skills needed to play, they underwent countless hours of practice, hard work, a lot of patience and perseverance, willingness to meet failures, and determination to achieve their goal.
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Likewise, you cannot master mathematics by reading about it or by simply watching your professor do it in class; you have to get involved and stay involved by doing it every day, just as skill is acquired in a sport. You can learn mathematics only in small, progressive steps, building on skills you have already mastered. Remember the saying: Rome wasn't built in a day. Keep using the vocabulary and facts you have already studied. They must be fresh in your mind; review them every week.
A Few Suggestions for Learning Mathematics
9 Read a few sections before each class. You might not fully understand the material, but you'll follow it far better when your professor discusses it in class. In addition, you will be able to ask more questions in class and answer more questions. 9 Whenever you study the book, make sure you have a pencil and enough paper to write down definitions, theorems, and proofs, and to do the exercises. 9 Return to review the material taught in class later in the same day. Read actively; do not just read as if it was a novel or a newspaper. Write down the definitions, theorems, and properties in your own words, without looking in your notes or the book. Good notetaking and writing aid retention. Rewrite the examples, proofs, and exercises done in class, all in your own words. If you find them too challenging, study them again and try again; continue until you succeed. 9 Always study the relevant section in the text and do the examples there; then do the exercises at the end of the section. Since the exercises are graded in order of difficulty, do them in order. Don't skip steps or write over previous steps; this way you'll progress logically, and you can locate and correct your errors. If you can't solve a problem because it involves a new term, formula, or some property, then restudy the relevant portion of the section and try again. Don't assume that you'll be able to do every problem the first time you try it. Remember, practice is the only way to
Success.
Solutions Manual
The Student's Solutions Manual contains additional helpful tips for studying mathematics, and preparing for and taking an examination in mathematics. It also gives detailed solutions to all oddnumbered exercises and a comprehensive list of references for the various exploratory writing projects.
A Final Word
Mathematics is no more difficult than any other subject. If you have the motivation, and patience to learn and do the work, then you will enjoy
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the beauty and power of discrete mathematics; you will see that discrete mathematics is really fun. Keep in mind that learning mathematics is a stepbystep process. Practice regularly and systematically; review earlier chapters every week, since things must be fresh in your mind to apply and build on them. In this way, you will enjoy the subject, feel confident, and to explore more. The name of the game is practice, so practice, practice, practice. I look forward to hearing from you with your comments and suggestions. In the meantime, enjoy the beauty and power of mathematics. Thomas Koshy
Chapter 1
The L a n g u a g e of Logic
Symbolic logic has been disowned by many logicians on the plea that its interest is mathematical and by many mathematicians on the plea that its interest is logical.
  A . N. W H I T E H E A D
ogic is the study of the principles and techniques of reasoning. It originated with the ancient Greeks, led by the philosopher Aristotle, who is often called the father of logic. However, it was not until the 17th century that symbols were used in the development of logic. German philosopher and mathematician Gottfried Leibniz introduced symbolism into logic. Nevertheless, no significant contributions in symbolic logic were made until those of George Boole, an English mathematician. At the age of 39, Boole published his outstanding work in symbolic logic, A n Investigation of the L a w s of Thought. Logic plays a central role in the development of every area of learning, especially in mathematics and computer science. Computer scientists, for example, employ logic to develop programming languages and to establish the correctness of programs. Electronics engineers apply logic in the design of computer chips. This chapter presents the fundamentals of logic, its symbols, and rules to help you to think systematically, to express yourself in precise and concise terms, and to make valid arguments. Here are a few interesting problems we shall pursue in this chapter: 9 Consider the following two sentences, both There are more residents in New York City head of any resident. No resident is totally sion: Is it true that at least two residents hairs? (R. M. Smullyan, 1978) true: than there are hairs on the bald. What is your concluhave the same number of
L
9 There are two kinds of inhabitants, "knights" and "knaves," on an island. Knights always tell the truth, whereas knaves always lie. Every inhabitant is either a knight or a knave. Tom and Dick are two residents. Tom says, "At least one of us is a knave." What are Tom and Dick?
Chapter I The Language of Logic
A r i s t o t l e (384322 B.C.), o n e of the greatest philosophers in Western culture, was born in Stagira, a small town in northern Greece. His father was the personal physician of the king of Macedonia. Orphaned young, Aristotle was . ~,; ~" i. ~'' % ~ ,,. "":. ;'i ' . ' ~ ' raised by a guardian. At the age of 18, Aristotle entered Plato's Academy in Athens. He was the "brightest and most learned student" at the Academy which he left when Plato died in 34 7 B.C. About 342 B.C., the king of Macedonia invited him to supervise the education of his young son, Alexander, who later became Alexander the Great. Aristotle taught him until 336 B.C., when the youth became ruler following the assassination of his father. ...*:4 ......}"~:~i:~:.r Around 334 B.C., Aristotle returned to Athens and founded a school called the Lyceum. His philosophy and followers were called peripatetic, a Greek word meaning "walking around," since Aristotle taught his students while walking with them. The Athenians, perhaps resenting his relationship with Alexander the Great, who had conquered them, accused him of impiety soon after the Emperor's death in 323 B.C. Aristotle, knowing the fate of Socrates, who had been condemned to death on a similar charge, fled to Chalcis, so the Athenians would not "sin twice against philosophy." He died there the following year.
W h a t are t h e y if T o m says, " E i t h e r I ' m a k n a v e or Dick is a k n i g h t " ? (R. M. S m u l l y a n , 1978) 9 Are t h e r e positive i n t e g e r s t h a t can be e x p r e s s e d as t h e s u m of t w o d i f f e r e n t cubes in two d i f f e r e n t ways? 9 Does t h e f o r m u l a E ( n ) = n 2  n + 41 yield a p r i m e n u m b e r for e v e r y positive i n t e g e r n?
A d e c l a r a t i v e s e n t e n c e t h a t is e i t h e r t r u e or false, b u t not both, is a p r o p o s i t i o n (or a s t a t e m e n t ) , w h i c h we will d e n o t e by t h e l o w e r c a s e l e t t e r p, q, r, s, or t. T h e v a r i a b l e s p, q, r, s, or t are b o o l e a n v a r i a b l e s (or l o g i c variables). T h e following s e n t e n c e s are propositions: (1) (2) (3) (4) S o c r a t e s w a s a G r e e k philosopher. 3+4=5. 1 + 1 = 0 a n d t h e m o o n is m a d e of g r e e n cheese. If i = 2, t h e n roses are red.
T h e following s e n t e n c e s a r e not propositions: 9 Let m e go! 9 x+3=5 (exclamation) (x is an u n k n o w n . )
1.1
Propositions
B a r o n Gottfried Wilhelm Leibniz (16461716), an outstanding German mathematician, philosopher, physicist, diplomat, and linguist, was born into a Lutheran family. The son of a professor of philosophy, he "grew up to be a genius with encylopedic knowledge." He had taught himself Latin, Greek, and philosophy before entering the University of Leipzig at age 15 as a law student. There he read the works of great scientists and philosophers such as Galileo, Francis Bacon, and Rend Descartes. Because of his youth, Leipzig refused to award him the degree of the doctor of laws, so he left his native city forever. During 16631666, he attended the universities of Jena and Altdorf, and receiving his doctorate from the latter in 1666, he began legal services for the 7 Elector of Mainz. After the Elector's death, Leibniz pursued scientific studies. In 1672, he built a calculating machine that could multiply and divide and presented it to the Royal Society in London the following year. In late 1675, Leibniz laid the foundations of calculus, an honor he shares with Sir Isaac Newton. He discovered the fundamental theorem of calculus, and invented the popular notationsd/dx for differentiation and f for integration. He also introduced such modern notations as dot for multiplication, the decimal point, the equal sign, and the colon for ratio. From 1676, until his death, Leibniz worked for the Duke of Brunswick at Hanover and his estate after the duke's death in 1680. He played a key role in the founding of the Berlin Academy of Sciences in 1700. Twelve years later, Leibniz was appointed councilor of the Russian Empire and was given the title of baron by Peter the Great. Suffering greatly from gout, Leibniz died in Hanover. He was never married. His works influenced such diverse disciplines as theology, philosophy, mathematics, the natural sciences, history, and technology.
, .......... Lb~,
9 Close the door! 9 Kennedy was a great president of the United States. 9 What is my line?
Truth Value
(command) (opinion) (interrogation)
1
The truthfulness or falsity of a proposition is called its t r u t h v a l u e , denoted by T(true) and F(false), respectively. (These values are often denoted by 1 and 0 by computer scientists.) For example, the t r u t h value of statement (1) in Example 1.1 is T and that of statement (2) is F. Consider the sentence, This sentence is false. It is certainly a valid declarative sentence, but is it a proposition? To answer this, assume the sentence is true. But the sentence says it is false. This contradicts our assumption. On the other hand, suppose the sentence is false. This implies the sentence
Chapter I The Language of Logic
George Boole (18151864), the son of a cobbler whose main interests were mathematics and the making of optical instruments, was born in Lincoln, England. Beyond attending a local elementary school and briefly a commercial school, Boole was selftaught in mathematics and the classics. When his father's business failed, he started working to support the family. At 16, he began his teaching career, opening a school of his own four years later in Lincoln. In his leisure time, Boole read mathematical journals at the Mechanics Institute. There he grappled with the works of English physicist and mathematician Sir Isaac Newton and French mathematicians PierreSimon Laplace and JosephLouis Lagrange. In 1839, Boole began contributing original papers on differential equations to The Cambridge Mathematics Journal and on analysis to the Royal Society. In 1844, he was awarded a Royal Medal by the Society for his contributions to analysis; he was elected a fellow of the Society in 1857. Developing novel ideas in logic and symbolic reasoning, he published his first contribution to symbolic logic, The Mathematical Analysis of Logic, in 184 7. His publications played a key role in his appointment as professor of mathematics at Queen's College, Cork, Ireland, in 1849, although he lacked a university education. In 1854, he published his most important work, An Investigation to the Laws of Thought, in which he presented the algebra of logic now known as boolean algebra (see Chapter 12). The next year he married Mary Everest, the niece of Sir George Everest, for whom the mountain is named. In addition to writing about 50 papers, Boole published two textbooks, Treatise on Differential Equations (1859) and Treatise on the Calculus of Finite Differences; both were used as texts in the United Kingdom for many years. A conscientious and devoted teacher, Boole died of pneumonia in Cork.
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I
is true, which again contradicts our assumption. Thus, if we assume t h a t the sentence is true, it is false; and if we assume t h a t it is false, it is true. It is a meaningless and selfcontradictory sentence, so it is not a proposition, but a p a r a d o x . The t r u t h value of a proposition may not be known for some reason, b u t t h a t does not prevent it from being a proposition. For example, around 1637, the F r e n c h m a t h e m a t i c a l genius PierreSimon de F e r m a t conjectured t h a t the equation x n + yn = z n has no positive integer solutions, where n >_ 3. His conjecture, known as F e r m a t ' s L a s t " T h e o r e m , " was one of the celebrated unsolved problems in n u m b e r theory, until it was proved in 1993 by the English m a t h e m a t i c i a n Andrew J. Wiles (1953) of Princeton University. Although the t r u t h value of the conjecture eluded m a t h e m a t i c i a n s for over three centuries, it was still a proposition! Here is a n o t h e r example of such a proposition. In 1742 the P r u s s i a n m a t h e m a t i c i a n Christian Goldbach conjectured t h a t every even integer greater t h a n 2 is the sum of two primes, not necessarily distinct. For example, 4  2 + 2, 6  3 + 3, and 18 = 7 + 11. It has been shown true for every
1.1 Propositions
F e r m a t (16011665) was born near Toulouse as the son of a leather merchant. A lawyer by profession, he devoted his leisure time to mathematics. Although he published almost none of his discoveries, he did correspond with contemporary mathematicians. Fermat contributed to several branches of mathematics, but he is best known for his work in number theory. Many of his results appear in margins of his copy of the works of the Greek mathematician Diophantus (250 A.D. ?). He wrote the following about his famous conjecture: "I have discovered a truly wonderful proof, but the margin is too small to contain it."
C h r i s t i a n Goldbach (16901764) was born in K6nigsberg, Prussia. He studied medicine and mathematics at the University of K6nigsberg and became professor of mathematics at the Imperial Academy of Sciences in St. Petersburg in 1725. In 1728, he moved to Moscow to tutor Tsarevich Peter H and his cousin Anna of Courland. From 1729 to 1763, he corresponded with Euler on number theory. He returned to the Imperial Academy in 1732, when Peter's successor Anna moved the imperial court to St. Petersburg. In 1742, Goldbach joined the Russian Ministry of Foreign Affairs, and later became privy councilor and established guidelines for the education of royal children. Noted for his conjectures in number theory and work in analysis, Goldbach died in Moscow.
even integer less than 4 • 1014, but no one has been able to prove or disprove his conjecture. Nonetheless, the Goldbach conjecture is a proposition. Propositions (1) and (2) in Example 1.1 are s i m p l e p r o p o s i t i o n s . A compound proposition is formed by combining two or more simple propositions called c o m p o n e n t s . For instance, propositions (3) and (4) in Example 1.1 are compound. The components of proposition (4) are I = 2 and Roses are red. The truth value of a compound proposition depends on the truth values of its components. Compound propositions can be formed in several ways, and they are presented in the rest of this section.
Conjunction
The conjunction of two arbitrary propositions p and q, denoted by p A q, is the proposition p a n d q. It is formed by combining the propositions using the word and, called a connective.
Chapter I The Language of Logic
Consider the s t a t e m e n t s
p: S o c r a t e s w a s a G r e e k p h i l o s o p h e r q: E u c l i d w a s a C h i n e s e m u s i c i a n .
and
Their conjunction is given by
p A q: S o c r a t e s w a s a G r e e k p h i l o s o p h e r a n d E u c l i d w a s a Chinese musician.
m
To define the t r u t h value of p A q, where p and q are a r b i t r a r y propositions, we need to consider four possible cases: 9 p is true, q is true. 9 p is true, q is false. 9 p is false, q is true. 9 p is false, q is false. (See the t r e e d i a g r a m in Figure 1.1 and Table 1.1.) If both p and q are true, t h e n p A q is true; i f p is t r u e and q is false, t h e n p A q is false; i f p is fhlse and q is true, t h e n p A q is false; and if both p and q are false, t h e n p A q is also false.
F i g u r e 1.1
Truth value ofp
Truth value ofq T
T F F
T a b l e 1.1
P
q
P^q
T T F F
T F T F
This information can be s u m m a r i z e d in a table. In the third column of Table 1.1, enter the t r u t h value ofp A q c o r r e s p o n d i n g to each pair of t r u t h values of p and q. The resulting table, Table 1.2, is the t r u t h t a b l e for
pAq.
1.1 Propositions Table 1.2
T r u t h table for p A q
p
T T F F
q
T F T F
pAq
T F F F
Expressions t h a t yield the value t r u e or false are b o o l e a n e x p r e s s i o n s , and they often occur in both m a t h e m a t i c s and computer science. For instance, 3 < 5 and 5 < 5 are boolean expressions. Ifstatements and whileloops in computer programs often use such expressions, and their values determine w h e t h e r or not ifstatements and whileloops will be executed, as the next example illustrates. Determine w h e t h e r the assignment s t a t e m e n t , sum _ 5)] _ then
SOLUTION: The s t a t e m e n t x ~ x + 1 will be executed if the value of the boolean expression ~ l ( a < b) v (b > _5)1 is true. By De M o r g a n ' s law, _ ~ l ( a < b) v (b > _5)1  ~ ( a < b) A ~(b > _5) _ _  (a > b) A (b < 5) Sincea7andb4, botha>_bandb < 5 are true; so, (a > _b ) A(b < 5) _ is true. Therefore, the a s s i g n m e n t s t a t e m e n t will be executed, m One of the elegant applications of the laws of logic is employing t h e m to simplify complex boolean expressions, as the next example illustrates. Using the laws of logic simplify the boolean expression (p A ~q) v q v (~p Aq). SOLUTION" [The justification for every step is given on its r i g h t  h a n d  s i d e (RHS). ] (p A ~q) v q v (~p A q)  [(p A ~q) v q] v (p A q) [qV(pA~q)]V(~pAq) assoc, law comm. law
1.2 Logical Equivalences
 [(q v p) A (q v ~q)] v (~p A q)  [(q V p) A t] V (~p A q) = (q V p) V (~p A q) = (~p
A
25
dist. law
qv~qt rAtr
q)
v
(p
v
A
q) [q v (p v q)] [q v (p v q)]
v
comm. law dist. law assoc, law
~p vpt tvqt tArr
 [~p v (p v q)] = [(~p v p) v q] _ (t
v
A
q)
A
[q
v
(p
q)]
= t A [q V (p Vq)]  q V (p V q)
=_qV(qVp) =(qVq) =_qvp pVq Vp
comm. law assoc, law idem. law comm. law
For any propositions p, q, and r, it can be shown t h a t p ~ (q v r) (p A~q) ~ r (see Exercise 12). We shall employ this result in Section 1.5. Here are two e l e m e n t a r y but elegant applications of this equivalence. Suppose a and b are any two real n u m b e r s , and we would like to prove the following theorem: I f a . b = O, then either a = 0 or b = 0. By virtue of the above logical equivalence, we need only prove the following proposition: I f a . b = 0 a n d a V: O, t h e n b = 0 (see Exercise 43 in Section 1.5). Second, suppose a and b are two a r b i t r a r y positive integers, and p a prime number. Suppose we would like to prove the following fact: Ifp[ab,* then either p la or p lb. Using the above equivalence, it suffices to prove the following equivalent s t a t e m e n t : I f plab a n d p Xa, t h e n p]b (see Exercise 37 in Section 4.2). We shall now show how useful symbolic logic is in the design of switching networks.
Equivalent Switching Networks (optional)
Two switching n e t w o r k s A and B are equivalent if they have the same electrical behavior, either b o t h open or both closed, symbolically described by A  B. One of the i m p o r t a n t applications of symbolic logic is to replace an electrical network, w h e n e v e r possible, by an equivalent simpler network to minimize cost, as illustrated in the following example. To this end,
*xly m e a n s
"x is a f a c t o r o f y . "
26
Chapter 1 The Language of Logic
let A be any circuit, T a closed circuit, and F an open circuit. T h e n A A T A, A A A'  F, A v T = T, and A v A'  T (see laws 3 t h r o u g h 8). Likewise, laws 1 t h r o u g h 11 can also be extended to circuits in an obvious way. ~ Replace the switching n e t w o r k in Figure 1.5 by an equivalent simpler network.
F i g u r e 1.5
 9
@
SOLUTION: The given network is r e p r e s e n t e d by (A A B') v [(A A B) v C]. Let us simplify this expression using the laws of logic. (The reason for each step is given on its RHS.) (A A B') v I(A A B) v CI = I(A A B') v (A A B)I v C  [A A (B' v B)I v C  (A A T) v C _=AvC assoc, law dist. law B'vBT AAT=A
Consequently, the given circuit can be replaced by the simpler circuit in Figure 1.6.
F i g u r e 1.6
 9
We close this section with a brief introduction to fuzzy logic.
Fuzzy Logic (optional)
"The binary logic of m o d e r n computers," wrote Bart Kosko and Satoru Isaka, two pioneers in the development of fuzzy logic systems, "often falls short when describing the vagueness of the real world. Fuzzy logic offers more graceful alternatives." Fuzzy logic, a b r a n c h of artificial intelligence, incorporates the vagueness or value j u d g e m e n t s t h a t exist in everyday life, such as "young," "smart," "hot," and "cold." The first company to use a fuzzy system was F. L. Smidth and Co., a cont r a c t i n g company in Copenhagen, Denmark, which in 1980 used it to r u n a
1.2 LogicalEquivalences
27
B a r t Kosko holds degrees in philosophy and economics from the Universityof Southern California, an
M.S. in applied mathematics, and a Ph.D. in electrical engineering from the University of California, Irvine. Currently, he is on the faculty in electrical engineering at the University of Southern California.
S a t o r u I s a k a received his M.S. and Ph.D. in systems science from the University of California, San Diego. He specializes in fuzzy information processing at Omron Advanced Systems at Santa Clara, and in the application of machine learning and adaptive control systems to biomedical systems and factory automation.
cement kiln. Eight years later, Hitachi used a fuzzy system to r u n the subway system in Sendai, Japan. Since then Japanese and American companies have employed fuzzy logic to control hundreds of household appliances, such as microwave ovens and washing machines, and electronic devices, such as cameras and camcorders. (See Figure 1.7.) It is generally believed t h a t fuzzy, commonsense models are far more useful and accurate t h a n standard mathematical ones.
F i g u r e 1.7
JuST TeEw~y (;INA LIKESIT
JUST THE WHY TEl) LIKES ir
dUSTTHEW Y A THE FEDEX 6UT LIKES IT
In fuzzy logic, the t r u t h value t(p) of a proposition p varies from 0 to 1, depending on the degree of its truth; so 0 _< t(p) _< 1. For example, the s t a t e m e n t "The room is cool" may be assigned a t r u t h value of 0.4; and the s t a t e m e n t "Sarah is smart" may be assigned a t r u t h value of 0.7.
28
(~hapter 1 The Language of Logic
Let 0 < x , y _< 1. T h e n t h e o p e r a t i o n s A, V, a n d ' a r e defined as follows" x A y  min{x,y} x v y  max{x,y} x 1x w h e r e min{x,y} d e n o t e s t h e m i n i m u m o f x a n d y, a n d max{x,y} d e n o t e s t h e m a x i m u m of x a n d y. N o t all p r o p e r t i e s in p r o p o s i t i o n a l logic are valid in fuzzy logic. F o r i n s t a n c e , t h e l a w o f e x c l u d e d m i d d l e , p v ~ p is true, does n o t hold in fuzzy logic. To see this, let p be a simple p r o p o s i t i o n w i t h t(p) = 0.3. T h e n t(p') = 1  0 . 3 = 0. 7; so t ( p v p ' ) = t ( p ) v t ( p ' ) = 0 . 3 v 0 . 7 = m a x { 0 . 3 , 0.7} = 0 . 7 # 1. T h u s p v p ' is not a t a u t o l o g y in fuzzy logic, l In p r o p o s i t i o n a l logic, t(p v p') = 1; so p v p' is a t a u t o l o g y . T h i n k of 1 r e p r e s e n t i n g a T a n d 0 r e p r e s e n t i n g an F. I Likewise, t(p A p') = t(p) A t(p') = 0.3 A 0. 7 = min{0.3, 0.7} = 0 . 3 # 0; so p A p ' is not a c o n t r a d i c t i o n , u n l i k e in p r o p o s i t i o n a l logic. N e x t we p r e s e n t briefly an i n t e r e s t i n g application* of fuzzy logic to decision m a k i n g . It is based on t h e Y a g e r m e t h o d , developed in 1981 by R o n a l d R. Yager of Iona College, a n d e m p l o y s fuzzy i n t e r s e c t i o n a n d i m p l i c a t i o n ~, defined by p ~ q = u p v q.
!
Fuzzy Decisions
S u p p o s e t h a t from a m o n g five U.S. c i t i e s   B o s t o n , Cleveland, M i a m i , N e w York, a n d San D i e g o   w e would like to select t h e b e s t city to live in. We will use seven c a t e g o r i e s C I t h r o u g h C7 to m a k e t h e decision; t h e y are climate, cost o f h o u s i n g , cost o f living, o u t d o o r activities, e m p l o y m e n t , crime, a n d culture, respectively, a n d are j u d g e d on a scale 0  6 : 0 = terrible, 1 = bad, 2 = pool', 3 = average, 4 = fairly good, 5 = very good, a n d 6 = excellent. T a b l e 1.15 shows t h e relative i m p o r t a n c e of each c r i t e r i o n on a scale 0  6 a n d t h e r a t i n g for e a c h city in each category. T a b l e 1.15
Category
C1 C2 C3 C4 C5 C6 C7
Importance
6 3 2 4 4 5 4
Boston
3 1 3 5 4 2 6
Cleveland
2 5 4 3 3 4 3
Miami
5 4 3 6 3 0 3
New York
1 0 1 2 4 1 6
San Diego
6 1 5 6 3 3 5
*Based on M. Caudill, "Using Neural Nets: Fuzzy Decisions," A/Expert, Vol. 5 (April 1990), pp. 5964.
1.2 Logical Equivalences
29
The ideal city to live in will score high in the categories considered m o s t i m p o r t a n t . In order to choose the finest city, we need to evaluate each city by each criterion, weighing the relative i m p o r t a n c e of each category. Thus, given a p a r t i c u l a r category's i m p o r t a n c e , we m u s t check the city's score in t h a t category; in o t h e r words, we m u s t c o m p u t e the t r u t h value of i + s  ~ i v s for each city, w h e r e i denotes t h e i m p o r t a n c e r a n k i n g for a p a r t i c u l a r category and s the city's score for t h a t category. Table 1.16 shows the r e s u l t i n g data. Now we take the conjunction of all scores for each city, using the m i n function (see Table 1.16). The lowest combined score d e t e r m i n e s t h e city's overall ranking. It follows from the table t h a t San Diego is clearly t h e winner.
T a b l e 1.16
Category
C1 C2 C3 C4 C5 C6 C7
~i
s
Boston
~ivs
Cleveland
s ,~ivs s
Miami
,.~ivs
N e w York
s ,~ivs
San D i e g o
s ~ivs
0 3 4 2 2 1 2
3 1 3 5 4 2 6
3 3 4 5 4 2 6 2
2 5 4 3 3 4 3
2 5 4 3 3 4 3 2
5 4 3 6 3 0 3
5 4 4 6 3 1 3 1
1 0 1 2 4 1 6
1 3 4 2 4 1 6 1
6 1 5 6 3 3 5
6 3 5 6 3 3 5 3 winner
Intersection
next best choices
Finally, suppose we add a sixth city, say, Atlanta, for consideration. T h e n the Yager m e t h o d e n s u r e s t h a t the revised choice will be the existing choice (San Diego) or Atlanta; it c a n ' t be any of the others. T h u s the p r o c e d u r e allows i n c r e m e n t a l decision making, so m a n a g e a b l e subdecisions can be combined into an overall final choice.
Exercises 1.2
Give the t r u t h value o f p in each case. 1. p  q, and q is not true. 2. p  q, q  r, and r is true.
Verify each, w h e r e f denotes a contradiction. (See Table 1.14.)
3. ~ ( ~ p ) =p
6. p A q =   q A p
4. p Ap p
7. p V q = q V p
5. p Vp = p
8. " ~ ( p V q ) =   ~ p A '  q
9. ~ ( p > q)  p
A ~q
30
Chapter I
The Language of Logic
10. p ~ q = ( p A ~ q )   ~ f 12. p ~ ( q V r )   ( p A ' ~ q ) ~ r
11. p A ( q A r ) =   ( p A q ) A r 13. (p v q) ~ r  (p ~ r) A (q ~ r)
U s e De M o r g a n ' s laws to e v a l u a t e e a c h b o o l e a n e x p r e s s i o n , w h e r e x = 2, y = 5, a n d z = 3.
14. ~ [ ( x < z ) 16. A(y p. W h e n x is divided by each of the primes 2, 3, 5 , . . . ,p, we get 1 as the remainder. So x is not divisible by any of the primes. Hence either x m u s t be a prime, or if x is composite t h e n x is divisible by a prime q # Pi. In either case, there are more t h a n k primes. But this contradicts the a s s u m p t i o n t h a t there are k primes, so our a s s u m p t i o n is false. In other words, t h e r e is no largest prime number, m Now we t u r n to yet a n o t h e r proof technique.
52
Chapter I
The Language of Logic
Proof by Cases
Suppose we would like to prove a t h e o r e m of the form H1 v H2 v . . . v
Hn + C. Since H I v H2 v . . . v H a + C  ( H I > C) A (H2 > C) A . .  A (Hn ~ C), the s t a t e m e n t H1 v H2 v . . . v Hn ~ C is t r u e if a n d only if each i m p l i c a t i o n Hi > C is true. Consequently, we need only prove t h a t each
implication is true. Such a proof is a p r o o f b y c a s e s , as i l l u s t r a t e d in the following example, due to R. M. Smullyan. Let A, B, and C be three i n h a b i t a n t s of the island described in Example 1.32. Two i n h a b i t a n t s are of the s a m e type if they are b o t h k n i g h t s or both knaves. Suppose A says, "B is a knave," and B says, "A a n d C are of the same type." Prove t h a t C is a knave.
PROOF BY CASES
Although this t h e o r e m is not explicitly of the form II1 v H2 v . . . v IIn ~ C, we artificially create two cases, namely, A is a k n i g h t and A is a knave.
Case 1
Suppose A is a knight. Since k n i g h t s always tell the t r u t h , his s t a t e m e n t t h a t B is a knave is true. So B is a knave and hence B's s t a t e m e n t is false. Therefore, A and C are of different types; t h u s C is a knave. Suppose A is a knave. T h e n his s t a t e m e n t is false, so B is a knight. Since k n i g h t s always tell the t r u t h , B's s t a t e m e n t is true. So A and C are of the same type; t h u s C is a knave. T h u s in both cases, C is a knave, m
Case 2
Existence Proof
Finally, t h e o r e m s of the form (~x)P(x) also occur in m a t h e m a t i c s . To prove such a theorem, we m u s t establish the existence of an object a for which P(a) is true. Accordingly, such a proof is an e x i s t e n c e p r o o f . T h e r e are two kinds of existence proofs: the c o n s t r u c t i v e e x i s t e n c e proof and the n o n c o n s t r u c t i v e e x i s t e n c e p r o o f . If we are able to find a m a t h e m a t i c a l object b such t h a t P(b) is true, such an existence proof is a c o n s t r u c t i v e p r o o f . The following example elucidates this method. Prove t h a t there is a positive integer t h a t can be expressed in two different ways as the sum of two cubes.
CONSTRUCTIVE PROOF
By the discussion above, all we need is to produce a positive integer b t h a t has the required properties. Choose b  1729. Since 1729  13 + 123 = 93 + 103, 1729 is such an integer.* m
*A fascinating anecdote is told about the number 1729. In 1919, when the Indian mathematical genius Srinivasa Ramanujan (18871920) was sick in a nursing home in England, the eminent
1.5 ProofMethods
53
A n o n c o n s t r u c t i v e existence proof of the t h e o r e m (3x)P(x) does not provide us with an e l e m e n t a for which P(a) is true, b u t r a t h e r establishes its existence by an indirect m e t h o d , usually contradiction, as i l l u s t r a t e d by the next example. Prove t h a t t h e r e is a p r i m e n u m b e r > 3.
NONCONSTRUCTIVE PROOF Suppose t h e r e are no primes > 3. T h e n 2 and 3 are the only primes. Since every integer >_ 2 can be expressed as a p r o d u c t of powers of primes, 25 m u s t be expressible as a p r o d u c t of powers of 2 and 3, t h a t is, 25  2 i 3 j for some integers i and j. But n e i t h e r 2 nor 3 is a factor of 25, so 25 c a n n o t be w r i t t e n in the form 2 i 3 j , a contradiction. Consequently, t h e r e m u s t be a p r i m e > 3. II
We invite you to give a constructive proof of the s t a t e m e n t in the example. We conclude this section with a brief discussion of counterexamples.
Counterexample
Is the s t a t e m e n t E v e r y g i r l is a b r u n e t t e t r u e or false? Since we can find at least one girl who is not a b r u n e t t e , it is false! More generally, suppose you would like to show t h a t the s t a t e m e n t (Vx)P(x) is false. Since ~[(Vx)P(x)] _= (3x)[~P(x)] by De M o r g a n ' s law, the s t a t e m e n t (Vx)P(x) is false if t h e r e exists an item x in the UD for which the predicate P(x) is false. Such an object x is a c o u n t e r e x a m p l e . Thus, to disprove the proposition (Vx)P(x), all we need is to produce a c o u n t e r e x a m p l e c for which P(c) is false, as the next two examples d e m o n s t r a t e . N u m b e r theorists d r e a m of finding formulas t h a t g e n e r a t e p r i m e n u m b e r s . One such f o r m u l a was found by the Swiss m a t h e m a t i c i a n L e o n h a r d E u l e r (see C h a p t e r 8), namely, E ( n )  n 2  n + 41. It yields a p r i m e for n = 1, 2, . . . , 40. Suppose we claim t h a t the f o r m u l a g e n e r a t e s a p r i m e for every positive integer n. Since E(41) = 412  41 + 41 = 412 is not a prime, 41 is a counterexample, t h u s disproving the claim. II A r o u n d 1640, F e r m a t conjectured t h a t n u m b e r s of the form f ( n )  22'' + 1 are prime n u m b e r s for all n o n n e g a t i v e integers n. For instance, f(0) = 3, f(1) = 5, f(2) = 17, f(3) = 257, and f(4) = 65,537 are all primes. In 1732, however, E u l e r established the falsity of F e r m a t ' s conjecture by producing a counterexample. He showed t h a t f(5)  22'~ + 1  641 • 6700417, a composite n u m b e r . (Prime n u m b e r s of the form 22'' + I are called F e r m a t primes.) I1 English mathematician Godfrey Harold Hardy (18771947) visited him. He told Ramanujan that the number of the cab he came in, 1729, was "a rather dull number" and hoped that it wasn't a bad omen. "No, Hardy," Ramanujan responded, "It is a very interesting number. It is the smallest number expressible as the sum of two cubes in two different ways."
54
Chapter I The Language of Logic
Exercises 1.5
D e t e r m i n e if each implication is vacuously t r u e for the indicated value of n. 1. If n > 1, t h e n 2 n > n; n  0 2. If n > 4, t h e n 2 n >_ n2"n,  O, 1, 2, 3 D e t e r m i n e if each implication is trivially true. 3. If n is a p r i m e n u m b e r , t h e n n 2 + n is an even integer. 4. If n > 41, t h e n n 3  n is divisible by 3. Prove each directly. 5. The s u m of a n y two even i n t e g e r s is even. 6. The s u m of a n y two odd i n t e g e r s is even. 7. The s q u a r e of an even i n t e g e r is even. 8. The product of any two even i n t e g e r s is even. 9. The s q u a r e of an odd integer is odd. 10. The product of any two odd i n t e g e r s is odd. 11. The product of any even i n t e g e r a n d any odd integer is even. 12. The s q u a r e of every integer of t h e form 3k + 1 is also of t h e s a m e form, w h e r e k is an a r b i t r a r y integer. 13. The s q u a r e of every integer of t h e form 4k + 1 is also of t h e s a m e form, w h e r e k is an a r b i t r a r y integer. 14. T h e a r i t h m e t i c mean ~~ of a n y two n o n n e g a t i v e real n u m b e r s a
and b is g r e a t e r t h a n or equal to t h e i r g e o m e t r i c m e a n j~ab. IHint" consider (v/a v/b)2 > 0.] Prove each u s i n g the law of the contrapositive. 15. If the s q u a r e of an integer is even, t h e n the integer is even. 16. If the s q u a r e of an integer is odd, t h e n the i n t e g e r is odd. 17. If the product of two integers is even, t h e n at least one of t h e m m u s t be an even integer. 18. If the product of two integers is odd, t h e n both m u s t be odd integers. Prove by contradiction, w h e r e p is a p r i m e n u m b e r . 19. j ~ is an irrational n u m b e r . 21. v/~ is an irrational n u m b e r . 20. v ~ is an i r r a t i o n a l n u m b e r . *22. log102 is an i r r a t i o n a l n u m b e r .
Prove by cases, w h e r e n is an a r b i t r a r y integer and Ixl denotes t h e absolute value of x.
1.5 ProofMethods
23. n 2 + n is a n e v e n i n t e g e r .
55
24. 2n 3 + 3n 2 + n is a n e v e n i n t e g e r .
25. n 3  n is divisible by 3.
3k, 3k + 1, or 3k + 2.)
( H i n t : A s s u m e t h a t e v e r y i n t e g e r is of t h e f o r m
26. ]  x [ = [ x [
27. [ x . y [ = [ x [ . [ y ]
28. [x +y] _< [x[ + [y[
P r o v e by t h e e x i s t e n c e m e t h o d . 29. T h e r e a r e i n t e g e r s x s u c h t h a t x 2 = x. 30. T h e r e a r e i n t e g e r s x s u c h t h a t [x] = x. 31. T h e r e a r e i n f i n i t e l y m a n y i n t e g e r s t h a t c a n be e x p r e s s e d as t h e s u m o f t w o c u b e s in t w o d i f f e r e n t ways. 32. T h e e q u a t i o n x 2 + y2 = z 2 h a s infinitely m a n y i n t e g e r s o l u t i o n s . Give a c o u n t e r e x a m p l e to d i s p r o v e e a c h s t a t e m e n t , w h e r e P(x) d e n o t e s a n arbitrary predicate. 33. T h e a b s o l u t e v a l u e of e v e r y real n u m b e r is positive. 34. T h e s q u a r e of e v e r y real n u m b e r is positive. 35. E v e r y p r i m e n u m b e r is odd. 36. E v e r y m o n t h h a s exactly 30 days. 37. (3x)P(x) ~ (3!x)P(x)
38. (~x)P(x) ~ (Vx)P(x) 39. F i n d t h e flaw in t h e following "proof": L e t a a n d b be real n u m b e r s s u c h t h a t a = b. T h e n a b = b 2. Therefore, a 2  ab = a 2  b 2 F a c t o r i n g , a ( a  b)  ( a + b ) ( a  b) C a n c e l a  b f r o m b o t h sides:
a=a+b
Since a  b, t h i s yields a  2a. C a n c e l a f r o m b o t h sides. T h e n we get 1 = 2. L e t a, b, a n d c b e a n y real n u m b e r s . T h e n a < b if a n d only if t h e r e is a positive real n u m b e r x s u c h t h a t a + x  b. U s e t h i s fact to p r o v e each. 40. If a < b a n d b < c, t h e n a < c. ( t r a n s i t i v e 41. I f a < b , thena+c_ 3. D e t e r m i n e the t r u t h value of each. 57. (Vx)lP(x) A Q(x)l 60. (Sz)lP(z)vQ(z)l 58. (Vx)[P(x) v Q(x)] 61. (Vx)[~P(x)] 59. (3y)[P(y) A Q(y)] 62. (3z)[~Q(z)]
Prove each, where a, b, c, d, and n are any integers. 63. The product of two consecutive integers is even. 64. n 3 + n is divisible by 2. 65. n 4  n 2 is divisible by 3. 66. I f a < b a n d c < d , 67. I f a + b thena+c6.
> 12, t h e n e i t h e r a > 6 o r b
68. I f a b  ac, t h e n either a  0 or b  c. [ H i n t : p > (qvr)  (p A~q) ~ r.] 69. If a 2  b 2, t h e n either a  b or a   b .
[ H i n t : p > (q v r) ( p A ~ q ) + r.]
70. Give a c o u n t e r e x a m p l e to disprove the following s t a t e m e n t : If n is a positive integer, t h e n n 2 + n + 41 is a prime n u m b e r . [Note: In 1798 the e m i n e n t F r e n c h m a t h e m a t i c i a n AdrienMarie Legendre (17521833) discovered t h a t the formula L ( n )  n 2 Zr n + 41 yields distinct primes for 40 consecutive values of n. Notice t h a t L ( n )  E (  n ) ; see Example 1.45. ]
The propositions in Exercises 7181 are fuzzy logic. 
Let p , q , t(r)0.5.
and r be simple propositions with t ( p )
1, t ( q ) 
0.3,
and
C o m p u t e the t r u t h value of each, where s' denotes the negation of the s t a t e m e n t s. 71. p A ( q v r ) 73. ( p A q ) V ( p A r ) 75. p'
A
72. p V ( q A r ) 74. ( p V q ) A ( p V r ) 76. (p
v
q'
q')
v
(p
A
q)
77. (p v q')' v q
78. (p A q)' A (p V q)
79. Let p be a simple proposition with t ( p ) = x and p' its negation. Show t h a t t ( p v p') = 1 if and only if t ( p ) = 0 or 1. Let p and q be simple propositions with t ( p ) 0 a , t h e n x <  a o r x > a .
3. (p *5. (p
v A
~q) ~q)
A
~(p (~p
A A
q)
v
*4. [p v q v (~p A ~q)] v (p A ~q) (~p
A
v
q)
~q)
*6. (p V q) A "~(p A q) A (~p V q)
7. Let p  q a n d r  s. D e t e r m i n e i f p + (p A r)  q + (q A S). N e g a t e each proposition, w h e r e U D = set of real n u m b e r s .
8. (VX)(3y)(xy > _1) _
10. (Vx)(Vy)(3z)(x + y = z) P r o v e each.
9. (Vx)(Vy)(xy = yx)
11. (Vx)(3y)(3z)(x + y = z)
12. T h e e q u a t i o n x 3 + y3 = z 3 h a s infinitely m a n y i n t e g e r solutions. "13. Let n be a positive integer. T h e n n(3n 4 + 7n 2 + 2) is divisible by 12. "14. Let n be a positive integer. T h e n n(3n 4 + 13n 2 + 8) is divisible by 24. "15. In 1981 O. H i g g i n s discovered t h a t t h e f o r m u l a h(x) = 9x 2 471x + 6203 g e n e r a t e s a p r i m e for 40 c o n s e c u t i v e v a l u e s of x. Give a c o u n t e r e x a m p l e to show t h a t not every value of h(x) is a p r i m e . "16. T h e f o r m u l a g ( x ) = x 2  2999x + 2248541 yields a p r i m e for 80 consecutive v a l u e s ofx. Give a c o u n t e r e x a m p l e to disprove t h a t every value o f g ( x ) is a prime. In a t h r e e  v a l u e d l o g i c , developed by t h e Polish logician J a n L u k a s i e w i c z (18781956), t h e possible t r u t h values of a p r o p o s i t i o n a r e 0, u, a n d 1, w h e r e 0 r e p r e s e n t s F, u r e p r e s e n t s u n d e c i d e d , a n d 1 r e p r e s e n t s T. T h e logical c o n n e c t i v e s A, V, ', ~, a n d o are defined as follows: A 0
u
0 0
0
U 0
u
v 0
u
0 0
u
u u
u
!
0
u
1
u
1
0 + 0 u 1
u 0 1 u 0 u 1 1 u
1
1
1
0 0 u 1 1 u 0 u u 1 u
1
0
Let p a n d q be a r b i t r a r y p r o p o s i t i o n s in a t h r e e  v a l u e d logic, w h e r e r' d e n o t e s t h e n e g a t i o n of s t a t e m e n t r a n d t(r) d e n o t e s t h e t r u t h value of r.
Chapter Summary
17. If t(p v p') = 1, show t h a t t(p) = 0 or 1. 18. Show t h a t p A q ~ p v q is a threevalued tautology. 19. Show t h a t (p + q) ~ (p' v q) is not a threevalued tautology. 20. Show t h a t (t9 ~ q) ~ (~q ~ ~p) is a threevalued tautology. 21. D e t e r m i n e if [p A (p ~ q)] ~ q is a threevalued tautology. Verify each. 22. (p A q)' =_p' v q' 23. (p V q)'  p' A q'
63
[Hint: Show t h a t t((p A q)') = t(p') V t(q').] [Hint: Show t h a t t((p v q)') = t(p') A t(q').]
Computer Exercises
Write a p r o g r a m to perform each task. C o n s t r u c t a t r u t h table for each proposition. 1. ( p v q ) A ~ q
4. (p ~ q) ~ ( ~ p v q)
2. p N A N D q
5. (p + q) ~ r
3. p N O R q
6. (p + q) a} is denoted by [a, oc) using the i n f i n i t y s y m b o l o~. Likewise, the set {x ~ R Ix < a} is denoted by (  ~ , a]. Next we present two interesting paradoxes related to infinite sets and proposed in the 1920s by the G e r m a n m a t h e m a t i c i a n David Hilbert.
The Hilbert Hotel Paradoxes Imagine a grand hotel in a major city with an infinite n u m b e r of rooms, all occupied. One m o r n i n g a visitor arrives at the registration desk looking for a room. "I'm sorry, we are full," replies the manager, "but we can certainly accommodate you." How is this possible? Is she contradicting herself?. To give a room to the new guest, Hilbert suggested moving the guest in Room 1 to Room 2, the guest in Room 2 to Room 3, the one in Room 3 to Room 4, and so on; Room 1 is now vacant and can be given to the new guest. The clerk is happy t h a t she can accommodate him by moving each guest one room down the hall. The second paradox involves an infinite n u m b e r of conventioneers arriving at the hotel, each looking for a room. The clerk realizes t h a t the hotel can make a fortune if she can somehow accommodate them. She knows she can give each a room one at a time as above, but t h a t will involve moving each guest constantly from one room to the next, resulting in total chaos and frustration. So Hilbert proposed the following solution: move the guest in Room 1 to Room 2, the guest in Room 2 to Room 4, the one in Room 3 to Room 6, and so on. This puts the old guests in evennumbered rooms, so the new guests can be checked into the oddnumbered rooms. Notice t h a t in both cases the hotel could accommodate the guests only because it has infinitely m a n y rooms.
2.1 The Concept of a Set
75
A third paradox: Infinitely m a n y hotels with infinitely m a n y rooms are leveled by an earthquake. All guests survive and come to Hilbert Hotel. How can they be accommodated? See Example 3.23 for a solution. We close this section by introducing a special set used in the study of formal languages. Every word in the English language is an a r r a n g e m e n t of the letters of the alphabet {A, B , . . . ,Z, a, b , . . . , z}. The alphabet is finite and not every a r r a n g e m e n t of the letters need make any sense. These ideas can be generalized as follows.
Alphabet
A finite set Z of symbols is an alphabet. (E is the uppercase Greek letter sigma.) A w o r d (or s t r i n g ) o v e r E is a finite a r r a n g e m e n t of symbols from E. For instance, the only alphabet understood by a computer is the b i n a r y a l p h a b e t {0,1 }; every word is a finite and unique a r r a n g e m e n t of O's and l's. Every zip code is a word over the alphabet {0,... ,9}. Sets such as {a, b, c, ab, bc} are not considered alphabets since the string ab, for instance, can be obtained by juxtaposing, that is, placing next to each other, the symbols a and b.
Length of a Word
The l e n g t h of a word w, denoted by llwli, is the n u m b e r of symbols in it. A word of length zero is the e m p t y w o r d (or the null w o r d ) , denoted by the lowercase Greek letter ~ ( l a m b d a ) ; It contains no symbols. For example, llabll = 2, llaabba]l  5, and IIs = 0. The set of words over an alphabet E is denoted by Z*. The empty word belongs to Z* for every alphabet Z. In particular, if Z denotes the English alphabet, then Z* consists of all words, both meaningful and meaningless. Consequently, the English language is a subset of Z*. More generally, we make the following definition.
Language
A l a n g u a g e over an alphabet Z is a subset of E*. The following two examples illustrate this definition. The set of zip codes is a finite language over the alphabet E  { 0 , . . . , 9}. m Let E  {a, b}. Then E*  {~, a, b, aa, ab, ba, bb, aaa, aab, aba, abb, b a a , . . . }, an infinite set. Notice that {aa, ab, ba, bb} is a finite language over E, whereas {a, aa, aba, bab, aaaa, a b b a , . . . } is an infinite language, m Words can be combined to create new words, as defined below.
76 Concatenation
Chapter2 The Language of Sets
The c o n c a t e n a t i o n of two words x a n d y over an a l p h a b e t , d e n o t e d by x y , is obtained by a p p e n d i n g t h e word y at t h e end of x. T h u s if x = x l . . . X m a n d y = Yl...Yn, xy  Xl . . .XmYl . . .Yn. For example, let E be t h e E n g l i s h alphabet, x = CAN, a n d y = ADA; t h e n x y = CANADA. Notice t h a t c o n c a t e n a t i o n is n o t a c o m m u t a t i v e operation; t h a t is, x y 7/= y x . It is, however, associative; t h a t is, x ( y z ) = ( x y ) z = x y z . Two i n t e r e s t i n g p r o p e r t i e s are satisfied by t h e c o n c a t e n a t i o n operation: 9 The c o n c a t e n a t i o n of a n y word x with )~ is itself; t h a t is, ~x  x  x~ for every x e E*. 9 Let
x,y ~
E*. T h e n
[]xY]l 
][xll +
ilYi]. (See Section 5.1 for a proof.)
For example, let E = {a,b }, x = aba, and y  bbaab. T h e n x y  a b a b b a a b and [[xyl[  8  3 + 5  IIxl[ + IlyJl. A useful notation: As in algebra, t h e e x p o n e n t i a l n o t a t i o n can be employed to e l i m i n a t e the r e p e a t i n g of symbols in a word. Let x be a symbol and n an i n t e g e r >_ 2; t h e n x n denotes the c o n c a t e n a t i o n x x . . . x to n  1 times. U s i n g this compact notation, the words a a a b b a n d a b a b a b can be a b b r e v i a t e d as a3b 2 a n d (ab) 3, respectively. Notice, however, t h a t (ab) 3 = ababab ~: a3b 3  aaabbb.
Exercises 2.1
Rewrite each set using the listing method. 1. The set of m o n t h s t h a t begin with the l e t t e r A. 2. The set of letters of the word GOOGOL. 3. The set of m o n t h s with exactly 31 days. 4. The set of solutions of t h e e q u a t i o n x 2  5x + 6  0. Rewrite each set u s i n g the setbuilder notation. 5. The set of integers b e t w e e n 0 and 5. 6. The set of J a n u a r y , F e b r u a r y , May, a n d July. 7. The set of all m e m b e r s of t h e U n i t e d Nations. 8. {Asia, Australia, Antarctica} D e t e r m i n e if t h e given sets are equal.
9. {x,y,z}, {xlx 2 {x,z,y} x},
10.
{xlx 2
= 1}, {xlx 2  x}
11.
{0, 1}
12. {x, {y}}, {{x},y}
M a r k each as t r u e or false.
13. a E {alfa}
14. b _ {a,b,c}
15. {x} _ {x,y, z}
2.1 The Concept of a Set 16. { 0 }  0 19. {0} = 0
22. {xlx ~ x }  0
77 17. 0 ~ 0 20. 0 __ 0
23. {x,y}  {y,x}
18. { 0 }  0 21. 0 ~ {0}
24. {x} ~ {{x},y}
25. 0 is a subset of every set.
26. E v e r y set is a subset of itself.
27. Every n o n e m p t y set has at least two subsets. 28. The set of people in the world is infinite. 29. The set of words in a dictionary is infinite. Find the power set of each set. 30. 0
31. {a}
32. {a,b,c}
33. Using Exercises 3032, predict the n u m b e r of subsets of a set with n elements. In Exercises 3437, n denotes a positive integer less t h a n 10. Rewrite each set using the listing method. 34. {nln is divisible by 2} 36. {nln is divisible by 2 and 3} 35. {nln is divisible by 3} 37. {nln is divisible by 2 or 3}
Find the family of subsets of each set t h a t do not contain consecutive integers. 38. {1,2} 39. {1,2,3}
40. Let an denote the n u m b e r of subsets of the set S  {1, 2 , . . . , n} t h a t do not contain consecutive integers, where n > 1. F i n d a 3 and a4. In Exercises 4146, a language L over E  {a, b} is given. Find five words in each language. 41. L  {x e E ' I x begins with and ends in b.} 42. L  {x ~ E*lx contains exactly one b. } 43. L  {x E E*fx contains an even n u m b e r of a's. } 44. L  {x e E ' I x contains an even n u m b e r of a ' s followed by an odd n u m b e r of b's. } C o m p u t e the length of each word over {a, b }. 45. aab 47. ab 4 46. aabbb 48. a3b 2
A r r a n g e the b i n a r y words of the given length in increasing order of magnitude. 49. L e n g t h two. 50. L e n g t h three.
78
Chapter 2 The Language of Sets A t e r n a r y w o r d is a word over the alphabet {0, 1, 2}. A r r a n g e the t e r n a r y words of the given length in increasing order of magnitude. 51. Length one. Prove each. *53. The empty set is a subset of every set. (Hint: Consider the implication x e ~ ~ x e A.) *54. The empty set is unique. (Hint: Assume there are two empty sets, 01 and ~2. T h e n use Exercise 53.) *55. Let A, B, and C be a r b i t r a r y sets such t h a t A c B and B c C. T h e n 52. Length two.
AcC.
(transitive property)
*56. If E is a n o n e m p t y alphabet, t h e n E* is infinite. (Hint: Assume Z* is finite. Since Z # 0, it contains an e l e m e n t a. Let x e E* with largest length. Now consider xa.)
J u s t as propositions can be combined in several ways to construct new propositions, sets can be combined in different ways to build new sets. You will find a close relationship between logic operations and set operations.
Union
The u n i o n of two sets A and B, denoted by A u B, is obtained by merging them; t h a t is, A u B  {xl(x e A) v (x e B)}. Notice the similarity between union and disjunction.
~
Let A  {a, b, c}, B  {b, C, d , e } , a n d C  { x , y } . T h e n A u B  { a , b ,
C, d , e } 
BUAandBUC
{b,c,d,e,x,y}CUB.
m
The shaded areas in Figure 2.4 r e p r e s e n t the set A u B in t h r e e different cases.
Intersection
The i n t e r s e c t i o n of two sets A and B, denoted by A N B, is the set of elements common to both A and B; t h a t is, A 5 B  {xl(x e A) v (x e B)}.
2.2 Operationswith Sets F i g u r e 2.4
U
79
U
A UB
A U B, where A and B are disjoint
AUB=B
Notice the relationship between intersection and conjunction.
Let A {Nov, Dec, Jan, Feb}, B  {Feb, Mar, Apr, May}, and C  {Sept, Oct, Nov, Dec}. T h e n A n B  {Feb}  B N A a n d B n C  0  C A B . (Notice t h a t B and C are disjoint sets. More generally, two sets are disjoint if and only if their intersection is null.) m
F i g u r e 2.5
O9 O9
Berkeley Street intersection
Figure 2.5 shows the intersection of two lines and t h a t of two streets, and Figure 2.6 displays the set A n B in three different cases.
F i g u r e 2.6
U
U
G
ANB ANB=O ANB=A
Let A  {a, b, c, d, g}, B and (A u B) N (A U C).
{b, c, d, e, f}, and C 
{b,c,e,g,h}.FindAu(BnC)
80
Chapter 2 The Language of Sets
SOLUTION: (1)
BNC
= {b,c,e} {a,b, c, d, e, g}
AU(BNC)
(2)
A UB = { a , b , c , d , e , f,g}
AuC{a, (AuB) N(AuC)
b, c, d, e, g, h} {a,b,c,d,e,g}
= A u (B n C) See the Venn diagram in Figure 2.7. F i g u r e 2.7
m A third way of combining two sets is by finding their difference, as defined below. Difference The d i f f e r e n c e of two sets A and B (or the r e l a t i v e c o m p l e m e n t of B in A), denoted by A  B (notice the order), is the set of e l e m e n t s in A t h a t are not in B. T h u s A  B = {x ~ AIx r B}. L e t A  { a , . . . , z , 0 , . . . , 9 } , and B  {0,...,9}. T h e n A  B  { a , . . . , z } and
BA=O. The shaded areas in Figure 2.8 r e p r e s e n t the set A  B in t h r e e different cases.
F i g u r e 2.8
U
U
AB
A  B =A
AB
m F o r any set A r U, a l t h o u g h A  U  ~, the difference U  A ~: ~. This shows yet a n o t h e r way of obtaining a new set.
2.2 Operationswith Sets Complement
81
T h e difference U  A is the (absolute) c o m p l e m e n t of A, d e n o t e d by A' (A prime). T h u s A ' = U  A = {x ~ U Ix r A}. I F i g u r e 2.9 r e p r e s e n t s t h e c o m p l e m e n t of a set A. ( C o m p l e m e n t a t i o n c o r r e s p o n d s to negation.)
F i g u r e 2.9
U
A'
~
{ a , . . . , Z }. F i n d the c o m p l e m e n t s of the sets A  {a, e, i, O~U} a n d Let U B = {a, c, d, e, . . . , w}. T h e n A' = U  A = set of all c o n s o n a n t s in t h e alphabet, and B' = U  B = {b, x, y, z}. m Let A = { a , b , x , y , z } , B F i n d (A u B)' and A' N B'. SOLUTION: (1) {c, d, e, x, y, z}, and U {a,b,c,d,e,w,x,y,z}.
A U B = {a, b, c, d, e, x, y, z}
(A u B)' = {w}
(2)
A' = {c, d, e, w}
B' = {a, b, w}
A' N B' = {w} = (A U B)' See F i g u r e 2.10.
F i g u r e 2.10
U
m
Since as a rule, A  B r B  A , new set. by t a k i n g t h e i r u n i o n we can form a
82
Chapter2 The Language of Sets Symmetric Difference
The s y m m e t r i c d i f f e r e n c e of A and B, denoted by A @ B, is defined by A @ B  (A  B) U (B  A ) . LetA  {a,...,z,0,...,9} andB  {0,...,9,+,,.,/}. ThenAB { a , . . . , z } and B  A = {+,,.,/}. SoA@B  (AB) u (BA) { a , . . . , z, + ,  , . , / } . a
The s y m m e t r i c difference of A and B is pictorially displayed in Figure 2.11 in three different cases.
Figure 2.11
A
AOB=AUB
A
=AB
Set and Logic Operations
Set operations and logic operations are closely related, as Table 2.1 shows.
Table 2.1
Set operation
AuB ANB A' A@B
Logic operation
pvq pvq ~p
pXORq
The i m p o r t a n t properties satisfied by the set operations are listed in Table 2.2. (Notice the similarity between these properties and the laws of logic in Section 1.2.) We shall prove one of them. Use its proof as a model to prove the others as routine exercises. We shall prove law 16. It uses De M o r g a n ' s law in symbolic logic, a n d the fact t h a t X = Y if and only if X ___ Y and Y _CX. __
PROOF:
In order to prove t h a t (AUB)'  A' NB', we m u s t prove two parts" (A UB)' c A' N B' and A' n B' c (A u B)'. 9 To prove t h a t (A u B)' c_ (A' n B')Let x be an a r b i t r a r y element of (A u B)'. T h e n x r (A u B). Therefore, by De M o r g a n ' s law, x r A and x r B; t h a t is, x e A' and x e B'. So x e A' NB'. T h u s every element of (A UB)' is also an element ofA' NB'; t h a t is, (A u B)' ___A' n B'.
2.2 Operations with Sets
83
T a b l e 2.2
Laws of Sets
Let A, B, and C be any three sets and U the universal set. Then:
1. A u A = A
3. A u O = A
I d e m p o t e n t laws 2. A n A = A I d e n t i t y laws 4. A n U = A Inverse laws
5. A u W = U 7. A u U = U 9. A u B = B u A
11. (At)t = A
6. A n A I = O
D o m i n a t i o n laws
8. A n O = O
C o m m u t a t i v e laws
10. A n B = B n A
Double c o m p l e m e n t a t i o n law A s s o c i a t i v e laws
12. A u ( B u C )  ( A u B ) u C 14. A u ( B n C ) = ( A u B ) n ( A u C ) 13. A n ( B n C )  ( A n B ) n C
D i s t r i b u t i v e laws
15. A n ( B u C )  ( A n B ) u ( A n C )
De Morgan's laws
16. (A u B)'  A' n B' 18. A u ( A n B )  A 20. I f A c _ B , t h e n A n B = A . 22. I f A c _ B , t h e n B tC_A t. 24. A  17. (A n B)'  A' u B'
A b s o r p t i o n laws
19. A n ( A u B ) = A (Note: The following laws have no names.) 21. I f A c _ B , t h e n A u B = B . 23. A  B  A n B t
9 To p r o v e t h a t A' c~ B' c_ (A u B ) " Let x be a n y e l e m e n t of A' n B'. T h e n x E A' a n d x ~ B'. T h e r e f o r e , x r A a n d x r B. So, by De M o r g a n ' s law, x r (A u B). C o n s e q u e n t l y , x ~ (A w B)'. T h u s , since x is a r b i t r a r y , A' • B' c (A u B)'. T h u s , (A u B)'  A' n B'. See t h e V e n n d i a g r a m s in F i g u r e 2.12 also.
Note" L a w 23 is a v e r y u s e f u l r e s u l t a n d will be u s e d in t h e n e x t
section.
A few words of explanation" T h e c o m m u t a t i v e laws i m p l y t h a t t h e o r d e r in w h i c h t h e u n i o n (or i n t e r s e c t i o n ) of t w o sets is t a k e n is i r r e l e v a n t . T h e associative laws i m p l y t h a t w h e n t h e u n i o n (or i n t e r s e c t i o n ) of t h r e e or m o r e
84
Chapter 2 The Language of Sets
F i g u r e 2.12
(A U B)' = shaded area
A' n B' = crossshaded area
I
sets is taken, the way the sets are grouped is immaterial; in other words, such expressions without p a r e n t h e s e s are perfectly legal. For instance, A U B u C  A u (B U C) = (A u B) U C is certainly valid. The two De M o r g a n ' s laws in propositional logic play a central role in deriving the corresponding laws in sets. Again, as in propositional logic, p a r e n t h e s e s are essential to indicate the groupings in the distributive laws. For example, if you do not parenthesize the expression A N (B U C) in law 15, t h e n the LHS becomes A N B U C = (A N B ) U C = (A U C) N (B U C) r (A N B ) U (A N C).
Notice the similarity between the set laws and the laws of logic. For example, properties 1 t h r o u g h 19 and 22 have their c o u n t e r p a r t s in logic. Every corresponding law of logic can be obtained by replacing sets A, B, and C with propositions p, q, and r, respectively, the set operators N, U, a n d ' with the logic operators A, v, and ~ respectively, and equality (=) with logical equivalence (=_). Using this procedure, the absorption law A w (A N B) = A, for instance, can be t r a n s l a t e d as p v (p A q)  p, which is the corresponding absorption law in logic. J u s t as t r u t h tables were used in Chapter I to establish the logical equivalence of compound statements, they can be applied to verify set laws as well. The next example illustrates this method. Using a t r u t h table, prove t h a t (A u B)'  A' N B'. SOLUTION. Let x be an a r b i t r a r y element. T h e n x may or may not be in A. Likewise, x may or may not belong to B. E n t e r this information, as in logic, in the first two columns of the table, which are headed by x E A and x e B. The table needs five more columns, headed by x ~ (A u B), x ~ (A u B)', x ~ A', x ~ B', and x ~ (A' n B') (see Table 2.3). Again, as in logic, use the entries in the first two columns to fill in the r e m a i n i n g columns, as in the table.
2.2 Operations with Sets
85
T a b l e 2.3
x e A
x e B
x e (A U B )
x e (A u B)'
x e A'
x 9 B'
x 9 (A' n B')
T T F F
T F T F
T T T F
F F F T
F F T T
F T F T
F F F T
Note: The shaded columns are identical Since the columns headed by x e (A u B)' and x e (A' n B') are identical, it follows t h a t (A u B)' = A' n B'. I Using t r u t h tables to prove set laws is purely mechanical a n d elem e n t a r y . It does not provide any insight into the d e v e l o p m e n t of a m a t h e m a t i c a l proof. Such a proof does not build on previously k n o w n set laws, so we shall not r e s o r t to such proofs in s u b s e q u e n t discussions.
. . . . . .
J u s t as the laws of logic can be used to simplify logic expressions a n d derive new laws, set laws can be applied to simplify set expressions a n d derive new laws. In order to be successful in this art, you m u s t k n o w the laws well and be able to apply t h e m as needed. So, practice, practice, practice.
Using set laws, verify t h a t (X  Y)  Z = X  (Y u Z).
PROOF(X  Y)  Z  (X  Y) n Z'
AB AB =AnB' =AnB'
= (X n Y') n Z' = X n (Y' n Z') = X n (Y u Z)'
=X(YuZ)
associative law 13 De M o r g a n ' s law 16
AB =AnB'
I
Simplify the set expression (A n B') U (A' n B) U (A' n B'). SOLUTION: (You m a y supply the justification for each step.) (A n B') u (A' n B) u (A' n B')  (A n B') u [(A' n B) u (A' n B')] = ( A n B') u [A' n (B u B')] = ( A n B') u (A' n U) = (A riB') u A '
86
(Jhapter2 The Languageof Sets
= A' u (A N B') = (A' u A) n (A' u B') = U n (A' u B')
=A'uB'
m
Often subscripts are used to n a m e sets, so we now t u r n o u r a t t e n t i o n to such sets.
Indexed Sets
Let I, called the i n d e x s e t , be the set of subscripts i used to n a m e the sets Ai. T h e n the u n i o n of the sets Ai as i varies over I is d e n o t e d b y . u Ai. Similarly,
tEI iEI n U
n Ai denotes the i n t e r s e c t i o n of the sets Ai as i r u n s over I. In p a r t i c u l a r ,
n iEI n n Ai i=I
cx.)
let I  {1, 2 , . . . , n }. T h e n u Ai  A1 uA2 u . . . UAn, which is often w r i t t e n as
i=1
Ai or simply U Ai. Likewise, N 1
iEb~
iEI
Ai 
n ? Ai  A1 n A2 N...
('x.)
NAn.
I f I  N, the expression
u Ai is w r i t t e n as u Ai  ~ Ai, u s i n g the infinity
/el
symbol oc; similarly, i ?b~ A i  i=ln A i  N1 A i . Before we proceed to define a new b i n a r y o p e r a t i o n on sets n , we define an o r d e r e d set. i=1
Ordered Set
Recall t h a t the set {al, a 2 , . . . , a , } is an u n o r d e r e d collection of elements. Suppose we assign a position to each element. T h e r e s u l t i n g set is an o r d e r e d s e t with n e l e m e n t s or an n  t u p l e , d e n o t e d by ( a l , a 2 , . . . , an). (Notice the use of p a r e n t h e s e s versus braces.) The set (al, a2) is an o r d e r e d pair. Two ntuples are e q u a l if and only if t h e i r c o r r e s p o n d i n g e l e m e n t s are equal. T h a t is, (al, a 2 , . . . , a,,) = (bl, b 2 , . . . , b,) if a n d only if ai = bi for every i.
•
Every n u m e r a l and word can be considered an ntuple. F o r instance, 345 = (3, 4, 5) l t $ ones tens hundreds ASCII* code for letter K EBCDIC** code for letter K
c o m p u t e r = (c, o, m, p, u, t, e, r) 1001011 = (1, 0, 0, 1, 0, 1, 1) 11010010 = (1, 1, 0, 1, 0, 0, 1, 0)
m
*American Standard Code for Intbrmation Interchange. **Extended Binary Coded Decimal Interchange Code.
2.2 Operationswith Sets
87
~Z ( .::
Rend D e s c a r t e s (15961650) was born near Tours, France. At eight, he entered the Jesuit school at La Fleche, where because of poor health he developed the habit of lying in bed thinking until late in the morning; he considered those times the most productive. He left the school in 1612 and moved to Paris, where he studied mathematics for a brief period. After a short military career and travel through Europe for about 5years, he returned to Paris and studied mathematics and philosophy. He then moved to Holland, where he lived for 20years writing several books. In 1637 he wrote Discours, which contains his contributions to analytic geometry. In 1649 Descartes moved to Sweden at the invitation of Queen Christina. There he contracted pneumonia and died.
We are now ready to define the next and final operation on sets.
Cartesian Product
The c a r t e s i a n p r o d u c t of two sets A and B, denoted by A x B, is the set of all ordered pairs (a,b) with a 9 A and b 9 B. T h u s A x B = {(a,b)l a 9 A A b 9 B}.A x A is denoted b y A 2. It is n a m e d after the F r e n c h philosopher and m a t h e m a t i c i a n Ren~ Descartes. Let A {a, b } and B {x, y, z}. T h e n
A x B = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)} B z A = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)} A 2  A x A = {(a, a), (a, b), (b, a), (b, b)} (Notice t h a t A x B r B x A.)
m
The various elements of A x B in Example 2.22 can be displayed in a r e c t a n g u l a r fashion, as in F i g u r e 2.13, and pictorially, using dots as in Figure 2.14. The circled dot in row a and column y, for instance, r e p r e s e n t s the element (a, y). The pictorial r e p r e s e n t a t i o n in F i g u r e 2.14 is the g r a p h ofA x B .
F i g u r e 2.13
a Elements of A b
(a, x) (b, x) x
(a, y) (b, y) y Elements of B
(a, z) (b, z) z
88
Chapter 2
2.14 a 9
The Language of Sets
Figure

9
Pictorial representation ofA •
x
y
z
Figure 2.15 shows the graph of the infinite set 1~ 2   1~ x N. The circled dot in column 4 and row 3, for instance, represents the element (4,3). The horizontal and vertical dots indicate that the pattern is to be continued indefinitely in both directions.
Figure 2.15
9
9
9
},
1
2
3
4
.
.
.
More generally, R 2   R • R consists of all possible ordered pairs (x,y) of real numbers. It is represented by the familiar x y  p l a n e or the c a r t e s i a n p l a n e used for graphing (see Figure 2.16).
Figure
2.16
The cartesian plane R 2"
(3,4)
(0,3)
(5,0) The following example presents an application of cartesian product. ~ Linda would like to make a trip from Boston to New York and then to London. She can travel by car, plane, or ship from Boston to New York, and by plane or ship from New York to London. Find the set of various modes of transportation for the entire trip.
SOLUTION:
Let A be the set of means of transportation from Boston to New York and B the set from New York to London. Clearly A  {car, plane, ship} and B  {plane, ship}. So the set of possible modes of transportation is given by
2.2 Operationswith Sets F i g u r e 2.17
89
London N Boston e w ~
plane
.~"~'~ship
A x B  {(car, plane), (car, ship), (plane, plane), (plane, ship), (ship, plane), (ship, ship)}. See Figure 2.17. I The definition of the product of two sets can be extended to n sets. The c a r t e s i a n p r o d u c t o f n s e t s A 1 , A 2 , . . . ,An consists of all possible n tuples (al, a 2 , . . . , an), where ai ~ Ai for every i; it is denoted by A1 x A2 x ... x An. If all Ai's are equal to A, the product set is denoted by An. LetA{x},B  {y,z}, and C {1,2,3}. Then
A x B x C
{(a,b,c)la ~ A , b ~ B, a n d c ~ C}
= {(x, y, 1), (x, y, 2), (x, y, 3), (x, z, 1), (x, z, 2), (x, z, 3)} Finally, take a look at the map of the continental United States in Figure 2.18. It provides a geographical illustration of partitioning, a concept that can be extended to sets in an obvious way.
F i g u r e 2.18
I
Partition
Consider the set S = {a, b, c, d, e, f, g, h, i} and the subsets $1 = {a, b }, $ 2  {c}, $3 = {d, e, f}, $4  {g, h}, and $5  {i}. Notice t h a t these subsets have three interesting properties: (1) They are nonempty; (2) they are pairwise disjoint; that is, no two subsets have any common elements; (3) their union
90
Chapter 2 The Language of Sets is S. (See Figure 2.19.) T h e set P  { S 1 , $ 2 , $ 3 , $ 4 , $ 5 } of S. is called a p a r t i t i o n
F i g u r e 2.19
S1
More generally, let I be an index set and P a family of s u b s e t s Si of a n o n e m p t y set S, w h e r e i ~ I. T h e n P is a p a r t i t i o n of S if: 9 Each set Si is n o n e m p t y . 9 The subsets are pairwise disjoint; t h a t is, Si n S j  ~,~ if i ~=j. 9 The union of the subsets Si is S; t h a t is, u Si = S.
i~I
(Each subset Si is a b l o c k of the partition.) T h u s a p a r t i t i o n of S is a collection of n o n e m p t y , pairwise disjoint subsets of S whose u n i o n is S. ~ Let Z,. denote the set of integers which, w h e n divided by 5, leave r as the r e m a i n d e r . T h e n 0 < r < 5 (see Section 4.1):
w
Z0 = { . . . ,  5 , 0 , 5 , . . . } Zl = { . . . ,  4 , 1 , 6 , . . . } Z 2  { . . . ,  3 , 2 , 7,...} Z3 = { . . . ,  2 , 3 , 8 , . . . } Z4 = { . . . ,  1 , 4 , 9 , . . . } P = {Z0, Zl, Z2, Z3, Z4} is a partition of the set of integers. See F i g u r e 2.20. (This example is discussed in more detail in Section 7.4.) m
F i g u r e 2.20
Set of integers Z.
2.2 Operationswith Sets
91
The sports pages of newspapers provide fine examples of partitions, as the next example illustrates. ~ In 2003, the set of teams S in the National Football League was divided into two conferences, American and National, and each conference into four divisions m East, South, North, and West. Let El, $1, N1, and W1 denote the set of teams in East, South, North, and West Divisions in the American Conference, respectively, and E2, $2, N2, and W2 the corresponding sets in the National Conference. Then: E1  {Buffalo, Miami, New England, NY Jets} $1  {Indianapolis, Tennessee, Houston, Jacksonville} N1  {Baltimore, Cincinnati, Cleveland, Pittsburgh} W1  {Denver, Kansas City, Oakland, San Diego} E2 = {Washington, Philadelphia, Dallas, NY Giants} $2  {Atlanta, Tampa Bay, Carolina, New Orleans} N2 = {Chicago, Detroit, Minnesota, Green Bay } W2 = {Arizona, Seattle, St. Louis, San Francisco } Clearly, P  {E1,S1,N1, W 1 , E 2 , S 2 , N 2 , W2} is a partition of S. We close this section with a brief introduction to fuzzy sets.
Fuzzy Sets (optional)
Fuzzy sets, a generalization of ordinary sets, were introduced in 1965 by Lotfi A. Zadeh of the University of California at Berkeley. They have applications to h u m a n cognition, communications, decision analysis, psychology, medicine, law, information retrieval, and, of course, artificial intelligence. Like fuzzy logic, they model the fuzziness in the natural language for example, in terms like young, healthy, wealthy, and beautiful. In fuzzy set theory, every element x in the universal set U has a certain degree of membership du(x), where 0 < du(x) < 1;du(x) indicates the degree of fuzziness. Accordingly, a fuzzy set S is denoted by listing its elements along with their degrees of membership; an element with zero degree of membership is not listed. For example, let Ube the fuzzy set of wealthy people and S  {Tom 0.4, Dick 0.7, Harry 0.6}. Then Harry belongs to S with degree of membership 0.6; ds(Harry)  0.6 measures Harry's degree of wealthiness. The concept of an ordinary subset can be extended to fuzzy sets also.
Fuzzy Subset
Let A and B be fuzzy sets. Then A is a f u z z y s u b s e t of B if A _ _B and _ dA(x) < dB(x) for every element x in A.
92
Chapter2 The Language of Sets
Lotfi A. Z a d e h (1921) was born in Baku, Azerbaijan. An alumnus of the University of Tehran (1942) and the Massachusetts Institute of Technology (1946), he received his Ph.D. from Columbia University in 1949 for his dissertation on frequency analysis of timevarying networks. He began his professional career in the Department of Electrical Engineering at Columbia. In 1959, he joined the Department of Electrical Engineering and Computer Science at the University of California, Berkeley, serving as its chair during the years 19631968. Currently, he is a professor at Berkeley and Director of Berkeley Initiative in Soft Computing. Zadeh's earlier "work was centered on systems analysis, decision analysis, and information systems. Since then his current research has shifted to the theory of fuzzy sets and its applications to artificial intelligence (AI). His research interest now is focused on fuzzy logic, soft computing, computing with words, and the newly developed computational theory of perceptions and precisiated natural language, "according to the University of California Web site. A truly gifted mind and an expert on AI, Zadeh has authored about 200journal articles on a wide variety of subjects relating to the conception, design, and analysis of information~intelligent systems. He serves on the editorial boards of more than 50 journals and on the advisory boards of a number of institutions related to AI. Zadeh is a recipient of numerous awards and medals, including the IEEE Education Medal, IEEE Richard W. Hamming Medal, IEEE Medal of Honor, the A S M E Rufus Oldenburger Medal, B. Bolzano Medal of the Czech Academy of Sciences, Kampe de Feriet Medal, AACC Richard E. Bellman Central Heritage Award, the Grigore Moisil Prize, Honda Prize, Okawa Prize, AIM Information Science Award, IEEESMC J. P. Wohl Career Achievement Award, SOFT Scientific Contribution Memorial Award of the Japan Society for Fuzzy Theory, IEEE Millennium Medal, and the ACM 2000 Allen Newell Award. He has received honorary doctorates from many universities from around the world.
9 2 "~
~
":t .... ."'
" ....
For example, letS = {Betsey 0.6, Mat 0.5} and T  {Betsey 0.8, J o n a t h a n 0.3, Mary 0.5, Mat 0.7} by fuzzy sets of smart people. Then S is a fuzzy subset of T. Operations on ordinary sets can be extended to fuzzy sets as well. Operations on Fuzzy Sets Let A and B be any fuzzy set. The u n i o n of A and B is AUB, where dA uB(x) = max{dA(x), ds(x)}; their i n t e r s e c t i o n is A N B, where dANB(X) = m i n { d A ( x ) , d s ( x ) } ; and the c o m p l e m e n t of A is A', where dA,(X) = 1  dA(x); in A' only the degrees of membership change. Using the sets S and T above, S u T = {Betsey 0.8, J o n a t h a n 0.3, Mary 0.5, Mat 0.7 } S n T = {Betsey 0.6, Mat 0.5} S' = {Betsey 0.4, Mat 0.5} Additional opportunities to practice the various operations are given in the exercises.
2.2
Operations with Sets
93
Exercises 2.2
L e t A = {a, e, f, g,i}, B = {b, d, e, g, h}, C = {d, e, f, h, i}, and U = { a , b , . . . ,k}. Find each set. 1. C l
5. (B n C)'
2. B Q C '
3. C Q A '
4. ( A U B ) ' 8. A ~ B 12. (A N B)  C
6. (A u C')' 10. A  ( B  C )
7. (B N C')' 11. (A u B)  C
9. ( A  B )  C
Using the Venn d i a g r a m in Figure 2.21 find each set. 13. ( A U B ) N C 16. ( A @ B ) U C F i g u r e 2.21
a
14. A n (B U C) 17. A Q ( B O C )
15. A 
(BC)
18. A  ( B @ C )
U A w f
Let A = {b, c}, B = {x}, and C = {x,z}. Find each set. 19. A x B 23. A x ( B u C ) 20. B x A 24. A x ( B N C ) 21. A x 0 25. A x B x C 22. A x B x O 26. A x C x B
M a r k each as true or false, where A, B, and C are a r b i t r a r y sets and U the universal set. 27. A  ~ ) = A
30. A  A = O
28. 0  A =  A 31. A  B = B  A 34. (A N B ) ' = A' NB' 37. A c A N B
29. O  O = 0 32. A  A ' = O 35. (A U B)' = A' U B' 38. B D (A  B ) = O
33. (A')' = A
36. A c A u B
Give a counterexample to disprove each proposition. 39. ( A  B )  C = A  ( B  C ) 41. A u ( B @ C ) = ( A u B ) G ( A u C ) 40. A U ( B  C ) = ( A U B )  ( A U C ) 42. A O ( B N C ) = ( A O B ) N ( A G C )
D e t e r m i n e if each is a partition of the set { a , . . . , z, 0 , . . . , 9}. 43. { { a , . . . , z } , { 0 , . . . , 9 } , 0 } 44. { { a , . . . , j } , { i , . . . , t } , { u , . . . , z } , { 0 , . . . , 9 } }
94
Chapter 2 The Language of Sets 45. {{a,... ,1}, { n , . . . , t } , {u,...,z}, {0,...,9}} 46. {{a,...,u}, {v,...,z}, {0,3}, { 1 , 2 , 4 , . . . , 9 } Prove each, where A, B, and C are any sets. 47. (A')' = A 49. A N ( A u B ) = A 51. A @ A = O 53. A @ B = B @ A 55. (A u B u C)' = A' n B' n C' Simplify each set expression. 57. A A (A  B)
60. 63. (A u B) u (A A B')' (A n B)' u (A u B')
48. A u (A n B ) = A 50. (A N B)' = A' u B' 52. A @ U = A ' 54. A  B = A A B '
56. (A A B n C)' = A' u B' u C'
58. (A  A') u (B  A)
61. 64. (A u B)  (A n B)' (A u B')' A (A' c~ B)
59. 62. 65.
(A  B')  (B  A') (A U B)' N (A n B') (A' u B')' u (A' ~ B)
*66. State De Morgan's laws for sets Ai, i ~ I. (I is an index set.) *67. State the distributive laws using the sets A and Bi, i ~ I. The s u m of two fuzzy sets A and B is the fuzzy set A  B, where dA ~ B(X) = 1A IdA (x) + d B ( x ) l; their d i f f e r e n c e is the fuzzy set A  B , where dAB(X) = 0 v IdA(x)  dB(x)i; and their c a r t e s i a n p r o d u c t is the fuzzy set A x B, where dA xB(X,Y) = dA(x) AdB(x). Use the fuzzy s e t s A = {Angelo 0.4, Bart 0.7, Cathy 0.6} and B = {Dan 0.3, Elsie 0.8, F r a n k 0.4} to find each fuzzy set. 68. A U B
72. A N B '
69. A A B
73. A ~ A '
70. A'
74. A 
71. A u B '
75. A  B
76. B  A
77. A x B
78. B x A
79. A x A
Let A and B be any fuzzy sets. Prove each.
*80. (A u B)' = A' • B' "81. (A • B)' = A' u B'
Sets and the various set operations can be implemented in a c o m p u t e r in an elegant manner.
Computer Representation
Although the elements of a set have no i n h e r e n t order, when the set is represented in a computer, an order is imposed upon t h e m to p e r m i t
2.3 ComputerOperations with Sets (optional)
95
implementation. The universal set U with n elements is represented as an array with n cells, each containing a 1: n1 U ] 1[ 1] ... 2 1 0
]1111111]
The elements are represented by the binary digits (or bits) 0 and 1 in the righttoleft fashion. Subsets of U are represented by assigning appropriate bits to the various cells. A bit 1 in a cell indicates the corresponding element belongs to the set, whereas a 0 would indicate the element does n o t belong to the set. ~ UsingU={a,b,...,h},representthesetsA={a,b,g}andB{c, 8bit strings. e, h } a s
SOLUTION: Remember, the elements are represented in the righttoleft order. Thus:
h
g
f
e
d
c
b
a
lllllrll ll llll AlOlllOlOlOlOlllll lllOlOlllOlllOlOl
m
Next we discuss how the various subsets of a finite set can be found methodically.
T a b l e 2.4
Subset
Bit String
0 {x} {y} {x,y} {z} {x,z} {y,z} {x,y,z}
000 001 010 011 100 101 110 111
Interestingly enough, there is a close relationship between sets and bit strings. Table 2.4, for instance, lists the various subsets of the set {x, y, z }. Notice that the table contains all possible threebit strings and their decimal
96
Chapter 2
The Language of Sets
values increase from 0 to 7. (See Section 4.3 for a discussion of nondecimal bases.) Next we present a systematic procedure to find the bit string of the subset that "follows" a given subset with bit string b 2 b l b o . Such a recipe for solving a problem in a finite number of steps is called an algorithm.*
NextSubset Algorithm
Take a good look at each string in Table 2.4. Can you find a rule to obtain each, except 000, from the preceding string? It is fairly simple: From right to left, locate the first 0. Change it to 1 and the l's to its right to 0's. For example, suppose you would like to find the subset following {x, y} with bit string b 2 b l b o  011. From right to left, the first 0 is b2. Change it to 1, and bl and b0 to 0's. The resulting string is 100 and the corresponding subset is {z }. This rule can be generalized and translated into an algorithm. See Algorithm 2.1. Use it to find the subsets following {z} and { y, z}.
Algorithm nextsubset (bnlbn2 ... bo) (* This algorithm finds the b i t string of the subset that follows a given subset of an nelement set S. *) Begin (* nextsubset *) find the f i r s t 0 from the r i g h t change i t to I replace the bits to i t s r i g h t with O's End (* nextsubset *) Algorithm 2.1
The nextsubset algorithm can be employed to find all subsets of a finite set S. Algorithm 2.2 shows the steps involved. Use it to find the subsets of {x,y,z}.
Algorithm subsets (S) (* Using the nextsubset algorithm, this algorithm finds the b i t representations of all subsets of an nelement set S. *) Begin (* subsets *) bnlbn2 . . . b o 4000
m
/ 75 +
*WordPerfect is a wordprocessing program marketed by Corel Corporation. **A character within single quotes indicates a literal character.
122
Chapter 3 Functions and Matrices
~
Let A  {0, 1 , . . . , 127}, the set of ordinal n u m b e r s in ASCII. Let f 9 A * ASCII be defined by n o n p r i n t a b l e control c h a r a c t e r uppercase letter lowercase letter other printable c h a r a c t e r if 0 < n < 31 or n  127 if 65 < n < 90 i f 9 7 < n < 122 otherwise
m
f(n)
Clearly, f is defined piecewise.
F u n c t i o n s defined piecewise are w r i t t e n as i f  t h e n  e l s e s t a t e m e n t s in most p r o g r a m m i n g languages. For example, the function in E x a m p l e 3.6 can be w r i t t e n as follows:
(x _> O) and (x _< 4000) then f ( x ) ~ 50 else f ( x ) ~ 50 + O.O06(x  4000) if
The geometrical r e p r e s e n t a t i o n of a function, called a g r a p h , is often used to study functions. Remember, a picture is w o r t h a t h o u s a n d words. Since every function f : X . Y is a set of ordered pairs (x,y), the g r a p h of f consists of points corresponding to the ordered pairs in f, as the next example illustrates. ~ G r a p h each function. (1) Let f 9 Z ~ Z defined by f(x) (2) Let g 9 R * R defined by
x 2.
g(x)
X 1 3X + 4
ifx > 0 if2 _x. The floor ofx r o u n d s d o w n x while the ceiling ofx r o u n d s _ up. Accordingly, if x r Z, the floor of x is the nearest integer to the left o f x on the n u m b e r line and the ceiling of x is the nearest integer to the right o f x, as shown in Figure 3.11. The floor f u n c t i o n f(x)  lxJ and the
F i g u r e 3.11
lxJ [xl1 x
[xJ+ 1 Ix]
32 SpecialFunctions
127
...., ,~," ,.2 ,, . '~,~ ,.
,.,..:.
a. So k + 1 > a/b or a/b  1 < k. So a a l n. T h e n obviously at least two pigeons m u s t roost in the same pigeonhole (see F i g u r e s 3.20 and 3.21). This property, called the pigeonhole principle, can be s t a t e d in t e r m s of functions, as the next t h e o r e m shows.
F i g u r e 3.20
J J
PROOF"
J
dS f J
F i g u r e 3.21
( T h e P i g e o n h o l e P r i n c i p l e ) L e t f" X ~ Y, where X and Y are finite sets, IX I  m, I Y I  n, and m > n. T h e n t h e r e exist at least two distinct e l e m e n t s Xl and x2 in X such t h a t f ( x l )  f(x2). Let X  {Xl,... ,Xm}. Suppose f is injective. T h e n f ( x l ) , . . . , f ( x m ) are distinct elements in Y. So m _< n. But this contradicts the a s s u m p t i o n t h a t m > n. Therefore, f is not injective and there m u s t be at least two distinct elements x l and x2 such t h a t f ( x l )  f(x2). Hence the theorem, m The pigeonhole principle is a simple but i m p o r t a n t c o u n t i n g principle t h a t we shall use in C h a p t e r s 4, 7, and 8. The pigeonhole principle, which can be applied in a variety of situations, can be restated as follows: If m objects are placed into n boxes, t h e n at least one box m u s t contain two or more objects, where m > n. Accordingly, the pigeonhole principle is also called the D i r i c h l e t B o x P r i n c i p l e after the G e r m a n m a t h e m a t i c i a n P e t e r G u s t a v Lejeune Dirichlet, who used it extensively in his work on n u m b e r theory. Although the principle looks simple and straightforward, to apply it successfully you m u s t choose the pigeons and pigeonholes appropriately, as the next few examples illustrate. Suppose we select 367 s t u d e n t s from campus. Show t h a t at least two of t h e m m u s t have the same birthday.
SOLUTION:
The m a x i m u m n u m b e r of days in a year is 366, and this occurs in a leap year. T h i n k of s t u d e n t s as pigeons and days of the year as pigeonholes. Let A be the set of s t u d e n t s and B the set of days, where IA[ = m = 367 and ]B] = n = 366. Let f : A ~ B defined by f ( a ) = b i r t h d a y of s t u d e n t a.
3.4 The Pigeonhole Principle
145
Gustav Peter Lejeune Dirichlet (18051859) was born in Duren, Germany. The son of a postmaster, he first attended a public school and then a private school that emphasized Latin. After attending the Gymnasium in Bonn for 2 years, Dirichlet entered a Jesuit college in Cologne where he received a strong background in theoretical physics under the physicist Georg Simon Ohm. In May 1822, he moved to the University of Paris. In 1826, Dirichlet returned to Germany and taught at the University of Breslau. Three years later, he moved to the University of Berlin where he spent the next 27 years. Dirichlet's primary interest in mathematics was number theory, inspired by Gauss' masterpiece, Disquisitiones Arithmeticae (1801). He established Fermat's Last Theorem for n = 14. Among the many results he discovered include the proof of a theorem presented to the Paris Academy of Sciences on algebraic number theory in 1837: The sequence {an + b} contains infinitely many primes, where a and b are relatively prime. In 1855, when Gauss died, Dirichlet moved to the University of GSttingen. Three years later, he went to Montreaux, Switzerland, to deliver a speech in honor of Gauss. While there, he suffered a heart attack and was barely able to return home. During his illness his wife succumbed to a stroke, and Dirichlet died.
Since m > n, by the pigeonhole principle, there should be at least two students al and a2 such t h a t f(al)  f(a2); t h a t is, at least two students have the same birthday, m The next example* is geometric, d e m o n s t r a t i n g t h a t the pigeonhole principle can pop up in seemingly unusual situations. Suppose five l a t t i c e p o i n t s , t h a t is, points with integer coordinates, are selected on the cartesian plane and each pair of points is joined by a line segment. Show t h a t at least one of the line segments m u s t contain a lattice point between its endpoints. SOLUTION: The set of lattice points can be partitioned into four n o n e m p t y disjoint classes according to the p a r i t y (evenness or oddness) of their coordinates: (odd,odd), (odd,even), (even,odd), and (even,even). Since there are five points (pigeons) and four classes (pigeonholes), by the pigeonhole principle, at least two of t h e m   s a y , A(a,b) and B(c,d)must belong to the same class. By the m i d p o i n t f o r m u l a in analytic geometry, the midpoint M of the line segment AB is (a~c, b+d)2 . Since the sum of any two odd or even integers
/
is an even integer, it follows t h a t M is also a lattice point. Thus AB contains a lattice point M different from its endpoints, m *Based on C. T. Long, "On Pigeons and Problems," Mathematics Teacher, Vol. 81 (January 1988), pp. 2830, 64.
146
Chapter 3
Functions and Matrices
It is well known that the decimal expansions of rational n u m b e r s are periodic. Using the pigeonhole principle, we shall establish this, but first a few words of explanation may be helpful. Using the familiar long division method, you may verify that 4111 = 0. 12345345345345... 33300 Although the decimal expansion is nonterminating, it is p e r i o d i c ; t h a t is, a certain block of digits, namely, 345, gets repeated. Accordingly, the expansion is usually written as 0.12345, using a bar over the first repeating block. The number of digits in the smallest repeating block is the p e r i o d of the expansion; here it is 3. We are now ready to prove the above proposition. ~ Prove that the decimal expansion of a rational number is periodic.
PROOF"
a
Consider, for convenience, a positive rational number ~, where 0 < a < b. Let ~  O . d l d 2 d 3 . . . where, by the division algorithm (see Section 4.1), we have: 10a  b d l + rl 10rl  bd2 + r2 10r2  bd3 + r3 (3.2)
lOrj  bdj+ 1 + rj+ 1
a
and 0 _< ri < b for every i. (Note: The digits a l l , d 2 , . . , in the decimal expansion are the quotients when 10a, 10rl,... are divided by b. Since a remainder has only b choices, by the pigeonhole principle, two of the remainders r l , r 2 , . . . ,rb+l must be equal; that is, rj  rk for s o m e j and k, where 1 < j < k _< b + 1. Consequently, dk+l  dj+l, dk+2  d]+2,..., d2kj  d k , d 2 k  j + l  dj+l, and so on. Thus d j + l . . , dk is the smallest block getting repeated and the period of the decimal expansion is k  j . m The next example, a rather sophisticated application of the pigeonhole principle, is due to the Hungarian mathematician Paul ErdSs. ~ ( E r d S s T h e o r e m ) Ifn + 1 integers are selected from the set {1, 2 , . . . , 2n}, one of them divides another integer that has been selected.
PROOF"
Let al, a2,... ,an+l denote the integers selected. Write each of them as a product of a power of 2 and an odd integer; that is, ai  2e/bi, where 1 < i < n + 1 andei > 0. The integers bl, b2,... ,bn+l are odd positive integers < 2n.
3.4 The Pigeonhole Principle
147
P a u l E r d 6 s (19131996) was born in Budapest, Hungary. Except for about three years in schools, Erd6s (pronounced airdosh) was taught at home, mostly by his father, who had returned from a Siberian prison after 6 years. A child prodigy, Erd6s, at age 3, discovered negative numbers for himself. In 1930 Erd6s entered the Peter Pazmany University in Budapest. Three years later, he discovered a beautiful proof of the celebrated Chebyshev theorem that there is a prime between any positive integer n and 2n. In 1934 he received his Ph.D. from the university. ~i , ~ . , ~ ~ An author of about 1500 articles and coauthor of about 500, Erd6s was 9~ one of the mostprolific writers in mathematics. A tribute in 1983 described him as "the prince of problemsolvers and the absolute monarch of problemposers." As "the Euler of our time," he contributed extensively to number theory, combinatorics, function theory, complex analysis, set theory, group theory, and probability, the first two areas being closest to his heart. "Always searching for mathematical truths," he deemed worldly possessions a nuisance, so he never had a home, a car, checks, or even an address. Always traveling from meeting to meeting, carrying a halfempty suitcase, he would stay with mathematicians wherever he went and donate the honoraria he earned as prizes to students. A recipient of many honors, Erd6s died of a heart attack while attending a mathematics meeting in Warsaw.
....~"i~(,: . ~.~.~ ""'"'~.~"~.:~ t~"..~..:~'~ .. , " " .. . . Jo,~, ~ ~..'~. , f~t..~~.;~." ...... "~z;i! ~ ~:: ~ ....~,_,~:.~,_:~: ~ ......... " ~ ~is~~ ! ' .'~' "" ~ " '   ~ " ~,: " ~j N ~
.. (..~,.v,.o:
.r
Since t h e r e are exactly n odd positive i n t e g e r s < 2n, by t h e pigeonhole principle, two of t h e e l e m e n t s bl, b 2 , . . . , bn+l m u s t be equal, say, bi  bj. T h a t is, aj  2~1b/  2(:Jbi. Thus, if ei < e] t h e n ai aj, a n d if ei < ei t h e n aj [ai.* m T h e pigeonhole principle tells us t h a t if m pigeons are d i s t r i b u t e d into n pigeonholes, w h e r e m > n, at least two pigeons m u s t s h a r e t h e s a m e pigeonhole. In fact, if m o r e t h a n 2m pigeons are assigned to m pigeonholes, t h e n at least t h r e e pigeons m u s t s h a r e t h e s a m e pigeonhole. T h u s t h e pigeonhole principle can be generalized as follows. ( T h e G e n e r a l i z e d P i g e o n h o l e P r i n c i p l e ) If m pigeons are assigned to n pigeonholes, t h e r e m u s t be a pigeonhole c o n t a i n i n g at least [ ( m  1)/nJ + 1 pigeons.
P R O O F (by c o n t r a d i c t i o n ) :
Suppose no pigeonhole c o n t a i n s m o r e t h a n [(m  1)/nJ pigeons. Then: m a x i m u m n u m b e r of pigeons  n . [(m  1)/nJ
k 9 ]Y], and k c IN. T h e n t h e r e is an e l e m e n t t ~ Y such t h a t f  l ( t ) c o n t a i n s m o r e t h a n k elements. 18. Prove t h a t any set S of t h r e e integers contains at least two i n t e g e r s whose s u m is even. (Hint: Define a suitable function f 9 S ~ {0, 1} and use Exercise 17.) "19. U s i n g the pigeonhole principle, prove t h a t the c a r d i n a l i t y of a finite set is unique.
Besides adding and m u l t i p l y i n g functions, t h e r e is a very f u n d a m e n t a l way of c o n s t r u c t i n g new functions. Consider the f u n c t i o n s f , g : R ~ I~ defined b y f ( x ) = 2 x + 3 a n d g ( x ) = x 2. Let x be an input into f. T h e n f(x) = 2x + 3 is a real n u m b e r a n d hence can be considered an i n p u t into g. The r e s u l t i n g o u t p u t is g ( f ( x ) ) (see F i g u r e 3.24). T h u s the functions f and g can be employed to define a new function, called the c o m p o s i t e of f and g, as shown in F i g u r e 3.25.
F i g u r e 3.24
f(x)
g(f(x) )
F i g u r e 3.25
go/"
This leads us to the following definition.
3.5
Compositionof F u n c t i o n s
151
Composition
Let f :X ~ Y and g : Y ~ Z. The c o m p o s i t i o n o f f and g, denoted by g o f (notice the order of the functions), is a function from X to Z, defined by (g o f)(x) = g ( f ( x ) ) . Read g o f as g circle f or the composition o f f a n d g. [In general, dom(g) need not be the same as codom(f); all t h a t is needed is t h a t range (f) c_ dom(g).] L e t f , g " R ~ IRt defined b y f ( x ) ( f o g)(x). SOLUTION:
f ( x )  2x + 3
2x + 3 a n d g ( x )  x 2. Find ( g o f ) ( x ) a n d
Then
(g o f)(x)  g ( f (x))
= g(2x + 3)  (2x + 3) 2
D
g(x)

x 2
So
( f o g)(x)  f (g(x))
= f(x 2)  2(x 2) + 3
= 2x2+3
m
It follows from Example 3.31 that, in general, f o g r g o f; in other words, composition is not a commutative operation. For instance, putting clothes in a washing machine and t h e n in a dryer does not yield the same result as p u t t i n g t h e m in a dryer and then in a washing machine!
(optional) Composition is easily accomplished in computer science. To illustrate this, study the following algorithm fragment, where x ~ ]R:
i. 2. 3. 4. 5. 6. 7. 8. if if x < 4 then x a, t h e n P(k + 1) is also true. Prove t h a t P(n) is t r u e for every n > a.
T h e celebrated euclidean a l g o r i t h m can be used to find the g r e a t e s t c o m m o n divisor of two positive integers, b u t first a very few properties of p r i m e a n d composite n u m b e r s , and some divisibility properties. Let a and b (r 0) be any two integers. If t h e r e is an integer q such t h a t a  b q , we say b d i v i d e s a, b is a f a c t o r of a, a is d i v i s i b l e by b, or a is a multiple of b. We t h e n write b l a ; otherwise, b ya. (Again, the m e a n i n g of the vertical b a r should be clear from the context.) For instance, 316, 8124, b u t 6 y14. A positive factor b of a positive integer a is a proper factor of a if b ~: a. For example, the proper factors of 6 are 1, 2, and 3. T h e r e are positive integers with exactly two positive factors. Accordingly, we m a k e the following definition.
Prime Numbers and Composite Numbers
A positive integer > 1 is a prime n u m b e r (or simply a p r i m e ) if its only positive factors are 1 and itself. A positive integer > 1 is a c o m p o s i t e n u m b e r if it is not a prime. For example, 2 and 19 are primes, w h e r e a s 6 and 21 are composite n u m b e r s (why?). T h e r e is a systematic procedure for d e t e r m i n i n g w h e t h e r or not a positive integer n >_ 2 is a prime. It is based on the next theorem. Any composite n u m b e r n has a p r i m e factor _< L~/nJ.
PROOF (by contradiction):
Since n is composite, t h e r e are positive integers a and b such t h a t n = a b w h e r e 1 < a < n and 1 < b < n. Suppose a > j ~ and b > ~/n. T h e n n  a b > v/n 9 j ~  n, which is impossible. Therefore, either a < v/n or b < j ~ . Since both a and b are integers, it follows t h a t either a < / v Z n / o r
b_< [J~J.
By the f u n d a m e n t a l t h e o r e m of a r i t h m e t i c (see T h e o r e m 4.13), every positive integer has a prime factor. Any such factor of a or b is also a factor of a . b  n, so n m u s t have a p r i m e factor [vZnJ. I
It follows from T h e o r e m 4.2 t h a t if n has n o p r i m e factors < L~r~J, ] t h e n n is a prime; otherwise, it is a composite n u m b e r . This fact can be used to d e t e r m i n e w h e t h e r or not an integer n > 2 is a prime, as the next example illustrates.
190
Chapter 4 Inductionand Algorithms
D e t e r m i n e if 1601 is a p r i m e n u m b e r .
SOLUTION: F i r s t list all p r i m e s < Lx/1601J. T h e y are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and 37. N o n e of t h e m is a factor of 1601 (verify); so 1601 is a prime, m
An a l g o r i t h m for d e t e r m i n i n g the p r i m a l i t y of a positive i n t e g e r n >_ 2 is given in A l g o r i t h m 4.1.
Algorithm prime number(n)
(* This algorithm determines i f a positive integer n>__2 is prime or not using Theorem 4.2. *) Begin (* algorithm *) l i s t all primes < Lv/n] i f any of them is a factor of n then n is not a prime else n is a prime End (* algorithm *)
Algorithm 4.1
In the r e m a i n d e r of this section we discuss some useful divisibility properties. We begin with a simple and s t r a i g h t f o r w a r d property. If a and b are positive integers such t h a t a i b and b i a, t h e n a  b. i
Notice t h a t this t h e o r e m does n o t hold if a a n d b are a n y integers. F o r example, 3 i (  3 ) a n d (  3 ) 13, b u t 3 r  3 .
Let a, b, and c be a n y integers. Then: (1) (2) (3) (4) If a If a If a Ifa I b and b i c, t h e n a I c ( t r a n s i t i v e p r o p e r t y ) . i b and a I c, t h e n a I (b + c). i b and a I c, t h e n a I (b  c). I b, t h e n a I bc.
PROOF: We shall prove p r o p e r t i e s 1 and 2, and leave the o t h e r s as exercises.
(1) Since a lb, t h e r e exists an integer q l such t h a t b = a q ] . Similarly, t h e r e exists an i n t e g e r q2 such t h a t c = bq2. T h e n c = bq2 = ( a q l ) q 2 a ( q l q 2 ) . T h u s , t h e r e exists an integer q = q l q 2 such t h a t c = a q . Therefore, a lc.

(2) As above, we have b  a q l and c  aq3. T h e n b + c  a q l + aq3 = a ( q l + q3). Since ql + q3 is an integer, it follows t h a t a l(b + c). i
4.2 DivisibilityProperties The Greatest Common Divisor
191
A positive integer can be a factor of two positive integers a and b. Such a positive integer is a c o m m o n f a c t o r of a and b. The largest such common factor is the g r e a t e s t c o m m o n d i v i s o r (gcd) of a and b, denoted by gcd {a, b }. For instance, gcd{6, 9}  3, gcd{12, 24}  12, and gcd{6, 35} = 1. This definition of gcd, although simple and clear, is not practical, so we give an alternate, equivalent definition below.
An Alternate Definition of GCD
A positive integer d is the g c d of two positive integers a and b if:
9 d[aandd[b;and
9 if d' [a and d' [ b, then d' I d, where d' is a positive integer. Thus, d is gcd{a, b} if (1) d is a common divisor of both a and b; and (2) any common divisor of a and b is also a divisor of d. The next theorem, an extremely useful and powerful result, can be applied to develop an algorithm to compute gcd{a, b}. Let a and b be any positive integers, and r the remainder when a is divided by b. Then gcd{a, b}  gcd{b, r}.
PROOF Let gcd{a, b} = d and gcd{b, r} = d'. To prove t h a t d = d', it suffices to show that d i d ' and d ' l d . By the division algorithm, a unique quotient q exists such that
a = bq + r To show that d i d ' :
(4.1)
Since d = gcd{a, b}, d f a and d I b. Therefore, d f bq, by Theorem 4.4. Then d l(a  bq), again by Theorem 4.4. In other words, d lr, by Equation (4.1). Thus, d Ib and d I r. Therefore, d I gcd{b, r }; that is, d I d'. Similarly, it can be shown that d' Id. (See Exercise 33.) Thus, by Theorem 4.3, d = d'; that is, gcd{a, b} = gcd{b, r}. 1 Illustrate Theorem 4.5, using a = 108 and b  20. SOLUTION: gcd{108, 20} = 4 (verify). When 108 is divided by 20, the remainder is 8. gcd{20, 8}  4 (verify). Thus, gcd{108, 20} = gcd{20, 8}. 1
Euclidean Algorithm
Among several procedures for finding the gcd of two positive integers, one efficient algorithm is the e u c l i d e a n a l g o r i t h m , named after the
192
Chapter 4 Induction and Algorithms
.:
\
:.:..
~,"~'/ }~!! ..
,
~'~ ..... "": .....=~:
. .~ . 9
Little is known about Euclid's life. He taught at the University of Alexandria and founded the Alexandrian School of Mathematics. When the Egyptian ruler King Ptolemy I asked Euclid if there were an easier way to learn geometry than by studying The Elements, he replied, "There is no royal road to geometry." Euclid is called the father of geometry. No work, except for the Bible, has been more widely read, studied, or edited," according to J. E. Lightner of Western Maryland College, Westminister, Maryland. "More than 2000 editions of the work have appeared since the first printed one in 1482; however, no extant copy of The Elements dates from Euclid's own time."
Greek m a t h e m a t i c i a n Euclid (330?275 B.c.), who included it in his extrao r d i n a r y work The Elements. The algorithm repeatedly applies the division algorithm and T h e o r e m 4.5. Before formally discussing the algorithm, we illustrate it in the next example. ~ Find gcd{ 1976, 1776}.
SOLUTION: Apply the division a l g o r i t h m with 1976 (the larger of the two n u m b e r s ) as the dividend and 1776 as the divisor:
1976 = 1. 1776 + 200 Apply the division algorithm again with 1776 and 200, u s i n g 1776 as the dividend and 200 as the divisor: 1776 = 8. 200 + 176 Continue this procedure until a zero r e m a i n d e r is obtained: 1976 = 1. 1776 + 200 1776=8. 200=1. 176=7. 24=3. 200+176 176+24 24+ 8+0 in this procedure is the gcd. Thus II 8 last nonzero r e m a i n d e r
The last nonzero r e m a i n d e r gcd{1976, 1776} = 8.
Will this m e t h o d work for any two positive integers a and b? If a  b , t h e n g c d { a , b }  a . So assume, for convenience, a > b. (If this is not true,
4.2 Divisibility Properties
193
simply switch them.) Let ro  b. T h e n by successive application of the division algorithm, we get a sequence of equations:
a qoro + rl ~r2 0 r3 > ... _> 0
Since the remainders are nonnegative and getting smaller and smaller, this sequence m u s t eventually t e r m i n a t e with r e m a i n d e r rn  O. Thus, the last two equations in the above procedure arer n  2  q n  l r n  l F rn 0 < rn < r n  1
and
rn1  q n r n
It then follows t h a t gcd{a, b } = gcd{a, r0 }  gcd{r0, rl }  gcd{rl, r2 } . . . . . }  r n , the last nonzero remainder. (This can be established by using mathematical induction; see Exercise 56 in Section 4.4.)
gcd{rnl,rn
~
Apply the euclidean algorithm to find gcd{ 2076, 1024}.
SOLUTION: By the successive application of the division algorithm, we get"
2076 = 2. 1024 + 28 1024=36. 2 8 = 1. 16 = 1. 12=3. 28+16 1 6 + 12 12 + 4 4+0 < last nonzero r e m a i n d e r
Since the last nonzero r e m a i n d e r is 4, gcd{2076, 1024} = 4. The euclidean algorithm is formally presented in Algorithm 4.2.
I
Algorithm Euclid(x,y,divisor)
(* This algorithm returns gcd{x,y} in divisor, where x>__y>O.*) O. Begin (* algorithm *) i. dividend ~ x 2. d i v i s o r 1. 52. Let n be a positive integer. Prove t h a t (n + 1)! + 2, (n + 1)! + 3 , . . . , (n + 1)! + (n + 1) are n consecutive composite numbers.
In everyday life we use the decimal notation, base ten, to represent any real number. For example, 234  2(102) + 3 (101) + 4(10~ which is the d e c i m a l e x p a n s i o n of 234. Likewise, 23.45  2( 101 ) + 3( 10 ~ + 4( 10 1 ) + 5(102). Computers use base two ( b i n a r y ) , and very long binary n u m b e r s are often handled by h u m a n s (as opposed to computers) using bases eight (octal) and sixteen ( h e x a d e c i m a l ) . Actually, any positive integer b > 2 is a valid choice for a base. This is a consequence of the following f u n d a m e n t a l result. ~ Let b be a positive integer>_ 2. Then every positive integer a can be expressed uniquely in the form a  akb k + a k  1 b k  1 + . . . + a 1b + a0, where ao, a 1 , . . . , ak are nonnegative integers less t h a n b, ak # 0, and k > 0. m This leads us to the following definition.
Baseb Representation
The expression a k b k 4 a k  l b k  1 + . . . + a l b + ao is the b a s e  b e x p a n s i o n of the integer a. Accordingly, we write a = ( a k a k  1 " " a l a o ) b in base b. The base is omitted when it is 10. For example, 234 = 234ten and 22 = 10110two (see Example 4.9). When the base is greater t h a n 10, to avoid confusion we use the letters A, B, C , . . . to represent the d i g i t s 10, 11, 12, . . . , respectively. It is easy to find the decimal value of an integer from its baseb representation, as the next example illustrates. [ ~ ~ ~ ~ ~ Express 10110two in base 10.
SOLUTION:
10110two  1(24) + 0(23) + 1(22) + 1(21) + 0(2 ~ = 16+0+4+2+0 = 22 m < binary expansion
198
Chapter 4 Induction and Algorithms
Conversely, suppose we are given a decimal integer. How do we e x p r e s s it in a n o t h e r base b? By T h e o r e m 4.8, all we have to do is express it as a s u m of powers of b, t h e n simply collect t h e coefficients in t h e r i g h t order. Always r e m e m b e r to account for m i s s i n g coefficients.
A Brainteaser
T a k e a look at the tablets A, B, C, D, a n d E in Figure 4.2. A s s u m i n g you are u n d e r 32 y e a r s old, identify t h e t a b l e t s on which y o u r age a p p e a r s ; we can t h e n easily tell y o u r age. F o r example, if y o u r age a p p e a r s on tablets A, B, C, and E, t h e n you m u s t be 23. Can you explain h o w t h i s puzzle works?
F i g u r e 4.2
1 3 5 7 9 11 13 15
A 17 19 21 23 25 27 29 31 2 3 6 7 10 11 14 15
B 18 19 22 23 26 27 30 31 4 5 6 7 12 13 14 15
C 20 21 22 23 28 29 30 31 8 9 10 11 12 13 14 15
D 24 25 26 27 28 29 30 31 16 17 18 19 20 21 22 23
E 24 25 26 27 28 29 30 31
R e t u r n i n g to n o n d e c i m a l r e p r e s e n t a t i o n s , a simple a l g o r i t h m e x p r e s s e s an i n t e g e r a in a n y n o n d e c i m a l base b: divide a, a n d its successive quot i e n t s by b u n t i l a zero q u o t i e n t is reached, t h e n pick t h e r e m a i n d e r s in t h e reverse order. T h e s e steps can be t r a n s l a t e d into t h e e l e g a n t a l g o r i t h m given in A l g o r i t h m 4.3.
Algorithm nondecimal base(n,b)
(* This algorithm finds the baseb representation (amam_1...alao) b of a positive integer n. The variables q and r denote the quotient of the remainder of the division algorithm, and i is a subscript. *) Begin (* algorithm *) (* i n i t i a l i z e the variables q, r, and i *) q no." So be careful.
A variety of interesting examples will show how useful this i m p o r t a n t proof technique is. The next example gives a nice formula for computing the total n u m b e r of cans in the collection in Figure 4.15. Using PMI, prove that, for every positive integer n, 1+2+3+.+nP R O O F (by i n d u c t i o n ) : n n ( n + 1) Let P(n)" ~ i  ~ . i=l 2
Basis step
n(n + 1)
To verify that P(1) is true (Note" Here no  1)"
1 '
Whenn1
RHS
1 (21 + 1) _ 1 _ ~ i  L H S ; s o i=l
P(1) is true.
I n d u c t i o n s t e p Let k be an arbitrary positive integer. We would like to show that P(k) , P(k + 1)" Assume P(k) is true; that is,
k
i=l
i
k(k + 1) 2
<
inductive hypothesis
To establish that P(k) ~ P(k + 1) is true, that is,
Eik+l k
(k + 1)(k + 2) 2
we start with the LHS of this equation:
LHS
Ei
i1
 ~~i + (k + I)
i=1
()
E
i=1
Xi
 ote +
Xi
i=1
Xk + 1
k(k + 1)
+ (k + 1),
by the inductive hypothesis
(k + 1)(k + 2)
= RHS
4.4
Mathematical Induction
211
Thus, if P(k) is true, t h e n P(k + 1) is also true. Therefore, by PMI, P(n) is t r u e for every n >_ 1; t h a t is, the f o r m u l a holds for every positive integer n. m F i g u r e 4.17 provides a geometric proof of this f o r m u l a w i t h o u t words.
Figure 4.17
S
+
S
9", S
=

n(n+l) n(n+l) 2
The next example, again an application of induction, employs a divisibility property, so we follow it in some detail. Prove t h a t 2n 3 + 3n 2 + n is divisible by 6 for every integer n > 1. P R O O F (by PMI): Let P(n): 2n 3 § 3n 2 § n is divisible by 6.
Basis step When n  1, 2n 3 + 3n 2 + n  2 ( 1 ) + 3(1) + 1  6
is clearly
divisible by 6. Therefore, P(1) is true.
Induction step Assume P(k) is true, t h a t is, 2 k 3 + 3 k 2 + k is divisible by 6 for any k > 1. T h e n 2 k 3 + 3 k 2 + k  6 m for some integer m (inductive hypothesis). We m u s t show t h a t P(k + 1) is true; t h a t is, 2(k + 1)3 + 3(k + 1)2 + (k + 1) is divisible by 6. Notice t h a t
2(k + 1) 3 + 3(k + 1)2 + (k + 1) = 2(k 3 + 3k 2 + 3k + 1) + 3(k 2 + 2k + 1) + (k + 1) = (2k 3 + 3k 2 + k) + 6(k 2 + 2k + 1) = 6m + 6(k 2 + 2k + 1) = 6(m + k 2 + 2k + 1), which is clearly divisible by 6. T h u s P(k + 1) is true. Thus, by induction, the given s t a t e m e n t is t r u e for every n >_ 1. by the inductive hypothesis
m
Notice t h a t in the above examples, n o  1, b u t it need not always be 1, as the next example shows.
212
Chapter4 Inductionand Algorithms
J a c o b L B e r n o u l l i (16541705), a member of the most distinguished family of mathematicans (see the family tree in Section 9.1), was born in Basel, Switzerland. His grandfather, a pharmacist in Amsterdam, had become a Swiss through marriage, and his father was a town councilor and a magistrate. Bernoulli received his M.A. in philosophy in 1671 and a theological degree 5years later. During this time, he studied mathematics and astronomy against his father's will. He spent the next 2 years tutoring in Geneva. In 1687 he became professor of mathematics at the University of Basel, .... 9 :,: remaining there until his death. His brother Johann succeeded him at ~" .."" ,,';S" " .~ "'~:::''7~ :S'' Basel. In May 1690 he used the term integral in the calculus sense known today. Bernoulli's most famous work, Ars Conjectandi, was published posthumously in 1713. It contains significant contributions to probability theory, the theory of series, and gravitational theory.
.. ::'::,:...,,,
( B e r n o u l l i ' s I n e q u a l i t y ) Let x be a n y real n u m b e r g r e a t e r t h a n  1 . Prove t h a t (1 + x) n > 1 + nx for every n > 0.
P R O O F (by PMI)" Let x be any real n u m b e r >  1 . Let P(n)" ( l § n >_ l + n x . _ (Note" The induction is on the discrete variable n and not on the " c o n t i n u o u s " variable x.)
Basis step
To verify t h a t P(0) is true: Notice t h a t (1 + x) ~  1 > l+0x
So P(0) is true. (Note" H e r e no  0.)
I n d u c t i o n s t e p A s s u m e P(k) is true; t h a t is, (1 + x ) k > 1 + kx for an arbitrary integer k > 0. We need to show t h a t P(k + 1) is true; t h a t is,
(1 + x ) k+l > 1 + (k + 1)x. By the inductive hypothesis, we have (1 + x) k > 1 + kx. T h e n (1 + x) k+l  (1 + x)(1 + x) k,
>_ (1 + x)(1 + kx), = 1 + (k + 1)x + k x 2
by IH a n d since 1 + x > 0
> 1 + (k + 1)x,
since kx 2 > 0
Therefore, P(k + 1) is also true. Thus, by PMI, (1 + X ) n >_ 1 + nx for every n >_ 0.
m
T h e next e x a m p l e inductively establishes T h e o r e m 2.3 from C h a p t e r 2.
4.4 Mathematical Induction
213
~
A finite set A with n elements has exactly 2 n subsets. PROOF (by PMI)"
B a s i s s t e p When n = 0, A = r so A has exactly 1 = 20 subset. T h u s the result is t r u e when n  0. I n d u c t i o n s t e p Assume any finite set with k elements has 2 k subsets, where k > 0. Let A be a set with k + 1 elements. We would like to show t h a t A has 2 k+l subsets. To this end, let x e A. Let B  A  {x}. Since IB ]  k, B has 2 k subsets by the inductive hypothesis. Each of the subsets of B is a subset of A. Now add x to each of them. The resulting 2 k sets are also subsets of A. Since every subset of A either contains x or does not contain x, by the addition principle, A has 2 k + 2 k  2 k+l subsets. Thus, by the principle of induction, the result holds for every finite set. i Both the basis and the induction steps are essential in the principle of induction, as the next two examples illustrate.
Let g(n) denote the m a x i m u m n u m b e r of nonoverlapping regions formed inside a circle by joining n distinct points on it. Figures 4.184.22 show the cases n  1, 2, 3, 4, and 5, where the various regions are n u m b e r e d 1, 2, 3, etc. The results are s u m m a r i z e d in Table 4.1. F i g u r e 4.18
Figure 4.19
214 F i g u r e 4.20
Chapter 4 Inductionand Algorithms
F i g u r e 4.21
l
F i g u r e 4.22
1
F i g u r e 4.23
/
It appears from the table t h a t g ( n )  2 n1. T h e n g(1) = 20  1, which is true (basis step). Nonetheless, this does not g u a r a n t e e t h a t g ( n )  2 n1 for every n >_ 1. If the formula were true, there would be g(6)  25  32 nonoverlapping regions with six points. Unfortunately, there are only 31 such regions (see Figure 4.23) We shall derive the correct formula in Chapter 6.
4.4
Mathematical Induction
215
Table
4.1
Number of points n
1
2
3
4
5
6
Maximum number of nonoverlapping regions g(n)
1
2
4
8
16
?
We can conclude t h a t t h e t r u t h f u l n e s s of t h e basis step a n d an a p p a r e n t p a t t e r n do n o t e n s u r e t h a t P(n) is t r u e for every n. m The following example shows t h a t t h e validity of t h e i n d u c t i o n step is necessary, b u t not sufficient, to g u a r a n t e e t h a t P(n) is t r u e for all i n t e g e r s in the UD. Consider the " f o r m u l a " P(n) 9 1 + 3 + 5 + . . . + (2n  1)  n 2 + 1. Suppose k P(k) is true: ~ (2i  1)  k 2 + 1. Then:
i=l
k+l ~~(2i 1)i=l
k ~~(2i 1 ) + (2k + 1)
i=l
= (k 2 + 1) + (2k + 1)  (k + 1) 2 + 1 So if P(k) is true, P(k + 1) is true. Nevertheless, the f o r m u l a does not hold for a n y positive integer n. Try P(1) (see Exercise 5). m U s i n g induction, t h e next e x a m p l e "proves" t h a t every person is of t h e s a m e sex. " P r o v e " t h a t every person in a set of n people is of the s a m e sex. "PROOF"" Let P(n): E v e r y o n e in a set of n people is of t h e s a m e sex. Clearly, P(1) is true. Let k be a positive i n t e g e r such t h a t P(k) is true; t h a t is, everyone in a set of k people is of t h e s a m e sex. To show t h a t P(k + 1) is true, consider a set A  { a l , a 2 , . . . , a k + l } of k + 1 people. P a r t i t i o n A into two overlapping sets, B  { a l , a 2 , . . . , a k } a n d C  { a 2 , . . . , a k + l } , as in F i g u r e 4.24. Since I B I  k  I CI, by the inductive hypothesis, everyone in B is of the s a m e sex and everyone in C is of the s a m e sex. Since B a n d C overlap, everyone in B U C m u s t be of t h e s a m e sex; t h a t is, everyone in A is of the s a m e sex. Thus, by PMI, P(n) is t r u e for every n > 1. This concludes the "proof."
m
N o t e : T h e a s s e r t i o n t h a t everyone is of t h e s a m e sex is clearly false.
Can you find t h e flaw in t h e "proof"? See Exercise 46.
216 Figure 4.24
Chapter4 Inductionand Algorithms
Before discussing the second version of the principle of induction, we will look at a few applications of the formula in Example 4.15. First a definition.
Polygonal Number
A polygonal n u m b e r is a positive integer n that can be represented by n dots in a polygonal array in a systematic fashion. For example, the integers 1, 3, 6, 10, ... are t r i a n g u l a r n u m b e r s since they can be represented by triangular arrays, as shown in Figure 4.25; the number of pins in a bowling alley and that of balls in the game of pool are triangular numbers. Let t,z denote the nth triangular number. Then t~ = 1 + 2 + 3 + . . . + n =
n ( n + 1)
Figure 4.25
tl=l
t2=3
t3=6
t4=lO For example,
Triangular numbers manifest delightful properties. of this result. See Exercises 4750.
tn + t n  1  n2; Figures 4.26 and 4.27 provide a nonverbal, geometric proof
Figure 4.26
\
\
\
\
\
4.4 Mathematical Induction F i g u r e 4.27
217
The next example is a n o t h e r application of the formula in Example 4.15 and the generalized pigeonhole principle. Let a l , a 2 , . . ~ be the first n positive integers in some order. Suppose they are a r r a n g e d a r o u n d a circle (see Figure 4.28). Let k be any positive integer < n. Prove t h a t there exists a set of k consecutive elements in the a r r a n g e m e n t with a sum [ [ k n ( n + 1)  2]/2nJ, where [xJ denotes the floor ofx. F i g u r e 4.28
a.~ a2
an
PROOF: Consider the following sums:
S1  a l
+a2 + ... +ak + ak+l
$ 2  a2 + a3 + . . .
S n  a n + a l + "'" + a k  1
Each of the first n positive integers appears k times in this set of sums. Then
n
ESi
k
ai
k
i

, by Example 4.15
i=1
i=1
218
Chapter4 Inductionand Algorithms
Consider kn(n + 1)/2 pigeons. We would like to distribute t h e m among n pigeonholes, called $ 1 , $ 2 , . . . ,Sn. By the generalized pigeonhole principle, at least one of the pigeonholes Si m u s t contain more t h a n Lkn(n § 1)/2n  1/nJ = L[kn(n § 1  2)]/2nJ pigeons. In other words, si > Lkn(n § 1)  2/2n J, as desired, m In particular, if n u m b e r s 1 t h r o u g h 10 are r a n d o m l y placed a r o u n d a circle, at least three consecutive integers in the a r r a n g e m e n t m u s t have a sum exceeding L [ 3 . 1 0 . 1 1  2]/(2.10)J = 16. We now discuss the strong version of the principle of induction.
Strong Version of Induction
Sometimes the t r u t h of P(k) might not be enough to establish t h a t of P(k § 1). In other words, the t r u t h f u l n e s s of P(k + 1) may require more t h a n t h a t of P(k). In such cases, we have to assume a s t r o n g e r inductive hypothesis t h a t P(n0), P(no + 1), ... , P(k) are all true; t h e n verify t h a t P(k § 1) is also true. This strong version, which can be proved using the weak version (see Exercise 57), is stated as follows. ( T h e S e c o n d Principle of M a t h e m a t i c a l Induction) Let P(n) be a predicate satisfying the following conditions, where n is any integer: 9 P(n0) is true for some integer no. 9 If k is an arbitrary integer > no such t h a t P(n0) A P(n0 + 1) A . . . A P(k) is true, then P(k + 1) is also true. Then P(n) is true for every n > no. The next theorem illustrates this proof technique, m
( T h e F u n d a m e n t a l T h e o r e m o f A r i t h m e t i c ) Every positive integer n >_2 either is a prime or can be written as a product of primes. P R O O F ( b y strong induction): Let P(n) denote the given predicate.
Basis s t e p
Choose n o  2. Since 2 is itself a prime, P(2) is true.
Inductive step Let k be a positive integer > 2 such t h a t P(2), P(3), . . . , P(k) are true; t h a t is, assume t h a t integers 2 t h r o u g h k are primes or can be written as products of primes. We would like to show t h a t P(k + 1) is also true; t h a t is, integer k + 1 is a prime or can be expressed as a product of primes. If k + 1 is itself a prime, t h e n we are done. If k + 1 is not a prime, it m u s t be the product of two positive integers x and y, where 1 < x , y < k + 1. By the inductive hypothesis, both x and y are primes or products of primes. Therefore, k + 1  x • is also a product of two or more prime numbers. In other words, P(k + 1) also holds: Thus, by the strong version of induction, P(n) is true for every n > 2. m
4.4 Mathematical Induction
219
We now present an interesting application of the f u n d a m e n t a l t h e o r e m of arithmetic, which is the cornerstone of n u m b e r theory, and the floor function. Find the n u m b e r of trailing zeros in 123! SOLUTION: By the f u n d a m e n t a l t h e o r e m of arithmetic, 123! can be factored as 2a5bc, where c denotes the product of primes other t h a n 2 and 5. Clearly a > b. Each trailing zero in 123! corresponds to a factor of 10 and vice versa. .'. N u m b e r of trailing zeros _ f N u m b e r of products of the form~ \ 2 . 5 in the prime factorization ]  m i n i m u m of a and b b, We proceed to find b" N u m b e r of positive integers < 123 and divisible by 5  [123/5J  24 Each of t h e m contributes a 5 to the prime factorization of 123! N u m b e r of positive integers < 123 and divisible by 25  [123/25J  4 (See Figure 4.29.) Each of t h e m contributes an additional 5 to the prime factorization. Since no higher power of 5 contributes a 5 in the prime factorization of 123!, the total n u m b e r of 5's in the prime factorization equals 24 + 4 = 28. Thus the total n u m b e r of trailing zeros in 123! is 28.
Figure 4.29
sincea>b
each contributes a 5
123!= 1 2 3 . . . 1 2 0 . . . 1 1 5 . . . 1 0 0 . . . 9 5 . . . 7 5 . . . 5 0 . . . 2 5 . . . 1 0 . . . 5 . . . 1
each contributes an additional 5
m
The next example is a n o t h e r interesting application of the floor function. It employs the following facts from n u m b e r theory:
9 Every positive integer that is not a square has an even n u m b e r ofpositive factors. For example, 18 has six positive factors: 1, 2, 3, 6, 9, 18; 21 has four: 1, 3, 7, 21; 19 has two: 1, 19. 9 Every perfect square has an odd n u m b e r of positive factors. For example, 25 has three positive factors, namely, 1, 5, and 25; 64 has seven: 1, 2, 4, 8, 16, 32, and 64.
220
Chapter 4
Induction
and
Algorithms
9 T h e r e a r e [~/nJ p e r f e c t s q u a r e s 1. 5. ~ ( 2 i  1 ) = n i=1
n
2
6. ~~i2= i=1
n
(n + 1)(2n + 1) 6
a(rn_l) r1
n [ (n+l)~ 2 7. ~_~i3  ~ i=1 2 9. n 2 + n is divisible by 2.
n 8. ~~ a r i  l = i=1
(r#
1)
10. n 4 + 2n 3 + n 2 is divisible by 4.
11. The n u m b e r of lines formed by joining n (> 2) distinct points in a plane, no t h r e e of which being collinear, is n ( n  1)/2. 12. T h e n u m b e r of diagonals of a convex ngon* is n ( n  1)/2 > 3. 13. Let a be a positive integer a n d p a p r i m e n u m b e r such t h a t p [a n. T h e n p [ a , w h e r e n > 1. (Hint: Use Exercise 37 in Section 4.2.) *An ngon is a polygon with n sides. An ngon such that the line segment joining any two points inside it lies within it is a convex polygon.
222
Chapter4 Inductionand Algorithms 14. P r o v e t h a t 1 + 2 + . . . + n  n ( n + 1)/2 by c o n s i d e r i n g t h e s u m in t h e
r e v e r s e order.* (Do n o t u s e i n d u c t i o n . ) Evaluate each sum. 30 50 15. ~ (3k 2  1) 16. ~ (k 3 + 2) k=l k=l
n
n
17. ~ Li/2]
i=l
18. ~ [i/21
i=l
F i n d t h e v a l u e of x r e s u l t i n g f r o m e x e c u t i n g e a c h a l g o r i t h m f r a g m e n t .
19.
x ~0
for
20.
t o n do (2i 1)
x ~0
for
i=l
i=l
t o n do
x + x +
x _ 2) cents can be made using two and threecent stamps. (Hint: Use the division algorithm and induction.) *56. Let a and b be any two positive integers with a>__b. Using the sequence of equations in the euclidean algorithm prove t h a t god {a, b }  god {r~_ 1, rn }, n >_ 1. *57. Prove the strong version of mathematical induction, using the weak version. *58. Prove the weak version of induction, principle. using the wellordering
**59. Let Sn denote the sum of the elements in the nth set of the sequence of sets of squares { 1 }, {4, 9 }, { 16, 25, 36 }, . . . . Find a formula for Sn. (J. M. Howell, 1989) **60. Redo Exercise 59 using the sequence of triangular numbers {1}, {3,6}, {10,15,21}, . . . . (J.M. Howell, 1988)
Suppose we wrote an algorithm to solve a problem and translated the algorithm into a computer program. Since it is impossible to test the program for all sets of input values, we rely on a mathematical proof to ensure that the program will always yield the correct output. The principle of induction can certify the correctness of algorithms.
Correct P r o g r a m
A c o r r e c t program yields the correct result for all legal input values, assuming the program contains no compilation and execution errors.
4.5 AlgorithmCorrectness
225
Proving the correctness of a program, especially a complex one, is not at all an easy task. It consists of two steps: (1) Proving t h a t the program will always terminate; and (2) proving t h a t it will always produce the correct result. The second step constitutes the partial c o r r e c t n e s s of the program.
Loop Invariant
First, we will establish the partial correctness of simple w h i l e loops. Let n denote the n u m b e r of iterations of a w h i l e loop. Assume a predicate P(n). A relationship among the variables holds true before the loop is executed and after each iteration of the loop, no m a t t e r how large n is. As the algorithm execution progresses, the values of the variables in the loop may vary, but the relationship remains unaffected. Such a predicate is a loop invariant. To prove t h a t P(n) is a loop invariant, we apply PMI, as the next two examples demonstrate.
•
Algorithm 4.6 computes the product of two positive integers x and y. Notice t h a t the values of the variables x and y are not affected by the loop in lines 37. But the values of i and a n s w e r do get changed during each iteration of the loop.
Algorithm m u l t i p l i c a t i o n ( x , y ) (* This algorithm computes the product of the p o s i t i v e integers x and y, and p r i n t s the answer. *) O. Begin (* algorithm *) i. answer O.
Basis step W h e n n  0, f a c t ( 0 )  1  1! by line 1; so P(0) is true.
4.5 AlgorithmCorrectness Induction step
Assume P(k) is true: f a c t ( k )  k!. Then:
227
fact(k + 1) = fact(k) 9(k + 1), by line 6 = k!. (k + 1), by the inductive hypothesis = (k + 1)~ Therefore, P(k + 1) is true. Thus, by induction, P(n) holds t r u e for every n _ _O; t h a t is, P(n) is a loop > invariant and hence the algorithm correctly computes the value of n!, for every n > _O. _ I
Searching and Sorting Algorithms
The r e m a i n d e r of this section establishes the partial correctness of a few s t a n d a r d searching and sorting algorithms. We begin with two searching algorithms, linear and binary.
Linear Search Algorithm
L e t X = [ X l , X 2 , . . . , X n ] be an u n o r d e r e d list (also known as a onedimensional a r r a y or simply an array) of n distinct items. We would like to search the list for a specific item, called key. If key exists in the list, the algorithm should r e t u r n the location of key. We search the list from right to left for convenience. Compare Xn and key. If Xn  key, key occurs and location = n. Otherwise, compare Xn1 and key. If they are equal, we are done. Otherwise, continue the search until it is successful or the list is empty. This algorithm is the linear search
algorithm.
For example, let X = [Dallas, Boston, Nashville, Albany, Portland] and
key = Albany. T h e n key occurs in the list at location 4. In general, we cannot assume key occurs in the list. To make the search process always successful, we store key in location 0: xo ~ key. So if the search routine r e t u r n s the value zero for location, it implies key does not
occur in the list. An iterative version Algorithm 4.8. of the linear search algorithm is given in
Algorithm linear search (X,n,key,location)
(* This algorithm searches a l i s t by the l i n e a r search method f o r a key and returns i t s location in the l i s t . To make the search always successful, we store key in xo. I f the algorithm returns the value 0 for l o c a t i o n , key does not occur in the l i s t . *) O. Begin (* algorithm *) i. x0 0.
P R O O F (by PMI)" Let P(n)" The algorithm r e t u r n s the correct location for every list of size n>0.
D
B a s i s step When n = 0, the w h i l e loop is skipped. The algorithm r e t u r n s the value 0 in location by line 5, which is correct. So P(0) is true. I n d u c t i o n step Assume P(k) is true for an arbitrary integer k > 0; t h a t is, the algorithm works when the list contains k items. To show that P(k + 1) is true, consider a list X with k + 1 elements. Case 1 Ifxk+l = key in line 3, the w h i l e loop will not be entered and the algorithm r e t u r n s the correct value k + 1 for location in line 5. Case 2 Ifxk+l r key, i  k at the end of the first iteration. This restricts us to a sublist with k elements. By the inductive hypothesis, the algorithm works correctly for such a list. In both cases, P(k + 1) holds. Thus, by induction, P(n) is true for n >_0. In other words, the algorithm r e t u r n s the correct location for every list with n > 0 elements.
Binary Search Algorithm
The b i n a r y s e a r c h a l g o r i t h m searches for a given key if the list X is ordered. The technique employed is d i v i d e and c o n q u e r . First compute the m i d d l e (mid) of the list, where m i d = [(1 + n)/2J. The middle item is Xmid. Now partition the list into three disjoint sublists: [x 1 , . . . , Xmid1], [Xmid], and [Xmid+l,., Xn]. If Xmidkey, the search is successful and location = mid. If they are not equal, we search only the lower half or the upper half of the list. If key < Xmid, search the sublist [xl, . . . , Xmid1]; otherwise, search the sublist [Xmid+l,..., Xn]. Continue like this until the search is successful or the sublist is empty. ~ se the binary search algorithm to search the list X = {3, 5, 8, 13, 21, 34, 55, 89] for key = 5.
SOLUTION"
Let xi denote the ith element of the list X, where 1 < i < n and n  8.
4.5 Algorithm Correctness
229
Step 1
C o m p u t e m i d for t h e list X:
m i d  L(1 + n)/2]  L(1 + 8)/2]  4.
Therefore, t h e middle t e r m is Xmid  13. S t e p 2 C o m p a r e Xmid a n d key: Since x4 r 5, key, if it occurs, m u s t exist in t h e lower sublist [Xl,X2,X3]  [3, 5, 8] or in t h e u p p e r sublist [x5,x6,x7,x8] = [21, 34, 55, 89]. Since key < x4, search t h e first sublist a n d c o n t i n u e steps 1 a n d 2 u n t i l e i t h e r key is located or t h e sublist becomes empty. Step 3 C o m p u t e m i d for t h e list [Xl,X2,X3]"
midSo Xmid  x2  5.
[(1 + 3)/2]  2
S t e p 4 C o m p a r e Xmid a n d key: Since Xmid = key, the search is successful. Key occurs at location 2 and we are done. (As an exercise, use the a l g o r i t h m to search t h e list X with key = 23.) m The steps in this e x a m p l e can be t r a n s l a t e d into an a l g o r i t h m . See A l g o r i t h m 4.9.
Algorithm binary search(X,l,n,key,mid)
(* This algorithm searches an ordered l i s t X of n elements f o r a special item (key). I t returns the location of key i f the search is successful and zero otherwise. The variable mid returns such a value. The variables low and high denote the lower and upper indices of the l i s t being searched. *) O. Begin (* algorithm *) 1. low 0.
P R O O F (by strong induction): Let P(n)" The algorithm works for every ordered list of size n. B a s i s step When n  0, l o w  1 and h i g h  O. Since l o w < h i g h is false in line 3, the w h i l e loop is not executed. So the algorithm r e t u r n s the correct value 0 from line 14, as expected, and P(0) is true. I n d u c t i o n step Assume P(i) holds for every i < k, where k > 0; t h a t is, the algorithm r e t u r n s the correct value for any list of size i < k.
To show t h a t P(k + 1) is true, consider an ordered list X of size k + 1. Since h i g h = k + 1 > 1 = low, the loop is entered and the middle index is computed in line 5.
Case 1 If k e y   X m i d , we exit the loop (line 7) and the value of m i d is returned, so the algorithm works. Case 2 If key < Xmid, search the sublist x l, . . . , Xmid1; otherwise, search the sublist Xmid+l, ... , X,z. In both cases, the sublists contain fewer t h a n k + 1 elements, so the algorithm works in either case by the inductive hypothesis. Thus P(k + 1) is true. So, by PMI, P(n) is true for n >_ 0; t h a t is, the algorithm works correctly for every ordered list of zero or more items, i
Next we present two s t a n d a r d sorting algorithms and prove their correctness.
Sorting Algorithms
Suppose we are given a list ofn items and would like to sort t h e m in "ascending order." Several methods are available. Two algorithms t h a t can do the job are bubble sort and selection sort.
Bubble Sort
B u b b l e sort is a simple, elegant algorithm for sorting a list of n items. It "bubbles up" smaller items to the top and pushes larger items to the bottom: Compare consecutive elements, beginning with the first pair. Swap t h e m if they are out of order. Compare the next pair and swap t h e m if necessary. Continue like this to the end of the list. This ends the first pass. Now place the largest element at the end of the list. Repeat these steps with all but the largest element until the resulting sublist consists of one element. The list is now ordered.
4.5 AlgorithmCorrectness
The following example demonstrates this method. Using bubble sort, sort the list X = [34, 13, 21, 3, 89].
231
SOLUTION: Let xi denote the ith element in the list, where 1 < i < 5. The given list is 1 2 3 4 5
Step 1
Compare Xl and
X2.
Since Xl 1 2
> X2,
swap them. This yields the list 5
3
4
Now comparex2 andx3. Sincex2 > x3, interchangex2 andx3. This produces the list 1 2 3 4 5
Since
X3 > X4,
switch them, yielding the list 1 2 3 4 5
Compare X4 and x5. Since X4 < X5, they are in the correct order and no interchanging is needed. This completes the first pass. At the end of the first pass, the largest element in the list is placed in proper position: 1 X 2 3 4 5 in correct position
11312113134189] to b e s o r t e d
S t e p 2 In the second pass, compare the elements xl through X4 and swap them if necessary. This results in the two largest elements being placed correctly:
1 2 3 4 5
to be sorted
correctly sorted
232
Chapter 4 Induction and Algorithms
S t e p 3 The t h i r d pass involves t h e e l e m e n t s x l t h r o u g h this pass, the t h r e e largest e l e m e n t s are correctly placed: 1 2 3 4 5
X 3.
At t h e end of
~
9
y
,J
to be sorted
in correct order
Step 4
At the end of t h e f o u r t h pass the list is completely sorted:
1 2 3 4 5
all in correct order
Two i m p o r t a n t observations"
9 At the end of t h e i t h pass, t h e i largest e l e m e n t s are correctly placed at the end of t h e list, w h e r e 1 < i < n. So t h e (i + 1)st pass involves t h e e l e m e n t s Xl t h r o u g h Xn_ i . 9 Bubble sort t a k e s n  1 passes to sort a list of n items, even if the list becomes o r d e r e d at t h e end of the ith pass, w h e r e i < n  1. Once the list is sorted, it m a k e s no sense to go t h r o u g h t h e r e m a i n ing passes, so t h e additional passes can be avoided w i t h a boolean variable.
T h e various steps in E x a m p l e 4.30 can be developed into an a l g o r i t h m for bubble sort, as p r e s e n t e d in A l g o r i t h m 4.10.
Algorithm bubble sort(X,n)
(* This algorithm sorts a l i s t X of n elements using the bubble algorithm. *) O. Begin (* algorithm *) 1. for i   1 to n  1 do 2. for j = l to n  i do 3. i f Xj > Xj+ 1 then 4. swap xj and Xj+ 1 5. End (* algorithm *)
Algorithm 4.10
4.5 Algorithm Correctness
233
Establish the correctness of the bubble sort algorithm. P R O O F (by PMI)" Let P(n)" The algorithm sorts every list of size n > 1. B a s i s s t e p When n  1, the list contains just one element and hence is clearly sorted, so P(1) is true. I n d u c t i o n s t e p Assume P(k) is true; t h a t is, the algorithm sorts correctly every list of k (> 1) items. To show t h a t P(k + 1) is true, consider a list X  [ X l , X 2 , . . . , Xk+l]. Since k + 1 > 2, the f o r loop in line 1 is entered. When i  1, j runs from 1 t h r o u g h n  1. Lines 3 and 4 are executed: the consecutive elements xj and Xj+l are compared and swapped if out of order. The inner f o r loop places the largest of the elements Xl, x2,..., Xk+l in position k + 1. This leaves a sublist of k elements, [xl, x2, . . . , Xk]. By the inductive hypothesis, the algorithm correctly sorts it. It follows t h a t the algorithm correctly sorts the entire list X; t h a t is, P(k + 1) is true. Thus, by the principle of induction, P(n) is true for n > 1; that is, the bubble sort algorithm always works, m
Selection
Sort
Unlike bubble sort, s e l e c t i o n s o r t finds the largest element and swaps it with xn if xn is not the largest element. Find the largest of the remaining elements Xl, x2,..., Xn1, and switch it with Xn1 if it isn't xn1. Continue like this until the list is completely sorted. In each pass, unlike in bubble sort, if two elements are out of order, we do not swap t h e m right away but wait to find the largest element of the sublist. At the end of the ith pass, the largest of the elements xl, x 2 , . . . , X n  i + l is swapped with X n _ i + l , where 1 < i < n. This outline of the selection sort algorithm can be a bit refined. In the ith pass, initially assume Xni+l is the largest element. Find the largest of the elements x l, x 2 , . . . , Xni. Swap it with Xni+l if necessary. Algorithm 4.11 results.
Algorithm selection sort(X,n) (* This algorithm sorts a l i s t X of n items using the i t e r a t i v e version of selection sort. Maxindex denotes the index of the largest element in a given pass. *) O. Begin (* algorithm *) 1. i f n > i then(* l i s t contains at least two elements *) 2. for i = 1 to n  1 do
234
Chapter4 Inductionand Algorithms
3. 4. 5. 6. 7. 8.
g. I0. 11.
begin (* for *) maxindex ~ n  i + i (* assume Xn_i+ 1 is the largest element; save its index. *) for j1 to n  i do i f Xj > Xmaxindex, then (* update maxindex *) maxindex ~ j i f maxindex r i + i , then (* found a larger element; swap the corresponding elements *) swap Xmaxindex and Xn_i+ 1
endfor
End
(* algorithm *)
A1gori thm 4. i i
~
Establish the correctness of Algorithm 4.11.
P R O O F (by PMI): Let P(n)" The algorithm works correctly for every list of size n _> 1. B a s i s s t e p When n = 1, the list contains one element and is clearly sorted, so P(1) is true. I n d u c t i o n s t e p Assume P(k) is true; t h a t is, the algorithm sorts correctly every list of size k > _1. _
To show that P(k + 1) is true, consider a list X = x x, x2, . . . , Xk+l] with k + 1 elements, where k + 1 _> 2. Since k + I >_ 2, the condition in line 1 is satisfied, and we enter the loop in line 2. When i = 1, m a x i n d e x = (k + 1) 1 + 1 = k + 1. The f o r loop in lines 57 compares each of the elements Xl, x 2 , . . . , xh with Xmaxindex and updates it as needed. Line 8 updates m a x i n d e x if we have found an element larger than Xh+l. If maxindex ~: k + 1, then the elements xk+ 1 and Xmaxindex are swapped. This stores the largest of the k + 1 elements in position k + 1, leaving a sublist of k elements, namely, x 1, x2,.. 9 xh to be sorted. Therefore, by the inductive hypothesis, the algorithm sorts correctly the list X containing k + 1 elements. Thus, by induction, P(n) is true for every n >_ 1; t h a t is, the algorithm correctly sorts every list of size n. B These searching and sorting algorithms are pursued again in Section 4.7. Additional sorting algorithms appear in the exercises.
Exercises 4.5
Prove that the given predicate P(n) in each algorithm is a loop invariant.
1.
Algorithmexponential (x,n) (* This algorithm computes
xn, where xEIR + and n ~W. *)
2.
Algorithm divi sion(x,y) (* This algorithm computes
the quotient and the remainder when a positive
4.5 Algorithm Correctness
235
0. Begin (* algorithm *) i. answer + I 2. while n > 0 do 3. begin (* while *) 4. answer + answer, x 5. n y then
3. x~xy 4. else 5. y,yx 6. gcd + x 7. End (* a l g o r i t h m *) P(n): g c d { x n , Y n }    g c d { x , y } , where Xn and Yn denote the values of x and y at the end of n i t e r a t i o n s of the loop.
integer x is divided by a positive integer y using addition and subtraction. *) 0. Begin (* algorithm *) 1. dividend ~ x 2. divisor di vi sor do 6. begin (* while *) 7. di vi dend _ n], and ]f2(n)l _< C2[g2(n)] for n >_ n2. Let C  C1C2 and no  max{n],n2}. Then"
](f]. f 2 ) ( n ) i  ]f](n). f2(n)i = [fl(n)]. ]f2(n)]
_ no
T h u s (fl" f2)(n)  O(gl(n)g2(n)).
I
The next two examples employ this h a n d y t h e o r e m along with the earlier theorems. Let f ( n ) = 6n 2 + 5n + 7 lg n!. E s t i m a t e the g r o w t h off(n). SOLUTION: Since 6n 2  O(n 2) and 5n  O(n), 6n 2 + 5n  O(n 2) by T h e o r e m 4.15. F u r t h e r m o r e , 7  O(1), and lg n!  O(n lg n) by Example 4.34. So 7 lg n! = O(1). O(n lg n) = O(1. n lg n), by T h e o r e m 4.16 = O(n lg n)
4.6 The Growthof Functions
243
Since lg n < n, n lg n < n 2 for n > 1 (see Figure 4.31), it follows by T h e o r e m 4.15 t h a t f ( n )  O(n 2) + O(n lg n)  O(n2). i Let f ( n )  (3n 2 + 4n  5) lg n. E s t i m a t e the g r o w t h off(n).
SOLUTION:
3n 2 + 4n  5  O(n2), by T h e o r e m 4.14 Clearly, lg n  O(lg n) So
f ( n )  (3n 2 + 4n  5) lg n
= O(n2) 9O(lg n)
= O ( n 2 lg n), by T h e o r e m 4.16
II
We now t u r n to the bigomega and the bigtheta n o t a t i o n s for investigating the g r o w t h of functions.
The BigOmega and BigTheta Notations
The bigoh notation has been widely used in the study of the g r o w t h of functions; however, it does not give us an exact order of growth. F o r instance, f(n)O(g(n)) j u s t implies t h a t the function f does not grow any faster t h a n g. In other words, it simply provides an u p p e r b o u n d for the size of f ( n ) for large values of n, b u t no lower bound. W h e n we need the lower bound, we employ the bigomega notation. W h e n we need both b o u n d s to estimate the growth of f, we use the bigt h e t a notation. Both n o t a t i o n s were introduced in the 1970s by Donald K n u t h of Stanford University. We now p u r s u e the bigomega notation. As you could imagine by now, its definition closely resembles t h a t of the bigoh notation; it can be obtained by simply changing _< to >_.
The BigOmega Notation
Let f , g " • ~ R. Suppose t h e r e is a positive c o n s t a n t C and a positive integer no such t h a t If(n)] > _C i g ( n ) i for every n >_ no. T h e n f ( n ) is g2(g(n)); _ t h a t is, f ( n )  ~ ( g ( n ) ) . [As above, read this as f ( n ) is bigomega of g(n).] The following example illustrates this definition. Let f ( n ) C50 50n 3  6n + 23. W h e n n >_ 0, 50n 3  6n + 23 > 50n 3. So, with and g ( n )  n 3, it follows t h a t f ( n ) >__ C . g ( n ) for every n >_ 0. T h u s f ( n )  g2(n3). (Notice t h a t here no  0.) i
244
Chapter 4 Induction and Algorithms
D o n a l d E r v i n K n u t h (1938), a pioneer in the development of the theory of compilers, programming languages, and the analysis of algorithms, is also a prolific writer in computer science. He was born in Milwaukee, Wisconsin, where his father, the first college graduate in the Knuth family, taught bookkeeping at a Lutheran high school; his talent for mathematics and music played a significant role in the intellectual development and pursuit of the young Knuth. As a youngster, Knuth had a marvelous gift for solving complex problems. As an eighth grader, he entered the Ziegler's Candies Contest to find the number of words that can be formed from the letters in Ziegler's Giant Bar. Knuth listed 4500 such words, 2000 more than in Ziegler's master list. This won a television set for the school and enough Ziegler candy for the entire student body. In high school, Knuth entered the prestigious Westinghouse Science Talent Search (now Intel Science Talent Search) with his project, The Prtrzebie System of Weights and Measures, that would replace the cumbersome British system. His project won an honorable mention, and $25 from MAD Magazine for publishing it. When he graduated from high school, he was already an accomplished mathematician, musician, and writer. He majored in physics at the Case Institute of Technology (now Case Western Reserve University) and was introduced to an IBM 650 computer, one of the earliest mainframes. After studying the manual from cover to cover, he decided that he could do better and wrote assembler and compiler code for the school's IBM 650. In 1958, Knuth developed a system for analyzing the value of a basketball player, which the coach then used to help the team win a league championship. Newsweek wrote an article about Knuth's system and Walter Cronkite carried it on the CBS Evening News. In his sophomore year, Knuth switched his major to mathematics. His work at Case was so distinguished that when he was awarded his B.S. in 1960, the faculty made an unprecedented decision to grant him an M.S. concurrently. Knuth then entered the California Institute of Technology for graduate work and received his Ph.D. in mathematics 3 years later. He joined the faculty there, also consulting for the Burroughs Corporation writing compilers for various programming languages, including ALGOL 58 and F O R T R A N H. From 19681969, he worked at the Institute for Defense Analyses, Princeton, New Jersey. In 1969, Knuth joined the faculty at Stanford University. Knuth's landmark project, The Art of Computer Programming, was initiated by AddisonWesley Publishing Co. in early 1962, while he was still in graduate school. Dedicated to the study of algorithms, it would be a sevenvolume series when completed. A revered work, it was the pioneer textbook in the 1970s and continues to be an invaluable resource. Knuth developed two computer languages to deal with mathematics typography, TEX, a typesetting program, and Metafont, a program to develop the shapes of letters. He has received numerous honorary degrees from universities around the world: the Grace Murray Hopper Award (1971), the Alan M. Turing Award (1974), the Lester R. Ford Award (1975), the National Medal of Science (1979), the McDowell Award (1980), the Computer Pioneer Award (1982), and the Steele Prize (1987). An accomplished church organist and composer of music for the organ, Knuth retired from Stanford in 1992.
4.6 The Growth of Functions
245
We now m a k e an i n t e r e s t i n g observation. To this end, let f ( n ) = f~(g(n)); so If(n)[ >_ Clg(n)[ for n >_ no. T h e n Ig(n)l _< C'[f(n)[ for some positive c o n s t a n t C t = 1/C; so g ( n ) = O ( f ( n ) ) . Conversely, let g ( n ) = O(f(n)). By r e t r a c i n g these steps, it follows t h a t f(n) = ~ ( g ( n ) ) . T h u s f(n) = f~(g(n)) if and only i f g ( n )  O(f(n)). We now define the bigtheta notation, using the bigoh and bigomega notations.
The BigTheta Notation
Let f , g 9 N + R such t h a t f ( n )  O(g(n)) and f ( n ) = gz(g(n)). T h e n f ( n ) is said to be of o r d e r g(n). We t h e n write f ( n ) =  read this as f(n) is bigtheta of g(n). The next two examples illustrate this definition. Let f ( n )  (3n 2 + 4n  5) lg n. By Example 4.37, f ( n )  O(n 2 lg n). W h e n n > 1, we also have: (3n 2 + 4n  5)lg n > 3n 2 lg n T h a t is,
f ( n ) > 3(n 2 lg n)
So
f ( n )  f2(n 2 lg n)
T h u s f(n)  O(n 2 lg n)  f2(n 2 lg n), so f(n)  
2 lg n).
II
Let f(n) show the n u m b e r of times the a s s i g n m e n t s t a t e m e n t x ~ x + 1 is executed by the nested for loops in Example 4.35. Recall t h a t f ( n ) n ( n + 1)/2  O(n2). Since n + 1 >_ n for every n >_ 1, it follows t h a t n ( n + 1)/2 >_ n2/2; so f(n)  f2(n2). T h u s f(n)   II
We now m a k e two i n t e r e s t i n g observations from Examples 4.39 and 4.40: 9 Iff(n) is a polynomial in n of degree m, t h e n f ( n ) =O(nm).
9 f(n)=  if and only if Alg(n)] _ 2 and select b + i integers at random. Find two integers in the list such t h a t their difference is divisible by b. 2. Read in an integer n >_ 2 and select n positive integers at random. Find a sequence of integers from the list whose sum is divisible by n. 3. Read in a positive integer > 2 and determine if it is a prime.
258
Chapter 4 Induction and Algorithms
4. Determine if each value of f ( n ) n 0.
m
67. A(2,2)
68. A(1, n)  n + 2
69. A(2, n)  2n + 3
"70. Predict a formula for A(3, n). "71. Prove the formula in Exercise 70, where n >_ 0.
The recursive definition of a function f does not provide us with an explicit formula for f ( n ) , but establishes a systematic procedure for finding it. This section illustrates the iterative method of finding a formula for f ( n ) for a simple class of recurrence relations.
5.2 SolvingRecurrence Relations
279
S o l v i n g t h e r e c u r r e n c e r e l a t i o n for a function f m e a n s finding a n explicit f o r m u l a for f ( n ) . T h e i t e r a t i v e m e t h o d of solving it involves two steps"
9 Apply t h e r e c u r r e n c e f o r m u l a i t e r a t i v e l y a n d look for a p a t t e r n to predict an explicit formula. ~ Use induction to prove t h a t t h e f o r m u l a does indeed hold for e v e r y possible value of t h e i n t e g e r n. The next e x a m p l e illustrates this m e t h o d . (The h a n d s h a k e p r o b l e m continued) By E x a m p l e 5.3, t h e n u m b e r of h a n d s h a k e s m a d e by n guests at a d i n n e r p a r t y is given by h(1)0
h(n)h(n
1)+ (n
1),n >_ 2
Solve this r e c u r r e n c e relation. SOLUTION: Step 1 To predict a f o r m u l a for h ( n ) "
h(n) h(n 
U s i n g iteration,
1) + (n  1) 2) + (n  2) + (n  1) 3) + (n  3) + (n  2) + (n  1)
= h(n = h(n 
= h(1) + 1 + 2 + 3 + . . . + (n  2) + (n  1) =0+ 1+2+3+... 1) 2
n(n ~ ,
+ (n
1)
n(n 
Step 2
To prove, by induction, t h a t h ( n ) W h e n n = 1, h(1) 
1)
2
w h e r e n > 1:
Basis step
1.0 2 = 0, which agrees with t h e initial
condition. So the f o r m u l a holds w h e n n  1.
Induction step
h ( k + 1) 
Assume h(k)
h(k) + k,
k(k
1)
for a n y k > 1. Then:
by t h e r e c u r r e n c e relation
280
Chapter 5 Recursion
k(k2 k(k+
1) 1)
+k,
by the induction h y p o t h e s i s
Therefore, if the f o r m u l a holds for n  k, it also holds for n  k + 1. Thus, by PMI, the r e s u l t holds for n >_ 1. More generally, u s i n g i t e r a t i o n we can solve the r e c u r r e n c e r e l a t i o n
an  a n  1 + f (n )
E
(5.5)
as follows:
an  a n  1 + f (n ) = l a n  2 + f ( n  1)] + f ( n )  a n  2 + f ( n  1) + f ( n )
= l a n  3 + f ( n  2)] + f ( n  1) + f ( n ) = a n  3 + f ( n  2) + f ( n  1) + f ( n )
lz
= ao +
r(i)
i=1
(5.6)
You can verify t h a t this is the actual solution of the r e c u r r e n c e r e l a t i o n (5.5). For example, in t h e h a n d s h a k e p r o b l e m f ( n )  n  1 a n d h(0) = 0, so the solution of the r e c u r r e n c e relation is
r/ ?/
h ( n )  h(O) + E f ( i )
i=1 n1
 0 + E(i
i=1
m
 1)

~
i=1
i =
n(n
1) , n>__l
2
which is exactly the solution o b t a i n e d in the example. ~ Solve t h e r e c u r r e n c e relation in E x a m p l e 5.6. SOLUTION: Notice t h a t an can be redefined as
n ( n + 1) an1 +   , 2
an
n > 1
5.2
Solving Recurrence Relations
281
w h e r e a0 
0. C o m p a r i n g this w i t h r e c u r r e n c e r e l a t i o n (5.5), we h a v e
f(n)

n(n + 1) T h e r e f o r e , by E q u a t i o n (5.6)
2
n
an  ao + ~~f (i)
i=1
n
i(i + 1)
2
n
= ao + ~~ i=1 = 2
_ 1 [n(n
1 E(i2
+ i)
i=l
i
i2 + i=1
+
1 ) ( 2 n + 1)
2
L
6
4 n(n2+ 1)]
n(n+1)(2n+12 6fi) n ( n2 nl + 4" 6 + ) 2 n(n + 1)(n + 2)
, n>_ 0 m
T h e following i l l u s t r a t i o n of t h e i t e r a t i v e m e t h o d b r i n g s us a g a i n to t h e T o w e r of B r a h m a puzzle. ~ Recall from E x a m p l e 5.4 t h a t t h e n u m b e r of moves n e e d e d to t r a n s f e r n disks from peg X to peg Z is given by bl1
b n  2bn1 + 1, n >__2
Solve this r e c u r r e n c e relation. SOLUTION: S t e p 1 To predict a f o r m u l a for U s i n g iteration,
bn"
b n  2bn1 + 1
= 212bn_2 + 1] + 1  22bn_2 + 2 + 1 = 2212bn_3 + 1] + 2 + 1  23bn_3 + 22 + 2 + 1
 2 n  l b l + 2 n2 +'''+ 2 2 A 2 + 1 __ 2 n1 jr_ 2n2 + . . . + 2 + 1 = 2 n  1, by Exercise 8 in Section 4.4.
282
Chapter 5 Recursion
Step 2
You m a y prove by i n d u c t i o n t h a t b n  2 n  1, w h e r e n > 1.
m
M o r e generally, y o u m a y verify t h a t t h e solution of t h e r e c u r r e n c e r e l a t i o n a n  C a n  1 + 1, w h e r e c is a c o n s t a n t (r 1), is
Cn 
1
an

c n aO
+
c1
F o r i n s t a n c e , in E x a m p l e 5.12, b0  0 a n d c  2, so 2 n  1
bn 2n "O + ~ =2 n 1
21 as expected. Let us p u r s u e E x a m p l e 5.12 a bit f u r t h e r . S u p p o s e t h e r e a r e 64 d i s k s at peg X, as in t h e original puzzle, a n d it t a k e s 1 second to m o v e a disk f r o m one peg to a n o t h e r . T h e n it t a k e s a t o t a l of 264  1 s e c o n d s to solve t h e puzzle. To get a n idea how incredibly large t h i s total is, notice t h a t t h e r e a r e about 365.24.60.60  31,536,000 seconds in a year. T h e r e f o r e , T o t a l t i m e t a k e n  264  1 seconds 1. 8 4 4 6 7 4 4 0 7 x 1019 seconds 5 . 8 4 9 4 2 4 1 7 x 1011 y e a r s 600 billion years! I n t r i g u i n g l y , a c c o r d i n g to s o m e e s t i m a t e s , 18 billion y e a r s old. t h e u n i v e r s e is only a b o u t
Exercises 5.2
U s i n g t h e i t e r a t i v e m e t h o d , p r e d i c t a s o l u t i o n to each r e c u r r e n c e r e l a t i o n s a t i s f y i n g t h e given initial condition. 1. s 0  1
Sn 2Sn1, n > 1
2. a l = l
an an1 + n, n > 2
3. a 0  1
an an1 + n, n > 1
4. a 1 = 1
a n  a n  1 +
(2n  1), n >_ 2
5.
aoO an an1 + 4n, n > 1
6. s l = l
Sn   ~ S n 
1 + n3, n _> 2
5.2 Solving Recurrence Relations
283 8. a 1   1
an  2 a n  1 + (2 n 
7. s l = l
8n = S n  1 + n 2 , n > 2
1), n >_ 2
9  1 6 . Using induction, verify the solutions to Exercises 18. 17. Using the data in Example 5.2, show t h a t the c o m p o u n d a m o u n t J u d y will receive at the end o f n years is given b y A ( n ) = 1000(1.08) n, w h e r e
n>0.
Use the recursive definition of fn in E x a m p l e 5.5 to a n s w e r Exercises 18 and 19. 18. Predict a formula for fn. 19. Prove t h a t the formula holds for n > 1. 20. Using induction, establish the explicit f o r m u l a for bn in E x a m p l e 5.12. Using induction, prove t h a t each is a solution to the c o r r e s p o n d i n g recurrence relation, where c is a c o n s t a n t and f ( n ) a function of n.
n
21.
an

ao + ~ f(i),
i=1
an
 an1
+
f (n)
+ l ( a s s u m e c ~: l )
22. a n  c n a o
+
cn  1 ~ anCan_l cl'
i=1
n 23. an  c n a o + ~ c n  i f (i), an  C a n  1 b f (n)
Let an denote the n u m b e r of times the s t a t e m e n t x ~ x + 1 is executed by the following loops.
for i = 1 to n do
f o r j = 1 to [ i / 2 ] x~x+l
do
24. Define an recursively. 0 25. Show t h a t an =
an1 + n/2 anl+(n1)/2
if n  1 if n > 1 and even if n > l a n d o d d
26. Solve the recurrence relation satisfied by an. Let an denote the n u m b e r of times the s t a t e m e n t x ~ x + 1 is executed by the following f o r loops:
for i = 1 t o n do
f o r j = 1 to [ i / 2 ] do x,x+1
27. Define an recursively. 1 28. Show t h a t an an1 + n/2 a n  1 + (n + 1)/2
if n  1 if n > 1 and even if n > l a n d o d d
284
Chapter 5 Recursion 29. Solve the recurrence relation satisfied by an. Let an denote the number of times the statement x ~ x + 1 is executed by the nested f o r loops in Exercise 35 in Section 4.4. 30. Define an recursively. 31. Solve the recurrence relation satisfied by an. 3233. Redo Exercises 30 and 31 using the loops in Exercise 36 in Section 4.4. 3435. Redo Exercises 30 and 31 using the loops in Exercise 37 in Section 4.4. 3637. Redo Exercises 30 and 31 using the loops in Exercise 38 in Section 4.4. Let tn denote the nth triangular number. 38. Define tn recursively. 39. Find an explicit formula for tn. 40. Prove that 8tn + 1 is a perfect square. The nth p e n t a g o n a l n u m b e r Pn is obtained from its predecessor by adding three rows of dots plus one. The first four pentagonal numbers are represented pictorially in Figure 5.11.
Figure 5.11
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
9
9
9
9
9
9
9
9
9
pl1
P2 = 5
P 3 = 12
P4 = 22
41. Represent P5 pictorially. 4243. Redo Exercises 38 and 39 usingpn. The nth h e x a g o n a l n u m b e r hn is obtained from its predecessor by adding four rows of dots plus one dot. The first four hexagonal numbers are shown pictorially in Figure 5.12. 4446. Redo Exercises 4143 using h n . 47. Prove that hn  Pn + tn  n, using the explicit formulas for Pn and tn. 48. Prove that hn  Pn + tn  n, using the recurrence relations for Pn and tn.
5.2 SolvingRecurrence Relations
Figure 5.12
285
9
9
9
9
9
9
9
Q
hl = 1
h2 = 6
h 3 = 15
h 4 = 28
Triangular pyramidal n u m b e r s Tn (or t e t r a h e d r a l numbers) a r e
positive i n t e g e r s t h a t can be r e p r e s e n t e d by t r i a n g u l a r p y r a m i d a l s h a p e s . T h e first f o u r t e t r a h e d r a l n u m b e r s a r e 1, 4, 10, a n d 20; see F i g u r e 5.13.
Figure 5.13
T1 1
T2 = 4
T 3 = 10
T 4 = 20
49. Define Tn r e c u r s i v e l y . 50. C o n j e c t u r e a n explicit f o r m u l a for Tn. 51. E s t a b l i s h t h e f o r m u l a in E x e r c i s e 50.
Square pyramidal n u m b e r s Sn a r e positive i n t e g e r s t h a t c a n be repr e s e n t e d by p y r a m i d a l s h a p e s , w h e r e t h e b a s e is a s q u a r e . T h e first f o u r s q u a r e p y r a m i d a l n u m b e r s a r e 1, 5, 14, a n d 30; see F i g u r e 5.14.
5254. Redo Exercises 4951 with
Sn.
L e t an d e n o t e t h e n u m b e r of s u b s e t s of t h e set S  {1, 2, . . . , n } t h a t c o n t a i n no c o n s e c u t i v e i n t e g e r s , w h e r e n > 0. W h e n n  0, S  O.t C o m p u t e each. 55. a0 56. a l 57. a2 58. a3
t Proposed by Irving Kaplansky of The University of Chicago.
286 F i g u r e 5.14
Chapter 5 Recursion
a
? ....
~

e
9
. . . .
~
  
9
.
.
.
.
S1 = 1
S 2 =
5
S 3 = 14
S 4 =
30
59. Define
an
recursively.
60. Solve the recurrence relation satisfied by an. Suppose we introduce a mixed pair of lmonthold rabbits into a large enclosure on the first day of a certain month. By the end of each month, the rabbits become m a t u r e and each pair produces k  1 mixed pairs of offspring at the beginning of the following month. ( N o t e : k >_ 2.) For instance, at the beginning of the second month, there is one pair of 2monthold rabbits and k  1 pairs of 0montholds; at the beginning of the third month, there is one pair of 3montholds, k  1 pairs of lmontholds, and k ( k  1) pairs of 0montholds. Assume the rabbits are immortal. Let a,~ denote the average age of the rabbit pairs at the beginning of the nth month. (P. Filipponi, 1990) * ' 6 1 . Define a , recursively. **62. Predict an explicit formula for a,~. **63. Prove the formula in Exercise 64. 64. (For those familiar with the concept of limits) Find lim
n  ~
an.
Unfortunately, the iterative method illustrated in the preceding section can be applied to only a small and simple class of recurrence relations. The present section develops a method for solving two large, i m p o r t a n t classes of recurrence relations.
5.3 SolvingRecurrence Relations Revisited Linear Homogeneous Recurrence Relations with Constant Coefficients (LHRRWCCs)
A kthorder
287
l i n e a r h o m o g e n e o u s r e c u r r e n c e r e l a t i o n w i t h cons t a n t c o e f f i c i e n t s is a r e c u r r e n c e r e l a t i o n of t h e f o r m
an  Clan1 4 C 2 a n  2 4 " ' " 4 C k a n  k
(5.7)
w h e r e Cl, C2,... ,Ck E R a n d ck r 0. First, a few w o r d s of e x p l a n a t i o n : T h e t e r m l i n e a r m e a n s t h a t e v e r y t e r m on t h e R H S of E q u a t i o n (5.7) c o n t a i n s at m o s t t h e first p o w e r of a n y p r e d e c e s s o r a i . A r e c u r r e n c e r e l a t i o n is h o m o g e n e o u s if e v e r y t e r m on t h e R H S is a m u l t i p l e of s o m e a i ; in o t h e r w o r d s , t h e relation is satisfied by t h e s e q u e n c e {0}; t h a t is, a n  0 for e v e r y n. All coefficients ci a r e c o n s t a n t s . Since a n d e p e n d s on its k i m m e d i a t e predecessors, t h e o r d e r of t h e r e c u r r e n c e r e l a t i o n is k. Accordingly, to solve a k t h  o r d e r L H R R W C C , we will n e e d k initial conditions, say, a0  Co, a l  e l , . . . , a k  1   C k  1 . T h e n e x t e x a m p l e i l l u s t r a t e s in detail t h e v a r i o u s t e r m s in t h i s definition. 9 T h e r e c u r r e n c e r e l a t i o n Sn  2Sn1 is a L H R R W C C . Its o r d e r is one. 9 T h e r e c u r r e n c e r e l a t i o n a n  n a , ~ _ l is l i n e a r a n d h o m o g e n e o u s . B u t t h e coefficient on t h e R H S is not a c o n s t a n t . T h e r e f o r e , it is not a LHRRWCC.
9 hn h n  1 + ( n  1) is a l i n e a r r e c u r r e n c e relation. B u t it is not h o m o g e n e o u s b e c a u s e of t h e t e r m n  1.
2 9 The recurrence relation an  an_ 1 + 3an2 not l i n e a r since t h e p o w e r of an1 is 2.
9 an  an1
4 2 a n  2
is h o m o g e n e o u s . B u t it is m
4 3 a n  6
is a L H R R W C C of o r d e r six.
Before we discuss solving s e c o n d  o r d e r L H R R W C C s , notice t h a t t h e solution of t h e r e c u r r e n c e r e l a t i o n Sn  2 S n  1 , w h e r e so  1, is Sn  2 n, n >_ 0 (see Exercise 1 in Section 5.2). M o r e generally, you m a y verify t h a t t h e solution of t h e r e c u r r e n c e r e l a t i o n a n  o t a n _ l , w h e r e a0  c, is a n  c a n, n>0. We now t u r n o u r a t t e n t i o n to t h e s e c o n d  o r d e r L H R R W C C
an  aan_l ~ b a n  2
(5.8)
c a n,
w h e r e a a n d b a r e n o n z e r o c o n s t a n t s . If it h a s a n o n z e r o solution of t h e f o r m t h e n c a n  acot n  1 + b c o t n  2 . Since ca r 0, t h i s yields c~2  a a + b; t h a t is, ot 2   a a  b  0, so a m u s t be a solution of t h e c h a r a c t e r i s t i c e q u a t i o n
X 2  ax 
b 
0
(5.9)
288
Chapter 5 Recursion
of the r e c u r r e n c e relation (5.8). The roots of E q u a t i o n (5.9) are t h e c h a r a c t e r i s t i c r o o t s of r e c u r r e n c e relation (5.8). T h e o r e m s 5.2 t h r o u g h 5.4 show how characteristic roots help solve LHRRWCCs. ~ Let a and fl be the distinct (real or complex) solutions of the e q u a t i o n
x 2  ax  b 
0, where a, b e R and b 4: 0. T h e n every solution of the LHRRWCC a n  a a n  1 + b a n  2 , where a0  Co and al  C1, is of the form an  A a n + B fl n for some c o n s t a n t s A and B.
PROOF"
The proof consists of two parts: 9 First, we will show t h a t a n = A(~ n + B f l n is a solution of the r e c u r r e n c e relation for any c o n s t a n t s A and B. 9 We will t h e n find the values of A and B satisfying the given initial conditions. First, notice t h a t since c~ and fl are solutions of equation (5.9), and ~ 2 __ a f t + b.
9 T o s h o w t h a t an  A a n + B f l n is a s o l u t i o n of the recurrence
~2 _
a~ + b
relation"
a a n  1 + b a n  2  a ( A ~ '~1 + B f l n  l ) + b ( A a n  2 + B f l n  2 ) = Ac~'~2(acv + b) + B f l n  2 ( a f l = Ac~,~2 . ol2 + B f l n  2 = Aol '~ + B f l n f12 + b)
T h u s an  A a n + B f l n is a solution of the recurrence relation (5.8). 9 Secondly, let an  A~x n + B f l n be a solution of (5.8). To find the values of A and B, notice t h a t the conditions a0  Co and a l  C1 yield the following linear system" Co  A + B
C1  A a + B f l
(5.10) (5.11)
Solving this system, we get (Verify.)
A C1  C o i l
and C0d  C1 (Remember, a 4= ft.)
~Z
at~
With these values for A and B, an satisfies the initial conditions and the recurrence relation. Since the recurrence relation and the initial conditions d e t e r m i n e a u n i q u e sequence, { a n } , an  A a n + B f l n is indeed the u n i q u e solution of the r e c u r r e n c e relation, m
5.3
Solving Recurrence Relations Revisited
289
A few interesting observations:
9 T h e s o l u t i o n s a a n d ~ a r e n o n z e r o , since a = 0, for i n s t a n c e , w o u l d i m p l y t h a t b = 0. 9 T h e o r e m 5.2 c a n n o t be applied if a  ft. H o w e v e r , it w o r k s even if a n d fl a r e c o m p l e x n u m b e r s . 9 T h e s o l u t i o n s otn a n d fin a r e t h e b a s i c s o l u t i o n s of t h e r e c u r r e n c e relation. In g e n e r a l , t h e n u m b e r of basic s o l u t i o n s e q u a l s t h e o r d e r of t h e r e c u r r e n c e relation. T h e g e n e r a l s o l u t i o n an  Aot n 4 B f l n is a l i n e a r c o m b i n a t i o n of t h e basic solutions. T h e p a r t i c u l a r solution is o b t a i n e d by selecting A a n d B in s u c h a w a y t h a t t h e initial c o n d i t i o n s a r e satisfied, as in T h e o r e m 5.2. T h e n e x t t h r e e e x a m p l e s i l l u s t r a t e h o w to LHRRWCCs using their characteristic equations. Solve t h e r e c u r r e n c e r e l a t i o n an  5 a n  1  6 a n  2 , SOLUTION:
9 To find the general solution of the recurrence relation:
solve
secondorder
w h e r e a0  4 a n d a l  7.
T h e c h a r a c t e r i s t i c e q u a t i o n of t h e r e c u r r e n c e r e l a t i o n is x 2  5x + 6  0; t h e c h a r a c t e r i s t i c roots a r e 2 a n d 3. T h e r e f o r e , by T h e o r e m 5.2, t h e g e n e r a l solution of t h e r e c u r r e n c e r e l a t i o n is an = A . 2 n + B . 3 n. (This solution is u s e d in E x a m p l e s 5.19 a n d 5.20.)
9 To find the values of A and B:
U s i n g t h e initial c o n d i t i o n s we find: a0 = A + B = 4 a l = 2 A + 3B = 7 Solving this l i n e a r s y s t e m yields A  5 a n d B   1 (Verify this.). T h u s t h e solution of t h e r e c u r r e n c e r e l a t i o n s a t i s f y i n g t h e given conditions is an  5 . 2 n  3 n, n >_ O. m T h e n e x t e x a m p l e finds a n explicit f o r m u l a for t h e n t h Fibonacci n u m b e r
F n , w h i c h we h a v e b e e n w a i t i n g for.
Solve t h e Fibonacci r e c u r r e n c e F1 = 1  F2.
relation Fn

Fn1
+ Fn2,
where
SOLUTION: T h e c h a r a c t e r i s t i c e q u a t i o n of t h e r e c u r r e n c e r e l a t i o n is x 2  x  1  O, a n d 1+~/5 1~/5 its solutions are a = 2 a n d fl  ~ . You m a y verify a + fl = 1 a n d aft   1. T h e g e n e r a l solution is F n  A a n + B f l n. To find A a n d B, we have:
F1  Aot + B fl 1 1
F2  A ot 2 }B fl 2 
290
Chapter 5 Recursi0n
Solving these two equations, we get (Verify)" Aa _ (1 + v/5)/2 (5 + vf5)/2 v~) 5 + 5~/5 v~25  5
v/~ (Verify this.).
1 + c~2
(1 + v ~ ) ( 5 (5 + v ~ ) ( 5  j 5 ) and similarly B fl 1 + ~2 1
_ 1 + j5 5 + ~/5 5 1 j~
T h u s t h e solution of t h e r e c u r r e n c e relation satisfying t h e given conditions is o,n _ _ ~n c~n _ _ ~n
an ~ 
which is the B i n e t f o r m for the n t h Fibonacci n u m b e r Fn. (See E x a m p l e 5.26 for a different method.) m The next example, proposed by Irving K a p l a n s k y of T h e U n i v e r s i t y of Chicago, also illustrates solving second order L H R R W C C s a n d is closely related to E x a m p l e 5.15. ~
{ 1, 2 , . . . n} t h a t do not Let an denote the n u m b e r of subsets of the set S contain consecutive integers, w h e r e n > _ 0. W h e n n  0, S  0 . Find an _ explicit f o r m u l a for a,,.
SOLUTION: To get an idea about a,,, let us find its value for n  0, 1, 2, 3, and 4 by c o n s t r u c t i n g a table, as in Table 5.3. It a p p e a r s from the table t h a t an is a Fibonacci n u m b e r and a,,  F,+2.
Table 5.3
S u b s e t s of S t h a t do n o t contain consecutive integers
an
D,
D,{I} D, {I}, {2} O, {i}, {2}, {3}, {1,3} O, {I}, {2}, {3}, {4}, {1,3}, {1,4}, {2,4}
Fn
+2
We shall, in fact, prove t h a t an  Fn+2 in two steps" F i r s t we shall define an recursively and t h e n solve the r e c u r r e n c e relation to obtain this explicit formula.
9 To define an recursively" F r o m Table 5.3, a0 = 1 and a l  2. So let n >_ 2. Let A be a subset of S t h a t does not contain two consecutive integers. T h e n e i t h e r n e A or nCA.
5.3
Solving Recurrence Relations Revisited
291
C a s e 1 Suppose n ~ A. T h e n n 1 r A. By definition, S* = {1, 2 , . . . , n  2} has a n  2 subsets not c o n t a i n i n g two consecutive integers. Add n to each of the subsets. T h e r e s u l t i n g sets are subsets of S satisfying t h e desired property, so S has a n  2 such subsets. C a s e 2 Suppose n r A. By definition, t h e r e are a n  1 such subsets of S h a v i n g the r e q u i r e d property. Since these two cases are m u t u a l l y exclusive, by t h e addition principle,
an  an1 zr a n  2 .
Thus
an
can be defined recursively as a0  1 , a l  2
an a n  1 + a n  2 ,
n>2.
9 To solve
the recurrence
relation"
This r e c u r r e n c e relation is exactly the s a m e as t h e Fibonacci one w i t h the initial conditions a0 = 1, a l = 2. So i n s t e a d of going t h r o u g h a complete solution, as in E x a m p l e 5.15, notice t h a t this definition yields the Fibonacci n u m b e r s 1, 2, 3, 5, 8, . . . . It follows t h a t an = F n + 2 , n > 0.
m
U s i n g the values of a and ~ from E x a m p l e 5.15,
an+2 _ fin+2
an
Fn+2

a~
,
n > 0 u
(Verify this. See Exercise 13.)
m
T h e o r e m 5.2 does not work if the c h a r a c t e r i s t i c roots a and fl are equal, t h a t is, if a is a root w i t h degree of multiplicity two. T h e following t h e o r e m , however, comes to our rescue. It shows t h a t , in addition to a n , n a n is a basic solution. Let a, b ~ I~ and b r 0. Let a be a real or complex solution of the e q u a t i o n
x 2 a x  b = 0 w i t h degree of multiplicity two. T h e n a n the general solution of the L H R R W C C a n  a a n _ 1 ~ b a n  2 . Aa n + Bnan
is
PROOF" Since a is a root of t h e e q u a t i o n x 2 city two,
x 2 axb =
ax 
b 
0
with degree of multipli
(xX 2 
a)2
2ax +
a 2
Therefore, a  2a and b
a 2
(5.12)
292
Chapter 5 Recursion
9 To show
that
an 
na n satisfies
the recurrence
relation:
Notice t h a t
aan1 + ban2 a[(n 
1)a n  l ] +
l)a nl]
b[(n

2)a n2]
= 2a[(n
~
(a2)[(n  2)a n2] by (5.12)
= an[2(n  1)  (n  2)]
 na n an
Therefore,
na n
is a solution of t h e r e c u r r e n c e relation.
T h e n a n  A a n + B n f l n is t h e general solution of t h e given r e c u r r e n c e relation, w h e r e A and B are selected in such a way t h a t t h e initial conditions are satisfied. (The values of A a n d B can be found u s i n g initial conditions, as in T h e o r e m 5.2.) n The next example i l l u s t r a t e s T h e o r e m 5.3. Solve the r e c u r r e n c e relation
an 6an 1 9an2,
w h e r e a0 = 2 a n d a 1
3.
SOLUTION: The characteristic e q u a t i o n of the r e c u r r e n c e relation is x 2  6x + 9  0; its solution is 3 with degree of multiplicity two. Therefore, by T h e o r e m 5.3, the general solution of the r e c u r r e n c e relation is a , , = A . 3 '2 + B 9 n 3 n . (We use this in E x a m p l e 5.21.) The initial conditions a0 = 2 and a l = 3 yield the e q u a t i o n s A.3 ~ +B. 0.3 ~  2 and A.3+B.1.33.
Solving these equations, we get A  2 and B   1 . (Verify). T h u s the solution of t h e r e c u r r e n c e relation satisfying t h e given conditions is a n  2 . 3 n  n . 3 n , n >_ O. m T h e o r e m s 5.2 and 5.3 can be combined to yield t h e following general result. ~ Let a be a c h a r a c t e r i s t i c root of t h e L H R R W C C (5.7). 9 If the degree of multiplicity of c~ is 1, t h e n a n is a basic solution of the LHRRWCC. 9 If t h e degree of multiplicity of a is m, t h e n a n, n a n , . . . , n m  l a n are basic solutions of t h e L H R R W C C . ( N o t e : A k t h  o r d e r L H R R W C C has k basic solutions.) 9 T h e general solution of the L H R R W C C is a linear c o m b i n a t i o n of all basic solutions, m The following e x a m p l e illustrates this general t h e o r e m .
5.3 Solving Recurrence Relations Revisited
293
Solve t h e r e c u r r e n c e r e l a t i o n an  7 a n  1  1 3 a n  2  3 a n  3 + 1 8 a n  4 , w h e r e a0  5 , a l  3 , a 2  6, a n d a3   2 1 . SOLUTION: T h e c h a r a c t e r i s t i c e q u a t i o n of t h e L H R R W C C i s x 4  7X 3 + 13X 2 +3X 18  0. Since x 4  7 x 3 + 13x 2 + 3x  18  (x + 1)(x  2)(x  3) 2, t h e c h a r a c t e r i s t i c r o o t s are:  1 a n d 2 w i t h d e g r e e of m u l t i p l i c i t y o n e e a c h and 3 w i t h d e g r e e of m u l t i p l i c i t y t w o
Since 3 is a r o o t w i t h d e g r e e of m u l t i p l i c i t y two, it y i e l d s t w o b a s i c s o l u t i o n s , 3 n a n d n 3 n. T h u s t h e g e n e r a l s o l u t i o n of t h e L H R R W C C is a l i n e a r c o m b i n a t i o n of t h e basic s o l u t i o n s (  1 ) n , 2 n , 3 n, a n d n3n; t h a t is, an = A (  1 ) n + B 2 n + C3 n ~ D n 3 n.
T o f i n d the v a l u e s o f A, B, C, a n d D: Since a0  5, a l  3, a2  6, a n d a3   2 1 , w e h a v e A+B+C
=5
 A + 2B + 3C + 3D = 3 A + 4B + 9C + 18D = 6 and A + 8B + 27C + 81D =  2 1
S o l v i n g t h i s l i n e a r s y s t e m , we g e t A  2  C, B  1, a n d D   1 (Verify this.). T h u s t h e s o l u t i o n of t h e L H R R W C C s a t i s f y i n g t h e i n i t i a l c o n d i t i o n s is an  2 (  1 ) n _+_ 2 n _+_2 . 3 n  n 3 n, n >_ O. m T h e t e c h n i q u e of s o l v i n g L H R R W C C s c a n n o t b e a p p l i e d to t h e s e e m i n g l y s i m p l e r e c u r r e n c e r e l a t i o n s fn  f n  1 + n ( E x a m p l e 5.5) a n d bn  2 b n  1 + 1 ( E x a m p l e 5.4), w h i c h a r e l i n e a r , b u t n o n h o m o g e n e o u s . So we n o w t u r n to solving l i n e a r n o n h o m o g e n e o u s r e c u r r e n c e r e l a t i o n s w i t h c o n s t a n t coefficients (LNHRRWCCs).
LNHRRWCCs
T h e g e n e r a l f o r m of a L N H R R W C C is
an  Clan1 + C2an2 + "'" + Ckank + f (n)
(5.13)
w h e r e cl, C 2 , . . . , Ck E R, Ck 7s O, a n d f ( n ) is n o t i d e n t i c a l l y zero. I t s s o l u t i o n d e p e n d s on t h a t of t h e a s s o c i a t e d l i n e a r h o m o g e n e o u s recurrence relation with constant coefficients (ALHRRWCCs)
an  Clan1 + C2an2 + "'" + Ckank
(5.14)
we s t u d i e d e a r l i e r .
294
Chapter 5 Recursion
Solving LNHRRWCCs
To solve the LNHRRWCCs (5.13), let a(n denote the general solution of the h) ALHRRWCCs (5.14). Suppose we know some solution a(np) of t h e r e c u r r e n c e relation (5.13); a(n is a p a r t i c u l a r s o l u t i o n of the L N H R R W C C s (5.13). p) T h e n the g e n e r a l s o l u t i o n of (5.13) is given by
ana (nh) + a (n p)
This fact is confirmed by the following theorem; we leave its proof as an exercise (see Exercise 44). Let a(nh) denote the general solution of the ALHRRWCCs (5.14) and a(n a p) particular solution of the L N H R R W C C (5.13). T h e n an  a(nh) + a(np) is the general solution of the L N H R R W C C s (5.13). I It follows from this t h e o r e m t h a t solving the L N H R R W C C s (5.13) depends on finding a particular solution a~n . Although no general algop r i t h m exists for solving an a r b i t r a r y LNHRRWCCs, two special cases can be handled fairly easily. When f ( n ) is a polynomial in n or is of the form Cot n, a particular solution can be extracted with ease, as the next two examples demonstrate, where C and a are constants. The techniques we employ are similar to those used to solve linear nonhomogeneous differential equations. Solve the LNHRRWCCs an al7.
5an1  6a~_2 + 8n 2, where a0 
4 and
SOLUTION: It follows from Example 5.14 t h a t the general solution of the ALHRRWCCs an  5 a n  1  6 a n  2 is given by a(n = A . 2 n + B . 3 n. Since f ( n )  8n 2 is h/ a quadratic polynomial in n, it seems reasonable to look for a particular solution of the same form, say, an  a n 2 + b n + c . T h e n the given recurrence relation yields
an 2 +bn +c =
5 1 a ( n  1) 2 + b ( n (8a)n 2 +(14ab)n
1)+c]61a(n2)
2 +b(n2)+c]+8n
2
19a+ 7bc
E q u a t i n g the coefficients of like terms, we get the linear systema8a b = 14ab
c = 19a + 7bc Solving the system, we get a  4, b  28, and c  60 (Verify). We now claim t h a t t(P)   4n 2 + 28n + 60 is a particular solution (Verify). tn
5.3
Solving Recurrence Relations Revisited
295
T h u s , by T h e o r e m 5.5, t h e g e n e r a l s o l u t i o n of t h e given r e c u r r e n c e r e l a t i o n is
ana (nh) + a (n p)
= A . 2n + B .
3 n + 4n 2 + 28n + 60
U s i n g t h e two given initial conditions, t h i s yields t h e l i n e a r s y s t e m :
A +B 
56
2A + 3B   8 5 T h i s yields A   8 3 a n d B  27 (Verify t h i s also.). T h u s t h e d e s i r e d solution is
an 
(  8 3 ) . 2 n + 2 7 . 3 n ~ 4n 2 A 28n + 60,
n > 0
m
f(n)
T h e n e x t e x a m p l e i l l u s t r a t e s h o w to solve t h e L N H R R W C C s (5.13) w h e n is of t h e f o r m C a n, w h e r e C a n d a a r e c o n s t a n t s .
5an1 6an2 +
Solve t h e L N H R R W C C s a n al=7.
3 . 5 n , w h e r e a0 
4 and
SOLUTION: As in E x a m p l e 5.19, t h e g e n e r a l solution of t h e A L H R R W C C s a n  5 a n  1 6 a n  2 is given by a(nh)  A . 2 n 4 B . 3 n. Since f ( n )  3 . 5 n, we s e a r c h for a p a r t i c u l a r solution of t h e f o r m a n = c 9 5 n. T h e n we m u s t h a v e c. 5 n  5(c. 5 n  l )  6(c. 5 n2) 4 3 . 5 n
Canceling 5 n2 f r o m b o t h sides, t h e r e s u l t i n g e q u a t i o n yields c  25/2. We n o w claim t h a t a n  (25/2)5 n is a p a r t i c u l a r solution of t h e r e c u r r e n c e r e l a t i o n (Verify this.). T h u s t h e g e n e r a l solution of t h e L N H R R W C C s is
a n  A . 2 n + B . 3 n +
(25/2)5 n
U s i n g t h e initial conditions, we get t h e l i n e a r s y s t e m : A + B  17/2 2A + 3B   1 1 1 / 2 Solving this s y s t e m , we get A  30 a n d B   7 7 / 2 (Verify this.). T h u s t h e solutions of t h e given r e c u r r e n c e r e l a t i o n a r e given by
an 
(30). 2 n  (77/2). 3 n + (25/2). 5 n,
n > 0 m
(Verify this also.)
296
Chapter 5 Recursion
A n i m p o r t a n t observation: In t h i s example, notice t h a t t h e 5 in f ( n ) is not a c h a r a c t e r i s t i c r o o t of t h e A L H R R W C C s . If it were, we w o u l d h a v e n e e d e d to m a k e a d j u s t m e n t s in o u r search for a p a r t i c u l a r solution, as in T h e o r e m 5.3. We shall p u r s u e this case shortly.
T h e following t h e o r e m justifies t h e t e c h n i q u e s d e m o n s t r a t e d in t h e s e two examples; we o m i t its p r o o f in t h e i n t e r e s t of brevity. In t h e L N H R R W C C s (5.13), s u p p o s e f (n) = (bk nk + b k _ l n k1 + . . . + b i n + bo)(~ n. If a is not a c h a r a c t e r i s t i c root of t h e A L H R R W C C s (5.14), t h e n a p a r t i c u l a r solution is of t h e f o r m (dk nk + d k _ l n k1 + ... + d i n 4 do)or n. If a is a c h a r a c t e r i s t i c root w i t h m u l t i p l i c i t y m, t h e n a p a r t i c u l a r s o l u t i o n is of t h e f o r m n m ( e k n k + ek_l n k  1 + . . . + e l n + eo)ot n. m We conclude this section w i t h t h e following example, w h i c h i l l u s t r a t e s this t h e o r e m w h e n a is a c h a r a c t e r i s t i c root of t h e A L H R R W C C s . Solve t h e L N H R R W C C s an  6an1  9an2 + 4(n + 1)3 n, w h e r e a0 = 2 a n d a l  3. SOLUTION: F r o m E x a m p l e 5.17, t h e general solution of t h e A L H R R W C C s is a ~ ) = A. 3 ~ + B. n3 n, w h e r e n >_ 0. Since 3 is a c h a r a c t e r i s t i c root w i t h m u l t i p l i c i t y 2, we search for a p a r t i c u l a r solution of the form n2(cn + d)3 n, w h e r e t h e c o n s t a n t s c a n d d are to be d e t e r m i n e d . T h e n we m u s t have
n2(cn + d ) 3 n  6{(n  1)21c(n  1) + d l 3 nl}
 9{(n  2)2[c(n  2) + dl3 n2 } + 4(n + 1)3 'z E q u a t i n g the coefficients of like t e r m s , this yields c  2/3 a n d d  4 (Verify); ( SO ~np ) _ 2n2(n + 6)3 n1
Thus the general soution of the recurrence relation is
an  A . 3 n + B . n3 n + 2n2(n + 6)3 n  l , U s i n g the initial conditions, this yields
an  (6  19n). 3 n1 + 2n2(n + 6)3 n1 ,
n>_0 i
n _> 0
(You can confirm this.)
E x e r c i s e s 5.3
D e t e r m i n e if each r e c u r r e n c e r e l a t i o n is a L H R R W C C .
1. Ln  Ln1 + L n  2 3. an  1 . 0 8 a n _ l 2. Dn  n D n  1 + (  1 ) n 4. bn = 2bn1 + 1
5.3
Solving Recurrence Relations Revisited
297
5. a n  a n  1 + n 7. a n  a n  1 + 2 a n  2 + 3 a n  5 Solve each LHRRWCC. 9. a n  a n  1 + 2 a n  2 , a o 10. 11.
an 5an1  3 , a l = 0 7
6. an  2 a n  1 + (2 n  1)
8. a n  a n  1 + 2 a n  3 + n 2
 6 a n  2 , a o = 4, a l 
an = an1 + 6an_2,ao = 2,al
= 5,al
= 8
= 0
12. an = 4an2,ao 13. an an1
+ an2,ao
= 1 , a l  2
1 4 . a n = a n  1 + a n  2 , ao  2, a l = 3 1 5 . Ln = L n  1 + L n  2 , L1 = 1, L 2 = 3 16.
an = 4an1
1
 4an_2,ao

= 3,al
= 10
~
17. an = 6an
9 a n  2 , a o = 2, a l
3
18. an = 3an1 + 4an2  12an_3,ao 19.
a n  8 a n  1  21an2 + 18an_3,ao
 3 , a l
 0 , a l
= 7,a2
= 7
 2 , a 2 = 13  5 , a 2  19 = 0,al = 16,a2  2,
20. a n  7 a n  1  1 6 a n  2 + 1 2 a n _ 3 , a o
21.
an = an1 + 16an2 + 4an3
 0 , a l
48an_4,ao
a3 = 1 4 2
22. a n = 1 3 a n  2  3 6 a n  4 , a0 = 7, a l = 6, a2 = 38, a3    8 4 23. a n = 9 a n  1  3 0 a n  2 + 4 4 a n _ 3  2 4 a n  4 , a o
a3 = 1 2 6 24.
an = 8an1 
= 5,al
=
12,a2
=
38,
24an2 + 32an3
 1 6 a n  4 , a o
=
1,al
=
4,a2
=
44,
a3 = 2 7 2 Find the general form of a particular corresponding to each function f(n). solution of the LNHRRWCCs (5.13)
25. f ( n ) = 28. f ( n ) =
Find the
n
26. f(n)= 29. f(n)form
1
n2 n
27. f(n)= 30. f(n)
3n 2 43n25 n
3n general
of a particular
a n  4 a n  1
 4an2 + f (n) c o r r e s p o n d i n g
solution of the LNHRRWCCs to each function f (n).
31. f (n) = 3  2 n
32. f ( n ) = n 2 n
34. (17n 31)2 n
33. f ( n )  2 3 n 2 2 n
Solve each LNHRRWCCs.
35. a n  2 a n  1
+ 1, a 0 
1
36. a n  7 a n  1  1 0 a n _ 2 + n 2, ao = 0, a l = 1
298
Chapter 5 Recursion
37. a n  7 a n  1  12an2 + 3n,a0  0 , a l = 2 38. a n  7 a n  1  12an2 + 3 n 4 n , a o
*39. *40.
an 
= 0,al  2
a n  1 ~ n , ao a n  1 + n


1
an

1,al  0
41. Let rn and Sn be two solutions of the r e c u r r e n c e relation (5.8). P r o v e t h a t a n   r n + S n is also a solution. 42. Let a be a solution of the e q u a t i o n x k  C l x k  1 t h a t c~n is a solution of L H R R W C C (5.7).
..... Ck
 0. S h o w
43. Let a be a characteristic root of the L H R R W C C a n = a a n _ 1 + b a n  2 +C a n  3 with degree of multiplicity three. Show t h a t a n, n(~ n, n 2 a n are solutions of LHRRWCC. 44. Let a(n denote the general solution of the A L H R R W C C s (5.14) a n d h) a particular solution of the L N H R R W C C s (5.13). P r o v e t h a t a n  a(nh) + a(np) is the general solution of the L N H R R W C C s (5.13).
a(np)
G e n e r a t i n g functions provide a powerful tool for solving L H R R W C C s , as will be seen shortly. They were invented in 1718 by the F r e n c h m a t h e matician A b r a h a m De Moivre, w h e n he used t h e m to solve the Fibonacci recurrence relation (see Example 5.26). G e n e r a t i n g functions can also solve combinatorial problems, as the next c h a p t e r shows. To begin with, notice t h a t the polynomial 1 + x + x ~ + x a + X 4 ~  X 5 can x6 1 be w r i t t e n as ~ . You may verify this by either crossmultiplication
x1
the familiar long division method, or Exercise 8 in Section 4.4. AccordX6 1
ingly, f ( x ) 
~ i s
x1
called the g e n e r a t i n g
f u n c t i o n of the sequence of
coefficients 1, 1, 1, 1, 1, 1 in the polynomial. More generally, we m a k e the following definition.
Generating Function
Let a0, al, a 2 , . . , be a sequence of real n u m b e r s . T h e n the function
g ( x )  ao + a I x + a2 x2 + . . . + a n x n + . . .
(5.15)
is the g e n e r a t i n g f u n c t i o n for the sequence { a n }. G e n e r a t i n g functions for the finite sequence a0, a l , . . . , a n can also be defined by letting ai = 0 for i > n; t h u s g ( x )  ao + a l x + a2 x 2 + . . . + a n x n is the g e n e r a t i n g function for the finite sequence a0, a l , . . . , a n .
5.4 GeneratingFunctions
299
A b r a h a m De M o i v r e (16671754), son of a surgeon, was born in VitryleFrancois, France. His formal education began at the Catholic village school, a n d then continued at the Protestant Academy at Sedan a n d later at S a u m u r . : , . , ,,' ..$ He did not receive good training in mathematics until he moved to Paris in 1684, where he studied Euclid's later books and other texts. A r o u n d 1686, De Moivre emigrated to England, where he began his lifelong profession, tutoring in mathematics, and mastered Newton's Principia Mathematica. In 1695 he presented a paper, his first, on Newton's theory of fluxions to the Royal Society of London a n d 2 years later he was elected a member of the Society. Unfortunately, despite his influential friends, he could not find an academic position. He h a d to earn a living as a tutor, author, a n d expert on applications of probability to g a m b l i n g a n d annuities. He dedicated his first book, a masterpiece, The Doctrine of Chances, to Newton. His most notable discovery concerns probability theory: The binomial probability distribution can
.' . .'.,~.'
be approximated by the normal distribution.
De Moivre died in London.
For example,
g ( x )  1 + 2x + 3x 2 + . . . + (n + 1)x n + . . .
is the g e n e r a t i n g function for the sequence of positive integers and
f ( x )  1 + 3x + 6x 2 + . . . + n ( n + 1)x2 + . . .
2
is the g e n e r a t i n g function for the sequence of t r i a n g u l a r n u m b e r s . Since
xn  1 x1 x n  1 g(x) = 1 + x +x2 ~... + x n1
x1
is the g e n e r a t i n g function for the sequence of n ones.
A w o r d o f caution" The RHS of E q u a t i o n (5.15) is a f o r m a l p o w e r
s e r i e s in x. The letter x does not r e p r e s e n t anything. T h e various powers
x n of x are simply used to keep t r a c k of the c o r r e s p o n d i n g t e r m s an of the sequence. In o t h e r words, t h i n k of the powers x n as placeholders.
Consequently, unlike in calculus, the convergence of the series is of no i n t e r e s t to us.
,, ,, , . . . . .
Equality of Generating Functions
Two g e n e r a t i n g functions f (x) an  bn for every n > _0. _ ~ a n x n and g ( x ) = ~ bnx n are e q u a l if
n =0
n =0
300
Chapter 5 Recurslon F o r example, let f ( x ) = 1 + 3x +
6X 2 q 10x 3 +  . .
and
2.3 3.4x2 4 . 5 x3 g(x)  1 +   ~ x +  ~ + ~ +....
T h e n f (x) = g(x).
A g e n e r a t i n g function we will use f r e q u e n t l y is
= 1 + ax + a 2 x 2 n . . . t[a n x n Jr " "
(5.16) (5.17)
Xax
Then
1 1  x = 1 +x +x 2 +...fX
n
Jr'''
Can we add a n d m u l t i p l y g e n e r a t i n g functions? Yes! Such o p e r a t i o n s a r e p e r f o r m e d exactly t h e s a m e way as polynomials are combined.
Addition and Multiplication of Generating Functions
CO CO
Let f (x) = ~ a n X n a n d g(x) n=0
~
n=0
bnx n
be two g e n e r a t i n g functions. T h e n
f (x) + g(x)  ~
n=0
(an
+ bn)x n
a n d f (x)g(x)  ~
n=0
~
i =0
aibn_
i
)
Xn
For example, 1 (1
 x) 2

1 1 x
Xi
1 x
Xi
)
n=O
1.1
)
xn
CX~
= E(n
n=O
+ 1)x n
= 1 + 2 x + 3x 2 + . . . + (n + 1)x n + . . . and 1 (lx) 3

(5.18)
1 1x
Xn
1 (lx) 2
(n
Ln =0
+ 1)X n
1.(n + 1i)
n=0
(X)
xn
= ~[(n
n0
+ 1) + n + . . . + 1 ] x n
5.4 Generating Functions
(x)
301
= E
n0
(n + 1)(n + 2)xn 2 (5.19)
= 1 + 3x + 6x 2 + 10x 3 + . . .
Before exploring how valuable g e n e r a t i n g functions are in solving LHRRWCCs, we illustrate how t h e t e c h n i q u e of p a r t i a l f r a c t i o n d e e o m p o s i t i o n , used in integral calculus, enables us to express t h e q u o t i e n t
p(x) of two polynomials p(x) a n d q(x) as a s u m of p r o p e r fractions, w h e r e q(x)
degp(x) < deg q(x). t For example, 6X+1 ( 2 X  1)(2X + 3) 1 2X 1 2 2X + 3
P a r t i a l F r a c t i o n Decomposition Rule for p ( x )
q(x)'
w h e r e d e g p ( x ) < deg q(x)
If q(x) has a factor of the form (ax + b) m, t h e n the decomposition contains a sum of the form
A1
ax + b
t
A2
(ax + b) 2
+'"+
Am (ax + b) m
~
w h e r e Ai is a r a t i o n a l n u m b e r . Examples 5.225.24 illustrate the partial fraction decomposition technique. We use t h e i r results to solve the r e c u r r e n c e relations in E x a m p l e s 5.255.27. x a Express (1  x)(1  2x) as s u m of partial fractions. SOLUTION: Since the d e n o m i n a t o r contains two linear factors, we let x (1  x)(1  2x) = A 1 x + B 1  2x
To find the c o n s t a n t s A and B, multiply b o t h sides by (1  x)(1  2x)"
x  A ( 1  2x) + B ( 1  x)
Now give c o n v e n i e n t values to x. S e t t i n g x  i yields A   1 and s e t t i n g x  1/2 yields B  1. (The values of A and B can also be found by e q u a t i n g tdeg f(x) denotes the degree of the polynomial f(x).
302
Chapter 5 Recursion
coefficients of like terms from either side of the equation resulting linear system.) x (1  x ) ( 1  2x) 1 1  x 1 1  2x
and solving the

..J_
(You m a y verify this by c o m b i n i n g t h e s u m fraction.) W e u s e t h i s r e s u l t in E x a m p l e 5.25.
x
on the RHS
into a single II
Express
1  x

X2 a s a s u m o f p a r t i a l f r a c t i o n s .
SOLUTION:
F i r s t , f a c t o r 1  x  x 2" 1  x  x 2 = (1  a x ) ( 1  f i x )
1F w/5 a n d fl w h e r e c~ = a  fl  v ~ . ) 2
~
1J5
.
( N o t i c e t h a t a + fl 
1, aft 
1, and
Let
x = 1  x  x 2 1 ax A t 1 fix B
Then
x A(1 fix) + B( 1 otx)
Equating
c o e f f i c i e n t s o f l i k e t e r m s , w e get"
A + B  O flA otB 1
Solving this linear system yields A Thus
1
,/g
= B
(Verify this.).
(1
x
x
2)
v/5
111
1]
1  fix
1c~x
W e u s e t h i s r e s u l t i n E x a m p l e 5.26. Express 2  9x 1  6x + 9x 2 as a sum of partial fractions.
m
SOLUTION:
Again, factor the denominator:
l  6X + 9X 2  (l  3X) 2
5.4
Generating Functions
303
By t h e d e c o m p o s i t i o n r u l e , let 2  9x 1  6x + 9x 2 Then 2  9x  A(1  3x) + B T h i s y i e l d s A  3 a n d B   1 (Verify this.). Thus 29x 1 6x + 9x 2 3 1  3x 1 (1  3x) 2
I
A = 1  3x t
B (1  3x) 2
W e u s e t h i s r e s u l t in E x a m p l e 5.27.
N o w we a r e r e a d y to u s e p a r t i a l f r a c t i o n d e c o m p o s i t i o n s a n d g e n e r a t i n g f u n c t i o n s to solve r e c u r r e n c e r e l a t i o n s in t h e n e x t t h r e e e x a m p l e s . U s e g e n e r a t i n g f u n c t i o n s to solve t h e r e c u r r e n c e r e l a t i o n bn  2 b n  1 + 1, w h e r e b l  1. SOLUTION: F i r s t , n o t i c e t h a t t h e c o n d i t i o n bl  i y i e l d s b0 = 0. To find t h e s e q u e n c e {bn} t h a t satisfies t h e r e c u r r e n c e r e l a t i o n , c o n s i d e r t h e c o r r e s p o n d i n g generating function
g ( x )  bo + b l x + b2 x2 + b3 x 3 + . . . + bnx n q...
Then
2xg(x) =
2 b l x 2 + 2b2x 3 + . 99+ 2bn_
1 xnJr"
99
Also, =l+x+
X 2 + X 3Jr" 9 9 ~X nJr" 9 9
1x Then
g(x)  2xg(x)
1x
= 1+
(bl  1)x + (b2  2bl  1)x 2 + . . .
Jr ' ' '
+ (bn  2 b n  1  1)x n
=1 since bl  I a n d bn  2 b n  1 + 1 for n >_ 2. T h a t is,
1
(1  2 x ) g ( x ) 
1 x
1
x
1 x
Then
g(x) =
(1  x ) ( 1  ~ c )
304
Chapter 5 Recursion 1 1x 1 12x'
+
=
§
by E x a m p l e 5.22
, by (5.16)
= 
xn
2nx n
: E(2 nn=O
OG
1)X n
But g(x) 
~
n:0
b n x n, so bn 
2n 
1, n > 1. (Notice t h a t t h i s is t h e s a m e II
solution o b t a i n e d in E x a m p l e 5.12.)
U s i n g g e n e r a t i n g functions, solve t h e Fibonacci r e c u r r e n c e r e l a t i o n F n F n  1 + F n  2 , w h e r e F1  1  F2.
SOLUTION: Notice t h a t t h e two initial c o n d i t i o n s yield F0  0. Let
g ( x )  Fo + F i x + F 2 x 2 4  . . . 4 F n x n 4  ' "
be t h e g e n e r a t i n g f u n c t i o n of t h e Fibonacci sequence. Since t h e o r d e r s of
F n  1 a n d F n  2 a r e 1 a n d 2 less t h a n t h e o r d e r of F n , respectively, we find xg(x) and x2g(x)" x g ( x )  F i x 2 4 F 2 x 3 4 F3 x4 4  . . . 4 F n  1 xn 4  ' " x2 g ( x )  F i x 3 4 F 2 x 4 4 F3 x5 4  . . . 4 F n _ 2 xn 4  . . .
Then
g ( x )  x g ( x )  x 2 g ( x )  F i x 4 (F2  F1)x 2 4 (F3  F2  F1)x 3 + ... + (Fn  F , _ 1  F n  2) xn 4 "'"
=X
since F2  F1 a n d Fn = F n  1 + F n  2 . T h a t is, (1  x  x 2 ) g ( x )  x
g(x) = 1 XX
2
111
14
2
1ax
1  fix
1]
' by E x a m p l e 5.23
where a =
1+V/5 2
a n d fl =
5.4 Generating Functions
305
Then
~ / 5 g ( x ) 1
1  otX
1
1  fiX
oo oo  E o l n x n   E ~ n x n  
(DO E(oln
 ~ n ) x n
n0 So
o4)
g(x) ~
n0
n=0
(Oln  ~ n ) x n
n=0
x/~
Therefore, by the e q u a l i t y of g e n e r a t i n g functions,
oln _ fin Fn,/g
otn _ fin
~  ~
(Recall t h a t this is t h e B i n e t f o r m of F n . ) We close this section with the following example. ~ U s i n g g e n e r a t i n g functions, solve the r e c u r r e n c e r e l a t i o n an 9 a n  2 , w h e r e ao  2 a n d a l  3. SOLUTION: Let
g ( x )  ao + a l x + a2 x2 + . . . + a n X n + . . . 6an1
m

Then
6 x g ( x ) = 6 a o x + 6 a l x 2 + 6a2x 3 + ... + 6 a n _ i X n + . . . 9x2g(x) 
9a0x 2 + 9 a l x 3 + 9a2x 4 + . . . + 9 a n _ 2 x n + . . .
Then
g ( x )  6 x g ( x ) + 9 x 2 g ( x )  ao + ( a l  6a0)x + (a2  6al + 9a0)x 2 + . . . + (an  6 a n  1 + 9 a n  2 ) x n + ' "
=29x u s i n g the given conditions. T h u s (1  6x + 9 x 2 ) g ( x ) = 2  9 x Therefore,
g(x) =
29x 1  6 x + 9x 2
306
Chapter 5 Recursion 3 1 = 1  3X  (1  3x) 2' by E x a m p l e 5.24
" 3
3nx n
nO
(n + 1)3nx n
(x:)
 E [ 3 n+l   ( n + 1 ) 3 n ] x n
n0
(x)
 E
n0
3n(2  n ) x n
Thus
an 
(2  n ) 3 n
n > 0
m
T h e following e x e r c i s e s p r o v i d e a m p l e p r a c t i c e in t h i s p r o b l e m  s o l v i n g technique.
Exercises 5.4 E x p r e s s e a c h q u o t i e n t as a s u m of p a r t i a l f r a c t i o n s . 1. x + 7 ( x  1)(x + 3) 5 l_x_6x 2 2.
4X 2  
3x
25
(x + 1 ) ( x  2)(x + 3) 2+4x l+8x+15x
2X 2 
3. 5.
4. 6.
2
x ( x + 2) (2 + 3x)(x 2 + 1)
X 3 ~  X 2 + X ~
2x + 2
( x  1)(x 2 + 2x)
X 3 + 2X 2 ~  X
7.
3
x 4 + 5x 2 + 6 3x 3  x 2 + 4x
X 4   X 3 ~ 2X 2   X ~ 1
8. x 4 + x 3 + x + 1 x 3 + x 2 + 5x  2 *10.
X4 X 2 +X1
9.
U s i n g g e n e r a t i n g f u n c t i o n s , solve e a c h L H R R W C C .
11. an = 2an_l, ao = 1
an = an1 +
12.
1,al = 1
= 1
13. a n = a n  l + 2 , a l
14. a n = a n  1 + 2 a n  2 , a o = 3, a l = 0 15. a n = 4 a n  2 , a o = 2, a l =  8 16.
an = an1
+ 6 a n  2 , a o = 5, a l = 0
5.5 RecursiveAlgorithms
307
17. a n = 5 a n  z 18. a n  a n  1
6an2,ao + an2,ao
= 4,a1 = 7 = 1,al = 2 = 2 , a l  3 = 1, L2 = 3
19. a n = a n  1 + a n  2 , a o 20. L n  L n  1 + L n  2 , L 1
21. 22.
an
=
4an1 6an1
 4 a n  2 , a0 = 3, a l = 10  9an2,ao
an

 2,a1  3
23. a n  3 a n  z + 4 a n  2  1 2 a n  3 , a0 = 3, a l =  7 , a2 = 7 24. a n  8 a n  1  2 1 a n  2 + 1 8 a n _ 3 , a o = 0, a l = 2, a2 = 13
2 5 . a n = 7 a n  1  1 6 a n  2 + 1 2 a n  3 , a0 = 0, a l = 5, a2 = 19 26. a n = 3 a n  1 + 4 a n  2  12an_3, a0 = 3, a l =  7 , a2 = 7 27. a n = 6 a n  1  1 2 a n  2 + 8 a n _ 3 , a o = 0,al = 2,a2 = 2 a2 = 38, a3 =  8 4 = 0, a l =  8 , a 2 = 4,a3 = 42
28.
a n  1 3 a n  2  3 6 a n  4 , a0 = 7, a l     6 ,
29. a n =  a n  z + 3 a n  2 + 5 a n  3 + 2 a n  4 , a o
f(n)
Recall t h a t t h e r e c u r s i v e d e f i n i t i o n o f t h e f a c t o r i a l f u n c t i o n f e x p r e s s e s in t e r m s of i t s e l f w i t h a s m a l l e r a r g u m e n t n  1. A c c o r d i n g l y , it c a n b e e m p l o y e d to w r i t e a s i m p l e a l g o r i t h m to c o m p u t e n! T h i s a l g o r i t h m h a s t h e i n t e r e s t i n g p r o p e r t y t h a t it i n v o k e s i t s e l f w i t h a s m a l l e r a r g u m e n t . S u c h a n a l g o r i t h m is a r e c u r s i v e a l g o r i t h m .
Recursive Algorithm
A n a l g o r i t h m is r e c u r s i v e if it i n v o k e s i t s e l f w i t h a s m a l l e r a r g u m e n t ; t h a t is, if it i n v o k e s a r e d u c e d v e r s i o n of itself. (See F i g u r e 5.1.) R e c u r s i v e d e f i n i t i o n s i n v a r i a b l y l e a d to r e c u r s i v e a l g o r i t h m s . T h i s sect i o n t r a n s l a t e s s o m e o f t h e e x a m p l e s d i s c u s s e d in S e c t i o n 5.1 i n t o r e c u r s i v e a l g o r i t h m s a n d p r e s e n t s a few n e w o n e s   g c d , b i n a r y s e a r c h , a n d m e r g e sort. ~ W r i t e a r e c u r s i v e a l g o r i t h m to c o m p u t e n!, w h e r e n >__0. SOLUTION: W h e n n = 0, t h e a l g o r i t h m m u s t t e r m i n a t e a n d y i e l d t h e v a l u e 1. W h e n n > 0, t h e r e c u r r e n c e r e l a t i o n f ( n ) = n . f ( n  1) m u s t be applied: t h e a l g o r i t h m m u s t i n v o k e i t s e l f w i t h n  1 as t h e n e w a r g u m e n t . T h e r e c u r s i v e a l g o r i t h m is g i v e n in A l g o r i t h m 5.1.
308
Chapter 5 Reeurslon
Algorithm factorial (n)
(* This algorithm computes n! using recursion *) O. Begin (* algorithm *) i. i f n = 0 then (* base case *) 2. factorial ~ I 3. else (* invoke the algorithm *) 4. factorial +  n 9 f a c t o r i a l ( n  1) 5. End (* algorithm *)
Algorithm 5.1 l
F i g u r e 5.15 shows t h e r e s u l t of invoking the factorial a l g o r i t h m w i t h n  3, w h e r e f m e a n s f a c t o r i a l .
Figure 5.15
~f(l) 9 / f recursive I y, gcd{x,y}  gcd{x  y , y } . (See Exercise 34 in Section 4.2.) We use this fact to write Algorithm 5.5.
AI gorithm gcd (x,y)
(* This algorithm computes the gcd of two p o s i t i v e integers x and y using recursion. *) O. Begin (* algorithm *) I. i f x > y then 2. gcd ~ gcd{x  y , y } 3. else i f x < y then 4. gcd 3. B a s i s s t e p Since the induction step below uses the r e c u r r e n c e relation Fk+i  Fk + Fh1, the basis step involves verifying t h a t both P(3) and P(4) are true.
9 To s h o w t h a t P ( 3 ) is true: W h e n n  3,
1+~/~
ot n  2   o~  
2 So P(3) is true.
9 T o s h o w t h a t P ( 4 ) is true:
<
1+3 2
= 2

F3
3+~/5
<
3+3
2
3F4
T h u s P(4) is also true.
5.7 Complexities of Recursive Algorithms (optional)
323
Induction
step
~ i  2 < Fi f o r 3 (~k1 < Fk+l"
A s s u m e P(3), P ( 4 ) , . . . , P(k) are true; t h a t is, a s s u m e < i _< k. We m u s t show t h a t P ( k + l ) is true; t h a t is,
We have
M u l t i p l y i n g both sides by
c~k  1
c~k  3 ,
  of k  2 + o~k  3
(Note: k  3 > 2. ) by t h e inductive hypothesis by t h e r e c u r r e n c e relation
< Fk + F k  1 ,
=
Fk+ I
,
T h u s P(k + 1) is true.
is,
Therefore, by the s t r o n g version of induction, P(n) is t r u e for n > 3; t h a t Of n2 < F n for every n > 3. m
Now we can e s t i m a t e the n u m b e r of divisions r e q u i r e d by the euclidean a l g o r i t h m to c o m p u t e gcd{a, b }. ( L a m ~ ' s T h e o r e m ) The n u m b e r of divisions needed to c o m p u t e g{a, b} by the euclidean a l g o r i t h m is no more t h a n five t i m e s the n u m b e r of decimal digits in b, w h e r e a > b > 2.
PROOF: Let Fn denote the n t h Fibonacci n u m b e r , a  r0, and b  rl. By the r e p e a t e d application of the division a l g o r i t h m we have"
ro  r l q l + r2 rl  r2q2 + r3 0 < r2 < rl _ 0 < r3 < r2 _
r n  2  r n  l q n  1 r n  1  rnqn
+ rn
0 _ 1 for 1 _< i _< n. In particular, since rn < r n  1 , qn >_ 2; SO rn >_ 1 a n d rn1 > 2  F3. Consequently, we have:
r n  2  r n  l q n  1 > _r n  1 + rn _ t rn
>F3+l
= F3 + F2  F4
324
Chapter 5 Recursion
r n  3  r n  2 q n  2
+ rn1
>__ r n  2 + r n  1
>_ F4 + F 3  F 5 C o n t i n u i n g like this,
r l  r 2 q 2 + r3 > r2 + r3
>_ Fn + Fn1  Fn+l
T h a t is,
b >_Fn+l
B y E x a m p l e 5.41, Fn+ 1 > c~n  l , w h e r e a =
1 + ~/5. Therefore, 2
b Then
>C~ n1
log b > (n  1) log Since a 1+v/5 1 1.618033989, log c~ ~ 0.2089876403 > ~. So log b > n1 5
Suppose b contains k decimal digits. T h e n b < 10 h. Therefore, log b < k n1 and hence k > ~ . T h u s n < 5k + 1 or n < 5k. T h a t is, the n u m b e r of 5 divisions needed by the a l g o r i t h m is no more t h a n five times the n u m b e r of decimal digits in n. n n1 Let us p u r s u e this example a bit further. Since log b > ~ , n 1 + 5 log b. Also, since b > 2, 5 log b >_ 5 log 2 >1 Thus n < 1 + 51ogb < 51ogb + 51ogb = 10 logb = O(log b) T h u s it takes O(log b) divisions to c o m p u t e gcd{a,b} by the euclidean algorithm. <
5.7 Complexities of Recursive Algorithms (optional)
325
Gabriel L a m d (17951870) was born in Tours, France. After graduating from the Ecole Polytechnique in 1817, he continued his studies at the E,cole des Mines, from which he graduated in 1820. The same year Lamd was appointed director of the School of Highways and Transportation in St. Petersburg, Russia. There he taught mathematics, physics, and chemistry and planned roads and bridges in and around the city. In 1832, he returned to Paris to form an engineering firm. Within a few months, however, he left it to become the chair of physics at the Ecole Polytechnique, where he remained until 1844. While teaching, he served as a consulting engineer, becoming the chief engineer of mines in 1836. He helped build the railroads from Paris to Versailles and to St. Germain. In 1844, Lamd became graduate examiner for the University of Paris in mathematical physics and probability, and professor 7 years later. In 1862, he became deaf and resigned his positions. He died in Paris. Although Lam4 did original work in number theory and mathematical physics, his greatest contribution was the development of the curvilinear coordinates and their applications. His work on the curvilinear system led him to number theory. In 1840, he proved Fermat's Last Theorem for n = 7. Gauss considered Lamd the foremost French mathematician of his time. French mathematicians, however, considered him too practical, and French scientists, too theoretical.
The next example, due to S. H. Friedberg, explores the n u m b e r of multiplications needed to compute the d e t e r m i n a n t of an n x n m a t r i x by cofactor expansion. (It may be omitted by those not familiar with d e t e r m i n a n t s and calculus.) (optional) Let fn denote the n u m b e r of multiplications needed to compute detA, the d e t e r m i n a n t of an a r b i t r a r y n x n m a t r i x A = (aij) by cofactor expansion. E s t i m a t e fn. SOLUTION: We estimate fn in three steps: 9 Define fn recursively. 9 Solve the recurrence relation. 9 Use the solution to estimate fn.
9 To define fn recursively: Let Cij denote the (n  1) x (n  1) d e t e r m i n a n t obtained from d e t A by deleting its ith row a n d j t h column. By expanding detA with respect to the first row, we have
n
detA
E(1)J+ZaljClj j=l
cofactor expansion by row 1
326
Chapter 5 Recursion
In particular let
f2
: [a
Clearly, two multiplications are needed to evaluate d e t A and hence  2. Also fl = 0. Suppose n _> 3. Then, by definition, it takes fn1 multiplications to compute Clj. Therefore, it takes fn1 + 1 multiplications to evaluate a l j C l j and hence n ( f n  i + 1) multiplications to c o m p u t e detA. Thus fn can be defined recursively as follows: fl0
fn  n ( f n  1 + 1),
n >_ 2
(5.20)
(This is a linear n o n h o m o g e n e o u s recurrence relation with n o n c o n s t a n t coefficients.)
9 To solve the recurrence relation (5.20)" Let fn = n!gn. Since fl  0, g l E q u a t i o n (5.20), we get n!gn = n [ ( n 

0. S u b s t i t u t i n g for fn in
1)!gn1 + 1]
= n!gn1 + n
So
(gn 
gn1)n! = n 1
gn  g n  1 
( n  1)!
(Note: g l  0. )
Solving this yields (see Exercise 64)
gn 
'••
11 ~, since gl = 0 k=l
II,.
So, nll)
fn = n ' g n   n ' ( k ~ _ _ l ~ .
=n!
Therefore,
fn 1. Then the middle t e r m is XL(n+l)/2j. Compare key to X [ ( n + l ) / 2 j . If they are not equal, search the lower sublist or the upper sublist, but not both. If n is even, L n + l j  L2J ; so the upper half contains 2 / n /\ n 9n 9 n /  [ ~ J elements and the lower half contains ~  1 . ( < L 2 A ) e l e m e n t s . n + 1 n + 1 On the other hand, if n is odd, then / 2 /  T ; ~ so both sublists contain n  2 1 = LnJ elements each. Thus, in any case, the m a x i m u m n u m b e r ~
C11
CnC[n/2J q 1,
of comparisons needed is Ckn/2] + 1. So
n >_ 2
(5.21)
328
Chapter 5 Recursion
To solve this r e c u r r e n c e relation, a s s u m e , for convenience, t h a t n is a p o w e r of 2, say n  2 k, w h e r e k >_ 0. Let Cn  a k . T h e n t h e r e c u r r e n c e r e l a t i o n (5.21) becomes ak  ak1 + 1, w h e r e a0 = 1. Solving t h i s r e c u r r e n c e r e l a t i o n yields ak  k + 1, k >_ 0 (Verify.). Since n = 2 k, k  l g n , so cn = 1 + lg n, n >_ 1. T h u s , if n is a p o w e r of 2, t h e n Cn  (O(lg n). S u p p o s e n is n o t a p o w e r of 2. T h e n , by induction, it can be s h o w n t h a t cn  1 + llg n J, w h e r e n >_ 1 (see Exercise 44), so Cn  (')(lg n). Thus, in b o t h cases, t h e a l g o r i t h m t a k e s ()(lg n) e l e m e n t c o m p a r i s o n s in t h e w o r s t case. m The p r e c e d i n g e x a m p l e is a special case of t h e following t h e o r e m . Since the p r o o f is s o m e w h a t complicated, we skip it (see Exercises 65 a n d 66). Let a, b E 1~ a n d c, d ~ IR+ w i t h b > 2. Let f be a n o n d e c r e a s i n g function* such t h a t f ( n )  a f ( n / b ) + c a n d f(1)  d. T h e n O(lgn) O(nlogba) if a  1 otherwise
f(n)

m
F o r example, let f be a n o n d e c r e a s i n g f u n c t i o n such t h a t f ( n ) = + 5 a n d f(1)  8. T h e n , by T h e o r e m 5.7, f ( n )  O(nlgS). The n e x t t h e o r e m is a g e n e r a l i z a t i o n of T h e o r e m 5.7. We s t a t e it w i t h o u t proof (see Exercises 6 7  6 9 for special cases of t h e t h e o r e m ) a n d apply it in E x a m p l e 5.45.
3f(n/2)
Let a, b ~ N a n d c, d e IR+ w i t h b >_ 2. Let f be a n o n d e c r e a s i n g f u n c t i o n such t h a t f ( n )
 af(n/b) + c n d. T h e n
O(n d)
f(n) 
if a < b d ifab d
m
O(ndlgn) O(nlogba)
if a > b d
(optional) Let A

(aij) a n d B =
(bij) be two n x n m a t r i c e s . Let C =
n
a i h b k j . Since C h a s n 2 e n t r i e s a n d k=l each t a k e s n multiplications, t h e p r o d u c t C can be c o m p u t e d u s i n g n 3 = O(n 3) multiplications; in fact, it can be c o m p u t e d u s i n g O(n 3) c o m p u t a t i o n s (additions a n d multiplications), as Exercises 40 a n d 41 indicate, m ~
(cij) be t h e i r p r o d u c t w h e r e cij
We close this section w i t h an analysis of t h e m e r g e sort a l g o r i t h m , a d i v i d e  a n d  c o n q u e r strategy. ( m e r g e s o r t ) T h e m e r g e sort m e t h o d in A l g o r i t h m 5.7 s o r t s a list of n elements. Assume, for convenience, t h a t n is a p o w e r of 2, say, n  2 k, k >_ 0. *Let S c_ IR. A function f 9S ~ •+ is said to be nondecreasing ifx < y implies f(x) 2. The list is split into two, with each sublist containing n/2 elements. In the worst case, the n u m b e r of comparisons resulting from line 4 is Cn/2, as it is from line 5. W h e n the merge a l g o r i t h m is invoked in line 6, each sublist contains n/2 elements; so the m a x i m u m n u m b e r of c o m p a r i s o n s from line 6 is n  1. T h u s
c21
Cn  2Cn/2 + ( n 
1),
n>3
Let ak

Cn
where n  2 k, k _> 0. T h e n a11
ak   2 a k _ l + (2 k 
1),
k>2
This recurrence relation (see Exercise 8 in Section 5.2) yields ak  ( k 1)2 k + 1, k>l
= k . 2k  2k + 1 Thus c,~  (lg n)n  n + 1 _2
fnfnl + (2n1),
fn = f n  1 + 4 n , n > 1
6. f
2
7. f
1
fn = f n  l + n , n > 2
fn 2fn1 + (2 n  1), n >_ 2
Find the number of comparisons needed to search for key  13 in each ordered list using the recursive binary search algorithm in Example 5.33. 8. 1 , 2 , 3 , 5 , 8 , 13 10. 3, 7, 8, 13, 21 9. 5,8, 13,21,34 11. 15, 16, 19, 21
Compute the maximum number of comparisons needed to search for a particular item in an ordered list containing the following number of items, using the recursive binary search algorithm.
12. 8 13. 20 14. 25 15. 31
Let bn denote the number of multiplications needed to compute n! using the recursive factorial algorithm in Example 5.1. 16. Define bn recursively. 17. Solve the recurrence relation satisfied by b,~. 18. Show that bn = O(n). 1922. Estimate the number of times a,~ the assignment statement, x x + 1, is executed by the nested f o r loops in Exercises 3538 of Section 4.4. Estimate the number a,~ of times the statement, x ~ x + 1, is executed by each nested f o r loop.
23. f o r i = 1 t o n do f o r j = 1 t o Li / 2 ]
x _ 0 . 3. Prove t h a t fn+ 1  f 2
_ 2fn + 2. (J. M. Schram, 1983)
4. Using Exercise 3, compute fl, f2, f3, and f4. 5. Let an be an infinite sequence with al  1, a5  5, a12  144, and an + an+3  2an+2. Prove t h a t an  Fn. (H. Larson, 1977)
1+4~
6. Let a
n,~
2 lim F~+I = ~.
gn
=
and Fn the n t h Fibonacci n u m b e r .
Prove t h a t
*7. Let S n denote the sum of the n u m b e r s in the n t h t e r m of the sequence of sets of p e n t a g o n a l n u m b e r s { 1}, {5, 12}, {22, 35, 51}, {70, 92,117, 145}, . . . . Find a formula for S n . *8. Let S n denote the sum of the n u m b e r s in the n t h t e r m of the sequence of sets of Fibonacci n u m b e r s {1}, {1,2}, {3,5,8}, {13,21,34,55}, . . . . Find a f o r m u l a for S n .
Chapter Summary
339
Describe the behavior of each n u m b e r sequence {an }, where a0  a, a l  b, and a2  c are positive numbers. (R. L. G r a h a m , 1991)
9. an+2  (1 + an+l)/an
10. an+3  (1 + an+l + an+2)/an
Let n E N and ~ Euler's phifunction. Define ~k _ ~gk1 o q), where ~1 _ and o denotes composition. Let f ( n )  ~(n) + ~2(n) + ~3(n) + . . . + q~(1). (D. L. Silverman, 1981) 11. Compute f(5) and f(8). 12. Prove t h a t if n  2 k, t h e n f ( n )  n. 13. Prove t h a t f ( n ) is even. [Hint: ~(n) is even for n > 2.] 14. Consider the sequence of right triangles Tn, n > 1, with legs An and Bn, and hypotenuse Cn such t h a t An+l  Bn and Bn+l  Cn. Compute lim Bn Cn n   * ~ An and n.eclim ~ . (M. Flavio, 1980) A set of integers A is f a t if each of its elements is > _ IAI. For example, _ {5, 7, 91} is a fat set, but {3, 7, 36, 41} is not. ~ is considered a fat set. Let fn denote the n u m b e r of fat subsets of the set {1, 2 , . . . , n}. (G. F. Andrews) "15. Define fn recursively. * 16. Find an explicit formula for fn.
Let f ( n , k ) denote the n u m b e r of kelement subsets of the set S {1, 2 , . . . ,n} t h a t do not contain consecutive integers. Let fn denote the total n u m b e r of subsets of S t h a t do not contain consecutive integers. (I. Kaplansky) "17. Define f(n, k) recursively. "18. Find an explicit formula for fn.
Computer Exercises
Write a p r o g r a m to perform each task. 1. Read in a positive integer n < 20, and p r i n t the various moves and the n u m b e r of moves needed to transfer n disks from peg X to peg Z, using the rules in Example 5.4. 2. Read in a positive integer n, and print the first n t r i a n g u l a r and t e t r a h e d r a l numbers. 3. P r i n t the t r i a n g u l a r n u m b e r s _< 1000 t h a t are perfect squares. 4. P r i n t the t r i a n g u l a r n u m b e r s < 1000 t h a t are primes. 5. There are eight palindromic t r i a n g u l a r n u m b e r s < 1000. Find them. 6. Search for two t r i a n g u l a r n u m b e r s tn such t h a t tn and n are palindromic, where 9 < n < 100. 7. Read in a positive integer n and print the first n Fibonacci numbers, using recursion and iteration.
340
Chapter 5 Recursion 8. Read in a positive integer n _< 20 and print the first n Lucas n u m b e r s . 9. Read in a positive integer n < 25 and print the values of Fn+l and
Fn
Ln+l correct to 10 decimal places, where Fn denotes the n t h Fibonacci Ln n u m b e r and Ln the n t h Lucas number.
Read in a list of n positive integers. Use recursion to print each. 10. Their sum, product, m a x i m u m , and m i n i m u m . 11. The list in the given order. 12. The list in the reverse order. 13. Read in a key and search the list for key. P r i n t the location if the search is successful; otherwise, print a suitable message. 14. Read in a key and a sorted list of n items; d e t e r m i n e if key occurs in the list using recursion and iteration. P r i n t the location of key if the search is successful. 15. Read in a list of n words and determine if each is a palindrome, using recursion. 16. Read in two lists of n integers. Determine if they are identical, using recursion. 17. Read in a nonnegative real n u m b e r x and a nonnegative integer n; compute the n t h power ofx. 18. Read in a positive integer n < 100 and a positive real n u m b e r x < 2. Use the binary r e p r e s e n t a t i o n of n and the technique of successive squaring to compute x n. P r i n t the n u m b e r of multiplications needed to compute it.
n
19. Read in a n u m b e r a, and a polynomial ~ aix ~(that is, coefficients and the corresponding exponents); print tt~e~ using H o r n e r ' s method. of the polynomial at a,
20. Read in n positive integers and print their m i n i m u m and m a x i m u m , using both iteration and recursion. 21. Read in a positive integer n < 10 and a r r a n g e the Stirling n u m b e r s of the second kind S(n, r) in a t r i a n g u l a r form, where 1 < r < n. 22. Read in n positive integers and sort t h e m using bubble sort, selection sort, and insertion sort. P r i n t the n u m b e r of elementcomparisons needed by each algorithm. 23. Read in n fourletter words. Sort them, using merge sort and quicksort. P r i n t the n u m b e r of element comparisons needed by each sort.
Chapter Summary Exploratory Writing Projects
341
Using library and Internet resources, write a team report on each of the following in your own words. Provide a welldocumented bibliography. 1. Describe the properties of Fibonacci numbers, their occurrences in nature, applications to various disciplines, and relationships to Lucas numbers. 2. Explain how the golden ratio is related to Fibonacci and Lucas numbers. Describe its various occurrences in nature. 3. Describe the various forms of Ackermann's function. Investigate its importance in the study of recursive functions and the analysis of algorithms. 4. Investigate the Josephus problem, named for the first century Jewish historian Flavius Josephus (37?100?). 5. Describe how, using Fibonacci numbers Fn (n >_ 2) as bases, nonnegative integers can be represented as binary numbers with no two adjacent l's. Express the integers 125 as such binary numbers. 6. Define continued fractions and describe their relationship to Fibonacci numbers. 7. Describe the Game of Life, invented in 1970 by British mathematician John H. Conway, now at Princeton University. 8. Describe the Game of Halma, invented in 1883 by George H. Monks, a Harvard Medical School graduate. 9. Examine the history of Catalan numbers and their properties and applications. Include a biography of E. C. Catalan. 10. Write an essay on the Tower of Brahma (Hanoi). 11. Write an essay on Quicksort. 12. Discuss the fifteen puzzle, invented by American puzzlist Samuel Loyd (18411911). 13. Discuss Markov chains, named after Russian mathematician Andrei A. Markov (18561922), who developed the theory of stochastic processes, and their applications to business.
Enrichment Readings
1. G. Brassard and P. Bratley, Algorithmics: Theory & Practice, PrenticeHall, Englewood Cliffs, NJ, 1986, pp. 2634, 4861.
342
Chapter5 Recursion 2. R. P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, 4th edition, AddisonWesley, Boston, MA, 1999, pp. 351403. 3. B. W. Jackson and D. Thro, Applied Combinatorics with Problem Solving, AddisonWesley, Reading, MA, 1990, pp. 226252. 4. T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley, New York, 2001. 5. C. Oliver, "The Twelve days of Christmas," Mathematics Teacher, Vol. 70 (Dec. 1977), pp. 752754. 6. S. Sahni, Concepts in Discrete Mathematics, 2nd ed., Camelot, Fridley, MN, 1985, pp. 205335. 7. R. Sedgewick, Algorithms, 2nd ed., AddisonWesley, Reading, MA, 1988, pp. 3189. 8. K. B. Strangeman, "The Sum of n Polygonal Numbers," Mathematics Teacher, Vol. 67 (Nov. 1974), pp. 655658. 9. C. W. Trigg, "Palindromic Triangular Numbers," J. Recreational Mathematics, Vol. 6 (Spring 1973), pp. 146147. 10. A. Tucker, Applied Combinatorics, Wiley, New York, 1984, pp. 222298. 11. H. S. Wilf, Algorithms and Complexity, PrenticeHall, Englewood Cliffs, NJ, 1986, pp. 2634, 4861.
Chapter 6
C o m b i n a t o r i c s and D i s c r e t e Probability
The theory of probability affords an excellent illustration of the application of the theory of permutations and combinations which is the fundamental part of the algebra of discrete quantity.
G. CRYSTAL
ombinatorics is a fascinating branch of discrete mathematics, which deals with the art of counting. Very often we ask the question, In how m a n y ways can a certain task be done? Usually combinatorics comes to our rescue. In most cases, listing the possibilities and counting them is the least desirable way of finding the answer to such a problem. Often we are not interested in enumerating the possibilities, but rather would like to know the total number of ways the task can be done. For instance, consider the following combinatorial problem: O n e t y p e of a u t o m o b i l e license p l a t e n u m b e r in M a s s a c h u s e t t s consists of one l e t t e r a n d five digits. F i n d t h e n u m b e r of s u c h license p l a t e n u m b e r s possible.
9 9 5H0515 MASSACHUSETTS ~ ~
C
Suppose you are willing to list all the possibilities and count them to find the answer. Assuming you know how to enumerate them systematically and that it would take a second to count each, it would take about 6 months to complete the counting alone. Obviously, this is an inefficient way to find the answer, especially when combinatorics can do the job in seconds. (See Example 6.6.) A few other interesting problems we examine in this chapter are: 9 A secretary types up 10 different letters and 10 envelopes. In how many different ways can she place each letter in an envelope so that no letter is placed in the correct envelope? 343
344
Chapter 6
Combinatorics and Discrete Probability
9 Eleven guests would like to order soft drinks with their dinner. There are five choices for a soft drink: Coke Classic, Diet Coke, root beer, Pepsi, and Sprite. Find the n u m b e r of different beverage selections possible. 9 What is the m a x i m u m n u m b e r of nonoverlapping regions formed inside a circle by joining n points on it? 9 In how m a n y ways can n married couples be seated at a r o u n d table so t h a t men and women sit on alternate chairs and no wife sits next to her husband? This chapter presents the f u n d a m e n t a l s of combinatorics.* In several instances, you will find recursion and generating functions useful in solving combinatorial problems, so review t h e m as needed.
This section, a natural continuation of Section 2.4, presents three fundamental principles that form the foundation of combinatorics. The addition and the inclusionexclusion principles discussed in Section 2.4 have fine applications to combinatorics, as will be seen shortly. Find the n u m b e r of ways of drawing a red queen or a black king from a standard deck of playing cards. SOLUTION: Let A denote the set of red queens and B the set of black kings. Clearly, ]Ai = 2 = IB]. Since A and B are disjoint sets, by the addition principle, IA u Bi = IA] + iBI = 2 + 2 = 4. Thus there are four different ways of drawing a red queen or a black king. m In this example, drawing a red queen can be considered a t a s k , say, task A. Likewise, drawing a black king can be considered task B. Since the two tasks cannot occur simultaneously, they are m u t u a l l y e x c l u s i v e . So finding the n u m b e r of ways of drawing a red queen or a black king is equivalent to finding the n u m b e r of ways task A or t a s k B can be performed. Accordingly, the addition principle can be restated in t e r m s of tasks as follows.
(Addition Principle) Let A and B be two mutually exclusive tasks. Suppose task A can be done in m ways and task B in n ways. T h e n task A or task B can take place in m + n ways. m
*The first book on combinatorics is Ars Conjectandi, written by the Swiss mathematician, Jakob Bernoulli (16541705) and published posthumously in 1713.
6.1 The Fundamental Counting Principles
345
The addition principle can be e x t e n d e d to a n y finite n u m b e r of pairwise m u t u a l l y exclusive tasks, u s i n g induction. For instance, let T 1 , . . . , Tn be n pairwise m u t u a l l y exclusive tasks. Suppose t a s k Ti can be done in mi ways, w h e r e 1 < i < n. T h e n t a s k T1, T 2 , . . . , or Tn can be done in m l § m2 § § mn ways, as the next e x a m p l e illustrates. ~ A f r e s h m a n has selected four courses and needs one m o r e course for t h e next t e r m . T h e r e are 15 courses in English, 10 in French, a n d 6 in G e r m a n she is eligible to take. In how m a n y ways can she choose t h e fifth course? SOLUTION: Let E be the t a s k of selecting a course in English, F the t a s k of selecting a course in F r e n c h , and G t h a t of selecting a course in G e r m a n . T h e s e t a s k s can be done in 15, 10, a n d 6 ways, respectively, a n d are m u t u a l l y exclusive, so, by the addition principle, the fifth course can be selected in IEI + IFI + I G I  15 + 10 + 6  31 ways. i Like the addition principle, the inclusionexclusion principle can be r e s t a t e d in t e r m s of t a s k s in an obvious way. It can also be e x t e n d e d to a finite n u m b e r of t a s k s (see T h e o r e m 6.19). ( I n c l u s i o n  E x c l u s i o n P r i n c i p l e ) Suppose a t a s k A can be done in m ways, t a s k B in n ways, and both can be accomplished in k different ways. T h e n t a s k A or B can be done in m + n  k ways. m The next p r o b l e m exemplifies this. In how m a n y ways can you deal a king or a black card from a s t a n d a r d deck of cards? SOLUTION: A king can be selected in four different ways and a black card in 26 different ways. These two t a s k s can be done s i m u l t a n e o u s l y in two ways, n a m e l y , by selecting a black king; so, by T h e o r e m 6.2, a king or a black card can be selected in 4 + 26  2  28 ways. m Before s t a t i n g the next c o u n t i n g principle, let us r e t u r n to E x a m p l e 2.23 in C h a p t e r 2. The t a s k of selecting a mode of t r a n s p o r t a t i o n for t h e trip from Boston to London via New York consists of two s u b t a s k s A a n d B" A is selecting a mode of t r a n s p o r t a t i o n from Boston to New Y o r k ~ c a r , plane, or ship ~ and B is selecting a mode of t r a n s p o r t a t i o n from New York to Lond o n ~ plane or ship. They can be done in ]Ai  3 a n d iBi  2 ways. Recall t h a t the trip can be m a d e in 6  IAI. iB] ways; t h a t is, IA • BI  6  IA]. IBI. More generally, we have t h e following result. ( M u l t i p l i c a t i o n P r i n c i p l e ) Suppose a t a s k T is m a d e up of two s u b t a s k s , s u b t a s k T1 followed by s u b t a s k T2. If s u b t a s k T1 can be done in m ] ways and s u b t a s k T2 in m2 different ways for each way s u b t a s k T1 can be done, t h e n t a s k T can be done in m l m 2 ways. m The next four examples i l l u s t r a t e this principle.
346
Chapter6 Comblnatoricsand Discrete Probability
F i n d t h e n u m b e r of t w o  l e t t e r words t h a t begin with a v o w e l   a, e, i,
or u.
O,
SOLUTION: The t a s k of f o r m i n g a t w o  l e t t e r word consists of two s u b t a s k s T1 a n d T2" T1 consists of selecting t h e first l e t t e r a n d T2 selecting t h e second letter, as F i g u r e 6.1 shows.
F i g u r e 6.1
number of choices
subtask T1
subtask T2
Since each word m u s t begin w i t h a vowel, T1 can be a c c o m p l i s h e d in five ways. T h e r e are no r e s t r i c t i o n s on t h e choice of t h e second letter, so T2 can be done in 26 ways (see F i g u r e 6.2). Therefore, by t h e m u l t i p l i c a t i o n principle, the t a s k can be p e r f o r m e d in 5 . 2 6 = 130 different ways. In o t h e r words, 130 t w o  l e t t e r words begin with a vowel.
F i g u r e 6.2
number of choices 5 subtask T 1 9 26 subtask T 2
m
T h e various twoletter words in this e x a m p l e can be e n u m e r a t e d systematically by c o n s t r u c t i n g a tree d i a g r a m , as in Figure 6.3. All desired words can be obtained by t r a v e r s i n g t h e various b r a n c h e s of the tree, as indicated.
F i g u r e 6.3
first element
~ a ~ a
second element
) 9 ~
word
aa 9
z a e ~ z
~ ~ ~ ~
az ea 9 ez
a
u
___.) ..)
ua
.
z
~
uz
6.1 The Fundamental Counting Principles
347
The multiplication principle can also be extended to any finite n u m b e r of subtasks. Suppose a task T can be done by n successive subtasks, T1, T2,..., Tn. If subtask Ti can be done in mi different ways after Ti1 has been completed, where 1 < i < n, t h e n task T can be done in m l m 2 . . . m n ways. The multiplication principle can be applied to prove t h a t a set with size n has 2 n subsets, as shown below. Show t h a t a set S with n elements has 2 n subsets. SOLUTION: Every subset of S can be uniquely identified by an nbit word (see Figure 6.4). The task of forming an nbit word can be broken down to n subtasks: selecting a bit for each of the n positions. Each position in the word
F i g u r e 6.4
number of choices
9
. .
9
.
9
. .
9
.
9
.
9
n bits has two choices, 0 or 1; so, by the multiplication principle, the total number of nbit words t h a t can be formed is 2 . 2 . . . . . 2  2 n (see Figure 6.5). In other words, S has 2 n subsets, n times
Y
F i g u r e 6.5
number of choices
2.2.2
..
2.2.2
II
We now t u r n to solving the problem posed at the beginning of the chapter. One type of automobile license plate n u m b e r in Massachusetts consists of one letter and five digits. Compute the n u m b e r of such license plate n u m b e r s possible. SOLUTION: For convenience, we decompose the task into three subtasks: 9 Choosing a letter. It can be done in 26 ways. 9 Choosing the position of the letter. It has six possible slots.
348
Chapter 6 Combinatorics and Discrete Probability
9 Choosing the five digits. T h e y can be selected in 1 0 . 1 0 . 1 0 . 1 0 . 1 0 100,000 ways.
=
Now we are ready to find the final answer. By t h e multiplication principle, the total n u m b e r of license plates is 2 6 . 6 . 100000 = 15,600,000. m The next example depends on the multiplication and addition principles. (optional) An identifier in a p r o g r a m m i n g language consists of a letter followed by a l p h a n u m e r i c characters.* F i n d the n u m b e r of legal identifiers of length at most 10. SOLUTION: Let Si denote the set of identifiers of length i, where i < i < 10. T h e n ISil 2 6 . 3 6 i1 (see Figure 6.6). Since the s u b t a s k s $ 1 , . . . ,$10 are m u t u a l l y F i g u r e 6.6 26 9 36 9 36 . . . . . 36
I
letter ~ (i  1) alphanumeric characters exclusive, by the addition principle, the total n u m b e r of identifiers of length < 10 is given by
10
JSit = ~ 2 6 "
36 i1  26 1)
36 i 1)
i=1
i=1  26. (36 l ~ 26(36 l ~ 35 36 1
 2,716,003,412,618,210 2.7 quadrillion! m
The final example in this opening section employs the multiplication and the inclusionexclusion principles. ~ An eightbit word is called a b y t e . Find the n u m b e r of bytes with t h e i r second bit 0 or the t h i r d bit 1. SOLUTION: N u m b e r of bytes with second bit 0  2 . 1 . 2 . 2 . 2 . 2 . 2 . 2 N u m b e r of bytes with third bit 1  2 . 2 . 1 . 2 . 2 . 2 . 2 . 2  27  27
Since these two s u b t a s k s are n o t m u t u a l l y exclusive, we c a n n o t add these two partial answers and claim t h a t the a n s w e r is 27 + 27  128 + 128  256. *An alphanumeric character is a letter or a digit.
6.1 The Fundamental Counting Principles
349
So, we m u s t find the n u m b e r of bytes t h a t have both properties. The number of bytes with second bit 0 and third bit 1 equals 2 . 1 . 1 . 2 . 2 . 2 . 2 . 2 26  64, so, by the inclusionexclusion principle, the n u m b e r of bytes with the given properties is 27 + 27  26  128 + 128  64  192. m
Exercises 6.1
Find the n u m b e r of positive integers < 1976 and divisible by: 1. 2 o r 3 . 2. 3 o r 5 . 3. 2,3, o r 5 . 4. 3,5, o r 7 .
5. In one version of BASIC, a variable name consists of a letter, or a letter followed by a digit, or the dollar sign ($). Find the total n u m b e r of possible variable names. Find the n u m b e r of terms in the expansion of each expression.
6. (a + b)(c + d + e)(x + y) 8. ( ~~ai)( ~ bi)( ~ci)
i=o i=l i=2
7. (b + c)(d + e + f)(x + y + z) 9. ( ~ el)( ~ bi)( ~ci)
i=2 i1 i=o
10. Find the n u m b e r of palindromes of length n over the English alphabet. 11. Find the n u m b e r of palindromic alphanumeric identifiers of length n. (See Example 6.7.) 12. Find the total n u m b e r of bytes. (See Example 6.8.) 13. A word over the alphabet {0, 1, 2} is called a t e r n a r y word. Find the n u m b e r of t e r n a r y words of length n that can be formed. A typical automobile license plate in New York contains three letters followed by three digits. Find the n u m b e r of license plates of this kind that: 14. Can be formed. 16. Begin with a vowel. 18. Repeat no letters or digits. 19. Contain the same letters and the same digits. 20. Have the property t h a t both words and n u m b e r s are palindromes. An old zip code in the United States consists of five digits. Find the total n u m b e r of possible zip codes that" 21. Have no repetitions. 23. End in K. 22. Begin with 0. 24. Are palindromes. 15. Begin with the word BAT. 17. Begin with the digit 6.
350
Chapter 6 Combinatorics and Discrete Probability
A zip code in C a n a d a consists of t h r e e letters and t h r e e digits. Each zip code begins with a letter. The letters and digits alternate; for instance, A1B2C3. Find the n u m b e r of zip codes that: 25. End in 6. 27. End in Z. 26. Begin with A and end in 3. 28. Are possible.
The password for a computer system consists of six a l p h a n u m e r i c characters and begins with a letter. Find the total n u m b e r of passwords that" 29. Are possible. 31. Contain the word BAT. 30. End in RED. 32. End in 2076.
33. Every radio and television station in the United States has a u n i q u e call name. Each contains three or four letters, beginning with K or W. For example, KEY and WASP are legal call names. Find the n u m b e r of possible call names. Find the n u m b e r of bytes that: 34. Begin with 101. 35. End with 110.
36. Begin with and end in the same bit. 37. Begin with and end in different bits. 38. Have the same third and fourth bits. 39. Have third bit or fifth bit 1. 41. Are palindromes. Find the n u m b e r of positive divisors of the following positive integers. 42. 2 i. 3 j . 5 k 43. 600 40. Have second bit 0 or third bit 1.
44. A numbertheoretic function used in the study of perfect n u m b e r s is the t a u f u n c t i o n v on N. (v is the Greek letter, tau.) v(n) denotes (?2 el, the n u m b e r of positive divisors of n ~ N. Let n  P l (" 1P2 "" "Pk', where P l , P 2 , . . . ,Ph are distinct primes and e l , e 2 , . . . ,e]~ ~ W. Find r(n). 45. Find the n u m b e r of n x n matrices t h a t can be constructed using bits. Find the n u m b e r of t e r n a r y words t h a t have: 46. Length at most 3. 48. Length at most n. 50. Length 4 and are palindromes. 47. Length at most 5. 49. Length 3 and are palindromes. 51. 4 < length < 6
In an alphabet of m characters, how m a n y words have: 52. Length 3?
6.2
Permutations
351
53. L e n g t h not m o r e t h a n 2? 54. L e n g t h at least 2, b u t not m o r e t h a n 4? 55. L e n g t h not m o r e t h a n n? Let A and B be two finite sets w i t h ]A] = m a n d ]B]  n. How m a n y : 56. F u n c t i o n s can be defined from A to B? 57. Bijections can be defined from A to B ( a s s u m e m  n)? 58. Invertible functions can be defined from A to B ( a s s u m e m  n)? 59. Injections can be defined from A to B ( a s s u m e m < n)? Let T denote t h e t a u function. Prove each. 60. T(n) is odd if a n d only if n is a square. "61. If m and n are relatively p r i m e n u m b e r s , t h e n ~'(mn)  ~:(m). T(n). *62. The h a r m o n i c m e a n m of t h e n u m b e r s a l , a 2 , . . . , an is the reciprocal of the a r i t h m e t i c m e a n of t h e i r reciprocals; t h a t is, 1 m 1~(1) n i=1
ai
Prove t h a t the h a r m o n i c m e a n of the positive factors of a perfect n u m b e r N is an integer. (Hint" I f d is a factor of N, t h e n so is N / d . ) (R. Euler, 1987)
The concept of ordered a r r a n g e m e n t s was familiar to Chinese m a t h e m a t i cians as early as 1100 B.C. This section i n t r o d u c e s t h e concept of ordered a r r a n g e m e n t s a n d shows how to find the n u m b e r of such a r r a n g e m e n t s . Suppose a p h o t o g r a p h e r would like to a r r a n g e 10 cats in a row for a television commercial. She can a r r a n g e t h e m in a n y order she likes. In how m a n y ways can she accomplish this? A l t h o u g h we could use the multiplication principle to arrive at an answer, we shall apply the concept of a p e r m u t a t i o n (see E x a m p l e 6.11). Recall t h a t t h e e l e m e n t s of an ordered set are assigned u n i q u e positions. For convenience, let us denote the ordered set (a,b,c) as the a r r a n g e m e n t abc. The words acb and bac are two different a r r a n g e m e n t s of the very s a m e letters. ( R e m e m b e r , the order m a k e s a difference.) E a c h of these a r r a n g e m e n t s is a p e r m u t a t i o n of the t h r e e l e t t e r s t a k e n all at a time, or
a 3permutation.
352
Chapter 6 Combinatorics and Discrete Probability
Permutation
A p e r m u t a t i o n of a set of n (distinct) elements t a k e n r (0 < r < n) at a time is an a r r a n g e m e n t of r elements of the set. For convenience, it is called an r  p e r m u t a t i o n . If r = n, t h e n the r  p e r m u t a t i o n is called simply a p e r m u t a t i o n . The n u m b e r of r  p e r m u t a t i o n s of a set of size n is d e n o t e d by P(n, r). We begin our discussion with a simple example. Find the n u m b e r of permutations; t h a t is, 3  p e r m u t a t i o n s of the e l e m e n t s of the set {a, b, c}. SOLUTION: By the multiplication principle, the n u m b e r of 3  p e r m u t a t i o n s of t h r e e elements is 3 92 9 1 = 6. Thus P(3, 3) = 6. m The various p e r m u t a t i o n s in Example 6.9 can be obtained systematically using a tree diagram, as Figure 6.7 shows. They are abc, acb, bac, bca, cab, and cba.
F i g u r e 6.7
first element
second element b c
a
third element c b
C
permutations ~ ~ ~ abc acb bac bca cab cba
12
a
~~x~~ x c ~ ~
a b
b a
Find the n u m b e r of 2permutations of the elements of the set {a, b, c}. SOLUTION" Again, by the multiplication principle, the n u m b e r of 2  p e r m u t a t i o n s is 3 . 2  6; t h a t is, P(3, 2) = 6. m The various 2permutations in Example 6.10 are ab, ac, ba, bc, ca, and cb. They can be obtained using the tree diagram in Figure 6.8. Examples 6.9 and 6.10 can be i n t e r p r e t e d as follows: Suppose you have three books in your hands and would like to a r r a n g e t h e m in a bookcase. If there is enough room for three books, they can be a r r a n g e d in 3 . 2 . 1  6 different ways. On the other hand, if there is room for only two books, t h e y can be a r r a n g e d in 3 . 2 = 6 different ways.
6.2 Permutations
353
F i g u r e 6.8
first element
second element b ~ ~ ~
'>
permutations
ab ac ba
be
a
~ c a
b
C
a
c ~
~ >
ca cb
b
M o r e g e n e r a l l y , we h a v e t h e following r e s u l t . ~ T h e n u m b e r of r  p e r m u t a t i o n s of a set of n (distinct) e l e m e n t s is g i v e n by n!
P(n,r) ~ . (nr)!
PROOF:
Since t h e r e a r e n e l e m e n t s , t h e first e l e m e n t can be c h o s e n in n ways. N o w n  1 e l e m e n t s a r e left; so t h e s e c o n d e l e m e n t c a n be c h o s e n in n  1 ways. C o n t i n u e like t h i s u n t i l t h e r t h e l e m e n t is r e a d y to be c h o s e n . At t h i s p o i n t t h e r e a r e n  r + 1 e l e m e n t s left. C o n s e q u e n t l y , t h e r t h e l e m e n t c a n be c h o s e n in n  r + 1 ways. T h u s , by t h e m u l t i p l i c a t i o n principle,
P(n,r) n(n n(n 
1)(n  2 ) . . . (n  r + 1) 1 ) . . . (n  r + 1)(n  r ) . . . 2 . 1 (n  r ) . . . 2 . 1
=
n! (n  r ) !
m
A l t h o u g h it is e a s y to r e m e m b e r t h e v a l u e of P ( n , r ) u s i n g t h i s form u l a , P ( n , r ) is o f t e n c o m p u t e d u s i n g t h e f o r m u l a P ( n , r ) = n ( n  1) (n  2 ) . . . (n  r + 1). T h e v a l u e s n! a n d (n  r)! m a y be too l a r g e e v e n for a c a l c u l a t o r to c o m p u t e . T h e n t h e v a l u e n ! / ( n  r ) ! m a y n o t be exact. You will find t h e fact t h a t n !  n ( n  1)! u s e f u l in c o m p u t i n g P ( n , r ) . F o r example, P(25, 5) = 25! ( 2 5  5)! = 25! 20! = 25.24.23.2221.20! 20!  6,375,600
 25.24.23.22.21
354
Chapter 6 Combinatorics and Discrete Probability
Suppose we let r = n in T h e o r e m 6.4. T h e n n? n? P ( n , n )  (n  n)?  O? n? 1  n?
Accordingly, we have the following result. ~ The n u m b e r of p e r m u t a t i o n s of a set of size n is given by P(n, n) = n!. T h a t is, n elements can be arranged in n? ways. m Interestingly enough, applications of this formula appear in the anonymous Hebrew book Sefer Yetzirah (The Book o f Creation), written between 300 and 600 A.D. The next two examples illustrate this theorem. ~ A p h o t o g r a p h e r would like to arrange 10 cats for a television commercial. How m a n y ways can she arrange t h e m in a row? SOLUTION: Since all the cats have to be in the commercial at the same time, r  n = 10. Therefore, the n u m b e r of possible a r r a n g e m e n t s is P(10, 10) = 107 = 3,628,800. m ~ Find the n u m b e r of words that can be formed by scrambling the letters of the word SCRAMBLE. (Remember, a word is j u s t an a r r a n g e m e n t of symbols; it need not make sense.) SOLUTION: The word SCRAMBLE contains eight distinct letters. Therefore, the number of words t h a t can be formed equals the n u m b e r of a r r a n g e m e n t s of the letters in the word, namely, P(8, 8) = 8? = 40,320. m The next example uses the Theorem 6.5. ~ A multiplication principle, as well as
salesperson at a computer store would like to display six models of personal computers, five models of computer monitors, and four models of keyboards. In how m a n y different ways can he a r r a n g e t h e m in a row if items of the same family are to be next to each other? SOLUTION: There are three types of items: personal computers, monitors, and keyboards. Think of the items in each family as tied together into one unit. These families can be arranged in P(3, 3) = 3! ways. Now the items within each family can be rearranged. The six models of personal computers can be arranged in P(6, 6)  6? ways, the monitors in P(5, 5)  5? ways, and the keyboards in P(4, 4) = 4? different ways. Thus, by the multiplication principle, the total n u m b e r of possible a r r a n g e m e n t s is 3? 6? 5? 4? = 12,441,600. m
6.2 Permutations
355
P e r m u t a t i o n s of e l e m e n t s a r r a n g e d in a circle w o r k s o m e w h a t differently from the linear s i t u a t i o n s we have i n v e s t i g a t e d up to this point.
Cyclic Permutation
In how m a n y different ways can you place five beads on a necklace? T h e a n s w e r is n o t 5! = 120, b u t far less, since it contains a lot of duplicate a r r a n g e m e n t s . For instance, t h e two circular a r r a n g e m e n t s s h o w n in F i g u r e 6.9 are identical. (Look at t h e relative positions of t h e beads p l t h r o u g h P5.) E a c h circular a r r a n g e m e n t is a cyclic permutation.
Figure 6.9
Pl
P5
Before we find the n u m b e r of cyclic p e r m u t a t i o n s of t h e five beads in E x a m p l e 6.14, the following general r e s u l t will be useful to prove. The n u m b e r of cyclic p e r m u t a t i o n s of n (distinct) items is (n  1)!.
PROOF:
To avoid duplicates, let us assign a fixed position to the first i t e m a 1 a r o u n d the circle (see Figure 6.10). Now n  1 positions are left. So t h e second i t e m a2 can be placed in any one of the n  1 positions. Now n  2 positions
Figure 6.10
a1
are left. Therefore, the t h i r d i t e m a3 can be placed in a n y of the n  2 positions. C o n t i n u e like this until all i t e m s have been placed. Thus, by t h e multiplication principle, the n u m b e r of cyclic p e r m u t a t i o n s is 1. (n  1) (n  2 ) .  . 2 . 1 = (n  1)! m The next example illustrates this result. ~ F i n d the n u m b e r of different ways five zinnias can be p l a n t e d in a circle.
356
Chapter 6 Combinatorics and Discrete Probability
SOLUTION: N u m b e r of ways of planting~ _ f N u m b e r of cyclic p e r m u t a t i o n s ) five zinnias in a circle ] ~of five items = ( 5  1 ) ~  24 m
Let us r e t u r n to "linear" p e r m u t a t i o n s . We would like to find the recurrence relation satisfied by P ( n , r). We will use a combinatorial a r g u m e n t to find it, leaving the s t r a i g h t f o r w a r d algebraic proof as a r o u t i n e exercise (see Exercise 45). The n u m b e r of r  p e r m u t a t i o n s of n distinct e l e m e n t s satisfies the recurrence relation P ( n , r)  P ( n  1, r) + r P ( n  1, r  1), where 0 < r < n. PROOF: Let X be a set with n elements and x an a r b i t r a r y element in it. The set of r  p e r m u t a t i o n s of X can be partitioned into two subsets: A, the set of p e r m u t a t i o n s n o t containing x, and B, the set of p e r m u t a t i o n s c o n t a i n i n g x.
9 To find the number o f e l e m e n t s i n A" Since no p e r m u t a t i o n s in A contain x, every element in A is an r  p e r m u t a t i o n of n  1 elements. The n u m b e r of such p e r m u t a t i o n s is P ( n  1, r).
9 To find the number
o f e l e m e n t s i n B: Since every p e r m u t a t i o n in B contains x, n  1 candidates are left in X for the r e m a i n i n g r  1 positions. They can be a r r a n g e d in P ( n  1, r  1) ways. Now the position ofx in a p e r m u t a t i o n has r choices. Therefore, by the multiplication principle, r P ( n  1, r  1) p e r m u t a t i o n s contain x.
Since A and B are disjoint sets, by the addition principle,
P(n,r) IAI + IBI
= P ( n  1, r) + r P ( n  1, r  1)
m
Note t h a t it's m u c h easier to c o m p u t e P ( n , r) u s i n g the explicit f o r m u l a in T h e o r e m 6.4 r a t h e r t h a n by u s i n g the recursive. T r y P(5, 3) b o t h ways to see the difference.
Fibonacci Numbers Revisited The following example p r e s e n t s an i n t e r e s t i n g confluence of p e r m u t a t i o n s and Fibonacci n u m b e r s . ~ Let Pn denote the n u m b e r of p e r m u t a t i o n s f of the set Sn   { 1 , 2 , . . . n } such t h a t li  f ( i ) l < 1 for all 1 < i < n, where n > 1. So Pn counts the n u m b e r of p e r m u t a t i o n s t h a t move each element no more t h a n one position from its n a t u r a l position.
cr~ cr~
iL
~j
i., o
~L
Jr
II
j~
~
L~
L~
o ~L pmL
= S~ 0
~.~
II
0
~'~.
o~.~
0
~ IIo 0
C~
~D
II
~D
II
o
IIo
o
II
L~ C~
h~
358
Chapter 6 Combinatorics and Discrete Probability
C a s e 1 L e t f ( n ) = n. T h e n t h e r e m a i n i n g n  1 e l e m e n t s can be u s e d to form P n  1 p e r m u t a t i o n s such t h a t ]i  f ( i ) ] < 1 for all i. C a s e 2 L e t f ( n ) V= n. T h e n f ( n )  n  1 a n d f ( n  1)  n. T h e r e m a i n i n g n  2 e l e m e n t s can be e m p l o y e d to form P n  2 p e r m u t a t i o n s w i t h t h e d e s i r e d property.
T h u s , by t h e a d d i t i o n principle, Pn  Pn1 tPn2, w h e r e P l  1 a n d P2 = 2. It now follows t h a t Pn = Fn+l, w h e r e n > 1, as conjectured, m Since t h e total n u m b e r of p e r m u t a t i o n s of Sn is n!, it follows from this e x a m p l e t h a t t h e r e are n !  F n + l p e r m u t a t i o n s f o f S n such t h a t l i  f ( i ) i > 1 for some i n t e g e r i, w h e r e 1 < i < n. T h u s t h e r e are n !  F n + l p e r m u t a t i o n s of Sn t h a t move at least one e l e m e n t of Sn by two spaces f r o m its n a t u r a l position. In p a r t i c u l a r , t h e r e are 3!  F4  3 such p e r m u t a t i o n s of t h e set {1, 2, 3}, as F i g u r e 6.12 depicts.
Figure 6.12 1 1 1 1
2
2
2
2
Exercises 6.2
E v a l u a t e each. 1. 5! 4~ 2. 10'. 3~7~ 3. P ( 5 , 3 ) 4. P ( 6 , 6 )
M a r k each s e n t e n c e as t r u e or false, w h e r e n is an a r b i t r a r y n o n n e g a t i v e i n t e g e r a n d 0 < r _< n. 5. 0 !  0 7. 5 . 4 ! = 5!
9. ( 2 + 3 ) ! = 2 ! + 3 !
6. 1 !  1
8. (m + n)!  m! + n!
10. (mn)! = m! n!
12.
n(n 
11. ( 2 . 3 ) !  2! 3! 13. P ( n , 0) = 0
15. P(n, r) = P ( n , n  r)
1 ) !  n!
14. P ( n , 1) = P ( n , n  1)
16. n! is divisible by 10 if n > 4.
17. F i n d t h e n u m b e r of twodigit n u m e r a l s t h a t can be f o r m e d u s i n g t h e digits 2, 3, 5, 6, and, 9 a n d t h a t c o n t a i n no r e p e a t e d digits. 18. F i n d t h e n u m b e r of t h r e e  d i g i t n u m e r a l s t h a t can be f o r m e d u s i n g t h e digits 2, 3, 5, 6, a n d 9, if r e p e t i t i o n s are not allowed.
6.2
Permutations
359
Find the n u m b e r of words t h a t can be formed by scrambling the letters in each word. 19. algorithm 20. word 21. computer 22. logic
The password for a computer system consists of eight distinct alphabetic characters. Find the n u m b e r of passwords possible that: 23. End in the string MATH. 24. Begin with the string CREAM.
25. Contain the word C O M P U T E R as a substring. 26. Contain the letters WORD together, but in any order. 27. Contain the string BLACK or the string WHITE. 28. Contain the strings BLACK and WHITE. 29. Do not contain the string SAMPLE. A botanist would like to plant three coleus, four zinnias, and five dahlias in a row in her front garden. How m a n y ways can she plant t h e m if: 30. They can be planted in any order. 31. Plants of the same family m u s t be next to each other. 32. The family of zinnias must be in between the other two families. Find the n u m b e r of ways seven boys and three girls can be seated in a row if: 33. A boy sits at each end of the row. 34. A girl sits at each end of the row. 35. The girls sit together at one end of the row. 36. Show t h a t P ( n , O) = 1. Using the recursive definition of P ( n , r), evaluate each.
37. P(5, 4) 38. P(6, 0)
39. P(3,2)
40. P(6,3)
Solve each equation. 41. P ( n , 1 ) = 6 43. P(n, n  1)  5040
42.
P(n, 2)
 42
44. P(5, r)  20
45. Using Theorem 6.4, prove t h a t P ( n , r)  P ( n  1, r) + r P ( n  1, r  1). Verify each. 46. ( n + l ) ! + n !  ( n + 2 ) n !
47. (n+l)!n!n(n!)
48. Prove by induction that 1.1! + 2 . 2 ! + . . . + n . n !  ( n + 1 ) !  1, n > 1. 49. Write an algorithm to compute P ( n , r), using Theorem 6.4.
360
Chapter6 Combinatorics and Discrete Probability
50. Write a recursive algorithm to compute P(n, r). "51. Show t h a t (n!)! > (2n)!, if n > 3.
At the beginning of the 18th century, the following problem was proposedi1! i iii
A secretary had written n different letters and addressed n different envelopes for them. Unfortunately, a wind storm mixed up the letters and the envelopes. After the storm was over, each letter was placed in an envelope. In how m a n y ways can the letters be placed in the envelopes, so t h a t every letter is in a w r o n g envelope?
i i i i ii i i i i ill
This problem has several variations. One involves n couples a t t e n d i n g a dance. In how m a n y ways can the men dance with women other t h a n their own wives? A second variation involves n guests checking in their coats at the coat room of a fancy r e s t a u r a n t . In how m a n y ways can the a t t e n d a n t r e t u r n their coats, so no person gets the right coat? Before answering these problems, we make the following definition.
Derangement
A p e r m u t a t i o n of n distinct items a 1, a 2 , . . . , a , in which no item ai appears in its original position i for any i, 1 _< i _< n, is called a d e r a n g e m e n t . We would like to find the n u m b e r of possible d e r a n g e m e n t s of n items, so we begin with an example. ~ Find the n u m b e r of d e r a n g e m e n t s of the elements 1, 2, 3, and 4. SOLUTION: There are nine d e r a n g e m e n t s of the four elements, namely: 2143 2341 2413 3142 3412 3421 4123 4312 4321
The p e r m u t a t i o n 2314 is not a d e r a n g e m e n t since 4 appears in its n a t u r a l position, m
6.3 Derangements
361
Let Dn denote the n u m b e r of d e r a n g e m e n t s of n items. T h e n Do  1, since t h e r e is one d e r a n g e m e n t with no elements. (This will be verified later.) T h e r e is no d e r a n g e m e n t with one element, so D1  0. T h e r e is exactly one d e r a n g e m e n t of the e l e m e n t s 1 and 2, namely, 21; therefore, D2  1. T h e r e are two d e r a n g e m e n t s of the elements 1, 2, and 3 : 2 3 1 and 312; therefore, D3  2. It follows by Example 6.15 t h a t D4  9.
To find an explicit formula for Dn, first we derive a r e c u r r e n c e relation satisfied by Dn. Let Dn denote the n u m b e r of d e r a n g e m e n t s of n distinct items. T h e n
Dn  ( n 
1)(Dn1 + Dn2),
n > 2
(6.1)
PROOF:
Let a l, a 2 , . . . , an denote the n items. I t e m an can be placed in a n y one of the positions 1, 2 , . . . , (n  1), so the location of an has n  1 choices. Suppose it is placed in position 1 (see Figure 6.13). Let us now consider the following two cases:
F i g u r e 6.13
an 1
. . . .
available
. . .
2
3
n
positions
C a s e 1 Suppose al is placed in position n (see Figure 6.14). T h e n n  2 items are left, namely, a 2 , a 3 , . . . ,an1 and also n  2 positions, namely, positions 2 t h r o u g h n  1. The n u m b e r of d e r a n g e m e n t s of n  2 items, by definition, is Dn2. Thus, if a l is placed in position n, t h e r e are Dn2 derangements.
F i g u r e 6.14
an
. . .
available
f
. .
~
.
.
,
,
a1
positions
362
Chapter6 Combinatoricsand Discrete Probability C a s e 2 Suppose a l is not placed in position n (see F i g u r e 6.15). T h e n a l m u s t occupy one of t h e positions 2 t h r o u g h n  1 a n d one of t h e i t e m s a2 t h r o u g h an1 m u s t occupy position n. Thus, we have n  1 i t e m s a n d n  1 positions, yielding Dn1 d e r a n g e m e n t s .
F i g u r e 6.15
available
not a 1
an
~~2",..~
9 S n
positions Thus, with an in position 1, a total of Dn1 +Dn2 d e r a n g e m e n t s are possible, by the addition principle. Since an m a y occupy a n y one of the n  1 positions, t h e total n u m b e r of d e r a n g e m e n t s is (n  1)(Dn1 + Dn2). T h u s Dn  ( n 1)(Dn1 +Dn2), n >_ 2. m U s i n g T h e o r e m 6.8, Dn can be defined recursively as follows.
A Recursive Definition of
Dn
DoDn1, (nD1  0 1)(D~_1 + Dn2), n > 2 (6.1)
T h e r e c u r r e n c e relation (6.1) can prove t h a t Do m u s t be 1. U s i n g t h e recurrence relation, D2 = (2  1)(D1 + Do). Since D2 = 1 a n d D1  0, this yields 1 = 1(0 + Do). F o r this to be true, Do m u s t be 1, as in t h e above definition. T h e next example illustrates this recursive definition. Use t h e recursive definition of Dn to c o m p u t e D3 and D4. SOLUTION: D3  (3  1)(D2 + D1) = 2(1 + 0) =2 D4  (4  1)(D3 + D2) = 3(2 + 1) =9 m
Notice t h a t these a n s w e r s agree w i t h those obtained earlier,
The above recursive definition can be used to derive an a l t e r n a t e definition of Dn.
6.3 Derangements
Notice t h a t the r e c u r r e n c e relation (6.1) can be r e w r i t t e n as
Dn  nDn1 :
363
(  1 ) [ D n  1  (n  1)Dn2]
dn =
To simplify this, we s u b s t i t u t e d n  D n  n D n  1 . T h e n it becomes dn1, w h e r e d l  D1  1D0 = 0  1(1)   1 . Solving this r e c u r r e n c e relation (see Exercise 9),
dn  (  1 ) n,
n _> 1
Thus
Dn  nDn_ l (  1)n
T h a t is,
Dn nDn_l
~ (1)n,
n >__ 1
(6.2)
Accordingly, D n can be defined recursively as follows: An Alternate Recursive Definition of Dn
Do1
D n  n D n  1 + (  1)n,
n>l
(6.3)
The next example uses this a l t e r n a t e definition. ~ With the a l t e r n a t e definition (6.3), c o m p u t e D 3 and D4. SOLUTION: D3 = 3D2 + (  1 ) 3 = 3(1) + (  1 ) =2 D4 = 4D3 + (  1 ) 4 = 4(2) + 1 =9
m
The r e c u r r e n c e relation (6.2) can be solved u s i n g the i t e r a t i o n m e t h o d (see Exercise 10). The solution is given in the following t h e o r e m . The n u m b e r of d e r a n g e m e n t s of n distinct e l e m e n t s is
1 1 1 1 ~.. + 2~  3~ + 4~ . . . . "
(  1 ) n II n > 0 
D n = n !
1
+
n!
A'
m
(_l)n
n!
r An i n t e r e s t i n g observation: It is s h o w n in calculus t h a t e1 = ~
n=O
so t h e expression inside t h e b r a c k e t s in t h e f o r m u l a is t h e s u m of t h e first (n + 1) t e r m s in t h e expansion of e 1. See Exercise 40.
364
Chapter6 Comblnatoricsand Discrete Probability The next example illustrates this theorem. ~ Using Theorem 6.9, compute D5. SOLUTION:
(
D5=57
1
1~q
1
2?
1
1
1)
5?
37 F 4?
(
=120 11t
1 1
2
1
1)
120
6 ~ 24
= 44 II Returning to the 18thcentury problem, n letters can be placed in wrong envelopes in Dn ways, where the value of Dn is given by Theorem 6.9. Exercises 6.3 Using the recursive definition (6.1), compute the number of derangements
Dn for each value of n.
1.5 2.6 3.7 4.10
Using Theorem 6.9, compute each. 5. D2 6. D4 7. D6 8. D7
9. Solve the recurrence relation dn   d n  1 , n >_2, where d l   1 . _ 10. Solve the recurrence relation (6.2). Prove each.
11. Dn is even if n is an odd integer.
12. Dn is odd if n is an even integer. Let bn denote the number of computations (additions and multiplications) needed to find Dn, using the recursive definition (6.1). Compute the following. 13. b2 14. b3 15. b4 16. b5
17. Define bn recursively. Using the recursive definition of bn, compute each. 18. b4 19. b5 20. b6 21. b8
22. Prove that bn is an even integer for every n >_ 0. Let Cn denote the number of computations (additions and multiplications) needed to find Dn, using the recursive definition (6.3). Compute each. 23. C2 24. c3 25. C4 26. c5
6.4 Combinations
365
27. Define
Cn
recursively.
Using the recursive definition of Cn, compute each.
28. c4 29. c5 30. c6 31. c8
32. Solve the recurrence relation obtained in Exercise 27. 33. Show that 34. Show that
Cn  O ( n ) .
D n  3 . 4 ..... n  4.5 ..... n ~ . . . q (  1 ) n  1 n + ( _ 1)n. Dn,
35. Let a n denote the n u m b e r of multiplications needed to compute using the formula in Exercise 34. Show t hat a n  O(n2). Letpn
Dn.
n! value of n.
36. 5
Compute the value
37. 6
ofpn
correct to six decimal places for each
38. 7 39. 10
*40. (For those familiar with the concept of a limit) Using Exercises 3639, predict lim P n .
n~
(1) n 41. Show t h a t p n n! 42. Using Exercise 41, derive the explicit formula for D n . ( H i n t : Solve the recurrence relation in Exercise 41.) The formula in Exercise 41 can be derived in a slightly different way also. Verify each. n1
43. Pn1 = ~ .
1
Pn1 +Pn2 n
44.
1
PnPn1 =
n
Pn
n
(Pn1Pn2)
(1) n
45.
Pn  Pn1 =
(Hint:
Let
n! gn = Pn  Pn 1
in Exercise 44.)
Dn.
46. Write a nonrecursive algorithm to compute 47. Write a recursive algorithm to compute
Dn.
Recall that a permutation is an ordered arrangement of elements in a set. Sometimes, however, the order of elements is irrelevant; only their membership is important. We will investigate such unordered arrangements in this section. For example, a committee such as A  {Costa, Shea,. Weiss, Hall, Chen} is just a set, and the order in which the names are listed is immaterial. Suppose we would like to form a subcommittee of A
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Chapter 6 Combinatorics and Discrete Probability
c o n s i s t i n g of t h r e e m e m b e r s . T h r e e s u c h s u b c o m m i t t e e s are: { C o s t a , S h e a , W e i s s } , {Costa, S h e a , Hall}, a n d { C o s t a , S h e a , C h e n } (see E x a m p l e 6.20). E a c h is a c o m b i n a t i o n of t h e five e l e m e n t s t a k e n t h r e e a t a t i m e , o r a
3combination.
More generally, we m a k e the following definition.
Combination
A n r  c o m b i n a t i o n of a s e t of n e l e m e n t s , w h e r e 0 < r < n, is a s u b s e t containing r elements. T h e n u m b e r o f r  c o m b i n a t i o n s o f a s e t w i t h n e l e m e n t s is d e n o t e d b y
C(n,r) o r ( n ) . * B o t h n o t a t i o n s f r e q u e n t l y a p p e a r in c o m b i n a t o r i c s . T h e
n u m b e r of c o m b i n a t i o n s is also c a l l e d t h e b i n o m i a l c o e f f i c i e n t * * for r e a s o n s t h a t will b e c l e a r f r o m S e c t i o n 6.6. B e f o r e d e r i v i n g a f o r m u l a for C(n, r), let u s s t u d y t h e f o l l o w i n g e x a m p l e . F i n d t h e n u m b e r of r  c o m b i n a t i o n s of t h e s e t {a, b, c}, w h e n r or3. SOLUTION: 9 E x a c t l y o n e s u b s e t c o n t a i n s z e r o e l e m e n t s : t h e n u l l set. N u m b e r of 0  c o m b i n a t i o n s  C(3, 0)  1. 9 T h r e e s u b s e t s c o n t a i n o n e e l e m e n t each" { a}, {b }, a n d { c }. N u m b e r of 1  c o m b i n a t i o n s  C(3, 1)  3. 9 T h r e e s u b s e t s c o n t a i n t w o e l e m e n t s each" {a, b }, { b, c }, a n d { c, a }. N u m b e r of 2  c o m b i n a t i o n s  C(3, 2)  3. 9 F i n a l l y , e x a c t l y o n e s u b s e t c o n t a i n s t h r e e e l e m e n t s : t h e s e t itself. N u m b e r of 3  c o m b i n a t i o n s  C(3, 3)  1. W e n o w d e r i v e a f o r m u l a for C(n, r). T h e n u m b e r of r  c o m b i n a t i o n s of a s e t of n e l e m e n t s is g i v e n b y C(n, r) n! 0 0. In particular, g5  C(5, 0) + C(5,2) + C(5, 4)  1 + 10 + 5  16 a n d g6  C(6, 0) + C(6, 2) + C(6, 4)  1 + 15 + 15  31, as expected. We shall p u r s u e this example a bit f u r t h e r in Section 6.6. m
Exercises 6.4 A c o m m i t t e e consists of nine m e m b e r s . F i n d t h e n u m b e r of s u b c o m m i t t e e s t h a t can be formed of each size. 1. Two 2. Five 3. Six 4. Seven
6.4
Combinations
373
5. Find the n u m b e r of ways a committee of three s t u d e n t s and five professors can be formed from a group of seven s t u d e n t s and 11 professors. 6. Find the n u m b e r of ways a committee of four students, four professors, and t h r e e a d m i n i s t r a t o r s can be formed from a group of six students, eight professors, and five administrators. 7. Find the n u m b e r of lines t h a t can be d r a w n using 10 distinct points, no three being collinear. 8. Find the n u m b e r of triangles t h a t can be drawn using 10 points, no three being collinear. 9. Solve the recurrence relation in Example 6.22.
10.
Prove the formula in Example 6.22, using induction.
Let A be a 10element subset of the set {1, 2 , . . . , 20}. 11. Determine if A has two fiveelement subsets t h a t yield the same sum of the elements. 12. Determine if A has two eightelement subsets t h a t yield the same sum of the elements. ( T w e l v e D a y s o f C h r i s t m a s ) Suppose t h a t on the first day of C h r i s t m a s you sent your love 1 gift, 1 + 2 gifts on the second day, 1 + 2 + 3 gifts on the third day, and so on. 13. Find the n u m b e r of gifts sent on the 12th day. 14. Find the total n u m b e r of gifts sent in 12 days. 15. Show t h a t the n u m b e r of gifts sent on the n t h day is C(n + 1, 2), where 1 n, the elements of each combination m u s t be repeated. Consequently, a combination may contain t h r e e a's, two a's and one b, one a and two b's, or three b's. Using the set notation, the 3combinations are {a,a,a}, {a,a,b}, {a,b,b} and {b,b,b}. So there are four 3combinations of a set of two elements, m Find the n u m b e r of 3combinations of the set {a, b, c}, where the elements may be repeated. SOLUTION: Again, using the set notation, the various 3combinations are:
{a,a,a} {b,b,b}
{c,c,c}
Thus the set
{a,a,b}
{a,a,c}
{a,b,b}
{a,b,c}
{a,c,c}
{b,b,c} {b,c,c}
{a,b,c} has 10 3combinations.
m
Before developing a formula for the n u m b e r of combinations with repetitions, let us r e t u r n to the beverage problem posed earlier. Five friends would like to order beverages with their d i n n e r at a local r e s t a u r a n t t h a t serves iced tea, hot tea, or coffee. Find the n u m b e r of beverage orders possible. SOLUTION: A convenient notation will prevent confusion.
380
Chapter 6
Combinatorics and Discrete Probability
Denote each type of beverage by a dash and separate t h e m u s i n g two slashes, as shown below: / iced tea hot tea / coffee
M a r k each person's selection by an X in the appropriate area. For instance, the distribution XX / X / XX indicates t h a t two people selected iced tea, one selected hot tea, and two selected coffee; t h e distribution XXX / / XX means, three people selected iced tea, none ordered hot tea, and two selected coffee. T h u s the n u m b e r of possible beverage orders equals the n u m b e r of p e r m u t a t i o n s of seven items (five X's and t w o / ' s ) of which five are alike (X's) and the other two are alike (/'s): 7? =21 5?2? This solution strategy produces the following theorem. The n u m b e r of rcombinations with repetitions from a set of n e l e m e n t s is
C(n + r PROOF"
m
1, r).
Each rcombination with repeated elements from a set of n e l e m e n t s can be considered a string of r X's and (n  1) slashes, as in Example 6.32. Each string contains r + n  1  n + r  1 symbols, of which r are alike (X's) and n  1 are alike (slashes). Therefore, by T h e o r e m 6.13, the n u m b e r of such strings, t h a t is, rcombinations, equals
(n + r  1)?
r?(n1)?
=C(n+rl,r)
m
This theorem helps solve the next example. There are five types of soft drinks at a fast food r e s t a u r a n t : Coke Classic, Diet Coke, root beer, Pepsi, and Sprite. Find the n u m b e r of beverage orders 11 guests can make. SOLUTION: Since there are five types of soft drinks, n = 5. Each beverage order is a selection containing 11 items, t h a t is, an l 1combination with r e p e a t i n g elements. Therefore, by T h e o r e m 6.14, the n u m b e r of possible beverage orders equals
C(n + r 
1, r)  C(5 + 11  1, 11) = C(15, 11) = 157 = 1365 11747
m
6.5 Permutations and Combinations with Repetitions
381
This problem has a nice i n t e r p r e t a t i o n . Let x i denote the n u m b e r of guests ordering soft d r i n k i, w h e r e 1 < i < 5. T h e n xl +x2 +x3 +x4 +x5 = 11, where x i > O. T h e n u m b e r of n o n n e g a t i v e integer solutions of this e q u a t i o n is the same as the n u m b e r of possible beverage orders, so the n u m b e r of integer solutions of this e q u a t i o n is C(5 + 11  1, 11)  C(15, 11)  1365. The next t h e o r e m generalizes this result, simply r e s t a t i n g T h e o r e m 6.14. Let X l , X 2 , . . . ,Xn be n n o n n e g a t i v e integer variables and integer. The e q u a t i o n X l + x 2 + . . . + X n  r has C ( n + solutions, F i n d the n u m b e r of solutions of the equation Xl ~x2 + x3  5 where Xl,X2, and x3 are n o n n e g a t i v e integer variables. SOLUTION: H e r e r  5 and n  3. By T h e o r e m 6.15, the n u m b e r of solutions is
C(n + r r a r
nonnegative 1, r) integer m
(6.6)
1, r)  C(3 + 5  1, 5) = C(7, 5)  21
(Can you list all the solutions? See E x a m p l e 6.35 also.)
m
T a k i n g this example a step further, suppose you would like to find all solutions of E q u a t i o n (6.6), w h e r e Xl, x2, x3 > 1. M a k e the s u b s t i t u t i o n Y i  x i  1, 1 _< i _< 3. Clearly, Y i >_ O. E q u a t i o n (6.6) becomes Yl +Y2 +Y3  2 By T h e o r e m 6.15, this equation has C ( n + r  1, r) = C(3 + 2  1, 2) C(4, 2)  6 solutions: (0, 1, 1), (1, 0, 1), (1, 1, 0), (2, 0, 0), (0,2, 0), and (0, 0,2). Consequently, E q u a t i o n (6.6) with x i > 1 has six solutions" (1, 2, 2), (2, 1,2), (2,2, 1), (3, 1, 1), (1,3, 1), and (1, 1,3).
F o r Loops Revisited
for
The following two examples provide applications of both these formulas to loops.
Find the n u m b e r of times the a s s i g n m e n t s t a t e m e n t x ~ x + 1 is executed by the following nested f o r loops:
For i = 1 t o n do For j = 1 t o i do
1 x ~x+
382
Chapter6 Combinatoricsand Discrete Probability SOLUTION: Notice t h a t t h e a s s i g n m e n t s t a t e m e n t is e x e c u t e d for each p a i r (j, i), w h e r e 1 < j _< i < n. F o r example, t h e s t a t e m e n t is e x e c u t e d w h e n j  3, i  5; j  3, i  3; b u t not w h e n j  3 a n d i  2. T h u s t h e n u m b e r of e x e c u t i o n s equals t h e n u m b e r of pairs (j, i) w i t h r e p e t i t i o n s allowed; in o t h e r w o r d s , it equals t h e n u m b e r of 2selections of t h e set {1, 2, 3 , . . . , n}. T h e n u m b e r of such 2selections is C(n + 2  1,2), so t h e a s s i g n m e n t is e x e c u t e d in C(n + 2  1, 2)  C(n + 1, 2)  tn different ways. m
We now p u r s u e this e x a m p l e w i t h one m o r e level added to t h e for loops. F i n d t h e n u m b e r of t i m e s t h e a s s i g n m e n t s t a t e m e n t x ~ x + 1 is e x e c u t e d by t h e following n e s t e d for loops"
For i = 1 to n do i do j do For j For = 1 to
k = I to
x ~  x + l
SOLUTION: T h e a s s i g n m e n t s t a t e m e n t is e x e c u t e d for each t r i p l e t (k, j , i), w h e r e 1 _~ k _ 0 a n d i + j § k  5. Let x i denote t h a t iced t e a was ordered by i c u s t o m e r s . Since 0, 1, 2, 3, 4, or 5 people can order iced tea, we use t h e polynomial 1 § x § x 2 § x 3 § x 4 § x 5 to r e p r e s e n t t h e various possibilities. (Notice t h e exponents.) Both hot tea a n d coffee can also be ordered by 0, 1, 2, 3, 4, or 5 people; correspondingly, t h e polynomials 1 + y § y2 § y3 § y4 § y5 a n d 1 § z § z 2 § z 3 § z 4 § z 5 r e p r e s e n t these possibilities. C o n s e q u e n t l y , we w a n t p r o d u c t s of the form xiyJz k, w h e r e i § j § k  5 in t h e p r o d u c t
t x/timothy!
i=o = k=O For convenience, let x  y  z. T h u s the total n u m b e r of beverage orders
possible is the coefficient o f x 5 in the product ( ~ xi) a. i=0 You m a y verify t h a t it is 21. Table 6.2 shows the various possible beverage orders.
T a b l e 6.2
Iced tea 0 0 0 0 0 0 1 1 1 1 1
Hot tea
Coffee
Iced tea
Hot tea
Coffee
0 1 2 3 4 5 0 1 2 3 4
5 4 3 2 1 0 4 3 2 1 0
2 2 2 2 3 3 3 4 4 5
0 1 2 3 0 1 2 0 1 0
3 2 1 0 2 1 0 1 0 0
The next p r o b l e m r e s e m b l e s E x a m p l e 6.34, w i t h additional c o n s t r a i n t s on the variables. F i n d the n u m b e r of solutions of t h e e q u a t i o n xl + xl > 7, 1 < x2, x3 < 3, and 0 < x4 < 3.
X2 +
X3 +
X4  
11, w h e r e
384
Chapter 6 Combinatorics and Discrete Probability
SOLUTION: Again, g e n e r a t i n g functions can do t h e job for us. Since xl > 7 , x l can be 7, 8, 9, 10, or 11, r e p r e s e n t e d by the polynomial x 7 + x 8 + . . . + x 11 T h e polynomials c o r r e s p o n d i n g to the c o n s t r a i n t s 1 _< x 2 , x 3 ~ 3 are x + x 2 + x 3 and x + x 2 + x 3. For 0 < x4 < 3, we have t h e polynomial 1 + x + x 2 + x 3. T h u s the n u m b e r of solutions is t h e coefficient of x 11 in t h e p r o d u c t
t h a t is, t h e coefficient of x 2 in the p r o d u c t
( xi) ( xi) ( xi) (ioxi)
You can verify t h a t it is 10. T h u s the e q u a t i o n has 10 solutions satisfying t h e given conditions. [The various solutions ( i , j , k , l) can be o b t a i n e d by picking the e x p o n e n t s in the p r o d u c t s xixJxkx l t h a t yield x 11 in t h e product (6.7).] m We close this section with an e x a m p l e t h a t is closely r e l a t e d to E x a m p l e 6.38. In how m a n y ways can 11 cookies be d i s t r i b u t e d a m o n g four c h i l d r e n m Amy, Betsy, Carol, and D a i s y   s o t h a t A m y gets at least seven cookies, both Betsy and Carol get at least one cookie each but not m o r e t h a n t h r e e , and Daisy gets no more t h a n t h r e e cookies? SOLUTION" Since A m y gets at least seven cookies, this case yields the p o l y n o m i a l x 7 + x s + x 9 + x 1~ + x 11. Similarly, the o t h e r t h r e e c o n s t r a i n t s yield t h e polynomials x + x 2 + x 3, x + x 2 + x 3, and 1 + x + x 2 + x 3. T h u s the n u m b e r of ways of d i s t r i b u t i n g 11 cookies u n d e r t h e given conditions is the coefficient of x 11 in the product (6.7) above, namely, 10. m Exercises 6.5 Find t h e n u m b e r of distinct words t h a t can be formed by s c r a m b l i n g t h e letters in each word. 1. C A L C U L U S Find the n u m b e r of bytes that: 3. C o n t a i n exactly two 0's. 5. C o n t a i n at least five 0's. 7. C o n t a i n exactly eight 0's. 4. C o n t a i n exactly five 0's. 6. C o n t a i n not m o r e t h a n two 0's. 8. C o n t a i n exactly nine 0's. 2. T A L L A H A S S E E
6.5
Permutations and Combinations with Repetitions
385
Find the num ber of ternary words over the alphabet {0, 1, 2} that are of length four and: 9. Contain exactly three O's. 11. Contain at most two 0's. 12. Contain two O's, three l's, and three 2's. In Exercises 1316, use Figure 6.17 to find the num ber of possible routes from A to the given point, traveling easterly or northerly for the given number of blocks. 13. Point F and 5 blocks. 15. Point D and 9 blocks. List the 4combinations of each set. 17. {a} 18. {a,b} 14. Point C and 8 blocks. 16. Point E and 11 blocks. 10. Contain at least six O's.
19. There are five types of desserts available at a restaurant. Find the number of ways eight people can select them, if order does not matter. 20. A restaurant offers six choices for the main dish. How many ways can a group of nine women select the main dish? Assume that order does not matter. 21. In how many ways can 10 quarters in a piggy bank be distributed among 7 people? Find the number of solutions to each equation, where the variables are nonnegative integers. 22. Xl +x2 +x3 = 3 24. xl +x2 +x3 +x4 = 10 23. Xl +x2 +x3 +x4 = 7 25. Xl +x2 +x3 +x4 +x5 = 11
Find the number of solutions to each equation, where xi > 1. 26. x l + x 2 + x 3 + x 4 = l l 27. x l + x 2 + x 3 + x 4 + x 5 = 1 3
Use generating functions to solve the following counting problems.
28. Use generating function (6.7) to find the various solutions of the equation in Example 6.38. Find the number of solutions to each equation.
29. x l + x2 + x3
=
10, Xl > _3, 1 < X2 _< 3, X3 > 5 _
30. X l + X 2 + X 3 = 1 2 , Xl, X 2 > 5 , 1 < X 3 < 4 31. X l + X 2 + X 3 + X 4 = 1 0 , 32. X l + X 2 + X 3 + x 4 = l l , Xl, X 2 > 2 , X 3 > 0 , X 4 > 5 x1, x 2 > 2 , 2 < X 3 < 4 , X4>3
3334. Find the solutions to the equations in Exercises 29 and 30.
386
Chapter 6
Combinatorics and
Discrete Probability
35. Find the number of ways 10 quarters can be distributed among three peopleAaron, Beena, and C a t h y   s o that both Aaron and Beena get at least one quarter, Beena gets no more than three, and Cathy gets at least two. 36. Find the number of ways 11 raisins can be distributed among four childrenDaisy, Emily, Francis, G a l l   s o that Daisy, Emily, and Francis get at least two raisins, Francis gets no more than four, and Gail gets at least three.
The binomial coefficients satisfy a vast array of properties. We shall visit a few of them shortly.
Pascal's Triangle
The various binomial coefficients ( n ) , where 0 _< r _< n, can be arranged in the form of a triangle, called P a s c a l ' s t r i a n g l e , * as shown in Figures 6.20 and 6.21.
Figure 6.20
(0) (1) (~1) (20) (21)0 (0)(]) (~)0 (4)0 (~)(~)(4)
1 1 1 2 3 ~, 4 6 3 4 1 1 1 1
e r o w 0 e r o w l ( r o w 2 e r o w 3 ( r o w 4
Figure 6.21
e row0 ( r o w 1 e row2 e r o w 3 e r o w 4
Pascal's triangle has many intriguing properties: 9 Every row begins with and ends in 1. This is no coincidence, since
C ( n , O) = 1 = C ( n , n ) .
*Although Pascal's triangle is n am e d after Pascal, it appeared in a 1303 work by the Chinese m a t h e m a t i c i a n Chu ShiKie.
6.6
The Binomial Theorem
387
9 Pascal's triangle is symmetric about a vertical line through the middle. This is so since C(n, r) = C(n, n  r). 9 Any interior number in each row is the sum of the numbers immediately to its left and to its right in the preceding row. This is so by virtue of Pascal's identity. 9 Recall that, by Example 6.22, the maximum number of places fn into which a pizza can be divided with n distinct cuts is given by fn = C(n, O) + C(n, 1) + C(n, 2). It is the sum of the first three numbers in row n of Pascal's triangle (see Figure 6.22).
Figure 6.22
fl~ f2~ f3~
(1
(1 4 3 6) 3)
1)
1 4 1
f4 ~ (1
9 By Example 6.26, the maximum number of nonoverlapping regions formed inside a circle by joining n distinct points on it is given by gn = C(n, O) + C(n, 2) + C(n, 4). Using Pascal's identity, this formula can be rewritten a s g n = C ( n  1,0) + C ( n 1,1)+C(n1,2)+ C ( n  1, 3 ) + C ( n  1, 4). (Verify this.) Consequently, the value of gn can be obtained by adding the first five numbers in row n  1 of Pascal's identity (see Figure 6.23).
Figure 6.23
gl> g3 g4 g5 g6 ~ (1 ~ (1 ~ ~ 4 ~ ~'1 5
(1) 2 3 6 10 10 3
0 1)
Pascal's Triangle and Abracadabra
Next we show how Polya's abracadabra problem is related to Pascal's triangle. Beginning with a 1 at the apex and using Pascal's identity, build the rhombic array in Figure 6.24. Each entry in Figure 6.24 gives the number of paths from the apex to the corresponding location. So there are 252 = C(10, 5) possible paths from the apex to the bottommost point in the array.
388 Figure 6.24
Chapter6 Combinatoricsand Discrete Probability
1 1 1 1 1 1 6 21 56 126 252 5 15 35 70 126 4 10 20 35 56 3 6 10 15 21 2 3 4 5 6 1 1 1 1 1
Next we show how Catalan n u m b e r s 1, 1, 2, 5, 14, 4 2 , . . . can be extracted from Pascal's triangle.
Pascal's Triangle and Catalan Numbers
In C h a p t e r 2 we defined the n t h
Catalan number C,,
(2n)!
n >_0
as
C, = ( n + l ) ! n ! '
(2n)!
Since n i n i =
(2n)
n
, this can be r e w r i t t e n as
n)
n+l '
n~0 
Consequently, C,z can be obtained by dividing the
central binomial coeffi
cient(2;)byn+l.
1(48 )  70 = 14; t h u s C4 is obtained by dividing by For example, C4  ~ 5 5 the central element 70 in row 8 in Pascal's triangle. Figure 6.25 shows the first five central binomial coefficients; we can use t h e m to c o m p u t e the first five Catalan n u m b e r s . Catalan n u m b e r s , like Fibonacci and Lucas n u m b e r s , have a propensity to appear in quite unexpected places. We shall p u r s u e a few such delightful occurrences later in this Section and in Section 9.6. The next t h e o r e m shows how the binomial coefficients and T h e o r e m 6.13 are useful in finding the binomial expansion of (x + y)n. We shall prove it using a combinatorial a r g u m e n t (see Exercise 43 for an algebraic method).
6.6 The Binomial Theorem
389
F i g u r e 6.25
The central binomial coefficients. 1 1 1 1 1 1 8 7 28 6 21 56 5 15 35 O 4 10 Q 35 56 1 1
1
3

3 Q 10
1
1 4 5 15 21 28 6 7 8 1 1 1 1 1
( T h e B i n o m i a l T h e o r e m ) * Let x a n d y be a n y real n u m b e r s , a n d n a n y n ( n) n o n n e g a t i v e integer. T h e n (x + y)n _ ~ xnryr.
r0
PROOF: Since (x + y)n _ (x + y)(x + y)... (x + y) to n factors, (x + y)n is e x p a n d e d by m u l t i p l y i n g an x from some of t h e factors on t h e R H S a n d a y f r o m t h e r e m a i n i n g factors. T h a t is, every t e r m is o b t a i n e d by selecting an x f r o m a n y of t h e n  r factors a n d a y from t h e r e m a i n i n g r factors. Thus, every t e r m in t h e e x p a n s i o n is of the form Cxnry r, w h e r e C d e n o t e s t h e coefficient a n d 0_ 1,
rl
0  I1 + (  1 ) i n  ~~ ( n ) l n  r (  1 ) r
r0
(o)
T h a t is,
I
We now m a k e an i n t e r e s t i n g observation. Recall t h a t the binomial coefficient C(n,r) denotes the n u m b e r of subsets of size r of a set with n elements, so the L H S of E q u a t i o n (6.8) r e p r e s e n t s the total n u m b e r of subsets with an even n u m b e r of elements and the RHS r e p r e s e n t s t h a t with an odd n u m b e r of elements. Since the total n u m b e r of subsets is 2 n, each equals 2 n  l , by E q u a t i o n (6.8).
6.6 The Binomial Theorem
393
Next we t u r n to two i n t e r e s t i n g occurrences of Catalan n u m b e r s .
Catalan Numbers Revisited
The great Swiss m a t h e m a t i c i a n L e o n h a r d E u l e r (see C h a p t e r 8) discovered Catalan numbers. He found t h e m in his study of triangulations of convex polygons, t h a t is, dividing the interior of a convex polygon into t r i a n g u l a r areas by drawing nonintersecting diagonals. Let Cn denote the n u m b e r of triangulations of convex (n + 2)gon, where n > 1. It follows from Figure 6.26 t h a t C1  1, C2 = 2, C3  5, and C4 = 14. More generally, Euler established t h a t
Ca
2 96 . 1 0 . . . ( 4 n  2) (n + 1)!
F i g u r e 6.26
Triangulations of convex (n + 2)gons.
CI= 1 C2=2
C35
C5=14
In 1759, the G e r m a n m a t h e m a t i c i a n J o h a n n Andreas von Segner (17071777), a contemporary of Euler, established a recursive procedure to compute Cn"
Cn CoCn1 + C1Cn2 '~ " " ~ C n  I C o
where Co  1. For example, C4  1 . 5 + 1 . 2 + 2 . 1 + 5 . 1 = 14.
394
Chapter6 Combinatoricsand Discrete Probability Catalan's Parenthesization Problem
Interestingly, Euler's triangulation problem is essentially the same as Catalan's parenthesization problem, which he solved in 1838: using n pairs of left and right parentheses, how many different ways we can parenthesize a sequence of n + 1 symbols for a binary nonassociative operation? For example, with two symbols, there is one possibility: (ab); with three symbols, there are two possibilities: ((ab)c) and (a(ab)); and with four symbols, there are five ways: ((ab) (cd)), (((ab)c)d), (a(b(cd))), (a((bc)d), and ((a(bc))d); they yield the Catalan numbers 1, 2, and 5; and so on. In 1961, H.G. Forder of the University of Auckland, New Zealand, showed that every triangulation of convex polygon yields a correctly parenthesized expression, and vice versa. This close relationship becomes clear if we consider the triangulation of the hexagon in Figure 6.27. We have labeled five of its sides a through e, leaving the base unlabeled. Label every diagonal spanning two adjacent sides with the concatentation of the corresponding labels in parentheses. Continue this algorithm until the base gets a label, as in Figure 6.28. Thus the triangulation in Figure 6.27 yields the correctly parenthesized expression (((ab)c)(de)). Retracing the steps, we can recover the triangulation from the expression.
Figure 6.27
Triangulation of a hexagon. b L
C
d
Figure 6.28
Parenthesized triangulation of a hexagon.
C
d
e
(((ab)c) (de))
Triangulations, Parenthesized Expressions, and Binary Numbers Each triangulation of a convex polygon and by extension the corresponding parenthesized expression, can be uniquely represented by a binary number. For instance, consider the expression (((ab)c)(de)) in Figure 6.28. If we replace each left parenthesis with a 1 and each letter with a 0 and ignore all right parentheses, the expression yields the binary number 111000100. We do not need to store the right parentheses, because if we know the left parentheses, we can insert the matching right parentheses correctly.
6.6 The Binomial Theorem
395
T h u s such a b i n a r y n u m b e r is a compact way of r e p r e s e n t i n g the expression and hence the triangulation. We shall revisit this b i n a r y designation in Section 9.6 on b i n a r y trees. Before closing this section, we derive the explicit f o r m u l a for C n . T h e proof employs the following g e n e r a t i n g function. (1  4x) 1/2  1  2 ~ 1 ( 2 n n   } ) x n n=l n
An Explicit Formula for the nth Catalan Number (optional)
Let P n denote the n u m b e r of ways of p a r e n t h e s i z i n g n + 1 symbols u s i n g n pairs of left and right p a r e n t h e s e s . T h e n P1  1. So, let n > 2. T h e first i symbols can be p a r e n t h e s i z e d in P i ways and the next n  i symbols in P n  i ways, where I _< i _ <  1. Using the multiplication and addition principles, %n we have the following r e c u r r e n c e relation for Pn"
P n  P 1 P n  1 + P2Pn2 +"" + PnIP1
(It follows from S e g n e r ' s formula t h a t P n  C n  1 . ) Now consider the g e n e r a t i n g function
f (x)  P l x + P 2 x2 + . . . + P n x n + . . .
Then If(x)] 2  p 2 x 2 + (P1P2 + P 2 P 1 ) x 3 + ' "
+ " "" + PnIP1)x n + """ = X 2 + P 3 x3 + . . . +Pnx n + . . . +
(P1P,~I + P 2 P n  2
=f(x)x
So
[f(x)] 2  f ( x ) + x  0 Solving, 1+ x/lf(x) 
4x
Since P n > 0 for every n, we t a k e the m i n u s sign, so
1~/i4x
f(x) 
396
Chapter 6 Combinatorics and Discrete Probability
Using the above power series expansion, this yields
f(x) = ~ 
n=l n
xn
Thus
Pnl_n (2:~2) Cn1
So
Cn as desired.
Exercises 6.6
n+l
1(2;)
'
n>0 
Find the coefficient of each.
1. xny 5 in the expansion of (x + y)8 2. x4y 6 in the expansion of ( x  y)10 3. x2y 6 in the expansion of (2x + y)8 4. x4y 5 in the expansion of (2x  3y) 9
Using the binomial theorem, expand each. 5. (x +y)4 6. ( x  y ) 5 7. ( 2 x  1 ) 5
8. (x + 2y) 6
Find the middle term in the binomial expansion of each.
9.
(1)
11.
(
x+
4
x
10.
(1)
x
2x+
x
14. (x + y)6
12.
( )1o
x2 + ~5
16. (x + y)8
x

6
Find the largest binomial coefficient in the expansion of each.
13. (x + y)5 15. (x + y)7
17. Using Exercises 1316, predict the largest binomial coefficient in the expansion of (x + y)n. Use Pascal's triangle in Figure 6.21 to answer Exercises 18 and 19. 18. Find the sum of the numbers along the northeast diagonals.
19. What do you notice about them?
6.6 The Binomial Theorem
397
The nth Fibonacci number Fn is given by the sum of the numbers along the nth northeast diagonal of Pascal's triangle; that is,
[(n1)/2J
Fn "
E
(ni1)~.
i=0
Using this formula, compute each Fibonacci number.
20. F1 21. F2
22. F5
23. F6
The Bell n u m b e r s Bn, named after the English mathematician Eric T. Bell (18831960) and used in combinatorics, are defined recursively as follows: B0=I
Bn=~(nl) i Bi,
n>l
i=0
Compute each Bell number. 24. B2 25. B3
26. B4
27. B5
Using the binomial theorem, prove each. 28. 24n + 3n  1 is divisible by 9. (Hint: 2 = 3  1.) 29. 42n + 10n  1 is divisible by 25. (Hint: 4 = 5  1.) 30. 31.
Y~ r=0 2r r=l n (;) }~2r 3 n r=0 2r 1
n(2n) (2n )int:UseheoremO18
32. r~0 (;)(n nr)(22)
[Hint: Consider (1 + x) 2n = (1 + x)n(1 + x) n. Equate the coefficients of x n from either side.]
33. i~1 (in l ) ( ? )
 (n2+nl)
[Hint: Consider (1 + x) 2n = (x + 1)n(1 + x) n. Equate the coefficients of x n+l from both sides.]
Evaluate each sum. 34. 1 ( 1 ) + 2 ( 2 )+3(3 )+...+n(:)
(Hint: Let S denote the sum. Use S and the sum in the reverse order to compute 2S.)
398
Chapter Combinatorics DiscreteProbability 6 and
35. a ( ; ) + ( a + d ) ( n l ) + ( a + 2 d ) ( 2
(Hint:
)+...+(a+nd)(n)
Use the same hint as in Exercise 34.) where
n+l 36. Show that C ( n , r  1) < C ( n , r ) if and only if r < ~ , 2 0 < r < n.
37. Using Exercise 36, prove that the largest binomial coefficient C ( n , r ) occurs when r = [n/2J. Using induction, prove each. 38" ( ; ) + (
(Hint:
n + l l ~/+(n+2)2 +  . . + ( n + r ) r Use Pascal's identity.) = n 2 n1 ( n ) 2 = (2n)
= (n+r+l)r
39. 1 ( 1 ) + 2 ( 2 ) + . . . + n ( n ) 40. ( ; ) 2 + ( 1 ) 2 + ( 2 ) 2 + . . . +
From the binomial expansion (1 + x) '~  ~ calculus that n(1 + x) "1  ~ 41.
,~ (n)
x r,
it can be shown using
,~ (n)
r0
rx '  1 .
Using this result, prove each. n1 . . . . n2 n2
r=l
1(1 ) +2(:)+3(3 ) +...+n( n)n2
42. 1 ( 1 ) + 3 ( 3 ) + 5 ( 5 ) + . . . . 2 ( 2 ) + 4 ( 4 ) + 6 ( 6 ) +
43. Prove the binomial theorem, using mathematical induction. *44. Using a combinatorial argument prove that
(n) (m)= (n)(n r)
(Hint:
(Newton's identity)
Select an relement subset of an nelement set in two ways.)
45. Prove the result in Exercise 44 algebraically. The following result is known as Vandermonde's identity, after the German mathematician AbnitTheophile Vandermonde (17351796):
(m :nl  (0) (nl + (~) (r n l)~ (~) (r n2)~ ~ (mr)(0)
6.7 The Generalized InclusionExclusion Principle (GIEP) (optional)
399
*46. Prove V a n d e r m o n d e ' s identity, using a combinatorial a r g u m e n t . (Hint: Consider the ways of selecting r people from a group of m m e n and n women.) 47. Prove V a n d e r m o n d e ' s identity algebraically. [Hint: Consider (1 + x)m(x + 1) n = (1 + X) m+n. ]
n (:)
48. Find a formula for ~ .
i=2
49. Using induction, establish the formula guessed in Exercise 48. 50. Find a formula for ~ .
i=3
51. Using induction, establish the formula guessed in Exercise 50. 52. Using Exercises 4851, predict a formula for ~ i=k .
The generalized version of the inclusionexclusion principle has interesting applications to n u m b e r theory, surjections, and d e r a n g e m e n t s , as will be seen shortly. ( G I E P ) Let A 1 , A 2 , . . . ,An be n finite sets. T h e n
i=1
U Ai
 E Inil E
1_ 1. So
(X2
S~  U Sn C S
n=l
m
Thus S ~ cS By assumption, R c S; so R ~ c S ~ , since every p a t h in R is also a p a t h in S. Therefore, R ~ _ S. Consequently, S c R ~ and R ~ __ S. Therefore, S = R ~ . In o t h e r words, t h e r e are no transitive relations in between R and R ~. So R ~ is the smallest transitive relation containing R. m It follows by T h e o r e m s 7.7 and 7.8 t h a t R*  R U R 2 u ... u R" and hence
M R ~  M R v MR.2 v . . . v MR,,
To illustrate this, using Example 7.36, the transitive closure of the relation R = {(a,b), (b,a), (b,c), ( c , d ) , (d,a)} on { a , b , c , d } is R* = R ~ = {(a,a), (a, b), (a, c), (a, d), (b, a), (b, b), (b, c), (b, d), (c, a), (c, b), (c, c), (c, d), (d, a), (d, b),
(d,c),(d,d)}.
Since R ~ = R*, the connectivity relation algorithm can be used to compute M R , , but it is not efficient, especially when M R . is fairly large. A b e t t e r m e t h o d to find R* is W a r s h a l l ' s a l g o r i t h m , n a m e d in h o n o r of S t e p h e n Warshall, who invented it in 1962.
Warshall's Algorithm
Let a  x l  x 2 . . . . . xmb be a p a t h in a relation R on a set A  { a l , a 2 , . . . ,an}. The verticesxl, x2,... ,Xm are the i n t e r i o r p o i n t s of the path. For instance, vertices c and d are the interior points on the p a t h acdb of the d i g r a p h in Figure 7.5. The essence of Warshall's a l g o r i t h m lies in c o n s t r u c t i n g a sequence of n boolean matrices W 1 , . . . , Wn, b e g i n n i n g with Wo  M R . Let Wk  (w/j), where 1 < k < n. Define wij  1 if a p a t h r u n s from ai to aj in R whose interior vertices, if any, belong to the set {al, a 2 , . . . , ak}. Since the i j t h
478
Chapter 7 Relations
element of Wn equals 1 if and only if a path exists from ai to aN whose interior points belong to the set {a l, a 2 , . . . , an }, Wn = WR*. In fact, the m a t r i x Wk = (WIN) can be constructed from its predecessor W k  1 = (vij) as follows. When can wij = 1? For WiN = 1, t h e r e m u s t be a path from ai to aj whose interior vertices belong to the set {a l, a 2 , . . . , ak }.
Case 1
If ak is not an interior vertex, all interior vertices m u s t belong to the set {hi, a 2 , . . . , ak1 }, so vij : 1.
C a s e 2 Suppose ak is an interior vertex (see Figure 7.35). If a cycle exists at ak, eliminate it to yield a shorter path. (This g u a r a n t e e s t h a t the vertex ak occurs exactly once in the path.) Therefore, all interior vertices of the paths ai . . . . . ak and ak . . . . . aj belong to the set {a l, a 2 , . . , ak1}. In other words, Vik = 1 and Vkj = 1.
Figure 7.35
ak
/ /
ai
a/
Consequently, WiN = 1 only if v,ii = 1, or vik = 1 and vkj = 1. This is the crux of Warshall's algorithm. Thus the i j t h element of Wk is 1 if: 9 The corresponding element of Wt,._1 is 1 or 9 Both the i k t h element and the k j t h element of Wk1 are 1; t h a t is, the ith element in column k of Wk1 and t h e j t h element in row k of Wk1 are 1. Use this property to construct W1 from W0 = M R , W2 from W1,..., and Wn from Wn1. Since Wn = MR*, the actual elements of R* can be read from Wn. The next two examples clarify this algorithm. Using Warshall's algorithm, find the transitive closure of the relation R = {(a, b), (b, a), (b,c)} o n A = {a,b,c}. SOLUTION: S t e p 1 Find W0.
Wo   M R 
[il!]
0
0
Step2
Find W1.
If the i j t h element of W0 is 1, the i j t h element of W1 is also 1. In other words, every 1 in W0 stays in W1. To find the r e m a i n i n g l ' s in W1, locate
7.7 Transitive Closure (optional)
479
the l ' s in column 1 (  k); there is just one 1; it occurs in position i  2. Now locate the l's in row 1 (  k). Again, there is just one 1, namely, in position j  2. Therefore, the ijth entry in W1 should be 1, where i  2 a n d j  2. Thus
W1
EoI!I
1 0 1 0
S t e p 3 Find W2. Again, all the l's in W1 stay in W2. To find the other l's, if any, locate the l's in column 2( k) and row 2 (  k). They occur in positions 1 and 2 of column 2 and in positions 1, 2, and 3 of row 2, so the ijth entry of W2 must be 1, where i = 1, 2 and j = 1, 2, 3. So change the O's in such locations of W1 to l's. Thus
W2
Ei x 1]
1 0 1 0
S t e p 4 Find W3. All the l's in W2 remain in W3. To find the remaining l's, if any, locate the l's in column 3   namely, positions 1 and 2 m and the l's in row 3. Because no l's appear in row 3, we get no new l's, so W3  W2. Since A contains three elements, WR. = W3. Thus,
WR,
I
1 1 0
1 1 0
1 0
11
which agrees with the transitive closure obtained in Example 7.37.
II
Using Warshall's algorithm, find the transitive closure of the relation R = {(a, a), (a, b), (a, d), (b, a), (c, b), (c, c), (d, b), (d, c), (d, d)} on {a,b,c,d}. SOLUTION: S t e p 1 Find W0.
1 1 0 0 1 0 1 1 0 0 1 1 1 0 0 1
Wo = MR 
480
Chapter 7 Relations
Step 2 Find W1. Locate the l ' s in column 1 and row 1; positions 1 and 2 in column 1; and positions 1, 2, and 4 in row 1. Therefore, W1 should contain a 1 in locations (1,1), (1,2), (1,4), (2,1), (2,2), and (2,4):
1 1 W1 ~'0 0 (All the l's in W0 remain in W1.) 1 1 1 1 0 0 1 1 1 1 0 1
Step 3 Find W2. Locate the l's in column 2 and in row 2; positions 1, 2, 3, and 4 in column 2, and positions 1, 2, and 4 in row 2. So W2 should contain a 1 in locations (1,1), (1,2), (1,4), (2,1), (2,2), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), and (4,4). Again, since all the l's in W1 stay in W2,
1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1
W2
Step 4 Find W3. The l's of column 3 occur in positions 3 and 4; those of row 3 in positions 1, 2, 3, and 4. Consequently, W3 should contain a 1 in locations (i,j) where i = 3,4 a n d j  1,2, 3,4:
1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1
W3
Step 5 Find W4. The l's of column 4 appear in positions 1, 2, 3, and 4; the l ' s of row 4 in positions 1, 2, 3, and 4. So W4 should contain a 1 in locations (i,j) where i = 1, 2, 3, 4 a n d j = 1, 2, 3, 4:
1 i 1 1 1 1 1 1 1 1 1 1 1 1
Since MR.  W4, this is the adjacency matrix of the transitive closure. (Finding the connectivity relation of R will verify this.) m
77 TransitiveClosure (optional)
481
Warshall's algorithm is presented in Algorithm 7.2. It is based on the discussion preceding Example 7.38.
Algorithm Warshal] (MR,W)
(* This algorithm employs the adjacency matrix of a relation R on f i n i t e set with n elements to find the adjacency matrix MR* of i t s t r a n s i t i v e closure. *) O. Begin (* algorithm *) (* I n i t i a l i z e W = (wij) *) I. W +MR 2. for k = i to n do (* compute W *) k 3. for i = i to n do 4. for j = I to n do
5. Wij + Wij V (Wik A Wkj ) (*compute the i j  t h 6. MR +W 7. End (* algorithm *)
element *)
Algorithm 7.2
A Comparison of Warshall's Algorithm with the Connectivity Algorithm
Why is this algorithm far more efficient than the connectivity relation algorithm? Notice that the number of boolean operations in line 5 is 2, so the total number of boolean operations in lines 2 through 5 (and hence in the algorithm) is 2. n. n. n  2n 3  ()(n3), whereas the connectivity algorithm takes ()(n4) bit operations.
Exercises 7.7
Find the transitive closure of each relation on A = {a, b, c}. 1. {(a,b), (b,a)} 3. {(b,a), (b,c), (c,b)} 2. {(a,b), (b,c), (c,a)} 4. {(a,a), (a,c), (b,c), (c,a)}
Find the transitive closure of each relation on A = {a, b, c, d}.
5. {(a,a), (a,b)}
6. {(a,b), (b,c), (c,a)}
In Exercises 79, find the adjacency matrix of the transitive closure of each relation R on {a, b, c} with the given adjacency matrix.
7.
0 0
1 0
8.
[i0 ]
1 1 0 1
9.
[!IZ]
1 0 find the transitive
1012. Using the connectivity relation algorithm, closure R* of each relation in Exercises 79.
482
Chapter 7 Relations 1315. For the relation R on {a,b,c} with each adjacency matrix in Exercises 79, compute the boolean matrix W1 in Warshall's algorithm. In Exercises 1618, the adjacency matrix of a relation R on {a, b, c, d} is given. In each case, compute the boolean matrices W1 and W2 in Warshall's algorithm.
Z0 i
0 1 1 0
0
1
0
0
1 0 0
0 1 1
0 0 0
0
i
18.
i010
0 0 0 0
0
1
0
1
1 1
1924. Using Warshall's algorithm, find the transitive closure of each relation in Exercises 79 and 1618. 2533. The r e f l e x i v e c l o s u r e of a relation on a set is the smallest reflexive relation t h a t contains it. Find the reflexive closures of the relations in Exercises 19. Find the reflexive closure of each relation on IR. 34. The lessthan relation. 35. The g r e a t e r  t h a n relation.
3644. The s y m m e t r i c c l o s u r e of a relation on a set is the smallest symmetric relation t h a t contains it. Find the symmetric closures of the relations in Exercises 19. Let R be any relation on a set A. Prove each. 45. R is reflexive if and only if A _ _R. _ 46. R U A is reflexive. *47. R u A is the smallest reflexive relation containing R . (Hint: Assume there is a reflexive relation S such t h a t R c S c R u A. Prove that S  R or S  R u A.) *48. R U R 1 is symmetric. [Hint: Consider (R u R  1) 1.]
*49. R u R 1 is the smallest symmetric relation t h a t contains R. (Hint: Suppose there is a symmetric relation S such t h a t R __ S c_
RuR1.)
Section 7.4 introduced relations t h a t are reflexive, symmetric, and transitive. Naturally we can now ask: Are there relations t h a t simultaneously manifest all three properties? The answer is yes; for instance, the relation is logically equivalent to on the set of
7.8 Equivalence Relations
483
propositions has all t h e s e properties. Such a r e l a t i o n is an equivalence relation.
Equivalence Relation
A relation on a set is an e q u i v a l e n c e r e l a t i o n if it is reflexive, s y m m e t r i c , a n d transitive. E x a m p l e s 7.407.42 explore equivalence relations. ~ The relation has the s a m e color h a i r as on the set of people is reflexive, s y m m e t r i c , and transitive. So it is an equivalence relation, m Let Z denote an alphabet. Define a relation R on E* by x R y if Ilxli  liYll, w h e r e Ilwll denotes t h e length of t h e word w. Is R an equivalence relation? SOLUTION: 9 Since every word has the s a m e length as itself, R is reflexive. 9 Suppose t h a t x R y . T h e n IIxll = Ilyll, so Ilyll = IIxll. Consequently, y R x . T h u s R is s y m m e t r i c .
9 I f x R y 4 and y R z , t h e n Ilxli = IlYl] and Ilyli = ilzil. Therefore, and hence x R y . In o t h e r words, R is transitive.
lixll = Ilzli
m
Thus, R is an equivalence relation,
(optional) Is the relation has the s a m e m e m o r y location as on the set of variables in a p r o g r a m an equivalence relation? SOLUTION: 9 Since every variable has the s a m e location as itself, the relation is reflexive. 9 If a variable x has the s a m e location as a variable y, t h e n y has the s a m e location as x, so t h e relation is s y m m e t r i c . 9 Suppose x has t h e s a m e location as y and y has t h e s a m e location as z. T h e n x has the s a m e location as z, so the relation is transitive. T h u s the relation is an equivalence relation, m
F O R T R A N provides an e q u i v a l e n c e s t a t e m e n t , so called since t h e relation has the s a m e location as is an equivalence relation. We can see this in the following F O R T R A N s t a t e m e n t :
EQUIVALENCE (A,B),(C,D,E),(F,G,H)
It m e a n s the variables A and B s h a r e the s a m e m e m o r y location; the variables C, D, and E s h a r e the s a m e m e m o r y location; a n d so do the variables F, G, and H. The congruence relation, an i m p o r t a n t relation in m a t h e m a t i c s , is a classic example of an equivalence relation. It is closely r e l a t e d to the equality relation a n d p a r t i t i o n s of a finite set, as will be seen shortly.
484
Chapter 7 Relations
K a r l F r i e d r i c h G a u s s (17771855), son of a laborer, was born in Brunswick, Germany. A child prodigy, he detected an error in his father's bookkeeping when he was 3. The Duke of Brunswick, recognizing his remarkable talents, sponsored his education. Gauss received his doctorate in 1799 from the University of Helmstedt. In his doctoral dissertation, he gave the first rigorous proof of the fundamental theorem of algebra, which states, "Every polynomial of degree n (> 1) with real coefficients has at least one zero." Newton and Euler, among other brilliant minds, had attempted to prove it, but failed. He made significant contributions to algebra, number theory, geometry, analysis, physics, and astronomy. His impressive work Disquisitiones Arithmeticae of 1801 laid the foundation for modern number theory. From 1807 until his death, he was the director of the observatory and professor of mathematics at the University of GSttingen. Called the "prince of mathematics" by his contemporary mathematicians, Gauss made the famous statement, "Mathematics is the queen of the sciences and the theory of numbers the queen of mathematics. "
The congruence symbol _ was invented a r o u n d 1800 by Karl Friedrich Gauss, the g r e a t e s t m a t h e m a t i c i a n of the 19th century.
Congruence Relation
L e t a , b , m ~ Z, w h e r e m > 2. T h e n a is c o n g r u e n t to b m o d u l o m , denoted by a = b(mod m), if a  b is divisible by m. The i n t e g e r m is the m o d u l u s of the c o n g r u e n c e r e l a t i o n . (This definition provides the basis of the r o o d o p e r a t o r we studied in C h a p t e r 3.) If a is not c o n g r u e n t to b modulo m, we write a ~ b (mod m). For example, since 5 1 ( 1 3  3), 13 = 3 (mod 5). Also,  5 ~ 3 (mod 4) since 41(5  3). But 17 ~ 4 (mod 6), since 6 ~ (17  4). The congruence relation has several useful properties, some of which are given below. Let a, b, c, d , m ~ Z w i t h m
> 2. Then:
(1) a = a (mod m). ( r e f l e x i v e p r o p e r t y )
(2) I f a  b (mod m), t h e n b = a (mod m). ( s y m m e t r i c p r o p e r t y ) (3) I f a = b (mod m) and b _= c (mod m), t h e n a _ c (mod m). ( t r a n s i t i v e
property)
(4) Let r be the r e m a i n d e r w h e n a is divided by m. T h e n a ~ r ( m o d m).
PROOF:
We shall prove p a r t 3 and leave the o t h e r p a r t s as exercises. (3) Suppose a _= b (mod m) and b =_ c (mod m). T h e n m i ( a  b) and m l ( b  c). Consequently, a  b  m q l and b  c  mq2 for some
7.8 EquivalenceRelations
integers q l and q2. T h e n ac( a  b) + ( b  c)
+ mq2 +
485
= mql = m(ql
q2)
Therefore, m l ( a  c) and a  c (mod m).
m
It follows by the t h e o r e m t h a t the congruence relation is an equivalence relation.
The Congruence Relation and the Mod Operator
Suppose a  r (mod b), w h e r e 0 < r < b. T h e n it can be shown t h a t r  a mod b. Conversely, if r  a mod b, t h e n a  r (mod b). T h u s a  r (mod b) if and only if r  a mod b, where 0 < r < b. See exercises 49 and 50. For example, 43  3 (mod 5) and 0 _< 3 < 5; clearly, 3  43 mod 5. Let us digress briefly to look at an i n t e r e s t i n g application of congruences*.
Fridaythe13th
Congruences can be employed to find the n u m b e r of F r i d a y  t h e  1 3 t h s in a given year. W h e t h e r or not F r i d a y  t h e  1 3 t h occurs in a given m o n t h depends on two factors: the day on which the 13th fell in the previous m o n t h and the n u m b e r of days in the previous month. Suppose t h a t this is a n o n  l e a p y e a r and t h a t we would like to find the n u m b e r of F r i d a y  t h e  1 3 t h s in this year. Suppose also t h a t we know the day the 13th occurred in December of last year. Let Mi denote each of the m o n t h s D e c e m b e r t h r o u g h N o v e m b e r in t h a t order and Di the n u m b e r of days in m o n t h Mi. T h e various values of Di a r e 31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, and 30, respectively. We label the days S u n d a y t h r o u g h S a t u r d a y by 0 t h r o u g h 6 respectively; so day 5 is a Friday. Let Di  di (mod 7), w h e r e 0 _< di < 7. T h e c o r r e s p o n d i n g values of di are 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, and 2, respectively. Each value of di indicates the n u m b e r of days the day of the 13th in m o n t h Mi m u s t be advanced to find the day the 13th falls in m o n t h Mi+l. For example, D e c e m b e r 13, 2000, was a Wednesday. So J a n u a r y 13, 2001, fell on day (3 + 3)  day 6, which was a S a t u r d a y .
*T. Koshy, Elementary Number Theory with Applications, Harcourt/Academic Press, Boston, MA, 2002.
486
Chapter 7 Relations i Let ti = ~ dj (mod 7), where 1 < i < 12. T h e n ti r e p r e s e n t s the j=l total n u m b e r of days the day of December 13 m u s t be moved f o r w a r d to d e t e r m i n e the day of the t h i r t e e n t h in m o n t h Mi. F o r example, t3  d l + d2 + d3 = 3 + 3 + 0  6 (mod 7). So, t h e day of December 13, 2000 (Wednesday) m u s t be advanced by six days to d e t e r m i n e the day of March 13, 2001; it is given by day (3 + 6) = day 2  T u e s d a y . Notice t h a t the various values of ti modulo 7 are 3, 6, 6, 2, 4, 0, 2, 5, 1, 3, 6, and 1, respectively; they include all the least residues modulo 7. Given the day of December 13, they can be used to d e t e r m i n e the day of the t h i r t e e n t h of each m o n t h Mi in a n o n  l e a p year. Table 7.5 s u m m a r i z e s the day of the 13th of each m o n t h in a nonleap year, corresponding to every choice of the day of December 13 of the previous year. You m a y verify this. Notice from the table t h a t t h e r e can be at most three F r i d a y  t h e  1 3 t h s in a n o n  l e a p year.
T a b l e 7.5
Day of the 13th in Each M o n t h in a N o n  l e a p Year.
~11 Jan. Feb. March April May June July Aug. Sept. Oct. Nov. Dec. Dec. 13 il 3 6 6 2 4 0 2 5 1 3 6 1
Sun Mon Tue Wed Thu Sat
II 3 II 4 II 5 II 6 It0
II
6 0 1 2 a
4
6 0 1 2 a
4
2 3 4 5 6
0
4 5 6 0 1 3
0 1 2 3 4 6
2 3 4 5 6
0
5 6 0 1 2 4
1 2 3 4 5 0
3 4 5 6 0
1
6 0 1 2 3
4
1 2 3 4 5 0
II 2
5
5
1
1
2
5
For a leap year, the various values of di are 3, 3, 1, 3, 2, 3, 2, 3, 3, 2, 3, and 2; and the c o r r e s p o n d i n g values of ti are 3, 6, 0, 3, 5, 1, 3, 6, 2, 4, 0, and 2. Using these, we can construct a similar table for a leap year. R e t u r n i n g to the congruence relation, we now explore a close relationship between equivalence relations and partitions; but first we m a k e the following definition.
Equivalence Class
Let R be an equivalence relation on a set A and let a ~ A. The e q u i v a l e n c e c l a s s of a, denoted by [a], is defined as [a] = {x ~ A I xRa}. It consists of all elements in A t h a t are linked to a by the relation R. If x ~ [a], t h e n x is a r e p r e s e n t a t i v e of the class [a]. The next two examples explore equivalence relations.
I !
The relation R  {(a, a), (a, b), (b, a), (b, b), (c, c)} on A  {a, b, c} is an equivalence relation. Find the equivalence class of each element in A.
7.8 Equivalence Relations
487
SOLUTION: (1) [ a ]  {x e AlxRa} = {a,b}
(2) [b] = {x e AlxRb} = {a,b} = [a]
(3) [c] = {x e AixRc} = {c}
Two distinct equivalence classes exist, [a] and [c]. Class [a] has two r e p r e s e n t a t i v e s and class [c] one representative. I The relation R on the set of words over the a l p h a b e t {a, b}, defined by xRy if ]]xli  Ilyli, is an equivalence relation (see E x a m p l e 7.41). Infinitely m a n y equivalence classes exist for R, such as {~ }, {a, b }, and {aa, ab, ba, bb}. l Find all equivalence classes of the congruence relation mod 5 on the set of integers. SOLUTION: Let r be the r e m a i n d e r w h e n an integer a is divided by 5. T h e n a  r ( m o d 5). Since the possible values of r, by the division algorithm, are 0, 1, 2, 3, a n d 4, t h e r e are five distinct equivalence classes: [0] = { . . . ,  1 0 ,  5 , 0 , 5 ,
[1] = { . . . ,  9 ,  4 ,
10,...}
1,6, 1 1 , . . . }
[2] = { . . . ,  8 ,  3 , 2 , [31 = { . . . ,  7 ,  2 , 3 , 8 ,
7, 12,...} 13,...}
I
[4] = { . . . ,  6 ,  1 , 4 , 9 , 1 4 , . . . }
These t h r e e examples lead us to the following observations: 9 Every e l e m e n t belongs to an equivalence class. 9 Any two distinct equivalence classes are disjoint. These results can be stated more formally as follows. Let R be an equivalence relation on a set A, with a and b any two e l e m e n t s in A. T h e n the following properties hold: (1) a ~ [a]. (3) If [a] r [bl, t h e n [ a ] n [b] = 0. (2) [a] = [bl if and only i f a R b .
PROOF: (1) Since R is reflexive, aRa for every a ~ A, so a e [a]. (2) Suppose [a]  [b]. Since a ~ [a] by p a r t (1), a ~ [b]. Therefore, by definition, aRb.
488
Chapter 7 Relations Conversely, let aRb. To show that [a] c_ [b]: Let x e [a]. Then xRa. Since x R a and aRb, x R b by transitivity. Therefore, x e [b] by definition. Thus [a] __ [b]. Similarly, [b] __ [a]. Thus, [a] = [b]. (3) We will prove the contrapositive of the given statement: If[a]u[b] r ~, then [a]  [b]. Suppose [a] u [b] r ~. Then an element x should be in [a] A [b]. Then x e [a] and x e [b]. Since x e [a], x R a and hence a R x by symmetry. In addition, since x e [b], xRb. Thus a R x and x R b . Therefore, a R b by transitivity. Thus [a] = [b], by part 2. This concludes the proof.
m
It follows by Theorem 7.10 that any two equivalence classes are either identical or disjoint, but not both. Notice that Example 7.43 has two disjoint equivalence classes, la] and [c]; their union is the whole set A. Therefore, {la], [c]} is a partition of A. In fact, every equivalence relation on a set induces a partition of the set, as given by the next theorem. ~ Let R be an equivalence relation on a set A. Then the set of distinct equivalence classes forms a partition of A. m The next four examples illuminate this theorem. ~ The relation belongs to the s a m e d i v i s i o n as is an equivalence relation on the set of teams in the American (National) League of majorleague baseball. Let x denote a certain team in the American League. Then the class [x l consists of all teams that belong to the same division as x. By Theorem 7.11, the set of teams in the league can be partitioned as {IYankees ], [White Sox], [Marinersl }. m By Example 7.41, the relation has the same length as on the set of words Z* over the alphabet E = {a, b} is an equivalence relation. Then the set of equivalence classes formed is {lZ 1, Ia 1, laa !, laaa 1,... }; it is a partition of E*. m (optional) Suppose a FORTRAN program contains the variables A through J and the equivalence statement:
• •
EQUIVALENCE (A,B),(C,D),(F,A,G),(C,J),(E,H)
By Example 7.42 the relation shares the s a m e m e m o r y location as is an equivalence relation on the set of variables V. Let V1 = {A,B,F,G}, V2 = {C,D,J}, V3 = {E,H}, and V4 = {I}. The partition of V induced by this relation is {V1, V2, V3, V4 }. See Figure 7.36.
7.8 EquivalenceRelations F i g u r e 7.36
Set of variables V.
489
m
By Example 7.45, the distinct equivalence classes formed by the c o n g r u e n c e relation modulo 5 on Z are [0], [1], [2], [3], and [4]. T h e y form a p a r t i t i o n of the set of integers, as shown in Figure 7.37.
F i g u r e 7.37
A partition of the set of integers Z.
m
Conversely, does every partition yield an equivalence relation? The next t h e o r e m shows t h a t every partition does. Every partition of a set induces an equivalence relation on it.
PROOF: Let P = {A1,A2,... } be a partition of a set A. Define a relation R on A as: aRb if a belongs to the same block as b. We shall show t h a t R is indeed an equivalence relation.
9 Since every element in A belongs to the same block as itself, R is reflexive. 9 Let aRb. T h e n a belongs to the same block as b. So b belongs to the same block as a. T h u s R is symmetric. 9 Let aRb and bRc. T h e n a belongs to the same block as b and b to the same block as c. So a belongs to the same block as c. Therefore, R is transitive. T h u s R is an equivalence relation, m
How can we find the equivalence relation c o r r e s p o n d i n g to a partition of a set? The next example d e m o n s t r a t e s how to accomplish this. Find the equivalence relation o n A
{a,b,c} c o r r e s p o n d i n g to the p a r t i t i o n
{{a,b},{c}}.
490
Chapter 7 Relations
SOLUTION: Define a relation R o n A as follows (see the above proof): xRy ifx belongs to the same block as y. Since a and b belong to the same block, aRa, aRb, bRa, and bRb. Similarly, cRc. T h u s R = {(a, a), (a,b,), (b, a), (b, b), (c, c)}. I
Example 7.50 can serve to develop an algorithm for finding the equivalence relation corresponding to a partition P of a finite set A. It is given in Algorithm 7.3.
Algorithm Equivalence Relation (P,A,R)
(* This algorithm determines the equivalence r e l a t i o n R corresponding to a p a r t i t i o n P of a f i n i t e set A. *) Begin (* algorithm *) while P # ~ do begin (* while *) e x t r a c t a block B pair each element in B with every element in B P 1582) is given by d = r + [2.6m  0.2J  2C + D + [C/4J + [D/4J (mod 7) where C = [y/100J and D = y mod 100; d = 0 denotes Sunday; and m  1 denotes March, m  11 J a n u a r y , and m  12 F e b r u a r y . This f o r m u l a is called Z e l l e r ' s f o r m u l a , after Christian J u l i u s J o h a n n e s Zeller ( 1 8 4 9 1899). Find the C h r i s t m a s day of each year. 43. 2000 44. 2020 45. 2345 46. 3000
Let a, b, c, d, m ~ Z with m > 2. Prove each. 47. If a _= b (mod m) and c _= d (mod m), t h e n a § c = b § d (mod m). 48. If a  b (mod m) and c _ d (mod m), t h e n ac =_ bd (mod m). 49. Let r be the r e m a i n d e r when a is divided by m. T h e n a ~ r (mod m). 50. If a  r (mod m) and 0 < r < m, r is the r e m a i n d e r w h e n a is divided by m. 51. Let rl and r2 be the r e m a i n d e r s w h e n a and b are divided by m, respectively. T h e n a  b (mod m) if and only if rl  r2 (mod m). 52. A positive integer N is divisible by 3 if and only if the s u m of its digits is divisible by 3. [Hint" 10  1 (mod 3).] 53. A positive integer N is divisible by 9 if and only if the s u m of its digits is divisible by 9. [Hint: 10  1 (mod 9).] Using the congruence relation, find the r e m a i n d e r w h e n the first integer is divided by the second. 54. 256, 3 55. 657, 3 56. 1976, 9 57. 389, 276, 9
(Hint" Use Exercise 52 or 53.)
58. The U n i t e d Parcel Service assigns to each parcel an identification n u m b e r of nine digits and a check digit. The check digit is the r e m a i n d e r mod 9 of the 9digit n u m b e r . C o m p u t e the check digit for 359,876,015.
7.9 Partialand Total Orderings
493
59. Every b a n k check has an 8digit identification n u m b e r d i d 2 . . , d8 followed by a check digit d given by d  ( d l d 2 , . . . , d s ) (7, 3, 9, 7, 3, 9, 7, 3) mod 10, where (Xl,X2,... ,Xn) 9 (Yl,Y2,..,Yn) n xiYi. (It is the dot p r o d u c t of the two ntuples.) C o m p u t e the i=1 check digit for 17,761,976. 60. Libraries use a sophisticated c o d e  a  b a r s y s t e m to assign each book a 13digit identification n u m b e r dl, d 2 . . . d13 and a check digit d. Let k denote the n u m b e r of digits a m o n g d l , d 3 , d5, d7, d9, d11, and d13 greater t h a n or equal to 5. T h e n d is c o m p u t e d as d  [  ( d l , d 2 , . . . ,d13) 9 (2, 1,2, 1,2, 1,2, 1,2, 1,2, 1,2)  k] m o d 10, where the dot indicates the dot product. C o m p u t e the check digit for 2,035,798,008,938. "61. ( T h e c o c o n u t s a n d m o n k e y p r o b l e m ) * Five sailors and a m o n k e y are m a r o o n e d on a desert island. D u r i n g the day they g a t h e r coconuts for food. They decide to divide t h e m up in the m o r n i n g and retire for the night. While the others are asleep, one sailor gets up and divides t h e m into equal piles, with one left over t h a t he t h r o w s out for the monkey. He hides his share, puts the r e m a i n i n g coconuts together, and goes back to sleep. L a t e r a second sailor gets up and divides the pile into five equal shares with one coconut left over, which he discards for the monkey. Later the r e m a i n i n g sailors repeat the process. Find the smallest possible n u m b e r of coconuts in the original pile.
J u s t as we used the concepts of reflexivity, s y m m e t r y , and t r a n s i t i v i t y to define equivalence relations, we can use reflexivity, a n t i s y m m e t r y , and transitivity to introduce a new class of relations: partial orders. We begin this section with an example. Building a house can be b r o k e n down into several tasks, as Table 7.6 shows. Define a relation R on the set of tasks as follows: Let x and y be any two tasks; t h e n x R y if x = y or m u s t be done before y. This relation is reflexive, antisymmetric, and transitive (verify). Such a relation is a partial order. *Writer Ben Ames Williams used this problem in a short story titled "Coconuts," which appeared in the October 9, 1926, issue of The Saturday Evening Post. The story concerned a contractor who wanted to bid on a large contract. Knowing of their competitor's strong passion for recreational mathematics, one of his employees gave him this problem. The competitor became so obsessed with solving the puzzle that he forgot to enter his bid before the deadline.
494 Table 7.6
Chapter 7 Relations
Task
(1) Building the foundation (t 1) (2) Framing (t 2) (3) Subflooring (t 3) (4) Partitioning into rooms (t4) (5) Roofing (t 5) (6) Plumbing (t6) (7) Wiring (t 7) (8) Siding (t8) (9) Flooring (t 9) (10) Interior painting (tlo) (11) Exterior painting (tll) (12) Carpeting (t12) ( 13 ) Installing fixtures (t 13 )
Requires the completion of
None tl
tl,t2 tl,t2,t3
tl,t2 tl,t2 tl,t2
tl,t2,t5,t9 tl,t2,t6
tl through t5, t 7
tl,t2,t8,t9
t 1 through t 7 , t 9 , t l o t I through t 11
Partial Order
A r e l a t i o n R on a set A is a p a r t i a l o r d e r if it is reflexive, a n t i s y m m e t r i c , a n d t r a n s i t i v e . T h e set A w i t h its p a r t i a l o r d e r R is a p a r t i a l l y o r d e r e d set (or poset), d e n o t e d by (A, R). W h e n t h e p a r t i a l order is clear from t h e context, call t h e poset A. T h e n e x t t h r e e e x a m p l e s i l l u s t r a t e t h e s e definitions. T h e r e l a t i o n < on R is reflexive, a n t i s y m m e t r i c , a n d t r a n s i t i v e , so < is a p a r t i a l order on R a n d (R, ___ 0. Let An denote the n u m b e r of i n d e p e n d e n t subsets of vertices in the path. W h e n n = 0, the p a t h Po consists of a single point vo, so t h e r e are two possible i n d e p e n d e n t subsets: O, {vo}. W h e n n = 1, the p a t h is vovl. T h e n t h e r e are t h r e e i n d e p e n d e n t subsets of {vo, Vl }, namely, O, {vo}, and {Vl}. W h e n n  2, the p a t h P2 contains t h r e e vertices" vo, vl, and v2. So, t h e r e are five i n d e p e n d e n t subsets: O, {vo}, {Vl}, {v2}, and {vo, v2}. *T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley, New York, 2001.
548
Chapter 8 Graphs
T h e s e d a t a are s u m m a r i z e d in Table 8.5. Clearly, a p a t t e r n e m e r g e s . It seems safe to conjecture t h a t A n  F n + 3 , w h e r e n > 0. We invite y o u to confirm this. T a b l e 8.5 P a t h Pn 9 v0 : : v0 Vl
: vo : Vl : v2
Independent subsets 0, O,
O,
An
2 3
5
{v0} {v0},{Vl}
{ v o } , { V l } , { v 2 } , { v o , v2}
. . . . v0 Vl v2
v3
O,
{v0}, {v1}, {v2}, {v3}
8
{v0,v2}, {v0, v3}, {Vl, v3} r~+3
m
A simple p a t h m a y r e p e a t vertices if t h e y be its endpoints. S u c h a p h e n o m e n o n carries a special label.
Cycle and Circuit A p a t h with e n d p o i n t s v0 and vn is c l o s e d if v0  v,z; otherwise, it is o p e n . A simple closed p a t h is a c y c l e ; a closed p a t h with no r e p e a t e d edges is a circuit. Table 8.6 s u m m a r i z e s and illustrates the basic t e r m s i n t r o d u c e d t h u s far in this section. Refer to it as often as needed.
T a b l e 8.6
Term Path Simple path Closed path Open path Cycle Circuit
Meaning Sequence v o  e l  v 1 . . . . . e n  v n , ei={vi_l,vi} ,l 3 vertices is Hamiltonian if deg (v) > ~ for every vertex v in G.
n
m
Notice that the condition in Theorem 8.9 does not apply to the graph G1 in Figure 8.73 and the 3cube in Figure 8.75. Nonetheless, both are Hamiltonian. So even if the condition is not satisfied, the graph may be Hamiltonian. The next theorem provides another sufficient criterion for a graph to be Hamiltonian. (Ore's T h e o r e m ) Let G be a simple connected graph with n > 3 vertices. If deg (u) + deg (v) > n for every pair of nonadjacent vertices u and v, then G is Hamiltonian. m You may verify that deg (u) + deg (v) > 5 for every pair of nonadjacent vertices u and v of the graph G3 in Figure 8.67. Consequently, by Theorem 8.10, G3 is Hamiltonian, as expected. The condition in the theorem is not satisfied by the 3cube in Figure 8.75; nonetheless, Q3 is Hamiltonian. Next we present the wellknown traveling salesperson problem, which employs Hamiltonian cycles and has interesting applications to the theory of communications.
8.5 Eulerianand HamiltonianGraphs
571
Gabriel A n d r e w D i r a c (19251984), a pioneer in graph theory, was born in Budapest. He moved to England in 1937 when his mother married Paul Adrien Maurice Dirac, a Nobel Laureate in physics. In 1942, he began his studies at Cambridge University. After a twoyear interruption by the war, he continued his studies at Cambridge and London, receiving his Ph.D. in mathematics from the University of London (1951). That same year he was awarded the Rayleigh Prize by Cambridge University. Dirac taught at the universities of London, Toronto, Hamburg, Wales, and Aarhus. He was a member of the editorial board of the Journal of Graph Theory and the European Journal of Combinatorics. Besides his work with graph theory, he made outstanding contributions to number theory and geometry. He was also a passionate art connoisseur.
Oystein Ore (18991968), a Norwegian mathematician, was born in Oslo. He received his Ph.D. from Oslo University in 1924 and taught there for 2 years. He joined Yale University in 1927 and taught there until his retirement in 1967, holding the chail ,:,fthe mathematics department from 1936 to 1945. Ore served on the board of American Relief for Norway from 1942 to 1947 and chaired the Relief Mission in 19451946. Recognizing his contributions to Norway, King Haakon VII decorated him Knight of St. Olaf in 1947. As an author of numerous articles and several books, Ore made outstanding contributions to graph theory, abstract algebra, number theory, and probability.
Traveling Salesperson Problem
Study the weighted graph in Figure 8.82. The vertices represent cities and the weights represent the distances between them. A salesperson assigned to city a would like to visit every other city exactly once and return to the home city so that the total distance traveled is a minimum. In other words, beginning at a, he would like to find a Hamiltonian cycle, so the sum of the weights along the cycle is a minimum. This is the celebrated t r a v e l i n g salesperson problem.
Figure 8.82
C
b , ~ ~ 1 0~  d a9 e
572
Chapter 8 Graphs When the graph contains only a few vertices, you can list all possible cycles, compute the sum of the weights along each, and find the shortest Hamiltonian cycle. For example, the graph in Figure 8.82, by the multiplication principle, has 4!/2 = 12 distinct Hamiltonian cycles, each containing five edges. You may verify that the cycle with least weight is adbcea; so the total distance traveled is 60 miles. Finding a Hamiltonian cycle in a complete graph Kn without searching through all distinct (n  1)!/2 Hamiltonian cycles remains an unsolved problem. We close this section with another graphtheoretic game. (optional) The game of H a m (for H a m p t o n Court) is a graphtheoretic game quite similar to SIM, invented by B. Recaman in 1977. Like SIM, Ham is also a twoperson game. Two players, R (for red) and B (for blue), take turns coloring an edge of the complete graph Kn. (A game on Kn is a game of o r d e r n.) Player R's objective is to obtain a Hamiltonian cycle made up entirely of red edges while B's goal is to prevent R from doing it. Whoever achieves his\her goal wins the game. The game of order 3 has a trivial winning strategy for player B. She can always block R from completing a Hamiltonian cycle. (So B wins.) See Figure 8.83, where solid edges indicate red edges and broken edges, blue.
Figure 8.83
iI ~
Player B wins.
J ~_
I
~k
Figure 8.84
Player B wins.
.
.
.
.
.
Figure 8.85
Player B wins. Player B can always win games of orders 4 and 5, as in Figures 8.84 and 8.85. The game of order 6, although a bit more complicated, can also be won by player B. After R has colored the first edge {x, y}, B colors an edge {u, v} not incident with x or y. Thereafter, B colors four more edges incident with x or y. At this stage, if R has at least two edges incident with both u and v, she can win (see Figure 8.86). But B can avoid this by coloring one of the two edges missing in the Hamiltonian cycle (see Figure 8.87). Is player B favored to win if the game is of order 7 or more? It is conjectured so.
8.5 Eulerianand Hamiltonian Graphs
573
F i g u r e 8.86
u
U
Player R wins.
", "b.~
',,V.:x.~
x 3'
~~/
Figure 8.87 Player B wins.
t
u
U
,," J.~{ z  . k \
" I \ i I ".
\
x
3'
I
Exercises 8.5
Determine if each graph is Eulerian.
2.fi
3.
0
5./%
7,, 'V" 
Construct an Eulerian circuit for each Eulerian graph. 7. b a ~ c e5 8. b a ~ c e8x x ~ / e5 d 9. ae b ~ e~ c
574
Chapter 8 Graphs 10.
a
11.
b
e2
~e 4 X x ~ e 6
d 12.
e2 C
e
/
e7
d
'Je4~ //~ 6= e8
d
13. Is the Petersen graph in Figure 8.28 Eulerian? 1419. Do the nonEulerian graphs in Exercises 16 have an Eulerian path? Find an Eulerian path in each graph, if possible.
20.
e 12
el
e4
21.
c
22.
b
a d
e
a d
e
Under what conditions will each graph be Eulerian? 23.
Kn
24.
Km,n
2 5  2 6 . Under what conditions will each graph in Exercises 23 and 24 contain an Eulerian path, but not an Eulerian circuit? 27. If a connected rregular graph is Eulerian, what can you say about r? 2 8  4 2 . Are the graphs in Exercises 112 and 2022 Hamiltonian? If one is not Hamiltonian, determine if it has a H a m i l t o n i a n path; if so, find it. 43. Under what conditions will the complete graph K,~ be Hamiltonian? 44. If G is a connected graph containing a vertex with degree 1, can it be Hamiltonian? Determine if each complete bipartite graph Km,n is Hamiltonian. If a graph is not Hamiltonian, does it contain a Hamiltonian path? 45. K2,3 46. K3,3 47. K2,4 48. K3,4
49. For what values of m and n will
Km,n be
Hamiltonian?
50. Is the Petersen graph in Figure 8.28 Hamiltonian?
8.,5 Eulerianand Hamiltonian Graphs
575
Consider the t r i a n g l e Figure 8.88.*
F i g u r e 8.88
graph
Tn, where T1, T2, T3 are shown in
w
A
T1
A
T2
/%
T3
51. When will Tn be Eulerian? Consider the s q u a r e Figure 8.89.
F i g u r e 8.89
v _
52. When will Tn be Hamiltonian? Sn, where $1, $2, $3 are pictured in
graph
S1
$2
$3
54. When will Sn be Hamiltonian?
53. When will S,, be Eulerian? Give an example of a graph that is: 55. Both Eulerian and Hamiltonian. 56. Eulerian, but not Hamiltonian. 57. Hamiltonian, but not Eulerian.
58. Neither Eulerian nor Hamiltonian. 59. Figure 8.90 shows five cities, a t h r o u g h e, and the distances between them. A salesperson based at a would like to visit each city exactly once and r e t u r n to a, covering the fewest miles. Find the route she should take and the m i n i m u m distance she would travel. 60. Display a Hamiltonian cycle for the 4cube. Find all power chains of order n, if they exist, for each value of n.
61. 6 62. 8 63. 9 64. 15
*Exercises 5154 are based on A. Guckin et al., The Euler Circuit Project, COMAP, Inc., Lexington, MA, 1989.
576 Figure 8.90
Chapter 8 Graphs
c
1
3 a
~ 5 e
6
*65. A power cycle of order n is a cyclic permutation of the first n (> 2) positive integers such that the sum of every pair of adjacent elements is a power. Find a power cycle of order 17. 66. Write an algorithm to determine if a connected graph is Eulerian, using its adjacency list representation. 67. Write an algorithm to determine if a connected graph contains an Eulerian path, using its adjacency matrix.
Take a good look at every graph presented so far. In some cases, the edges meet only at the vertices, whereas in the other cases the edges meet at nonvertices. For example, the edges of the K6nigsberg bridge model (Figure 8.2) meet only at its vertices, whereas the edges of the pentagram in Figure 8.5 intersect at points that are not necessarily vertices. However, the confusion can be resolved. The pentagram can be redrawn in such a way that its edges intersect only at its vertices, as Figure 8.91 shows. The graphs in Figures 8.5 and 8.91 are clearly isomorphic. You are invited to verify this.
Figure 8.91
0
3
2
w
From these observations arises the following definition.
Planar Graph
A graph is p l a n a r if it can be drawn in the plane, so its edges meet only at the vertices. Such a drawing is a p l a n a r r e p r e s e n t a t i o n of the graph.
8.6 Planar Graphs
577
The previous discussion indicates the pentagram is planar. The graph in Figure 8.92 is planar since it can transform to Figure 8.93; they are isomorphic graphs. The graph in Figure 8.94 is also planar; Figure 8.95 shows its planar representation.
F i g u r e 8.92 a b
d
c
F i g u r e 8.93
a
b
C v
wd
F i g u r e 8.94
b
c
d
F i g u r e 8.95
c)
Planarity is an important concept with fine, realworld applications. For example, it figures prominently in designing circuit boards. The following example answers a problem posed at the beginning of the chapter.
~
(The T h r e e H o u s e s  U t i l i t i e s P u z z l e ) A developer is building three new houses m A , B, and C r o o n one side of a street. She would like to connect three utilities m electricity (E), telephone (T), and water ( W )  to each house. This situation can be modeled by the complete bipartite graph K3,3 in Figure 8.96. Can the developer lay the utility lines without any crossings? To answer this, try to make a planar representation of K3,3. Draw eight of the nine edges in K3,3 (see the solid edges in Figure 8.97). How can we draw the edge from W to A? Notice that W lies on the circuits WCTBW and WCEBW. Consider the regions bounded by them. Since A lies outside both, any edge
578 F i g u r e 8.96
Chapter 8 Graphs A B C
E
T
W
from W to A m u s t cross a b o u n d a r y (see the broken edges in the figure). T h u s K3,3 is n o n p l a n a r and the puzzle has no solution. The utility lines will cross. (Example 8.41 will prove this algebraically.)
F i g u r e 8.97
"
B
RI",, ,,,~_~ 1 C
m
An a r g u m e n t like this geometric proof can show t h a t the complete graph /(,5 is also nonplanar. Notice t h a t any planar r e p r e s e n t a t i o n of a connected graph divides the plane into regions, including the u n b o u n d e d region. For instance, the planar graph in Figure 8.98 divides the plane into five regions, m a r k e d 1 t h r o u g h 5. The graph has six edges and and three vertices. Let r denote the n u m b e r of regions formed by e edges and v vertices. T h e n r  e  v + 2. In fact, this relationship is not a coincidence, but is always true.
F i g u r e 8.98
5
( E u l e r ' s f o r m u l a ) Let G be a connected planar graph with e edges and v vertices. Let r be the n u m b e r of regions formed by a planar r e p r e s e n t a t i o n of G. T h e n r  e  v + 2.
8.6 PlanarGraphs
579
P R O O F (by i n d u c t i o n on e)" Let P(n)" E u l e r ' s f o r m u l a holds for every c o n n e c t e d p l a n a r g r a p h w i t h n edges. B a s i s s t e p Clearly, P(0) is t r u e . S u p p o s e t h e g r a p h c o n t a i n s one edge. T h e n t h e r e are two possibilities: T h e edge m a y or m a y not be a loop (see F i g u r e 8.99). In t h e first case, e  1, v  1, a n d r  2. In t h e second case, e  1, v  2, a n d r  1. In b o t h cases, r  e  v + 2. C o n s e q u e n t l y , P(1) is also true.
F i g u r e 8.99
I n d u c t i o n s t e p A s s u m e P(k) is t r u e for some k > 0; t h a t is, a s s u m e t h e f o r m u l a holds for every c o n n e c t e d p l a n a r g r a p h w i t h k edges. C o n s i d e r a c o n n e c t e d p l a n a r g r a p h G w i t h k + 1 edges, v vertices, a n d r regions. G e i t h e r has a cycle or does not.
C a s e 1 Suppose G does not c o n t a i n a cycle. T h e n G m u s t c o n t a i n a v e r t e x u of degree 1. To find such a vertex, build a p a t h from a n y v e r t e x u n t i l you c a n n o t go a n y f u r t h e r . The last v e r t e x visited is such a v e r t e x u. (See F i g u r e 8.100.) Let a be t h e edge i n c i d e n t w i t h u. Let H be t h e g r a p h o b t a i n e d by deleting v e r t e x u a n d edge a from G. H c o n t a i n s e'  k edges a n d v' = v  1 vertices. Since G is connected, so is H. Therefore, by t h e i n d u c t i v e hypothesis, the f o r m u l a holds for H ; t h e n u m b e r of regions r' f o r m e d by H satisfies t h e f o r m u l a r' = e'  v' + 2. B u t r' = r. So r'T h a t is, ( e  1)  (v  1) + 2
rev+2
F i g u r e 8.100
/a
C a s e 2 Suppose G c o n t a i n s a cycle. Let a be an edge in the cycle (see F i g u r e 8.101). Let L be t h e s u b g r a p h o b t a i n e d by d e l e t i n g edge a from G. (Note: Deleting edge a m e r g e s regions R1 a n d R2.) This s u b g r a p h c o n t a i n s e' = k edges a n d v'  v vertices; t h e r e f o r e , by t h e i n d u c t i o n hypothesis, t h e n u m b e r of regions r' f o r m e d by L is given by r' = e'  v' + 2. B u t r'  r  1, e' = e  1, a n d v' = v. So r1(e1) v+2
580
Chapter 8 Graphs That is, rev+2
F i g u r e 8.101
By cases 1 and 2, P(k + 1) is true. Therefore, by PMI, the formula holds for every connected planar graph, n The next two examples illustrate Euler's formula. ~ Verify Euler's formula for the connected planar graph in Figure 8.102.
F i g u r e 8.102
SOLUTION: The graph contains 22 edges and 13 vertices, which form 11 regions. So e  v + 2  2 2  1 3 + 2  11  r. m
~ A connected planar graph has 17 edges, dividing the plane into nine regions. How many vertices does the graph have?
SOLUTION: Here e  17 and r  9. By Euler's formula,
ver+2179+210
m
Euler's formula can generate simple relationships in special cases. One such formula is given in the following theorem. If G is a simple, connected, planar graph t h a t contains e (>2) edges and v vertices, then e < 3v  6.
PROOF: Suppose the plane is divided into r regions by a planar r e p r e s e n t a t i o n of the graph. First, we shall show t h a t 2e > 3r. If e  2, then r  1. So 2e > 3r.
8.6 PlanarGraphs
581
Suppose e > 2. Let s be the total n u m b e r of edges along the boundaries of the various regions. Every edge appears on the b o u n d a r y of two regions, so every edge is counted twice in this sum and s  2e. Now every region, including the u n b o u n d e d region, is bounded by at least t h r e e edges. Therefore, s > 3r and 2e > 3r. By Euler's formula,
r+v2e
Then
3r + 3v  6 3 r + 3v  6  e 3e 2e
So
3r + 3v  6  e > 3r
T h a t is, e4. Figure 8.117 shows a coloring with exactly four colors. How do we interpret this result? Since the chromatic number is four, the final exams can be scheduled conflictfree using four time slots, as Table 8.8 shows.
Table 8.8
Block
Course(s)
Ibl
A,C
234
B,G D E,F
We close this section with the next example, which illustrates how graph coloring can be employed to design traffic light patterns at intersections. Figure 8.118 shows the intersection of two divided avenues, California and New York, where all left and right turns are permitted. The arrows indicate
8.7
GraphColoring
593
the traffic flows along each avenue. Assuming they are equally heavy in each direction, design a traffic signal pattern for the intersection.* F i g u r e 8.118 I I
i I
li il orniaAve Ca, i
'l
I t I
i I I
SOLUTION: Since the four right turns do not interfere with the other traffic flows, they can safely be dropped from our discussion. The remaining traffic directions are pictured in Figure 8.119 and are labeled A through H. We need to develop a traffic pattern so that vehicles can pass through the intersection without interfering with other traffic flows. F i g u r e 8.119
>
,H,~
~.I+I I I ~
California Ave. F
~i~I~
k
A
iill'i, ,
I ICID
*Based on J. Burling et al., "Using Graphs to Solve the Traffic Light Problem," F A I M Module, COMAP, Inc., Lexington, MA, 1989.
594 Figure 8.120
Chapter 8 Graphs
To this end, represent each traffic flow with a vertex. Two vertices are adjacent if the corresponding traffic flows cross each other. For instance, directions C and E intersect, so vertices C and E are adjacent. Figure 8.120 shows the resulting graph. Now color its vertices. Since the graph contains C3 as subgraph (verify), its chromatic number is > 3. Is it in fact three? Let us see.
Figure 8.121
H
A E
Suppose we color A red, C blue, and H yellow (see Figure 8.121). Then B cannot be blue or yellow, but can be red; so color it red. Then E must be yellow; this forces F to be blue. D, being adjacent to B, F, and H can't be red, blue, or yellow, so it must be assigned a new color, say, green. Thus the graph takes at least four colors. Figure 8.121 shows a coloring of the graph with exactly four colors, which depicts an efficient way of designing the traffic signal pattern. It consists of four phases: 9 Vertices B and F are green, so traffic flows B and F proceed, while others are waiting. 9 Vertices D and H are yellow; that is, only traffic directions D and H proceed simultaneously.
8.7 GraphColoring
595
9 Vertices A and E are red; t h a t is, traffic flows A and E continue at the same time, while others are stopped. 9 Vertices C and G are blue, so only traffic directions C and G proceed simultaneously. See Table 8.9 also.
T a b l e 8.9 Phase 1
Only B and F proceed.
Traffic light pattern Phase 2
Only D and H proceed.
Phase 3
Only A and E proceed.
Phase 4
Only C and G proceed.
II
Exercises 8.7
Find the chromatic n u m b e r of each map or graph.
0
0
3.
~
4.
596
Chapter 8 Graphs
.
Figure
8.122
i
Continental United States.
6. Figure 8.5
7. Figure 8.18
8. Figure 8.56 11. Figure 8.106
9. Graph G1 in Figure 8.73 10. Figure 8.102 12. Petersen graph 15. 3cube Q3 16. Characterize graphs with chromatic number 1.
13. The Star of David 14. Wheel graph
Wn
17. Let G be the union of two simple disconnected subgraphs H1 and H2 with chromatic numbers m and n, respectively. What can you say about the chromatic number c of G? In Exercises 18 and 19, schedule conflictfree committee meetings using the smallest number of blocks. Identify such a schedule.
18.
Committee
1 B
Committee
2 A
Committee
3 C
Committee
4 A
Committee
5 B
D E F
E G H
D F H
C E F
G H I
19.
Committee Committee Committee Committee Committee Committee
1 B 2 D 3 A 4 B 5 A 6 C
D E F
E G H
C D F H
C E H I
E G H
G H I
8.7
Graph Coloring
597
A zoo curator would like to add new open "islands" for seven species of animals to roam about freely. Unfortunately, some species prey on some others, as given by the incompatibility a r r a y in Table 8.10.
Table 8.10
Species
1
2
3
4
5
6
7
.
.
.
9
9
9
.
.
9
9
.
9
9
.
9
9
9
9
.
9
9
9
9
7
9
9
20. Determine the m i n i m u m n u m b e r of islands needed to keep them. 21. Find a possible way of accommodating the animals on those islands. A pet shop owner just received a shipment of 10 species of tropical fish. Since some species are incompatible, t h a t is, they fight with some other species, they cannot be kept in the same tank. Table 8.11 s u m m a r i z e s the incompatibility of the various species, where a dot in row i and column j indicates species i and j are incompatible.
Table 8.11
Species
1
2
3
4
5
6
7
8
9
10
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
.
9
.,
.
.
,
.
,
.,
9
9
9
9
9
10 9 9
22. Determine the m i n i m u m n u m b e r of t a n k s needed to store the fish. 23. Find a possible way of storing t h e m a m o n g those tanks. Figure 8.123 shows the traffic flows at an exit from a shopping center into a twoway street. (J. Williams, 1992) 24. Represent this information in a graph.
598
Chapter 8 Graphs 25. Develop a traffic light pattern so that traffic will flow smoothly at the exit.
Figure
8.123
D A
v
Shopping Center
I I I I
B
C
Figure
8.124
Shopping Center
C~ ~ A
F
j
E D
I
Figure 8.124 shows the traffic flows from a twoway street into a shopping center and from the shopping center into the street. 2627. Redo Exercises 8.24 and 8.25. *28. Let G be a simple graph with n vertices. Let k denote the maximum degree of any vertex in G. Prove that the chromatic number of G is 2 edges and v vertices. Then e _< 2v  4. Using this fact, show that each bipartite graph is nonplanar. 43. K3,3 44. K3,4 45. K3,5 46. K4,5
Supplementary Exercises
1. Find the number of vertices in C,,, W,~, and Qrz. (See Figures 8.16, 8.17, and 8.75.) 2. Find the number of edges in C,, W,~, and Qn. 3. For what value(s) of n is C, a regular graph? *4. Prove that if Wn is regular, then n = 3. A graph G with n vertices and e edges is g r a c e f u l if the vertices can be labeled 1, 2 , . . . , n in such a way that the values li  j ] are different for every pair of adjacent vertices i a n d j . For example, the graph in Figure 8.126 is graceful.
Figure 8.126
1
2
3
5
4
Are the following graphs graceful?
0.:
:
o
]
7]
]
w
Chapter Summary
605
8. Prove t h a t every open path with n vertices is graceful. 9. Prove t h a t Kl,n is graceful. A graph G  (V, E) is c o m p l e t e n  p a r t i t e if the following conditions are satisfied:
9 V  V1 [,J V2 u . . . [.J V n , where Vi r
0 for every i and Vi ~ ~  ~ for
i ~:j; and 9 If u ~ Vi and v e ~ , then {u, v} e E, where i r 10. What can you say about G if IVil  1 for every i? 11. Let JViJ  vi, 1 2) denote the n u m b e r of vertices in a selfcomplementary graph. Show t h a t n _ 0 (mod 4) or n  1 (mod 4). 18. Delete any vertex and edges incident with it in the Petersen graph. Show t h a t the resulting subgraph is Hamiltonian. 19. Find the n u m b e r of distinct H a m i l t o n i a n cycles in Kn, where n > 3. "20. Find the n u m b e r of distinct H a m i l t o n i a n cycles in Kn,n, where n > 2. "21. Find the n u m b e r of distinct Hamiltonian paths in Kn+l,n, where n>l. 22. Delete any edge from K5. Show t h a t the resulting subgraph is planar.
606
Chapter 8 Graphs 23. Delete any edge from g3, 3. Show t h a t the resulting subgraph is planar.
Computer Exercises
Let G be a graph with n vertices, labeled 1 t h r o u g h n, where 1 _< n < 10. Write a program to do each task. 1. Read in n and the various edges {i, j} of G, where 1 < i, j < n. 9 P r i n t the adjacency matrix. 9 Print the degree of each vertex. 9 Print the linked list representation of G. 9 Determine if G is a simple graph. 9 Determine if G is a complete graph. 2. Let G be a weighted graph. Read in n, the various edges {i, j} of the graph, and their weights w. 9 Print the weighted adjacency matrix. 9 Print the adjacency list representation. 3. Read in the adjacency matrix of G. Print its adjacency list representation. Use the adjacency list representation to print the adjacency matrix of G. 4. Read in the adjacency matrix of a simple graph and determine if it is rregular. 5. Read in two positive integers m and n, where m, n sets V1 and V2 of a bipartite graph G, where iVll and the edges {i, j} in G, where i, j < 10. Using the and adjacency list representation of G, determine if bipartite graph Km,n. < 10; the vertex m and IV2i  n; adjacency matrix it is the complete
6. Read in the adjacency matrix of G. Determine if G is connected. Find how many simple paths r u n from vertex i to vertex j, where 1 _< i, j < 10 and i r 7. Read in the adjacency matrix of a graph and determine if the graph is Eulerian. If it is not, see if it contains an Eulerian path. 8. Read in the edges {i, j} of a graph, where 1 < i, j < n. With the linked list representation of the graph, determine if it is Eulerian. If it is not Eulerian, determine if it contains an Eulerian path. 9. Let G be a complete weighted graph whose vertices and weights represent cities and distances between them, respectively. Read in the
Chapter Summary
607
weighted adjacency matrix of G. Find a Hamiltonian cycle so t h a t the sum of the weights along the cycle is a minimum. 10. Solve the knights puzzle in Example 8.26. List all moves. 11. Read in the various class lists in Example 8.46. Find a conflictfree final exam schedule for the courses. 12. Read in the adjacency matrix for the 48 states of the continental United States. Assign a coloring to them in such a way that adjacent states receive different colors.
Exploratory Writing Projects
Using library and Internet resources, write a team report on each of the following in your own words. Provide a welldocumented bibliography. 1. Explain the applications of graph theory to various disciplines. 2. Describe the history of the traveling salesperson problem. Name a few practical applications. 3. Give a historical background of the fourcolor problem. Include a biography of K. Appel and W. Haken. 4. Explain the Chinese Postman Problem. 5. Explain how de Bruijn sequences, named after Nicolaas G. de Bruijn, can be generated from Eulerian circuits. 6. Explain the relationship between Hamiltonian cycles and the Tower of Brahma. 7. Example 8.11 is an application of Ramsey's theorem, developed by the English logician Frank P. Ramsey (19031930). The theorem laid the foundation for a branch of combinatorics called Ramsey theory. Give a brief introduction to Ramsey theory. 8. Discuss the Instant Insanity Puzzle. 9. Discuss the relationship between ncubes and parallel computers. How can two (n  1)cubes be used to construct an ncube? Also, discuss the relationship between an ncube and Gray code. 10. Write an essay on graceful graphs, a term coined by S. W. Golomb of the University of Southern California. 11. Write an essay on the game of SIM. 12. Write an essay on the game DIM, a threedimensional variation of SIM invented in 1972 by D. Engel. 13. Describe the game of Hackenbush. 14. Write an essay on chromatic polynomials.
608
Chapter 8 Graphs 15. Write an essay on the Petersen graph. 16. Investigate rook polynomials. 17. Write an essay on matching theory. 18. Write an essay on parallel algorithms.
Enrichment Readings
0
K. Appel and W. Haken, "Every Planar Map is 4Colorable," Bulletin of the American Mathematical Society, Vol. 82 (1976), pp. 711712. J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, Elsevier, New York, 1976. R. A. Brualdi, Introductory Combinatorics, 3rd ed., PrenticeHall, Upper Saddle River, NJ, 1999. N. Cristofides, Graph Theory: An Algorithmic Approach, Academic Press, New York, 1975. B. W. Jackson and D. Thoro, Applied Combinatorics with Problem Solving, AddisonWesley, Reading, MA, 1990, pp. 134200. J. A. McHugh, Algorithmic Graph Theory, PrenticeHall, Englewood Cliffs, NJ, 1990. A. Ralston, "Debruijn Sequences   A Model Example of the Interaction of Discrete Mathematics and Computer Science," Mathematics Magazine, Vol. 55 (May 1982), pp. 131143. S. S. Skieno, Implementing Discrete Mathematics, AddisonWesley, Reading, MA, 1990. K. Thulasiraman and M. N. S. Swamy, Graphs: Theory and Algo
0
0
0
0
0
Q
0
0
rithms, Wiley, New York, 1992.
10. A. Tucker, Applied Combinatorics, 2nd ed., Wiley, New York, 1984, pp. 379, 389410. 11. D. West, Introduction to Graph Theory, 2nd ed., PrenticeHall, Upper Saddle River, NJ, 2001. 12. R. J. Wilson and J. J. Watkins, Graphs: An Introductory Approach, Wiley, New York, 1990.
Chapter 9
Trees
A n expert problem solver m u s t be endowed with two incompatible qualities m a restless imagination a n d a pertinent pertinacity.
HOWARD W. EVES
rees are the most important class of graphs and they make fine modeling tools. In 1847, the German physicist Gustav Robert Kirchoff used them to solve systems of linear equations for electrical networks. Ten years later, A r t hur Cayley studied the isomers of saturated hydrocarbons CnH2n+2 with them. Today trees are widely used in mathematics and computer science, as well as in linguistics and the social sciences. This chapter presents the concept of a tree, and two necessary and sufficient conditions for a graph to be a tree. In addition, it presents the concept of a spanning tree for a connected graph, algorithms for finding it, and the important class of rooted trees. Trees facilitate sorting and searching, representing and computing numeric expressions, and constructing efficient coding schemes and algorithms, as will be seen in Sections 9.5 and 9.6. Some of the interesting problems trees handle well are: 9 Can four queens be placed on a 4 • 4 chessboard, so they are not attacking each other? 9 A utility company would like to lay pipelines for natural gas between five towns, a through e, as in Figure 9.1. The weights of the edges represent the costs of building the various pipelines. How cheaply can the company build them, so natural gas can be sent from any town to any other town? 9 There are 64 entrants in a singles tennis tournament. The winner of each round is advanced to the next round and plays another. Find the number of matches and rounds played to determine the champion. 9 There are eight coins in a collection plate. Although they look identical, one of them is counterfeit and heavier. Identify it, using an equal arm balance and a m i n i m u m number of weighings.
609
T
610
(;hapter 9 Trees
~ ~ 1 1 ~
~"~ , " , / " 9
~'., ," ' . ~ ' ' ,,' "~n
Gustav R o b e r t K i r c h o f f (18241887), an outstanding German physicist and son of a lawyer, was born in K6nigsberg, Prussia. After graduating from the local gymnasium, Kirchoff entered the University of K6nigsberg at the age of 18 and received his doctorate 5 years later. In 1848, his teaching career began in Berlin. Two years later he joined the faculty of the University of Breslau, where he met his future research collaborator, the wellknown chemist Robert Bunsen. In 1854, they both moved to Heidelberg. While there, with Bunsen's collaboration, Kirchoff made his greatest contributions to science. In 1875, he accepted the chair of theoretical physics at the University of Berlin, a position he held with great distinction until his death. Although Kirchoff made significant contributions to every branch of physics, he is best known for his pioneering work in spectroscopy.
F i g u r e 9.1
c
7~~/~,,~/15
v
1010
12 e
a
As in graph theory, tree terminology is not standardized, so be aware of this when you refer to different texts on the topic.
Notice t h a t the graphs in Figures 9.2 and 9.3 are connected; each is a c y c l i c , m e a n i n g it does not contain a cycle. Such a graph is a tree.
Figure 9.2
Propane C3H8.
H H~C~C
H
H H
I I CI I HI HI H
a
F i g u r e 9.3
d
e
f
g
9.1 Trees
Tree
611
A connected, acyclic graph is a tree. Trees are everywhere. You have seen several examples of trees in earlier chapters. Figure 1.1, Figures 5.165.19, and Figures 6.7, 6.8, and 6.14 offer fine cases. The graph in Figure 9.4 is not a tree, since it contains a cycle. The graphs in Figures 9.2 and 9.3 are connected; each is acyclic, so each is a tree. The graph in Figure 9.5 is not a tree, either, because it is not connected. Nonetheless, it is a set of disjoint trees, called a forest.
F i g u r e 9.4
Ethylene C2H4.
H
H
H
H
F i g u r e 9.5
A forest.
Figure 9.6 shows the f a m i l y t r e e of the Bernoullis of Switzerland, the most distinguished family of mathematicians.
F i g u r e 9.6
The Bernoulli family. Jakob I
Nicolaus Nicolaus Nicolaus Johann Nicolaus Daniel Johann II Daniel II Christoph Figure 9.7 depicts a tree of a corporation's top management. The tree in Figure 9.8 shows a partial hierarchical chart for teams in the National Hockey League (NHL) during the 20022003 season. To qualify as a tree, a connected graph must fulfill path and edge requirements. First we present the path requirement. Jakob II
Johann III
I
612 Figure 9.7
Chapter 9 Trees
Chief Executive Officer
l
[President[
I
Vice President Finance Vice President Manufacturing
I
Vice President Marketing
I
Vice President Research and Development
Figure 9.8
NHL
Eastern Conference
Western Conference
Atlantic
Northeast
Southeast
Central
Northwest
Pacific
Devils
San ,Jose
~
A connected graph is a tree if and only if there is a unique, simple path between any two vertices.
PROOF: Let G be a tree, and v and w any two of its vertices. Since G is connected, by Theorem 8.3, a simple path m u s t run between them. If there are two distinct simple paths between them, then one path followed by the other in reverse order would form a cycle. This, however, is impossible since G is a tree; so G contains a unique, simple path between v and w. Conversely, let G be a graph with a unique, simple path between any two vertices. Clearly, G is connected. Suppose G contains a cycle, and v and w are two vertices in it. Then the cycle can be split into two distinct simple paths between v and w, a contradiction, so G is acyclic. Being connected and acyclic, G is a tree. m
You may verify t h a t the graph in Figure 9.2 is connected and contains a unique, simple path between any two vertices. Therefore, by Theorem 9.1, it is a tree. The same holds for Figure 9.3. Next we establish the edge r e q u i r e m e n t for a connected graph to be a tree.
9.1 Trees
613
A connected g r a p h w i t h n vertices is a tree if a n d only if it has exactly n  1 edges.
PROOF: F i r s t we prove by s t r o n g induction t h a t a tree w i t h n vertices has n  1 edges. Let P(n)" A tree T with n vertices has n  1 edges. B a s i s s t e p Suppose T contains one vertex. Since T is acyclic, it is loopfree; consequently, T contains no edges and P(1) is true. I n d u c t i o n s t e p Suppose t h e r e s u l t is t r u e for every t r e e with k or fewer vertices. Let T be a tree with k + 1 vertices and e an edge in T. Deleting e from T yields two connected disjoint graphs, T1 a n d T2. E a c h is a tree. Suppose T1 has p vertices; t h e n T2 has q  k + 1  p vertices. By the inductive hypothesis, T1 contains p  1 edges and T2 contains q  1 edges; so T contains (p  1)+ (q  1) + 1  (p  1) + (k  p ) + 1  k edges. T h u s by s t r o n g induction, P(n) is t r u e for every n _> 1. Conversely, let G be a connected g r a p h with n vertices a n d n  1 edges. If it were not a tree, it would have a cycle. Remove an edge from this cycle. The r e s u l t i n g g r a p h is still connected. If it is not acyclic, r e m o v e an edge from a cycle. C o n t i n u e this p r o c e d u r e to get an acyclic g r a p h H. T h u s H is connected and acyclic; it m u s t be a tree with n vertices. So, by the first part, H has n  1 edges and hence G has more t h a n n  1 edges, which is a contradiction. T h u s G is connected and acyclic, and hence a tree. m
For example, the g r a p h in F i g u r e 9.2 is connected and has 11 vertices, so it m u s t contain 10 edges to be a tree, which is true.
Exercises 9.1
D e t e r m i n e if each g r a p h is a tree.
l.d/~
2.
i 7. Is the g r a p h in Figure 8.4b a tree? If not, why?
//x,
"x
8. D e t e r m i n e if the P e t e r s e n g r a p h in Figure 8.28 is a tree. If it isn't, explain why.
614
Chapter 9 Trees
Let n denote t h e n u m b e r of vertices of a t r e e a n d e t h e n u m b e r of edges. Verify t h a t e  n  1 for each tree.
I I I I
/A
n i=1
12. How m a n y bonds does the h y d r o c a r b o n molecule CnH2n+2 have? A s s u m e a carbon molecule has degree four. 13. Let T be a tree with vertices v l , . . . , Vn. Show t h a t ~ deg ( v i )  2 n  2. 14. For w h a t values of n is K n a tree? D e t e r m i n e if each complete b i p a r t i t e g r a p h is a tree. 15. K1,2 16. K1,3 17. K2,2 18. K2,3
19. For w h a t values of m and n is K m , n a tree? 20. Let G be an r  r e g u l a r tree with n vertices. Prove t h a t n = 1 or n = 2. D r a w all n o n i s o m o r p h i c trees with the given n u m b e r of vertices n.
21. 2
22. 3
23. 4
24. 5
25. The e c c e n t r i c i t y of a vertex v in a tree is the length of the longest simple p a t h from v. Find the eccentricity of each vertex of t h e t r e e in Figure 9.9.
F i g u r e 9.9
c
e
f
h
a
d
i
26. T h e c e n t e r of a tree is a vertex w i t h t h e least eccentricity. F i n d t h e center(s) of the tree in F i g u r e 9.9. 27. U s i n g the adjacency m a t r i x of a connected g r a p h with n vertices, w r i t e an a l g o r i t h m to d e t e r m i n e if it is a tree.
All connected g r a p h s have trees t h a t span t h e m . We will discuss t h r e e ways to find t h e m , as well as solve the 4queens problem s t a t e d at the
9.2 Spanningtrees
615
beginning of the chapter, but first a more pressing practical job needs to be done. A county's five towns, A through E, are connected by roads (see Figure 9.10), but 2 feet of snow cover them. The county would like to plow as few roads as possible, so one can travel between the towns. Figure 9.11 displays one possible solution.
F i g u r e 9.10
Graph G.
C
B
D
A E
F i g u r e 9.11
Graph H.
C
B
A D
You may notice that the graph H in Figure 9.11 is a subgraph of G. In fact, it is a tree containing every vertex of G. It is called a spanning tree of G.
Spanning Tree
A subgraph H of a connected graph G is a s p a n n i n g t r e e of G if H is a tree containing every vertex of G. Using this definition, we can solve the snowplowing problem by finding a spanning tree for the graph G. Two additional spanning trees are given in Figure 9.12, indicating that the spanning tree of a graph need not be unique. C
B F i g u r e 9.12
o
B
o
A
E
E
Does every connected graph have a spanning tree? Theorem 9.3 affirms that it does.
616
Chapter 9 Trees
J o s e p h B e r n a r d K r u s k a l (1928) was born in New York City. He graduated from the University of Chicago in 1948 and received his Ph.D. from Princeton in 1954. After being an instructor at Princeton and at the University of Wisconsin, he became an assistant professor at the University of Michigan in 1958. The following year he joined the technical staff at Bell Telephone Labs, a position he still holds. Kruskal has served as visiting professor at Yale, Columbia, and Rutgers.
Every connected graph has a spanning tree.
PROOF:
Let G be a connected graph. If G is a tree, we are done. If G is not a tree, it must contain a cycle. Remove an edge from the cycle. The new graph is still connected. If it is acyclic, then it is a tree and hence a spanning tree. Otherwise, it m u s t have a n o t h e r cycle. Remove an edge from this cycle. Continue this procedure until a subgraph H is acyclic. Since H is both connected and acyclic, it is a tree. H also contains every vertex of G, so it is a spanning tree of G. I Three algorithms to find a spanning tree in a connected graph are given below: Kruskal's, the DepthFirst Search, and the BreadthFirst Search. Kruskal's algorithm represents a special case of an algorithm for a minimal spanning tree that we will examine in the next section. It was developed by the American m a t h e m a t i c i a n Joseph Bernard Kruskal in 1956.
Kruskars Algorithm for a Spanning Tree
Although the proof of Theorem 9.3 can find a spanning tree, Kruskal's algorithm takes advantage of Theorem 9.2. It is given in Algorithm 9.1.
Algorithm Spanning Tree (G,T)
(* G = (V, E) is a connected graph with n v e r t i c e s and T denotes a spanning t r e e . Numedges denotes the number of edges selected. Begin (* algorithm *) O~ T 3. Figure 9.47 shows the first six Fibonacci trees, T1 t h r o u g h T6.
9.5
Binary Trees
647
Figure 9.47
Fibonacci trees.
/%
T1 T2 T3 T4
A
T5
/%
T6
Out of curiosity, we now explore the n u m b e r of leaves ln, the n u m b e r of internal vertices in, the n u m b e r of vertices Vn, and the n u m b e r of edges en of the nth Fibonacci tree T n . To facilitate our investigation, study the trees in Figure 9.47, collect the needed data, and t h e n s u m m a r i z e t h e m in a table, as in Table 9.5.
Table 9.5
rl
1
2
3
4
5
6
...
n
v~
In in
en
1 1 0 0
1 1 0 0
3 2 1 2
5 3 2 4
9 5 4 8
15 8 7 14
... ... ... ...
? ? ? ?
Vn 
Using the table, we conjecture t h a t In  F n , in  In  1  F n  1, in + In  2 F n  1, and en  2in  2F,~  2. They are in fact true. We can confirm them; see Exercises 7477.
Binary Tree T r a v e r s a l s An important tree operation is tree traversal, visiting every vertex of a tree in a systematic way. Three elegant methods for traversing a n o n e m p t y binary tree exist: p r e o r d e r , i n o r d e r , p o s t o r d e r t r a v e r s a l s . Each of t h e m can be defined recursively: Preorder traversal 9 Visit the root. 9 Traverse the left subtree in preorder. 9 Traverse the right subtree in preorder. Inorder traversal 9 Traverse the left subtree in inorder.
648
Chapter' 9 Trees 9 Visit the root. 9 Traverse the right subtree in inorder.
Postorder traversal
9 Traverse the left subtree in postorder. 9 Traverse the right subtree in postorder. 9 Visit the root. The binary tree in Figure 9.48 has Tz and T2 as left and right subtrees. To traverse the tree in preorder, first visit the root v; traverse T1 in preorder and then traverse T2 in preorder. The other two traversals can be interpreted similarly.
F i g u r e 9.48
v
The next three examples illustrate these tree traversals stepbystep. Give the output from traversing the binary tree in Figure 9.49 in preorder.
F i g u r e 9.49
b d c /
SOLUTION: 1. First visit the root a and output a. 2. Traverse the left subtree rooted at b in preorder (see Figure 9.50). O u t p u t b. Now traverse the left subtree rooted at d in preorder (see Figure 9.51). O u t p u t d. Traverse the left subtree ofd; it is empty. So traverse its right subtree; it is also empty. Therefore, back up to b and traverse its right subtree (see Figure 9.52). Visit e. O u t p u t e. It has no subtrees. 3. Backtrack to a. Traverse its right subtree in preorder (see Figure 9.53). O u t p u t c. It has no left subtree, so traverse its right subtree (see Figure 9.54). O u t p u t f.
9.5 BinaryTrees Figure 9.50
649
a
c
Figure 9.51
a
f
Figure 9.52
b d
a
c f
Figure 9.53
a
Figure 9.54
a
Traverse its left subtree (see Figure 9.55). Output g. It has no subtrees; so backtrack to f and traverse its right subtree in preorder, which is empty. All vertices have been visited, so the traversal is done. The output is a,
b,d,e,c,f,g.
650 Figure 9.55
Chapter 9 Trees
a
b d
c
m
Give the o u t p u t from traversing the binary tree in Figure 9.49 in inorder. SOLUTION: 1. First traverse the left subtree in inorder (see Figure 9.56). Traverse its left subtree in inorder (see Figure 9.57). Again traverse its left s u b t r e e in inorder. It is empty, so visit the root d and o u t p u t it. Now traverse its right subtree. Since it is empty, back up to b and o u t p u t it. Traverse its right subtree (see Figure 9.58). Vertex e has no left subtrees, so o u t p u t e. It has no right subtrees.
Figure 9.56
a
f
Figure 9.57
a
f
Figure 9.58
a
d
f
2. Backtrack to a and o u t p u t a.
9.5 BinaryTrees
651
3. Now traverse the right subtree in inorder (see Figure 9.59). Since the left subtree of c is empty, output c. Traverse its right subtree (see Figure 9.60). Move to the left and traverse its left subtree (see Figure 9.61). Since g has no left subtree, output g. It also has no right subtree, so back up to f and output it. Since f has no right subtree, the traversal is over. The output is d, b, e, a, c,g, f . F i g u r e 9.59
a
d
F i g u r e 9.60
a
F i g u r e 9.61
a
m Find the output from traversing the binary tree in Figure 9.49 in postorder. SOLUTION: The various steps are summarized in Figures 9.629.67. 1. Traverse the left subtree in postorder (see Figures 9.629.64). The output from this traversal is d, e, b. 2. Traverse the right subtree in postorder (see Figures 9.659.67). This traversal yields the output g, f, c. 3. Backtrack to the root a. Output a. The output from the postorder traversal is d , e , b , g , f , c , a .
652 Figure 9.62
T r a v e r s e t h e left s u b t r e e of a in postorder.
Chapter 9 Trees
a
/
Figure 9.63
d has no subtrees; o u t p u t d.
a
/
Figure 9.64
Back up to b and t r a v e r s e its right subtree; e has no subtree, so o u p u t e. B a c k t r a c k to b; o u t p u t b. b
a
c
d
/
Figure 9.65
c has no left subtree; t r a v e r s e its right subtree.
a
a
Figure 9.66
T r a v e r s e f ' s left subtree. d
9.5 BinaryTrees F i g u r e 9.67
g has no subtrees; ouput g. Backtrack to f; it has no subtree. O u t p u t f. Backtrack to c; o u t p u t it. b d
a
653
c
m
Binary trees are fine tools for representing and evaluating algebraic expressions involving binary operators.
Infix, Prefix, and Postfix Notations
Before discussing how an algebraic expression with only binary operators can be represented by a binary tree, we should examine three s t a n d a r d forms for such expressions. In the familiar form, the operator appears between its two operands. This is the infix form. For instance, a + b is an infix expression. Two alternatives are the p r e f i x and p o s t f i x forms. In the prefix notation, also called the P o l i s h n o t a t i o n , the binary operator precedes its two operands; in the postfix notation, also known as the R e v e r s e d P o l i s h N o t a t i o n ( R P N ) , the binary operator follows the two operands. These two notations are attributed to the Polish logician J a n Lukasiewicz. The three general forms are summarized in Table 9.6. The prefixes pre, in, and post indicate the location of the operator with respect to the operands.
T a b l e 9.6
Notation
Infix Prefix Postfix
General form
< operand >< operator >< operand >< operator >< operand >< operand >< operand >< operand >< operator >
Example
a+b +ab ab+
Some calculators, such as HewlettPackard's, use the postfix notation to evaluate numeric expressions. The main advantage of the two Polish notations is that algebraic expressions can be rewritten w i t h o u t parentheses. Invalid expressions can be detected easily, too. Algebraic expressions involving the binary operators + (addition),  (subtraction), 9 (multiplication),/(division), and 1' (exponentiation) have the following order of precedence (from highest to lowest): Exponentiation Multiplication, division Addition, subtraction Certainly, parentheses can override this precedence to specify a n o t h e r order in which to perform operations, so parenthesized subexpressions have the highest priority. [Note t h a t a 1" b 1" c = a 1" (b 1" c).]
654
Chapter 9 Trees
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J a n L u k a s i e w i e z (18781956), a Polish logician and philosopher, was born in Lvov. His father was a captain in the Austrian army. After studying mathematics and philosophy, Lukasiewicz earned his Ph.D. in philosophy from the University of Lvov in 1902, where he taught for the next 5 years. In 1915, he accepted an invitation to teach at the University of Warsaw. In 1919, he served as the minister of education in independent Poland, and the following year returned to the University where he was a professor from 1920 to 1939, serving twice as its rector. In 1946, while exiled in Belgium, Lukasiewicz accepted a professorship at the Royal Irish Academy, Dublin, where he remained until his death. A "resourceful and imaginative scholar" and "a gifted and inspiring teacher," he was one of the founding fathers of the Warsaw School of Logic, a member of several scientific societies, and the recipient of numerous honors.
To convert an infix expression into prefix or postfix, look for the subexpression that must be evaluated first, and translate it into the desired form. Now translate the next subexpression that must be evaluated. Continue like this until the whole expression is converted into the desired form. The next example illustrates this technique. Rewrite the expression a 9 (b + c)/d  e in prefix and postfix forms. SOLUTION: 1. To convert the expression into prefix f o r m : 9 The subexpression b + c must be evaluated first, so translate it into prefix: +bc. 9 Scanning the expression from left to right indicates the next operation that must be performed is multiplication. The two operands of 9 are a and +bc. So translate this subexpression into prefix: , a + bc. 9 Now perform division. The two operands o f / are , a + bc and d; rewrite this subexpression in prefix: / 9 a + bcd. 9 Finally, perform subtraction. The operands o f  are / 9 a + bcd and e; convert this subexpression into prefix:  / 9 a + bcde. Since the expression is finished, we are done; so the desired prefix expression is  / 9 a + bcde. These steps are summarized in Figure 9.68. 2. To convert the expression into postfix form: Use essentially the same steps as above, but convert subexpressions into postfix form.
9.5 BinaryTrees F i g u r e 9.68
655
a * (b + c ) / d  e
I ]
+bc
I I
9a + bc
I I
/ *a + bcd
I I
 / * a + bcde
9 Convert b + c into postfix: b c + 9 Convert a 9 ( b c + ) into postfix (parentheses are used for readability):
abc+,
9 Convert ( a b c + , ) / d
into postfix: a b c + , d / into postfix: a b c + , d / e  . This is the desired
9 Convert ( a b c + , d / )  e postfix expression.
These steps are summarized in Figure 9.69. F i g u r e 9.69
a * (b + c ) / d
I I I
 e
bc+
I I
abc + *
I
abc + , d /
I J
abc + , d / e 
I
Since every binary operator has two operands, every algebraic expression containing only binary operators can be represented by a b i n a r y e x p r e s s i o n t r e e . In such a tree the leaves contain operands and the internal vertices contain operators. Since not all data stored in the vertices are of the same kind, it is a h e t e r o g e n e o u s tree. To construct a binary expression tree, store the operator in the root of the tree and represent the first operand by the left subtree and the second operand by the right subtree. Example 9.21 employs this technique. Represent the expression a 9 (b + c ) / d  e in a binary expression tree. SOLUTION: Build the tree from the bottom up, using the order in which the various operations are performed. Construct a binary tree for each expression b + c, a 9 (b + c), a 9 (b + c ) / d , and a 9 (b + c ) / d  e successively, as in Figures 9.709.73. F i g u r e 9.70 +
b
c
656 F i g u r e 9.71
Chapter 9 Trees
a
+
b
c
F i g u r e 9.72
*
a +
/
d
b
c
F i g u r e 9.73
Binary expression tree. *
a +
/
e
d
b
c
m
For a computer to evaluate a legal infix expression, it translates it into a Polish expression and then evaluates it. The following example illustrates the second half of this process. ~ Evaluate the prefix expression singledigit number. / 2 + 75429, where each operand is a
SOLUTION: In a prefix expression, the binary operator precedes its operands; so scan the expression from left to right until you encounter two successive operands. Then back up to the operand immediately preceding the operator and perform the operation. Repeat this procedure until the expression consists of a single operand. Figure 9.74 summarizes the various steps involved; so the value of the expression is 27.
F i g u r e 9.74
1"/,2+ 75429
I I
12
I I
24
I I
6
I I
36
I I
27
m
9.5 BinaryTrees
657
The steps illustrated in Example 9.22 can easily be adapted to evaluate postfix expressions. Binary expression trees can evaluate numeric expressions recursively: replace each operator vertex with the value of the expression tree rooted at the vertex. The next example illustrates this method. Evaluate the expression represented by the binary expression tree in Figure 9.75.
F i g u r e 9.75
/ 8 4
3
7
4
SOLUTION: First evaluate the subtree rooted at/, giving 2; replace this subtree with 2 (see Figure 9.76). Now evaluate the subtree rooted at 1", giving 8; replace the subtree rooted at 1" with 8 (see Figure 9.77). Continue like this until the root of the tree contains an operand. The remaining steps transpire Figures 9.78 and 9.79. The value of the expression is 24.
F i g u r e 9.76
3
F i g u r e 9.77
7
4
8 7
F i g u r e 9.78 9
4
8
F i g u r e 9.79
3 24
J
m
658
Chapter 9 Trees Suppose you traverse a binary expression tree in preorder. What can you say about the output? In a preorder traversal, the root precedes the subtrees. Since the root of a binary expression tree represents a binary operator and the subtrees represent its operands, the preorder traversal yields the prefix form of the expression. Similarly, the postorder traversal produces the postfix expression. For example, the binary tree in Figure 9.80 represents the infix expression [a + (b t c)] 9 [d/(e  f)]. You may verify that preorder traversal yields the prefix expression 9 + a ~ b c / d  e f and postorder traversal produces the postfix expression a b c "~ + def  / . .
F i g u r e 9.80
+ a b ? c d
/ e ]'
Traversing this tree in inorder yields a + b t c 9 d / e  f , the same as the infix expression except for parentheses. In general, the inorder traversal of a binary expression tree yields the infix expression without parentheses. Finally, binary expression trees can represent assignment statements and boolean expressions; the assignment operator ~ commands least priority among the operators. For instance, the statement x ~ w + y t z is represented by the tree in Figure 9.81. Such binary trees can find the value of the variable on the LHS of the assignment operator ~ . With the tree in Figure 9.82, you may verify that the value assigned to x by the statement x ~ (5 + 6) 9 (2 t 3) is 88.
F i g u r e 9.81
(.___
x
+
w y
? z
The trees in Figures 9.83 and 9.84 display the boolean expressions (a < b) v (c < d) and a 9 (b + c) < d t e  f, respectively. The r e l a t i o n a l o p e r a t o r s , =, and V: have nexttolast precedence among all operators.
9.5 BinaryTrees Figure 9.82
659
X
$
+ 5 62
1" 3
Figure 9.83
a
b
c
d
Figure 9.84
a b
+ c d
1" e
[
Next we show how parenthesized triangulations are closely related to binary trees.
Binary Trees and Parenthesized Triangulations
Recall from Section 6.6 t h a t the triangulation of a convex (n + 2)gon corresponds to correctly parenthisizing a sequence of n + 1 symbols and vice versa. Interestingly, each procedure can be represented in a binary tree. For example, let us r e t u r n to the parenthesized triangulation in Figure 6.28. Figure 9.85 shows the order of operation employed there; this leads to the binary tree in Figure 9.86 in an obvious way.
Figure 9.85
Lukasiewicz discovered an intriguing way to extract the binary n u m b e r corresponding to a polygonal dissection from its binary tree representation.
660 Figure 9.86
Chapter 9 Trees
a
b
clyde
d
(((ab)c)(de))
iii000100
First we label each leaf in Figure 9.86 with a 0 and each i n t e r n a l vertex with a 1, as in Figure 9.87. Next, beginning at the root, we traverse the entire tree. Reading each unvisited vertex, we get the same b i n a r y n u m b e r 111000100 as before.
Figure 9.87
Traversing the binary tree for its binary number.
' o 'I x .
tJ ~, \ : ~ / s ', \ ' / , '
,'o: S
Ill
s. .'/,'
SOl
._.
..
/o',/ ii
I, ml ~vi I I ~  7 "r~"U I
~,.'~k
Finally, we observe a close relationship between binary trees and Catalan numbers.
Binary Trees Meet Catalan Numbers
An interesting combinatorial problem is to find the n u m b e r of nonisomorphic binary trees we can draw with n vertices. For example, t h e r e is exactly one binary tree with n = 1; there are exactly two distinct b i n a r y trees with n = 2, exactly five with n = 3 (see Figure 9.88), and so on. In general, with n vertices we can draw exactly Cn nonisomorphic binary trees. This should be obvious, because we found earlier t h a t there is a bijection between triangulated convex polygons and binary trees.
Figure 9.88
/
n=l n=2 n=3
9.5 Binary Trees
661
Exercises 9.5
Give the output from traversing each binary tree in preorder.
1o a 2. a
b
c~[
d
0
a
b
c
h k 46. Give the output from traversing the binary trees in Exercises 13 in inorder. 79. Give the output from postorder traversing the binary trees in Exercises 13. 1012. Redo Exercises 46 using the DFS method. 1315. Redo Exercises 46 using the BFS method. Rewrite each infix expression in prefix form.
16. a + b , c / ( d  e) ~ f 18. (a + b 9 c ) / ( d  e / f ) "t g
1 7 . a "t (b ~ c) + d / e  f
19. a  (b 9 c + d ) / e 9 f  g "~ h
2023. Translate the infix expressions in Exercises 1619 into postfix expressions. 2427. Construct a binary expression tree for each infix expression in Exercises 1619. Rewrite each prefix expression in infix form, supplying parentheses when necessary. 28. + , 1" abc 9 de 30.  "t + a , bcd , e f 29. +a ~ /b  cde
31. * *  a + bcd ~ e  fg
Convert each postfix expression into infix form, supplying parentheses when necessary. 32. ab  cd  / e f ~ 9 33. abc + d 9 9 f 34. a b  c d  / e
~
35. a b / c d / e f g  + 9 +
662
Chapter 9 Trees Evaluate each binary expression, where each operand is a singledigit number. 36.  1 " 2 1 " 2 3 , + 8 5 7 38. 6 3 / 9 2 1 + / + 7 3  , 37. 3 7 , 4 + 5 / 2 1 " 39. 8 6 + 3 4 + / 5 t
Represent each binary expression in a binary expression tree. 40. a , b + c
"~ d
41. a , b + [ c  ( d  e ) ] 43. l(a + b) 9 c] 9 ld 1" (e 1" f)l
42. (a + b 9 c) t (d/e)
Evaluate each binary expression tree. 44.
///•,
9 5
45.
T
/ \1"
2
6
/ \
9 5 4
/k,
3
2
6 8
4647. Traverse the binary trees in Exercises 44 and 45 in preorder. 4849. Traverse the binary trees in Exercises 44 and 45 in postorder. 5051. Traverse the binary trees in Exercises 44 and 45 in inorder. (Supply parentheses when needed.) Represent each assignment s t a t e m e n t and boolean expression in a binary expression tree.
52. x ~ (x + y , z ) $ ab
53. x ~ x t (Y t z ) / x + ( y  z) 55. ( a < b ) v l ( b _ < c ) A ( d _ < e ) l 57. (x/y) , z + w > (w  y) t z
54. x + y
t z < w+y hit; so insert rat as the right child of hit (Figure 9.96).
F i g u r e 9.93
Insert hit.
hit
F i g u r e 9.94
Insert hat.
hit
hat /
F i g u r e 9.95
Insert cat.
hit
cat
666 Figure 9.96
Insert rat.
Chapter 9 Trees
hit hat//% cat Z rat
The next word sat is greater than both hit and rat; store it as the right child of rat (Figure 9.97). Now fat < hit and fat < hat, but greater t h a n cat; append it as the right child of cat (Figure 9.98). The remaining words similarly fall in place, as in Figures 9.999.101. The completed binary search tree stands in Figure 9.101.
Figure 9.97
Insert sat.
hit
cat
sat
Figure 9.98
Insert fat.
hit
cat ~,. fat
sat
Figure 9.99
Insert mat.
hit
cat ~,. fat
sat
9.6 Binary Search Trees F i g u r e 9.100 Insert p a t .
hit
667
cat x. tat " pat
sat
F i g u r e 9.101 Insert kat.
hit
cat fat" kat" " pat
sat m
The order in which data are inserted in the tree determines its shape. In other words, the same data entered differently will produce differently shaped trees. For instance, the data kat, rat, hat, cat, hit, pat, fat, sat, m a t produces the tree in Figure 9.102. F i g u r e 9.102
kat
cat fat" mat 9
sat
A n i m p o r t a n t o b s e r v a t i o n : If the data in Example 9.25 are stored as an
unordered linear list, linear search takes a maximum of nine comparisons to find a given key. On the other hand, searching the binary tree in Figure 9.101 (or 9.102) takes at most four comparisons (Why?), much less than nine.
,,,
More generally, a maximum of n comparisons are needed to locate an item in an unordered list of n items, but if they are stored in a binary search tree of height h, the worst case takes only h + 1 comparisons. So making the tree as b u s h y as possible minimizes the search time.
668
Chapter 9 Trees Binary search trees can also be used to eliminate duplicates from a homogeneous list. To accomplish this, build a binary search tree consisting of distinct elements, as the next example illustrates. Using a binary search tree, eliminate all duplicates in the list 11, 15, 8, 11, 8, 12, 15, 10, 12, 17, 6. SOLUTION: With each item, search the tree to determine if it already exists in the tree; if it does, then the item is a duplicate. If it is a new element, insert it in the tree. This procedure finds four duplicates: 11, 15, 8, and 12. The final binary search tree is displayed in Figure 9.103.
Figure 9.103
11
6
17
m
These two examples indicate how efficiently binary search trees can handle homogeneous lists.
Exercises 9.6 Construct a binary search tree for each set.
1. i , a , u , o , e 4. i, a, e, o, u 2. a , e , i , o , u 3. u , o , i , e , a 6. 5 , 2 , 13, 17,3, 11
5. 8, 5, 2, 3, 13, 21
7. do, re, me, fa, sol, la
8. inning, input, output, insect, inroad, inset, insole 9. order, ouch, outfit, outing, outcome, outlet, outcry
10. canna, coleus, balsam, celosia, dahlia, azalea, tulip
1120. Find the maximum number of comparisons needed to locate an item in the binary search trees of Exercises 110. Construct a binary search tree using the words in each phrase or sentence. 21. Fourscore and seven years ago. 22. Ask not what your country can do for you. 23. All that glitters is not gold. 24. Necessity is the mother of invention.
9.6 BinarySearch Trees
669
25. Write an algorithm to print the contents of a binary search tree in lexicographic order. 26. T o u r n a m e n t s o r t is a sorting technique t h a t reflects the s t r u c t u r e of a t o u r n a m e n t . For a list of n items where n is a power of 2, a full, complete binary tree springs from the leaves to the root. For example, consider the list 13, 8, 5, 1, 21, 3, 34, 2. Store the n u m b e r s as leaves in a binary tree (Figure 9.104). At each level move up the larger of the siblings to its p a r e n t (Figure 9.105). Now the root contains the largest element m. O u t p u t it and store 0 in the leaf t h a t contained m. Repeat this procedure until all elements are output. Give the final output. F i g u r e 9.104
13
8 5
1
21
3 34
2
F i g u r e 9.105
34
13
8 5
1
21
3 34
2
H e a p s o r t is also a sorting method based on binary trees. A h e a p is a balanced homogeneous binary tree such that: (1) all leaves in the lowest level are as far left as possible; (2) info(root) is greater t h a n both info(left child) and info(right child) if they exist; and (3) every subtree is also a heap. To sort a list by this version of heapsort, first store the items level by level from left to right. Beginning with the leftmost internal vertex, build a heap to move the largest element to the root. O u t p u t it. Store 0 in the root. Repeat the procedure until all vertices contain 0's. [Heapsort takes O(n log n) comparisons) in the worst case to sort a list o f n elements.] Use the list in Figure 9.106 for Exercises 27 and 28. *27. Build the heap from the tree. *28. Show how the tree looks after building the second heap.
670 F i g u r e 9.106
Chapter 9 Trees
13
3
34
One way binary trees are used in the field of communications is through the Huffman coding scheme presented below. The "Huffman Code is one of the fundamental ideas that people in computer science and data communications are using all the time," according to Donald E. Knuth of Stanford University. It is used in computer networks, highdefinition televisions, modems, and VCR Plus, a device that automatically programs a VCR. Suppose we would like to transmit a message over a certain alphabet. We would like to encode it in terms of bits, using an unambiguous coding scheme. The ASCII scheme (Appendix A. 1) could translate each letter into a binary word. In ASCII every character is represented by a 7bit word. Such a system is a f i x e d  l e n g t h code. Using ASCII the word GRAPH is encoded as 10001111010010100000110100001001000 and the message 101010010010001000101100111110100101011001 is decoded as THEORY. (You can verify both.) A fixedlength code has the advantage of being relatively easy to encode and decode, but characters are assigned the same length codes whether or not they appear frequently, which usually wastes both storage space and time. To rectify this, codes of variable length can be assigned to the symbols in the alphabet. This kind of technique is a v a r i a b l e  l e n g t h code. Characters that occur frequently receive shorter codes than those that occur infrequently. One such system, the H u f f m a n c o d i n g s c h e m e developed by D. A. Huffman, shortens the encoded messages considerably. For instance, ASCII encodes the word TREE as 101010010100101000101 1000101, 28 bits long, while the Huffman codes in Table 9.7 renders the same word as 111000, a saving of 22 bits.
T a b l e 9.7
Symbol
Code
E R T
0 10 11
9.7 Huffman Trees (optional)
671
t
D a v i d A l b e r t H u f f m a n (19251999) was born in Alliance, Ohio. After graduating from Ohio State University in electrical engineering at the age of 18, he joined the Navy. He received his M.S. in electrical engineering from Ohio State in 1949 and his D.Sc. from M I T 4 years later. In 1951, while at MIT, in a course on information theory, he and his classmates were given a choice of taking a final examination or writing a term paper on a coding problem. Huffman worked on it for months without much success, but j u s t as he decided to start preparing for the final, a solution came to his mind; the result was the Huffman Code. Huffman left M I T in 1967 to head the newly created computer science department at the University of California, Santa Cruz. Although best known for the H u f f m a n Code, he made significant contributions to switching theory, information theory, and picture analysis.
To verify this, scan the encoded message from left to right, using Table 9.7. If the bit you encounter is a 0, it represents the letter E. However, if it is a 1, look at the next bit; if 0, the character is R; otherwise, it is T. See Figure 9.107. Now try it yourself.
Figure 9.107
0 startl
E
R
T
Huffman Algorithm
To illustrate the H u f f m a n a l g o r i t h m , consider the message TERNARY TREE over the alphabet { ~, A, E, N, R, T, Y}, where the character [] indicates a blank space.
S t e p 1 Construct a frequency table for the various symbols in the message (Table 9.8).
T a b l e 9.8
Chapter Frequency
[] 1
A 1
E 3
N 1
R 3
T 2
Y 1
672
Chapter 9 Trees S t e p 2 Find two characters that have the least frequencies, say, El, and A. (Any two of the characters ~, A, N, and Y will do.) Concatenate them to form a new symbol DA. The frequency of a newly created symbol is the sum of the frequencies of its components. So the frequency of DA is 2 (Table 9.9). Figure 9.108 shows the resulting symbols and their frequencies. (If the frequencies of the symbols differ, the symbol with the smallest frequency is made the left child; otherwise, preserve the alphabetic order.)
F i g u r e 9.108
Table 9.9 Symbol Frequency
]I~AENRTY
2 3 1 3 2 1
S t e p 3 Again look for two symbols of lowest frequencies, namely, N and Y. Concatenate them; the frequency of NY is 2. See Figure 9.109. Five symbols remain in Table 9.10.
F i g u r e 9.109
Table 9.10
Symbol Frequency
sA 2
NY 2
E 3
R 3
T 2
S t e p 4 Continue like this until one symbol remains: Tables 9.119.14 summarize the steps.
TER~ANY.
9.7 HuffmanTrees (optional) T a b l e 9.11
673
S
bo,
E
R
T
Frequency II
4
3
3
2
T a b l e 9.12
Symbol Frequency
[]ANY 4
TE 5
R 3
T a b l e 9.13
Symbol Frequency
RsANY 7
TE 5
T a b l e 9.14
Symbol tl Frequency
TER[ANY 12
The steps of combining symbols to form new symbols can produce a binary tree, as in Figure 9.110. When we drop all symbols and their frequencies except the original characters, a H u f f m a n t r e e remains (Figure 9.111). In this full binary tree, the leaves represent the original characters.
F i g u r e 9.110
ER I]ANY, 12
R DANY, 7
(~ANY,g
674 F i g u r e 9.111
H u f f m a n tree.
Chapter 9 Trees
0
1
T
E
0
1
[]
A
N
Y
S t e p 6 Assign a 0 to each edge t h a t leads to a left child and a 1 to each edge t h a t leads to a right child. See Figure 9.111. Step 7 To find the H u f f m a n code for any character, traverse the path from the root to the corresponding leaf. The sequence of bits along the path is its Huffman code. The codes of the various characters are given in Table 9.15.
Table 9.15
Character Code
[]
1100
A
1101
E
01
N
1110
R
10
T
O0
Y
1111
According to Table 9.15, the Huffman code of the message TERNARY T R E E is 00011011101101101111110000100101, only 32 bits long, with a saving of 52 bits over ASCII. Interestingly enough, the code for any character does not appear as a prefix in the code for any other character. Such a nonrepetitive code is a p r e f i x c o d e . This prefix property g u a r a n t e e s t h a t every message has a unique H u f f m a n code and vice versa. We close this section with the next example, which illustrates decoding a message using a H u f f m a n code. Using Table 9.15, decode the message 110111101111110000100101. SOLUTION: Scan the given message from left to right. Since the first bit is 1, the first character can be [], A, N, R, or Y. The second bit is also a 1, so the first
9.7 HuffmanTrees (optional)
675
character must be D, A, N, or Y. With the third bit 0, the character must be [] or A. The next bit is 1, making the first character A. Since the fifth bit is 1, the second character must be [], A, N, R, or Y. The sixth bit is also 1: the character must be [], A, N, or Y. With the seventh bit 1 and eighth bit 0, the second character is N. Continuing like this, you can verify that the original message was ANY TREE. m Huffman codes offer a unique and expeditious transmission service, as the previous two examples suggest.
Exercises 9.7
With Table 9.15, encode each word. 1. EARN 2. EATEN 3. AERATE 4. TREATY
Using Table 9.15, decode each message.
5. 1 1 1 1 0 1 1 1 0 1 1 0 6. 0 0 1 1 0 1 0 0 0 0 0 0 1 0
7. 1101111011001101111000
Figure 9.112
0 1
8. 110111001001001001110100
0
1
c
f
e
d
Using the Huffman tree in Figure 9.112, find the Huffman code for the characters in Exercises 912.
9. e 10. u 11. f 12. d
With the Huffman tree in Figure 9.112, encode each word. 13. cud 14. decode
15. educed 16. deduced
Using the Huffman tree in Figure 9.112, decode each message.
17. 11110110 19. 01010011011110110
18. 10111111
20. 011110110111
21. Using the following frequency table, construct a Huffman tree for each
character in the alphabet {a, b, c, d, e, f}.
676
Chapter 9 Trees
Character Frequency
a 4
b 1
c 2
d 3
e 5
f 4
2 2 . Using the following frequency table, construct a H u f f m a n tree for the
alphabet {a, b, c, e, g, l, o, s, u }.
Character Frequency
a 4
b 3
c 2
e 3
g 1
l 2
o 4
s 1
u 5
23. Using Exercise 21, find the H u f f m a n code for t h e c h a r a c t e r s a t h r o u g h f. Using Exercise 21, encode each word. 24. babe
2 5 . bead
26. abba
27. ceded
Using Exercise 21, decode each message. 28. 1101010110 30. 010011010110 29. 00111010110 31. 10010100111110
With Exercise 22, find a Huffman code for each character.
32. a 33. c 34. o 35. s
Using Exercise 22, encode each word. 36. cabbage 37. babbage 38. calculus 39. caboose
Using Exercise 22, decode each message. 40. 001100110010101001100
42. 1010011011100010001
41. 1001100110111000001100
43. 001101000000001010000100
The w e i g h t w of a Huffman code, which measures its efficiency, is defined as follows. Let Cl, c 2 , . . . , Cn denote the characters in an alphabet; fl,f2,... ,fn their frequencies; and 11,12,... ,ln the lengths of their codes.
n
T h e n w = ~ fill. Compute the weight of the Huffman code in:
i=1
44. Table 9.15.
45. Exercise 21.
46. Exercise 22.
The wide application of mary trees embraces the general decisionm a k i n g process. Consider this guessing game: A n n a t h i n k s of a n u m b e r
9.8 Decision Trees (optional)
677
n < 1000; each t i m e B e e n a gives an incorrect response, A n n a says w h e t h e r B e e n a ' s guess is less t h a n or m o r e t h a n n; by m a k i n g at m o s t 10 guesses, B e e n a can find the n u m b e r n. Suppose A n n a chooses n  687. We shall employ t h e b i n a r y s e a r c h algor i t h m to find n. B e e n a ' s first guess is L(1 + 1000)/2J  500. Since 500 < n, A n n a says more. Since the n u m b e r m u s t lie b e t w e e n 500 a n d 1000, B e e n a c o m p u t e s L(500 + 1000)/2J  750 as h e r second guess. A n n a ' s response? Less. So B e e n a guesses L(500+ 750)j  625. A n n a ' s response? More. B e e n a ' s f o u r t h guess isL(625 + 750)/2]  687; she has m a d e the correct guess. Figure 9.113
< / \ ><
/
\
>
yes
T h e s e steps form the b i n a r y tree in Figure 9.113, a d e c i s i o n t r e e . At each i n t e r n a l vertex, we select one of two alternatives, t h e r e b y selecting a subtree. T h e next two examples provide additional decision trees. Sort t h r e e distinct e l e m e n t s a, b, a n d c. C a s e 1 Let a < b. T h e n c o m p a r e a a n d c. Ifc < a, c < a < b. On t h e o t h e r hand, if c > a, t h e n c o m p a r e b a n d c. If b < c, t h e n a < b < c; otherwise, a_ [lg n!l a n d the list will t a k e at least [lg n!l comparisons to sort. A m o r e d o w n  t o  e a r t h application follows.
678 Figure 9.114
Chapter 9 Trees
a?b}
< >
a
23
23
234 The g r a m m a r defined in this example is G = (N, T, P, a), where: 9 N = {, , , , (digit> }, 9 T = {+,,0,1,2,3,4,5,6,7,8,9}, 9 The production rules are: _ 1}. I The g r a m m a r G = (N, T , P , ~) in E x a m p l e 11.15 is contextfree, so L(G) {anb n In >_ 1} is a contextfree language. E x a m p l e 11.53 will d e m o n s t r a t e t h a t G is not regular. II A language L(G) m a y contain words derivable from a in more t h a n one way. Accordingly, we m a k e the following definition.
Ambiguous Grammar A g r a m m a r G is a m b i g u o u s ifa string in L(G) has more t h a n one derivation tree. The next two examples p r e s e n t a m b i g u o u s g r a m m a r s . The following g r a m m a r G defines the s y n t a x of simple algebraic expressions: (expression) ::= (expression) (sign> (expression) I (letter>
::= + l : : = a l b l c l . . . Iz
This g r a m m a r can produce the expression a  b + c two ways, as the derivation trees in F i g u r e 11.11 show. As a result, G is an a m b i g u o u s grammar. 1 o ~ (optional) T h e following are simplified production rules for an ifthen statement S: S ::= if (expression) t h e n I if (expression) t h e n (statement) else (statement) (expression) ::= E1 I E2 (statement) ::= S1 I $2 I if (expression> t h e n (statement) To see t h a t these rules produce an a m b i g u o u s g r a m m a r , notice t h a t the ifthen s t a t e m e n t I f E1 then i f E2 then $1 else $2 (11.3)
754 F i g u r e 11.11
Chapter I!
FormalLanguages and FiniteState Machines
(expre ssion) (let ;er)
(sign)
I
(expression) (letter)
(sign) +
(expression) c
I
I
(expression) ~ ~ (sign) (expression) (letter) (sign) + (expression) (letter) c
(expression) (letter)
I
I
I I b
I
can be interpreted in two ways: (i) If E1 then (if E2 then S 1 else $2), or (ii) If E 1 then (if E2 then S 1) else $2. Using indentation, these possibilities can be displayed as follows:
(i) i f El then i f E2 then Sl else $2 (ii) i f El then i f E2 then Sl else $2
Accordingly, statement ( 11.3) can be generated by two distinct derivation trees (see Figure 11.12). To avoid this confusion, each e l s e is paired with the nearest if. Consequently, statement (i) is the correct interpretation of statement (11.3). If you would like statement (11.3) to mean statement (ii), you have
11.2 Grammars F i g u r e 11.12
755
if
(expression>
then if
0 } 39. {amb n I m, n > 1} 37. { a n b l n > l } 38. { a n b a l n > l }
40. The set of palindromes.
Using Example 11.18, draw the derivation tree for each integer. 41. 234 42.  2 3 4
43. An identifier in Java is a letter, underscore, or $, followed by any number of alphanumeric characters. With BNF, define the g r a m m a r for a Java identifier. Use the g r a m m a r in Exercise 43 to see if each string is a valid J a v a identifier. 44. catch 22 45. 20/20 46. algorist 47. three roots
Construct a derivation tree for each identifier. 48. result2 49. value 50. R2D2 51. math
The production rules of a g r a m m a r for simple arithmetic expressions are: (expression) ::= (digit)I ((expression))I + ((expression)) I
 ((expression)) I (expression)(operator)(expression)
(digit) ::= 01 1 1 2 1 3 1 4 1 5 1 6 1 (operator) ::= + l  I * [ / i 1 "
71819
Use this g r a m m a r to a n s w e r Exercises 5 2  5 9 .
Determine if each is a valid arithmetic expression. 52. 2 , 3 + 4 53.  ( 3 , 4 1 " 5 ) 54. 3+ 1"7 55. 6 + 5 / 8 ,
Construct a derivation tree for each expression. 56. 3 + 5 , 6
57. 5 + ( 4 1 " 3 ) 58. ( 5 + 3 )  7 / 4 59.  ( 3 1" ( 5 + 2 ) )
A number in ALGOL (excluding the exponential form) is defined as follows: (number) ::= (decimal number)I (sign) (decimal number) (decimal number) :: (unsigned integer) I. (unsigned integer) i (unsigned integer). (unsigned integer)
758
Chapter II
FormalLanguages and FiniteState Machines
(unsigned integer) "' (digit) I (unsigned integer)(digit) (digit)::=01 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 ::= + l Use this g r a m m a r to answer Exercises 6067. Determine if each is a valid ALGOL number. 60. 234 61. 2.34 62. 234. 63..234
Draw a derivation tree for each ALGOL number. 64.  3 . 7 6 65. +376 66..376 67. 0.23
For Exercises 6873, use the following definition of a simple algebraic expression: (expression) ::= (term) I (sign) (term) I {expression) {adding operator){term) (sign) ::= + 1 (adding operator) ::= + l (term) ::= (factor) [ {term) {multiplying operator){factor) (multiplying operator) ::= * 1 / (factor) ::= (letter) I ( (expression) )l (expression) (letter/::=alblcl... Determine if each is a legal expression.
68. a + b , ( c / d ) 69. a + b + c 70.  a , b / c + d 71. ( ( a  b ) + c )
Iz
Construct a derivation tree for each expression. 72. ( a , b ) + c / d 73. a , ( b + c / d )
74. Use BNF to define a g r a m m a r for the language of wellformed parentheses (wfp). Use the g r a m m a r in Exercise 74 to see if each is a valid sequence of parentheses. 75. (()) 76. ()(()) 77. (()()) 78. ()()()
79. Figures 11.13 and 11.14 diagram the syntax for an unsigned integer and an unsigned number, respectively. Define the g r a m m a r for an unsigned number in BNF.
11.3 FiniteState Automata F i g u r e 11.13
759
unsigned integer:
[ ~
digit
]
F i g u r e 11.14
unsigned number: unsigned integer ~ 9 unsigned ~ integer ~ ~ E + unsigned integer
Using the g r a m m a r in Exercise 79, check if each is a valid unsigned number. 80. 177.76 81. .1776 82. 1776. 83. 17.76E2
This section presents an abstract model of a machine that accepts input values, but produces no output values. Often the question arises whether or not a word over an alphabet is acceptable. For example, is 2R2D an acceptable identifier or is 17.06 a valid real number in C + +? Finitestate automata can model the steps in determining if a given word exists in a language. Accordingly, finitestate automata, also known as l a n g u a g e r e c o g n i z e r s , play a central role in the development of compilers. Before we study the definition, we present a simple example of a language recognizer. Determining if an input string over the alphabet {a, b} contains abba substring involves the following five steps:
as
a
S t e p 0 If the first symbol in the string is a, move to step 1 and look for the character b. Otherwise, no progress has been made. S t e p 1 If the next character is b, the substring ab has occurred, so go to step 2 and look for another b. Otherwise, the symbol b is still missing, so stay in step 1. S t e p 2 If the next symbol is b, the substring abb exists; go to step 3; if a, return to step 1. S t e p 3 If the next symbol is a, the given input string contains the substring abba; otherwise, return to step 0 and start all over again.
760
Chapter II FormalLanguages and FiniteState Machines S t e p 4 Once the substring abba has occurred in the input string, any sequence of a's and b's may follow. These steps can be represented by a digraph (see Figure 11.15), each vertex representing a step. Exactly two edges, labeled a or b, leave each vertex.
F i g u r e 11.15
a
b
To determine the action required from a given step, simply follow the directed edges from the corresponding vertex. For example, at vertex s3 (step 3) if the next input symbol is a, move to vertex s4 (step 4); otherwise, return to vertex so (step 0). The other (labeled) edges are interpreted similarly. The digraph indicates a string contains abba as a substring if and only if the directed path the string determines terminates at vertex s4. The string abab determines the path sosls2sls2, which does not end at s4; consequently, abab is not acceptable. On the other hand, the string ababbab determines the path sosls2sls2s~s4s4, which terminates at s4; so the string does have the desired property, m The digraph in Figure 11.15 displays a f i n i t e  s t a t e a u t o m a t o n . (Automaton is the singular form of automata.) Its five vertices, so through s4, are the s t a t e s of the automaton. Since the whole process begins at so (step 0), so is the i n i t i a l s t a t e . A string is acceptable, that is, contains abba as a substring, if and only if its path ends at s4; accordingly, s4 is an a c c e p t i n g state. The digraph shows the transition of the machine between states. For example, if the automaton is at state s2 and the input symbol is a, the automaton switches its state to s 1. The digraph is the t r a n s i t i o n d i a g r a m of the finitestate automaton. The initial state is customarily identified by an arrow pointing to it and an accepting state by two concentric circles, as Figure 11.16 shows. The transition diagram appears in Figure 11.17.
F i g u r e 11.16
The initial state
An accepting state
!1.3 FiniteStateAutomata F i g u r e 11.17
761
a
a
a
b
Each state si and an input symbol d e t e r m i n e a u n i q u e state sj. So we can define a function f : S x I ~ S as follows, where S = {so, Sl, s2, s3, s4 }, the set of states, and I = {a, b}, the i n p u t alphabet:
f ( s o , a) = 81 f(s2, a) = s i f(s4, a)   8 4 f ( s o , b) = so
f(s2, b) = s 3 f(s4, b) = 8 4
f ( s l , a) = s i f(s3, a)   8 4
f(si,b) =s2 f(s3, b)   s o
The function f is the t r a n s i t i o n f u n c t i o n of the finitestate a u t o m a t o n . It can also be defined by the t r a n s i t i o n t a b l e in Table 11.1.
T a b l e 11.1
State a
s0 Sl s2 s3 s4 s1 81 81 s4 s4
Input symbol b
80 s2 s3 so s4
We are now ready to define a finitestate a u t o m a t o n .
FiniteState Automaton
A f i n i t e  s t a t e a u t o m a t o n (FSA), M, manifests five characteristics: 9 A finite set, S, of s t a t e s of the a u t o m a t o n . 9 A specially designated state, so, called the i n i t i a l state. 9 A subset A of S, consisting of the a c c e p t i n g s t a t e s (or final s t a t e s ) of the a u t o m a t o n . 9 A finite set, I, of i n p u t s y m b o l s . 9 A function f : S x I ~ S, called the t r a n s i t i o n f u n c t i o n or the n e x t state function. In symbols, M  ( S , A , I , f
, so).
762
Chapter II
FormalLanguages and FiniteState Machines
For instance, for the FSA in Example 11.26, S {80,s1,82,83,84}, A = {s4}, I  {a,b}, and the t r a n s i t i o n function f i s defined by Table 11.1. New York City subway c o m m u t e r s use an FSA everyday, as the next example shows.
=
A turnstile in the s u b w a y e n t r a n c e contains four a r m s at waist level (Figure 11.18). Initially, it is locked so t h a t the a r m s c a n n o t be moved. Depositing a token into the slot, however, unlocks it and allows the a r m s to r o t a t e t h r o u g h one q u a r t e r of a complete turn, so the c o m m u t e r passes t h r o u g h the turnstile.
F i g u r e 11.18
The t u r n s t i l e has two states: locked (l) and unlocked (u). Depositing a token (t) shifts the t u r n s t i l e from the locked state to the unlocked state and no m a t t e r how m a n y times the c o m m u t e r inputs t, the t u r n s t i l e r e m a i n s in the same state. P u s h i n g (p), the arms, takes the t u r n s t i l e back to the locked state. Once it is in the locked state, it r e m a i n s t h e r e regardless of how m a n y times the c o m m u t e r pushes the arms; t h a t is, regardless of the n u m b e r of times he inputs p into the device. The t u r n s t i l e exemplifies an FSA. Figure 11.19 shows its t r a n s i t i o n diagram.
F i g u r e 11.19
l
The next two examples draw t r a n s i t i o n d i a g r a m s of FSAs from their algebraic definitions. Draw the t r a n s i t i o n d i a g r a m of the FSA M  (S, A, I, f, so), where S { s o , s l , s 2 } , A = {s2}, I = {a, b}, and the t r a n s i t i o n function f is defined by
f(s0, a)
f(sl,
= Sl, b) = so,
f(s0, b) f(s2, a)
= so,
 s2,
f(sl, f(s2,
a)  s2, b)  so.
*Based on B. Hayes, "On the FiniteState Machine, A Minimal Model of Mousetraps, Ribosomes, and the Human Soul," Scientific American, Vol. 249 (Dec. 1983), pp. 2028, 178.
11.3 FiniteStateAutomata
763
SOLUTION: The FSA contains t h r e e states   so, S l , and s2   w i t h s2 t h e only accepting state. Since t h e r e are two i n p u t symbols, exactly two edges leave each vertex. D r a w a directed edge from s t a t e si to s t a t e sj if t h e r e is an i n p u t symbol x such t h a t f ( s i , x )  sj; t h e n label the edge x. F o r example, since f ( s l , b)  so, a directed edge r u n s from Sl to so labeled b. F i g u r e 11.20 shows the r e s u l t i n g t r a n s i t i o n diagram.
F i g u r e 11.20
~
b a
a
m
D r a w the t r a n s i t i o n d i a g r a m of the FSA M  (S, A, I, f, so), w h e r e S {so, s 1, s2, s3, s4}, A  {s2}, I  {a, b, c }, and f is defined by Table 11.2.
T a b l e 11.2
a
b
s2 s2 s4 s2 s4
c
s3 s3 83 s4 s4
s4
s3 s4
Sl s4
SOLUTION: The a u t o m a t o n contains five states, with s2 the only accepting one. Since t h e r e are t h r e e i n p u t symbols, t h r e e edges originate from every state. D r a w a directed edge from s t a t e si to s t a t e sj if t h e r e exists an i n p u t symbol x such t h a t f ( s i , x )  sj. For instance, f ( s l , c ) = s3, so a directed edge labeled c r u n s from state Sl to s t a t e s3. F i g u r e 11.21 displays the r e s u l t i n g t r a n s i t i o n diagram, where, for convenience, t h e t h r e e loops at s4 a p p e a r as a single loop with labels a, b, a n d c.
F i g u r e 11.21
c c
m
764
Chapter11 FormalLanguages and FiniteState Machines
Suppose a string is input into an FSA. If the path it determines ends at an accepting state, the string is a c c e p t e d (or r e c o g n i z e d ) by the automaton; otherwise, it is r e j e c t e d by the automaton. Determine if the strings a3b2ab and ab3a are accepted by the FSA in Figure 11.17. SOLUTION: First, find the path determined by the string and check if it terminates at s4, the accepting state. (Recall that a3b2ab  aaabbab.) Begin at the initial state, so. When a is input, move to state Sl. Every time a is input, remain there, so the path defined by a a a is sos 1s 1s 1. When b is input, transfer to state s2. The path obtained thus far is s o  s l  s l  s l  s 2 . Now b moves to s3 and a to s4, yielding the path 808181s1828384. Once in s4, remain there no matter what the input is. Thus the path determined by the given string is SOSl8181S2SnS4S4. Since it terminates at s4, the FSA accepts the given word. Notice that the path determined by the string ab3a is 808182838081, and it does not end at the accepting state s4; consequently, the automaton rejects the string, m Two different FSAs may accept the same language over an alphabet. This occurrence requires that we make a new definition.
Equivalent FiniteState Automata
The set of words accepted by an FSA, M, is the l a n g u a g e a c c e p t e d (or r e c o g n i z e d ) by M and is denoted by L ( M ) . Two finitestate automata, M and M', are e q u i v a l e n t if they recognize the same language: L ( M ) = L ( M ' ) . Identify the language L ( M ) accepted by the automaton M in Figure 11.20.
SOLUTION: Look for paths beginning at so and terminating at words over {a, b } that end in aa.
82. L ( M )
consists of all
1
By Example 11.31, the automaton in Figure 11.20 accepts the language of words over {a, b } ending in a a . You may verify that the FSA in Figure 11.22 accepts the same language. Consequently, the automata in Figures 11.20 and 11.22 are equivalent.
F i g u r e 11.22
b
v
1
11.3 FiniteStateAutomata
765
The next four examples build FSAs with desired properties, as Example 11.26 did. Design an FSA t h a t accepts words over I = {a, b} containing an even n u m b e r of a's. SOLUTION: Every word over I contains either an even n u m b e r of a ' s (E) or an odd n u m b e r of a's (O), so the a u t o m a t o n has two states, E and O, E being the accepting state. Initially, the n u m b e r of a ' s in the word is zero, an even integer; E is the initial state of the automaton. If the a u t o m a t o n is at E and an a is input, it moves to state O. If it is at O and an a is input, it moves to state E. Figure 11.23 shows the transition diagram of the FSA.
Figure 11.23
9 a
A word over I has even parity if it contains an even n u m b e r of a's and o d d p a r i t y if an odd number. Since the a u t o m a t o n in Example 11.33 determines w h e t h e r a word has even or odd parity, it is called a parity
c h e c k machine.
I
Design an FSA accepting words over {a, b } t h a t begin with a a and end in bb. SOLUTION: We build the a u t o m a t o n step by step:
Step 0
Initially, the a u t o m a t o n is at the initial state so.
Step 1 If the first symbol is a, move to state Sl from so and wait for the next symbol. But if the first symbol is b, the word is not acceptable (state s2 ). See Figure 11.24. Figure 11.24
Step 2
If the input symbol at s l is a, move to state s3 and d e t e r m i n e w h e t h e r the string ends with bb. On the other hand, if the input symbol at Sl is b, move to s2 to trap such unacceptable words. Once at s2, r e m a i n there no m a t t e r w h a t the input symbol is. See Figure 11.25.
766 F i g u r e 11.25
Chapter 11 FormalLanguages and FiniteState Machines
b
b
S t e p 3 Every word t h a t triggers a move from so to 8 3 begins w i t h aa. Any n u m b e r o f a ' s can follow it (see the loop at s3 in Figure 11.26). However, if b follows the word, move to a new state s4, as in Figure 11.26.
F i g u r e 11.26
a
a
b
b
b
S t e p 4 If the input symbol at 8 4 is a, r e t u r n to S 3 and look for the pair bb. But if it is b, move to a new state s5. See Figure 11.27.
F i g u r e 11.27
a
a
a,b
S t e p 5 Once at s5, any n u m b e r of b's may occur. However, if the input symbol at s5 is a, r e t u r n to s3 to look for bb. Since words e n d i n g in bb are acceptable, s5 is the accepting state. These six steps create the FSA in Figure 11.28.
F i g u r e 11.28
a
b b b
a
b
Y
m
! 1.3
FiniteState Automata
767
(optional) An identifier in a p r o g r a m m i n g language consists of a letter followed by any n u m b e r of alphanumeric characters (Section 11.1). Design an FSA that recognizes such legal identifiers. SOLUTION: Let I denote the set of all characters in the alphabet recognizable by a compiler. Let l denote a letter, d a digit, and n any n o n a l p h a n u m e r i c character. The a u t o m a t o n will have three states: so, s l, and s2. State s2 traps all invalid strings. (Accordingly, it is called a t r a p s t a t e or a d u m p s t a t e . ) The resulting a u t o m a t o n appears in Figure 11.29. F i g u r e 11.29
d, n
B
The FSA in Figure 11.29 can be translated into an algorithm which determines if a sequence of characters is a legal identifier. See Algorithm 11.1.
Algorithm i d e n t i f i e r (* This algorithm determines whether a sequence of characters is a v a l i d i d e n t i f i e r , using the FSA in Figure 11.29. All characters are read from the same input l i n e . SymboZ denotes an a r b i t r a r y character; stote denotes an a r b i t r a r y s t a t e ; stoteO, stotel, and store2 denote the various states of the FSA. stote2 is a dump state. *) Begin (* algorithm *) state ~ state0 (* i n i t i a l i z e state *) read (symboI ) while not at the end of the current l i n e begin case state of state0: i f symbol is a l e t t e r then state
('~ a/1 b/1
a/1
b/O '~ a I0 (~ alO a/O
15.
a/O b/1
b/1
778
Chapter 11 Formal Languages and FiniteState Machines
16.
a/1 b/O
.>
(~ b/O a/O
b/1
b/O
Using the FSM in Figure 11.33, find the output from each input string. 17. abba 18. baab 19. a2b3a 20. a3b2ab 3
Using the unit delay machine in Figure 11.35, find the output of each input string.
21. 1101 22. 1111 23. 0000 24. 101110
25. With a transition table, define the transition function f a n d the output function g of the FSM for binary addition in Figure 11.39. Using the FSM in Figure 11.39, compute the sum of each pair of binary numbers.
26. 1001 0110 27. 00111 28. 1011 0110 29.
10010
11011 10101
30. Redraw Figure 11.20 as the transition diagram of an FSM. 3134. Redraw the transition diagram of each automaton in Exercises 1720 of Section 11.3 as that of an FSM. Determine if the input string in Exercises 3538 is accepted by the FSM in Figure 11.42.
F i g u r e 11.42
('~ b/O
a/1
b/O
35. abba 36. aabb 37. a 3 38. b3a 4
39. Identify the language accepted by the FSM in Figure 11.42. Design an FSM accepting strings over {a, b} that: 40. Contain aa as a substring. 41. Contain exactly one a. 
With x an input symbol and s an arbitrary state of an FSM M (S,I, O , f , g , so), define g(s,x) in each case. 42. f(s, x) is an accepting state.
43. f(s, x) is a nonaccepting state.
1I.,5 Deterministic FiniteState Automata and Regular Languages
779
Is the language accepted by an FSA contextsensitive? Or is it contextfree, regular, or something else? This section provides a definitive a n s w e r to these questions. In an F S A M  ( S , A , I , f , so), where II[ = m, exactly m outgoing edges leave every state si, each labeled with a unique element of I. Besides, since f 9S x I ~ S, every s t a t e  i n p u t pair yields a unique state; in other words, every s t a t e  i n p u t pair uniquely determines the next state. For the a u t o m a t o n in Figure 11.15, the pair (s2, a) determines the state Sl, whereas the pair (s2, b) determines the state s3. Accordingly, the a u t o m a t a in Section 11.3 are called d e t e r m i n i s t i c f i n i t e  s t a t e a u t o m a t a (DFSA). This determinism suggests t h a t the language accepted by a DFSA is indeed regular, as the next example demonstrates. By Example 11.31 the language L ( M ) accepted by the DFSA in Figure 11.43 consists of words over {a, b } ending in a a . Employing it, a regular g r a m m a r G  ( N , T , P , c~) can be constructed. Choose {a, b} as the set of t e r m i n a l symbols: T  {a, b}. Choose the states as the n o n t e r m i n a l symbols: N = {so, s l, s2 }. Select the initial state so as the start symbol: a  so.
Figure 11.43
b a a a
Define the two productions rules: 9 If there is an edge labeledx from state si to state sj, define the production si ~ xsj. The various productions obtained this way are"
so ~ asl, So ~ s2 ~ bso, as2,
81 ~
as2, s2 > bso.
S l > bso,
and
9 If there is an edge labeled x from state si to an accepting state, induce the production si ~ x. Two additional productions can be obtained by this method: Sl > a and s2 > a
The g r a m m a r G  (N, T, P, a) where N, T, P, and a are defined as above is clearly regular, therefore L ( G ) is a regular language. You may verify t h a t L ( G ) consists of strings over T ending in a a . T h u s L ( M )  L ( G ) . m
780
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This example leads us to a f u n d a m e n t a l result whose proof resembles t h a t in Example 11.44. ~ The language accepted by a DFSA is regular.
PROOF:
Let M = (S,A, I, f, so) be a DFSA and L ( M ) denote the language accepted by the automaton. We shall construct a regular g r a m m a r G using the machine M and show t h a t L ( G )  L ( M ) . To construct the g r a m m a r G  (N, T , P , ~ ) , choose N  S as the set of states, T = I as the input alphabet, and a = so as the initial state. Define the productions P this way: Let si and sj be any two states, and x any input symbol. If f ( s i , x ) = sj, define the production si ~ xsj; if f ( s i , x ) = sj, an accepting state, include the production si > x. Clearly, G is a regular g r a m m a r .
T o p r o v e t h a t L ( M ) c_ L ( G ) :
Let x = X l X 2 . . . Xn be a string accepted by the a u t o m a t o n M; t h a t is, let x ~ L ( M ) . Then the transition diagram of the a u t o m a t o n contains a directed path sos 1s2 . . . . . Sn, where Sn is an accepting state. Correspondingly, these production rules follow:
so * X l S l
S 1 > X 2 8 2
(11.4)
Si1
> X i S o i
S n  1 > X n
(Note: Sn is an accepting state.)
and the derivation of the string x:
~, X l S 1 ~, x l X 2 8 2
(11.5)
~, X l X 2 . . . X n _ l S n _ l XlX2 . . . Xn_lXn
since S n  1 > X n . Thus x ~ L(G), so L ( M ) c_ L ( G ) . Conversely, let x = X l X 2 . . . X n E L ( G ) . Then it m u s t have a derivation of the form (11.4). Correspondingly, the transition diagram of the a u t o m a t o n M m u s t contain a directed path, sos lS2 . . . . . Sn. The string determined by this path is x = X l X 2 . . . Xn. Since the last production in the derivation
11.5 Deterministic FiniteState Automata and Regular Languages
781
(11.4) is S n 
1 > Xn, Sn
must be an accepting state, thus x ~ L ( M ) and hence In other words, the language accepted by the DFSA m
L ( G ) c_ L ( M ) . Thus L ( M ) = L ( G ) .
is regular,
This proof provides an elegant method for finding the regular language accepted by a DFSA. We demonstrate it again in the next example. ~ Find the g r a m m a r of the regular language accepted by the parity check machine in Example 11.33. SOLUTION: Using the transition diagram in Figure 11.23, N = {E, O}, T  {a, b}, S {E}, and the production rules are: EoaO, E~bE, O~aE, O+bO, Eob, and O+a
The regular g r a m m a r defined by the parity check machine M is G (N, T , P , S ) . [So L ( G ) = L ( M )  the set of strings over T containing an even number of a's.] n Finally, is the converse of Theorem 11.3 true? With G a regular grammar, does a DFSA exist such that L ( M ) = L(G)? The next two sections will give us an answer.
Exercises 11.5
Determine if each is a DFSA.
0
0
a a b
b
)
0
a
+
a
~..
782
Chapter II
Formal Languages and FiniteState Machines
0
a,b
a
a a,b
r
Write the regular g r a m m a r defined by the DFSA in each figure. 5. Figure 11.17 6. Figure 11.28
714. Construct the regular g r a m m a r defined by each DFSA in Exercises 1724 of Section 11.3. By m a k i n g a DFSA, define a regular g r a m m a r G = (N, T , P , ~ ) t h a t generates the language consisting of strings over {a, b } that: 15. Contain exactly one a. 17. Begin with aa. 19. Contain a b a as a substring. 21. Begin with a a or bb. 16. Contain at least one a. 18. End with bb. 20. Contain a a a as a substring. 22. Contain b a a b as a substring.
We ended the preceding section with a question: For a regular g r a m m a r G, is there a DFSA M such t h a t L ( G ) = L(M)? The obvious t e m p t a t i o n is to simply reverse the steps in Example 11.44 (or Theorem 11.3) to look for it. Let's see what happens if we do so. ~ With the regular g r a m m a r G  (N, T, P, ~), where N  {A, c~}, T = {a, b}, and P = {a * aa, a ~ aA, A ~ b}, let us see what happens if we reverse the steps in Theorem 11.3 in order to construct a D F S A M = ( S , A , I , f , so). Then I = T = {a, b} and so = a. Corresponding to the productions a  . aa and a * aA, there m u s t be two states, namely, a and A; besides, by virtue of the production A ~ b, an accepting state F must exist. T h u s S m u s t be {a,A, F}. Use the productions to draw the edges in the transition diagram of the automaton: If si ~ xsj, draw an edge from state si to sj and label it x; if si ~ x, draw an edge from si to the accepting state F and label it x. The diagram in Figure 11.44 results. Unfortunately, it is not a DFSA for two reasons: (1) A state, c~, has two outgoing edges with the same label a; (2) not every state, namely A and F, has two edges with different labels. T h u s reversing the steps illustrated in Example 11.44 does n o t yield a DFSA.
11.6 NondeterministicFiniteState Automata
783
F i g u r e 11.44
a
@
b
@
n
But, fortunately, we have a n o t h e r option. The a u t o m a t o n in F i g u r e 11.44 is a n o n d e t e r m i n i s t i c finitestate a u t o m a t o n . " N o n d e t e r m i n i s t i c " m e a n s t h a t each s t a t e  i n p u t pair m a y d e t e r m i n e more t h a n one state. F o r instance, the pair (a, a) d e t e r m i n e s two states, a and A. If a is i n p u t at state a, two choices exist for the next state: r e m a i n at a or move to A. We can now move to the following definition.
Nondeterministic FiniteState Automata
A n o n d e t e r m i n i s t i c f i n i t e  s t a t e a u t o m a t o n (NDFSA) M exhibits five characteristics: 9 A finite set S of states; 9 A specially designated state ~, called the i n i t i a l state; 9 A subset A of S consisting of the a c c e p t i n g s t a t e s (or final s t a t e s ) of the a u t o m a t o n ; 9 A finite set I of i n p u t symbols; 9 A function f : S • I , P(S), called the t r a n s i t i o n f u n c t i o n (or the n e x t  s t a t e function). [Note: P(S) denotes the power set of S.] In symbols, M  (S, A, I, f, c~). In an NDFSA, each s t a t e  i n p u t pair is linked with a set of states, not necessarily a u n i q u e state; it can be the null set. A N D F S A can be represented by a t r a n s i t i o n d i a g r a m and a t r a n s i t i o n table can define a t r a n s i t i o n function, as the next two examples illustrate. For the N D F S A in Figure 11.44, S  {a, A, F} a n d A table in Table 11.7 defines the t r a n s i t i o n function. {F}. The t r a n s i t i o n
Table 11.7
a
{c~,A} 0 0
0 {F} 0
m
The N D F S A M  (S, A, I, f, a), where S  { a , A , B , C } , A  {F}, I {a, b}, and f is defined by Table 11.8. Its t r a n s i t i o n d i a g r a m is given in Figure 11.45.
784 T a b l e 11.8
Chapter 11 FormalLanguages and FiniteState Machines
I
a
A
{a,A} {A} {F} O
O
{B,F}
{ } B O
F i g u r e 11.45
~a~b~a
~ ~
b
~~/.... I
The definition of a string accepted by an FSA can be extended to NDFSA as well.
Equivalent Nondeterministic FiniteState Automata
A string is a c c e p t e d or r e c o g n i z e d by a NDFSA M = (S, A, I, f, so) if a directed path runs from the initial vertex so to an accepting state t h a t generates the string. The language of all strings accepted by M is L ( M ) . Two NDFSAs are e q u i v a l e n t if they accept the same language. The next two examples illustrate the definition of ( L ( M ) ) . The word a3b is accepted by the NDFSA in Figure 11.44 since the corresponding path, ~~~AF, ends at an accepting state F. Notice t h a t
L ( M ) = {a nb I n >_ 1 }. I
The string a2b3a is accepted by the NDFSA in Figure 11.45. Two paths generate it, a  a  A  B  B  B  F and a  A  A  B  B  B  F . The a u t o m a t o n accepts strings amb and a mb ha, where m, n >_ 1. Thus L ( M )  {a rob, amb na ] m, n>l}. I The question we posed at the beginning of this section can be partially answered now. Every regular language is accepted by an NDFSA.
PROOF: Let G = (N, T, P, a) be a regular g r a m m a r . Through essentially the same steps as in Example 11.46, make a suitable N D F S A M = (S, A, I, f, so) such that L ( G ) = L ( M ) . Select I = T, so = {a }, and N as the set of nonaccepting states of M. Since the g r a m m a r contains productions of the form si x, introduce an accepting state F; choose S = N u {F} and A  {F}. Finally, since every production of G is si ~ xsj or si ~ x, the t r a n s i t i o n function f : S x I + P ( S ) follows: f ( s i , x )  {sj I si + x s j } U {F I si + x}.
11.6 Nondeterministie FiniteState Automata
785
As in Theorem 11.3, it can be shown t h a t L ( G ) = L ( M ) . (Complete the proof.) m Although nondeterministic finitestate a u t o m a t a have been defined, an explicit answer to the question posed earlier has yet to surface: Given a regular g r a m m a r G, does there exist a DFSA such t h a t L ( G ) = L(M)? We will answer this in the next section.
E x e r c i s e s 11.6
Draw the transition diagram of the NDFSA M = ( S , A , I , f , so), where:
1. S = {so,sl,s2), A = {s2}
S I a b 2. S 
{so,sl,s2}, A = {81}
J
I s2
a
b
80 81
.
{81} {81}
{80} {81,82}
{81} {82}
{80} {81,82}
O
. . .
0
0
0
3. S = { s o , s l , s 2 , s 3 ) , A = {s2}
S I a
4. S = {so, s l , s 2 , s 3 } , A = {s2}
S I a b
80 81 s2
{80,81} {81,82} {s2}
{83} {81}
80 81
{80, 81 } {81,82}
{83 } {80}
{s~}
{s3} {s3}
s2
0
{81}
0
{83}
5. S{8o,81,'32,83,84), A = {s2,s3}
a b
6. S  { 8 0 , 8 1 , 8 2 , s 3 , 8 4 , 8 5 ) , A = {s2,s5}
a b
so 81 82 83 84
{S0,Sl} {Sl,82} 0 0 {84}
{84} {81,S3} 0 0 {84}
s0 81 82 83 84 85
{80,81} {Sl,S2} {S2} {S3} {83} {s5}
{84} {83} {S2} {S3} {S4,S5} {s5}
Construct a transition table for each NDFSA.
a
, ~
b
786
Chapter 11 Formal Languages and FiniteState Machines
0
a
b
0
a
a
a
a
b
b
10.
a, b Does the NDFSA in Figure 11.45 accept each string? Identify a path defined by any accepted string. 11. ab2a 13. a3b 12. abab 14. ab2ab
Is each string accepted by the NDFSA in Exercise 1? Give a path for accepted strings. 15. a2b 17. a3b 3 16. ab2a 18. (ab) 3
Does the NDFSA in Exercise 10 accept each string? Show a path that defines any accepted string. 19. abba 21. a2b 2 20. (ab) 3 22. a4b2ab 3
Construct a NDFSA that accepts the language generated by the regular grammar G  (N, T, P, or), where: 23. N = { a , A , B } , T = { a , b } , a n d P = { a o a A , bB, A ~ a} A~aA, A~bB, B~
24. N = {a,A,B}, T = {a,b}, a n d P = {a ~ aA, a ~ bA, A  ~ aB, a b, B o b}
11.7 Automataand Regular Languages
787
25. N = {a, A, B, C, D}, T = {a,b}, and P = {cT ~ ba, a ~ aA, A ~ aA, A ~ bB, B ~ aA, B ~ bC, C ~ aD, C ~ ba, D ~ aD, D ~ bD, C
a}
26. N = { a , A , B , C } , T = {a,b}, and P = {a ~ bcr, a ~ A ~ bB, B o aA, B ~ bC, C ~ aA, C ~ bcr, B ~ b} aA, A ~ aA,
Create a N D F S A t h a t accepts the regular language over {a, b} of strings that: 27. Contain exactly one a. 29. Begin with aa. 31. Contain a b a as a substring. 33. Begin with a a or bb. *35. Begin with aa, but not end in bb. *36. Begin with aa and end in bb. 28. C o n t a i n at least one a. 30. E n d with bb. 32. Contain a 3 as a substring. 34. Contain ba2b as a substring.
The preceding two sections d e m o n s t r a t e d t h a t the language accepted by a DFSA is regular and t h a t every regular language is accepted by an NDFSA. This section shows t h a t every N D F S A is equivalent to a DFSA, which answers affirmatively our question about the existence of a possible DFSA M such t h a t L ( G ) = L ( M ) . Every regular language is, in fact, accepted by a suitable DFSA. The next two examples illustrate step by step how to construct a DFSA equivalent to a given NDFSA. Consider the regular g r a m m a r G  (N, T, P, a), where N  {A, a }, T = {a,b}, a n d P {a ~ aa, c~ ~ aA, A ~ b}. The N D F S A M = ( S , A , I , f , so) t h a t accepts L ( G ) is shown in Figure 11.46 (same as Figure 11.44). By Example 11.49, L ( M )  {a'~b In > 1}. U s i n g M, we shall construct the DFSA M'  (S',A', I', f', s 0) which accepts L(G)" '
F i g u r e 11.46
a r@ b ~@
S t e p 1 C h o o s e I '  I = {a,b}, s 0  {so}  {a}, a n d S '  P ( S ) . The various states in M' are subsets of S. If t h e r e are n states in M, there can be 2 n states in M', so the states of M' are: O, {or}, {A}, {F}, {~,A}, {cy,F}, {A,F}, and { ~ , A , F }
!
788
Chapter II
Formal Languages and FiniteState Machines
S t e p 2 The accepting states of M' are those states of M' t h a t contain an accepting state of M. They are {F}, { a , F } , {A,F}, and {a,A,F}. Step 3 Let X  {s l, s 2 , . . . , sin} be a state in M'. An input symbol x leads
m
from state X to state Y, where Y 
u f ( s i , x ) . In other words, an edge
i=l
m
labeled x r u n s from state X to state Y if Y = u f ( s i , x ) .
i=l
Figure 11.46 produces all possible transitions: f ( 0 , a) = 0 f ( A , a) = 0 f ( 0 , b) = 0 f ( A , b) = {F} f ( a , a)  {a,A} f ( F , a) = 0 f ( a , b) = 0 f ( F , b) = 0
Since f(O, a)  0  f ( 0 , b), edges r u n from 0 to itself labeled a and b. Since f ( a , a)  {~, A} and f ( a , b ) = O, an edge labeled a goes from {~ } to {a, A} and an edge b from {a } to 0. Similarly, t h e r e is an edge labeled a from {A} to 0, an edge b from {A} to {F}, and two edges a and b from {F} to O. Since f(~, a) u f ( A , a) = {~, A} u O = {~, A}, an edge labeled a r u n s from {a, A} to {a, A}. Also, f ( a , b ) u f(A,b) = 0 u {F} = {F}, so an edge b goes from {a,A} to {F}. Similarly, t h e r e are edges labeled a and b from {a, F} to {~, A} and 0, respectively; edges a and b from {A, F} to 0 and {F}, respectively; and edges a and b from {cr, A,F} to {or, A} and {F}, respectively. These results appear in the transition table in Table 11.9.
Table 11.9 a 0 b
{A} {F} {cy,A} {a,F} {A,F} {a,A,F}
0 {a,A} 0 0 {a,A} {a,A} 0 {a,A}
0 0 {F} 0 {F} 0 {F} {F}
Figure 11.47 shows the resulting DFSA. Since the states {A}, {or, F}, {A, F}, and {~, A, F} cannot be reached from the initial state {or }, they can be dropped out to yield the simplified DFSA M' in Figure 11.48. F r o m this transition diagram, L ( M ' )  {aanb In > 0}  { a n b l n > 1} L(G). T h u s the a u t o m a t a M and M' are equivalent, so the NDFSA is the same as the DFSA. m Construct a DFSA M ' (S',A' , I' , f', 8 0 equivalent to the N D F S A ') M  ( S , A , I , f , so) in Example 11.50. Recall t h a t L(M)  {amb, ambna[ m, n > 1 }. The key steps lie below. (Fill in the details.)
11.7 Automataand Regular Languages F i g u r e 11.47
789
a,b
a,b
b
F i g u r e 11.48
SOLUTION: S t e p 1 Select I' = I  {a,b}, s~  {so}  {a }, and S' = P(S). The states of M' are 0, {a}, {A}, {B}, {F}, {a,A}, {a,B}, {a,F}, {A,B}, {A,F}, {B,F}, {a,A,B}, {a,A,F}, {~,B,F}, {A,B,F}, and {~,A,B,F}. S t e p 2 The accepting states of M' are {F}, {a,F}, {A,F}, {B,F}, {a,A,F}, { a , B , F } , {A,B,F}, and {a,A,B,F}.
790
Chapter 11 FormalLanguages and FiniteState Machines Step 3 The transition table of the DFSA is Table 11.10.
Table 11.10 a b
0 {A} {B} {F} {c~, A} {a,B} {c~, F} {A, B} {A, F} {B,F} {c~, A, B} {c~,A, F}
0 {a,A} {A} {F} {a,A} {a, A, F} {or, A} {A,F} {A} {F} {c~,A, F} {a,A} {a, A, F} {A, F} {a, A, F}
0 0 {B, F} {B} {B,F} {B} 0 {B, F} {B, F} {B}
{B,F}
{B,F} {B} {B, F} {B,F}
{a,B,F} {A,B,F} {a,A,B,F}
S t e p 4 The table indicates the states {a, B}, {a, F}, {A, B}, {c~, A, B}, {a, B, F }, {A, B, F}, and {a, A, B, F } are not reachable from any state, so they are not the initial state {a }. Delete the corresponding rows from the table. It is now obvious from the table that the states {A}, {A, F}, {a, A, F} also cannot be reached from {a }; delete those rows also from the table. The resulting transition Figure 11.49.
F i g u r e 11.49
diagram
of the
DFSA M'
appears
in
:,b
a,b
~
a
~
b
L(M). T h u s M
From the diagram, it follows that L(M') = {amb,ambnaim, n >_ 1} = a n d M ' are equivalent automata. As in the previous example, we have shown that the equivalency between an NDFSA and a DFSA. i The techniques illustrated in the two previous examples can be generalized to arrive at the following result. (The proof is a bit complicated, so we omit it.)
11.7 Automataand Regular Languages ~ Every NDFSA is equivalent to a DFSA. The next theorem follows from Theorems 11.3, 11.4, and 11.5. ~ A language is regular if and only if it is accepted by a DFSA.
791 m
m
As Theorem 11.6 indicates, a DFSA can define a regular g r a m m a r and vice versa. Each is a characterization of the other. We now look for an example of a simplelooking language that is regular.
not
~
Show that the language L  {anb n I n > 1} is not regular. P R O O F (by c o n t r a d i c t i o n ) : Suppose L is regular. Then, by Theorem 11.6, a DFSA M exists such that L ( M )  L . Suppose M has m states. Since the string x  a m + l b m+ 1 E L , x is accepted by the DFSA. Let P be the path corresponding to x; it ends at an accepting state F.
F i g u r e 11.50
a
a
)
@a@a
~
,~
@b@
...
The path corresponding to the substring a m+l contains m + 1 states. But, since only m states exist, by the pigeonhole principle at least two of the m + 1 states, say, si and sj, where i < j, must be the same; consequently, there must be a directed cycle at si, each edge labeled a (see Figure 11.50). Let l be the length of the cycle. The path s o  s 1 . . . . . s i  s j + 1sj+2 . . . . . F generates the string x' = a m + l  l b m + l . Since this path ends at F (an accepting state), x' is accepted by the automaton; so x' ~ L. This is a contradiction, since x' does not contain the same number of a's and b's. Thus L is not a regular language, m It follows by this example that the set of wellformed nested parentheses is also n o t a regular language. (Why?) These discussions lead us to a powerful conclusion: Regular languages are accepted by DFSAs.
792
Chapter 11 FormalLanguages and FiniteState Machines
Exercises 11.7
16. Construct a DFSA equivalent to each N D F S A in Exercises 14, 7, 8 of Section 11.6. Eliminate all unreachable states. 78. Design DFSAs equivalent to the NDFSAs in Exercises 23 and 24 of Section 11.6. Eliminate all unreachable states. Let L be the language recognized by an FSA and L R   { X n . . . Xl [ X l . . . X n E L}. Construct an NDFSA t h a t accepts L R from each FSA in Exercises 916. ( H i n t : Reverse the directions of the edges; switch the roles of the initial state and the accepting states.) 9. Figure 11.20 11. Exercise 18 in Section 11.3 13. Exercise 37 in Section 11.3 15. Exercise 40 in Section 11.3 10. Exercise 17 in Section 11.3 12. Exercise 36 in Section 11.3 14. Exercise 38 in Section 11.3 16. Exercise 41 in Section 11.3
1724. Identify the language L ( M ) accepted by the FSA in Exercises 916. 2532. Construct a DFSA equivalent to the NDFSA in Exercises 916.
The abstract models of computing machines with limited capabilities are DFSA, FSM, and NDFSA. An a u t o m a t o n checks if a given input string has a desired property and produces no output values. An FSM, on the other hand, yields an o u t p u t value corresponding to each input.
Formal Language
9 A f o r m a l l a n g u a g e over an alphabet E is a subset of E*, the set of all possible words over Z (page 734). 9 The c o n c a t e n a t i o n of two languages A and B over Z consists of words ab with a e A and b E B (page 736).
9 A ~ = { a l a 2 . . . an l ai E A}, where A ~  A
cx$
(page 739). (page 739).
9 A* = u A n is the K l e e n e c l o s u r e of a language A
n:0
Grammar
9 A g r a m m a r G = (N, T , P , ~) consists of a finite set N of n o n t e r m i h a l s y m b o l s , a finite set T of t e r m i n a l s y m b o l s , a finite set P of p r o d u c t i o n r u l e s , and a s t a r t s y m b o l c~ (page 746).
Chapter Summary
793
9 A word w' = xfly is directly derivable from a word w = x a y if a p r o d u c t i o n a ~ / ~ exists; we t h e n write w ~ w'. A word Wn is d e r i v a b l e from wi if t h e r e exists a finite sequence of d e r i v a t i o n s , Wl =, w2, w2 w 3 , . . . , W n  1 ~ Wn. T h e l a n g u a g e derivable from cr is t h e l a n g u a g e generated by G, denoted by L ( G ) (page 746). 9 In BNF, each production w ~ w' is w r i t t e n as w ::= w'; each n o n t e r m i nal symbol w is enclosed by angle brackets, as in (w); a n d p r o d u c t i o n rules with t h e s a m e lefthand sides are c o m b i n e d w i t h vertical b a r s into a single rule (page 748). 9 A t y p e 0 g r a m m a r has p h r a s e  s t r u c t u r e form c~Ac~' ~ c~flc~'. (page 751 ). (page 751)
9 In a contextsensitive (type 1) g r a m m a r , every p r o d u c t i o n is of t h e
9 In a contextfree (type 2) g r a m m a r , every p r o d u c t i o n is of t h e form A ~ ~. (page 751). 9 In a regular (type 3) g r a m m a r , every p r o d u c t i o n is of the form A~ t or A ~ tB (page 751). 9 A language L ( G ) is contextsensitive, contextfree, or regular according as w h e t h e r G is contextsensitive, contextfree, or r e g u l a r (page 751). 9 An ambiguous language contains a word t h a t has more t h a n one derivation tree (page 753). 9 The l a n g u a g e accepted by a DFSA is r e g u l a r (page 780).
FiniteState Automaton (FSA)
9 A F S A M  ( S , A , I , f , so) consists of a finite set S of s t a t e s , a finite set A of a c c e p t i n g states, a finite set I of i n p u t s y m b o l s , a t r a n s i t i o n function f : S x I ~ S, and an initial state so. Every s t a t e  i n p u t pair yields a u n i q u e n e x t  s t a t e of the a u t o m a t o n (page 761). 9 A transition table defines t h e t r a n s i t i o n function. (page 761).
9 A transition diagram can r e p r e s e n t a DFSA. The initial s t a t e so is identified by d r a w i n g an a r r o w t o w a r d it; an accepting s t a t e by two concentric circles a r o u n d it (page 762). 9 An i n p u t s t r i n g is a c c e p t e d by an a u t o m a t o n M if a n d if the s t r i n g traces a p a t h t h a t ends at an accepting state. language L(M) accepted by M consists of all words recognized (page 9 Two a u t o m a t a M and M ' are equivalent if L ( M ) = L ( M ' ) only The by it 764 ).
(page 764).
794
Chapter II FormalLanguages and FiniteState Machines FiniteState Machine (FSM)
9 An F S M M = (S,I, O , f , g , so) consists of a finite set S of states, a finite set I of i n p u t s y m b o l s , a finite set O of o u t p u t s y m b o l s , a t r a n s i t i o n f u n c t i o n f : S x I ~ S, an o u t p u t f u n c t i o n g : S x I ~ O, and an i n i t i a l s t a t e so. Every s t a t e  i n p u t pair produces a n e x t  s t a t e and an o u t p u t value (page 772). 9 A t r a n s i t i o n t a b l e can define the t r a n s i t i o n and o u t p u t functions of an FSM (page 772). 9 A transition d i a g r a m also can define an FSM (page 772).
Nondeterministic FiniteState Automaton (NDFSA)
9 An N D F S A M  ( S , A , I , f , ~) consists of a finite set S of states, a subset A of S of accepting states, a finite set I of i n p u t symbols, a t r a n s i t i o n function f : S x I ~ P(S), and an initial state a. A s t a t e  i n p u t pair m a y be paired with zero, one, or more states (page 783). 9 Every regular language is accepted by a N D F S A 9 Every N D F S A is equivalent to a DFSA 9 Every regular language is accepted by a DFSA (page 784). (page 787). (page 787).
Review Exercises
Let A = {4, a, bc} and B = {a, ab}. Find each.
1. A B
2. B A
3. A 3
4. B 3
Find three words belonging to each language over {a,b,c }. 5. {a,b}{c}* 7. {ab}{ab}* 6. {a}b*{c}* 8. {b} {a,b,c}* {b}
A g r a m m a r G = (N, T , P , a ) has N = {{noun phrase), (verb}, (adjective), (noun}, (article) }, T = {a,the,chicken,wolf, cabbage,eats,walks,reliable, discreet, gracious}, a = (sentence), and the production rules are: (sentence) ~ (noun phrase) ~ {noun p h r a s e ) ( v e r b ) ( n o u n phrase) (article){noun) I (article){adjective)(noun)
{article) ~ the la (noun) ~ chicken I w o l f l c a b b a g e (adjective) ~ reliable I discreet I gracious (verb) ~ eats l w a l k s
Chapter Summary
D e t e r m i n e if each is a valid sentence in L ( G ) . 9. The gracious chicken walks the wolf. 10. The reliable wolf eats a chicken. Make a derivation tree for each sentence. 11. The discreet wolf eats the cabbage. 12. The reliable cabbage walks the gracious chicken.
795
Using the g r a m m a r G  (N, T, P, a) where N = {a, A, B}, T = {a,b}, and P = {a ~ ba, a > aA, A ~ aB, A ~ b a , B ~ aB, B > b B , A ~ a , B ~ a, B ~ b}, d e t e r m i n e if each string belongs to L ( G ) . 13. ab3a 14. (ab) 3 15. aba2b 16. ab2a 4
C o n s t r u c t a parse tree for each string. 17. ba2b 18. b2a3b 19. aba 2 20. a2b2a 2
Develop a g r a m m a r t h a t generates each language over {a,b }. 21. {b n in >_ 1~ 22. {anba'* In > 0} 23. {b 2n+l In >_0} 24. {ab'*aln >_ 0} With the g r a m m a r below, construct parse trees for the simple w h i l e s t a t e m e n t s in Exercises 25 and 26. (while s t a t e m e n t ) ' while (expression) do (statement)
(statement) " '  ( a s s i g n m e n t s t a t e m e n t ) f (while s t a t e m e n t ) I s {assignment s t a t e m e n t ) " '  ( v a r i a b l e ) = {expression) (variable/ "' a [b {c I . . . I z (expression) "  (variable)(sign)(variable) J (variable) (operator)(variable/ (operator) " '  =J ~:l < l < l > l > {sign)' +125. While x >_y do x "= y + z. 26. While x >_y do while y < z do a " b + c. 27. Draw the t r a n s i t i o n diagram of the D F S A M  ( S , A , I , f , so), w h e r e S  { s o , s l , s 2 , s 3 , s 4 } , A = {83} , I = {a,b}, and f is defined by Table 11.11. 28. Redo Exercise 27 w i t h A
= {81,83}
and f defined by Table 11.12.
C o n s t r u c t the t r a n s i t i o n table for each DFSA.
796
Chapter 11
Formal Languages and FiniteState Machines
T a b l e 11.11
sO Sl s2 s3 s4 Sl s4 s4 s3 s4 s4 s2 S3 s3 s4
T a b l e 11.12
a
Sl Sl s4 s3 s4
b
s2 Sl s3 s3 s4
s2 s3 s4
29.
b
30.
a;b
3 1  3 4 . Identify the language L ( M ) accepted by the a u t o m a t a in Exercises 2730. Design a DFSA t h a t accepts strings over {a, b } that: 35. Begin with aaa.
37.
36. Contain abb as a substring.
g
II
s0 81 82 s3 s0 s2 80 s3
f b a
38.
b
f
g
a
80 81 S2 S3 sO
b
Sl
a
0
b
0
sl 82 83
S2
1 0 1 1
0 0 0 0
s2 s3 s3
s3 s2 s2
0 1 1
1 1 0
Chapter Summary
Construct a transition table for each FSM. 39.
~
797
a/O
b/1 ~ a/O ~
O Jl
40.
a / O~~)
a / l~')
b/1
b/O
Using the FSM in Figure 11.33 (Example 11.38), find the output from each input string. 41. a2b2aba 42. aba2ba 43. ab3ab 44. a2b3a 2
4546. Redraw the DFSAs in Exercises 27 and 29 as FSMs. Design an FSM to accept string over {a, b } that: 47. Contain ab 2 as a substring. 48. Begin with a or b 2.
4950. Compose the regular g r a m m a r defined by the DFSA in Exercises 2728. Draw the transition diagram of the NDFSA M  ( S , A , I , f , {a,b } and:
51.
S{so,sl,s2),A
so), where I 
= {Sl}
52.
S 
{80, S l , 8 2 } , A 
{81 } . b {81} {80,82}
a
11
so II {sit
{s0t
si I] {s2} s2 {s2}
{sl,s2} {s2}
I
s2
{80,82} I] {Sl}
Construct a transition table for each NDFSA. Determine if the NDFSA in Exercise 52 accepts each input string. 53.
a,b
)
b
798
Chapter II
FormalLanguages and FiniteState Machines
54.
a
_...)
('~ a, b b
a, b ~")
55. a 3
56. ab2ab 4
57. a2b 3
58. a3b 4
5 9  6 2 . D e t e r m i n e if each i n p u t s t r i n g in Exercises 5558 is accepted by t h e N D F S A in Exercise 54. Create a N D F S A t h a t accepts t h e language L ( G ) g e n e r a t e d by t h e r e g u l a r g r a m m a r G = (N, T, P, a), where: 63. N = { a , A , B } , T = { A , B } , a n d P = {a ~ ba, a ~ aA, A ~ aB, A bA, A ~ b B , B ~ aB, B ~ b B , a ~ a , A ~ b} 64. N = { a , A , B } , T = {a,b}, a n d P = {a > aA, a ~ b a , A ~ a a , A ~ aB, A ~ bA, B > aB, B ~ b a , B ~ bA, a ~ a , A ~ b} 6 5  6 6 . C o n s t r u c t a DFSA e q u i v a l e n t to each N D F S A in Exercises 51 and 52. 6 7  6 8 . W h a t languages do the DFSAs in Exercises 65 a n d 66 accept? Let A and B be a n y languages over a finite a l p h a b e t E. Prove each. *69. (A u B*)* = (A* u B)*
Supplementary Exercises
**70. (AUB)* = (A*UB*)* = ( A ' B * ) *
Let m denote the n u m b e r of a ' s and n the n u m b e r of b's in a s t r i n g over {a,b }. Design an FSA t h a t accepts strings with t h e given properties. 1. m  l ( m o d 2) and n  l ( m o d 2).
2. m  0 ( m o d 3) and n  l ( m o d 3).
3. m  0(mod 2) and n  l ( m o d 2), or m  l ( m o d 2) a n d n  0(mod 2). 4. Design an FSA t h a t accepts positive integers n divisible by 3. *5. Using t h e s y n t a x d i a g r a m in Figure 11.51 for a real n u m b e r , design an FSA to recognize valid real n u m b e r s . *6. The R o m a n n u m e r a l s M, D, C, L, X, V, and I have values 1000, 500, 100, 50, 10, 5, and 1, respectively. In the strict additive n o t a t i o n no n u m e r a l with a s m a l l e r value precedes a n u m e r a l w i t h a larger value. For instance, 19 is w r i t t e n as XVIIII in lieu of t h e s h o r t e r
Chapter Summary F i g u r e 11.51
799
representation XIX and MMLXXVI, unlike MCMXCVI, is a wellformed sequence. Excepting M's, C, X, or I should not appear more than four times in the same sequence, and D, L, or V no more t h a n once. This makes sense since CCCCC  D, XXXXX  L, and so on. Design an FSA to recognize the language of such wellformed sequences of additive Roman numerals. Develop a g r a m m a r t h a t generates each language over {a,b }. *7. The set of words t h a t begin and end with a. *8. The set of words t h a t begin with aa and end with bb. 9. Using BNF, define a g r a m m a r for the language of wellformed nested parentheses. 10. Use productions instead of BNF to define the g r a m m a r in Exercise 9. A M o o r e m a c h i n e M = (S,I, O , f , g , so), named after Edward Moore who introduced it in 1956, is an FSM consisting of a finite set S of states, a finite set I of input symbols, a finite set O of o u t p u t symbols, a transition function f : S • I ~ S, an o u t p u t function g : S ~ O, and an initial state so. Draw a transition diagram for the Moore machine defined by each transition table.
f
input
$
input 1
s o
. . . .
1
11.
SO Sl S2 S3
SO 83 S2 80
Sl S2 S3 81
12.
SO
Sl
S3
0
Sl s2
83
Sl s3
Sl
82 s2
82
1 1
0
Construct a transition table for each Moore machine.
13.
1
1
0
1
1
800
Chapter II
Formal Languages and FiniteState Machines
F o r r e s t M o o r e (1925) was born in Baltimore, Maryland. He graduated from Virginia Polytechnic Institute in 1947 and received his Ph.D. in mathematics from Brown 3 years later. After teaching at the University of Illinois for a year, he joined the technical staff at Bell Telephone Labs. In 1966, he joined the faculty of the University of Wisconsin, Madison, and taught there until his retirement in 1985. Moore has made outstanding contributions to the logical design of switching circuits, automata theory, graph theory, and database management. Edward
14. 0 ~ 0 ~'~ Y ~ 0 ff v 1
1 The o u t p u t g e n e r a t e d by the Moore m a c h i n e M  (S,I, O , f , g , so) for the i n p u t s t r i n g a l a 2 . . . a m is g ( s o ) g ( s l ) . . . g ( s m ) , w h e r e si  f ( s i  l , a i ) a n d 1 _< i _< m. F i n d the o u t p u t produced by the m a c h i n e in Exercise 13 for each input. 15. 011 16. 1010 17. 10001 18. 1101101
19. L e t L be a r e g u l a r language. Prove t h a t L R  { x n . . . x ] I x l . . . x , , ~ L} is also regular. **20. An F S M M  (S,I, O , f , g , so) is s i m p l y m i n i m a l if no o u t p u t rows in its t r a n s i t i o n table are identical. If l S i  n, I I I  m, a n d I O i  P, how m a n y simply m i n i m a l F S M s are possible?
Computer Exercises Write a p r o g r a m to do each task, w h e r e E  {a,b }. 1. D e t e r m i n e if a s t r i n g over Z" 9 Begins w i t h aa. 9 E n d s with bb. 9 C o n t a i n s aba as a substring. 9 Has its n u m b e r of a ' s c o n g r u e n t to 1 mod 3. 9 C o n t a i n s exactly one a. 9 C o n t a i n s at least one a.
Chapter Summary
9 Has an even n u m b e r of a ' s a n d b's. 9 Has b o t h its n u m b e r of a's a n d b's c o n g r u e n t to 1 mod 3.
801
2. Let m d e n o t e the n u m b e r of a ' s a n d n t h e n u m b e r of b's in a s t r i n g over E. Read in a word over E and see if it has: 9 m  0(mod 5) 9 m _ 0(mod 3) or m  l ( m o d 3) 9 m0(mod3) andnl(mod3) 9 m _ 3(mod 5) 9 m  0(mod 3) or m  2(mod 3) 9 m0(mod3) andn=2(mod3)
3. For a DFSA with n (< 10) states, labeled 1 t h r o u g h n, r e a d in its n u m b e r of states n a n d t r a n s i t i o n table. Read in a sequence of i n p u t strings over Z and d e t e r m i n e if each is accepted by t h e DFSA. 4. I m p l e m e n t the u n i t delay m a c h i n e in E x a m p l e 11.41. 5. Read in two b i n a r y n u m b e r s and use the F S M in E x a m p l e 11.42 to c o m p u t e t h e i r sum. 6. By m e a n s of the s y n t a x d i a g r a m in F i g u r e 11.31, d e t e r m i n e if a s t r i n g of c h a r a c t e r s r e p r e s e n t s a valid integer. 7. Ascertain with the s y n t a x d i a g r a m in F i g u r e 11.32 w h e t h e r a s t r i n g of c h a r a c t e r s r e p r e s e n t s a valid real n u m b e r . (Excluding t h e exponential form.) 8. Read in the n u m b e r of s t a t e s n, the t r a n s i t i o n table, a n d a set of i n p u t strings for an FSM with n (< 10) states, labeled 1 t h r o u g h n. P r i n t the o u t p u t produced by each i n p u t string. 9. U s i n g a DFSA with n (< 10) states, labeled 1 t h r o u g h n, read its n u m b e r of s t a t e s n and t r a n s i t i o n table. D e t e r m i n e the c o r r e s p o n d i n g r e g u l a r grammar.
Exploratory Writing Projects
Using library a n d I n t e r n e t resources, write a t e a m r e p o r t on each of t h e following in y o u r own words. Provide a w e l l  d o c u m e n t e d bibliography. 1. Discuss how B N F rules are used to define p r o g r a m m i n g l a n g u a g e s such as C § § and Java. 2. Discuss T u r i n g m a c h i n e s and C h u r c h ' s thesis. 3. Explain how v e n d i n g m a c h i n e s , slot m a c h i n e s , a n d garage door openers can be modeled by FSAs. 4. Write an essay on Kleene closure. 5. Write an essay on different types of F S M s a n d t h e i r applications.
802
Chapter II
FormalLanguages and FiniteState Machines
Enrichment Readings
1. W. J. Barnier, "FiniteState Machines as Recognizers," The UMAP Module 671 (1986), pp. 209232. 2. B. Hayes, "On the FiniteState Machine, a Minimal Model of Mousetraps, Ribosomes, and the Human Soul," Scientific American, Vol. 249 (Dec. 1983), pp. 1928, 178. 3. J. E. Hopcroft and J. D. Ullman, Introduction to Automata Theory, Languages, and Computation, AddisonWesley, Reading, MA, 1979. 4. Z. Kohavi, Switching and Finite Automata Theory, 2nd ed., McGrawHill, New York, 1978. 5. P. Linz, An Introduction to Formal Languages and Automata, D. C. Heath, Lexington, MA 1990. 6. J. C. Martin, Introduction to Languages and the Theory of Computation, 2nd ed., McGrawHill, New York, 1997. 7. M. Sipser, Introduction to the Theory of Computation, PWS, Boston, 1997. 8. W.A. Wulfet al., Fundamental Structures of Computer Science, AddisonWesley, Reading, MA, 1981, pp. 164.
Chapter 12
B o o l e a n A l g e b r a and Combinatorial Circuits
Mathematics is music for the mind; music is mathematics for the soul.
ANONYMOUS
eorge Boole's classic An Investigation of the Laws of Thought, published in 1854, led to the development of two closely related areas of mathematics: symbolic logic and the mathematical system called boolean algebra. Chapter i of Boole's book demonstrated how valuable the symbols and the laws of logic are in investigating how we reason in order to reach conclusions. That material will help us in studying this chapter. Until the late 1930s boolean algebra did not seem to have many useful applications. In 1938, Claude E. Shannon, while working at the Massachusetts Institute of Technology, used boolean algebra to analyze electrical circuits, thus opening the door for a world of applications of boolean algebra. Since then, boolean algebra has played a central role in the design, analysis, and simplification of electronic devices, including digital computers. In this chapter we will address some of the interesting problems that boolean algebra handles well: 9 Three switches for a light fixture are in a hallway. If the light is on, it can be turned off by flipping one of the switches off. On the other hand, if it is off, flipping one of the tongues turns it on. What does the circuit look like? 9 Design a circuit to compute the sum of two 3bit numbers. 9 How can boolean algebra simplify a circuit while maintaining the circuit's capabilities? 9 Electronic devices display digits by lighting up a maximum of seven line segments in the adjacent configuration. What kind of circuit will accept the binarycoded decimal expansion of a decimal digit and light up a segment?
803
G
804
Chapter 12 Boolean Algebra and Combinatorial Circuits
C l a u d e E l w o o d S h a n n o n (19162001), a pioneer in artificial intelligence and information theory, was born in Gaylord, Michigan. After graduating from the University of Michigan in 1936, he earned his M.S. in 1937 and Ph.D. in applied mathematics in 1940 from MIT. Shannon was a research mathematician at the Bell Telephone Labs from 1941 to 1957 and became professor of electrical engineering at M I T in 1958. In 1948, he coined the word b i t for binary digit. He received honorary doctorates from many universities: Michigan, Pittsburgh, Princeton, Northwestern, Edinburgh, Oxford, CarnegieMellon, Tufts, and Pennsylvania. He also received numerous honors and awards, including the Nobel Peace Prize, Morris Liebmann Memorial Award, Ballantine Medal, Medal of Honor from the Institute of Electrical and Electronics Engineers (IEEE), and National Medal of Science. Shannon made outstanding contributions to the theory of switching circuits, information theory, cryptography, and artificial intelligence.
.~
~ ~~  ; ~ ~ ~ _ ~ ~ ~, ' , , ~  ~
A boolean algebra is a m a t h e m a t i c a l s y s t e m ; it consists of a n o n e m p t y set S with one or more operations defined on S, and a set of axioms t h a t the elements of S satisfy. A m a t h e m a t i c a l system can be t h o u g h t of as a skeleton, like a h u m a n skeleton. W h e t h e r people are black or white, Caucasian or Chinese, their skeletons have common characteristics. Likewise, concrete examples of a m a t h e m a t i c a l system share common properties. When we study m a t h e m a t ical systems, we need to study properties common to all examples of such systems. Real n u m b e r s form an i m p o r t a n t n u m b e r system. Before formally defining a boolean algebra, we examine two concrete examples. Look for the properties common to both because they will help us define a boolean algebra. Let U b e an a r b i t r a r y set and P(U) its power set. Let A , B , and Cbe any t h r e e elements of P(U). You may recall from Chapter 2 t h a t the union, intersection, and complementation operations on P ( U ) satisfy the following:
C o m m u t a t i v e properties 9 AUB=BUA Associative properties 9 AU(BUC)=(AUB)UC Distributive properties 9 AN(BuC) = (ANB)U(ANC) 9 Au(BNC) = (AuB)N(AuC) 9 AN(BNC)=(ANB)NC 9 ANB=BNA
12.1 Boolean Algebra
805
I d e n t i t y properties
9 AUO=A
9 AN U =A
C o m p l e m e n t properties 9 AUA'U
T h e n e x t e x a m p l e is a b i t s o p h i s t i c a t e d . ~ L e t D30 b e t h e s e t o f p o s i t i v e f a c t o r s o f 3 0 : D 3 0 D e f i n e t h r e e o p e r a t i o n s @, @, a n d '  {1, 2, 3, 5, 6, 10, 15, 30}.
9 ANA'O
I
o n D30 a s f o l l o w s :
a @ b  l c m {a, b} a  b  g c d {a,b} t at u 3O a
For instance, 2  3  lcm{2,3}  6  3  2; 6 @ 10  l c m { 6 , 10}  30 10 @ 6;
2 Q 3  gcd{2,3} 6'= 3O ~ =5; 6
1  3 E) 2;
6 G 10  g c d { 6 , 10}  2 = 10 E) 6; 5' 3O 5 = 6;
6 @ 6'  6 
5  l c m {6, 5}  30;
5 Q 5'  5 E) 6  g c d { 5 , 6 }  1.
These operations satisfy the following properties:
C o m m u t a t i v e properties 9 a@b=b•a Associative properties 9 aO(b@c)=(aGb)oc Distributive properties 9 a@(boc)=(aQb)@(a@c) I d e n t i t y properties
9 a@ 1 =a
9 aQbbQa
9 aQ(bQc)=(aQb)@c
9 a@(bQc)=(a@b)G(a@c)
9 aQ30a
tRecall that lcm {a,b} denotes the least common multiple of a and b, and gcd {a,b} the greatest common divisor of a and b.
806
Chapter 12 Boolean Algebra and Combinatorial Circuits
Complement properties
9 a@a ~=30 9 a(3a'=l I
Take a close look at both examples. W h a t do they have in common? In each case, a n o n e m p t y set B contains two special elements ~ ~ and U in Example 12.1; 1 and 30 in Example 12.2. F u r t h e r m o r e , both examples contain two binary operators and a u n a r y operator, satisfying the same 10 properties. These commonalities suggest a m a t h e m a t i c a l system called a boolean algebra, defined below.
Boolean Algebra
A b o o l e a n a l g e b r a consists of a n o n e m p t y set B containing two distinct elements 0 and 1, two binary operators + a n d . , and a u n a r y o p e r a t o r ' satisfying the following conditions for all x,y, and z in B: Commutative laws
9 x+y=y+x 9 x.y=y.x
Associative laws 9 x+(y+z)=(x+y)+z Distributive laws
9 x.(y+z)=(x.y)+(x.z) 9 x+(y.z)=(x+y).(x+z)
9 x.(y.z)=(x.y).z
Identity laws
9 x+0=x 9 x.l=x
Complement laws
9 x+x'=l 9 x.x'=O
In symbols, the boolean algebra is denoted by (B, +,., ', 0, 1) . We clarify a few points before we go any further: 9 The operations +,., a n d ' are called s u m , p r o d u c t , and c o m p l e m e n t a t i o n , respectively. (For instance, in Example 12.1, the binary operators are n and U; the u n a r y operator is complementation.) These operators are generic symbols: the operator + does not stand for addition nor the o p e r a t o r , for multiplication. 9 Since + a n d . are binary operators, both x + y and x .y belong to B for all x and y in B; since' is a u n a r y operator, x' belongs to B for every x ~ B. 9 The elements 0 and 1 are the z e r o e l e m e n t and the u n i t e l e m e n t , respectively. Again, they are generic symbols for the zero element and the unit element, respectively; they need not be the familiar n u m b e r s zero and one. In Example 12.1, the zero element is 0 and the unit element is U; in Example 12.2, they are 1 and 30, respectively.
12.1 BooleanAlgebra
807
9 When it is clear from the context, we designate the boolean algebra (B, +,, ', 0, 1) as the boolean algebra B for convenience. 9 The o p e r a t o r , in x . y is usually omitted for convenience; t h u s x y = x . y . No p a r e n t h e s e s appear when t h e r e is no danger of confusion. F o r instance, (xy) + (xz) = x y + x z and x + y + z = x + (y + z) = (x + y) + z. 9 Precedence rules govern evaluating expressions in boolean algebra: First, parenthesized subexpressions are evaluated. Complementation has the highest priority a m o n g the operators, followed b y . and then +. For example, x y + z x ' = (xy) + [z(x')]. 9 The 10 axioms are paired off in two columns. In each pair, an axiom can be obtained from the other by swapping + w i t h . , and 0 with 1. These are d u a l a x i o m s . For instance, the dual of x + x' = 1 (axiom 9) is x . x' = 0 (axiom 10). (In a boolean algebra, the dual of every t r u e s t a t e m e n t is also true. This property is the p r i n c i p l e o f d u a l i t y . ) According to the definition, a boolean algebra contains at least two elements, the zero and the unit. Consequently we can ask: Does any boolean algebra contain exactly two elements? The next example shows t h a t a n s w e r is yes. Let B  {0, 1 }. Define the operations +, 0+0=0 0.0=0 0'= 1 0+1=1 0.1=0 1'=0
" 3
and'
on
B
as
follows: 1+1=1 1.1=1
1+0=1 1.0=0
[Think of the operators + and 9 as the o r (v) and the a n d (A) operators, respectively.J By T h e o r e m 7.1, the commutative, associative, and distributive laws are satisfied. Clearly, x + 0 = x and x 9 1 = x, for every x e B; therefore, 0 is the zero element and 1 is the unit element. Besides, 0+0'=0+1=1 0.0'=0.1=0 and and
1+ 1'= 1+0=
1.1'= 1.0=0
1
Consequently, the complement laws are also satisfied. T h u s (B, +,., ', 0, 1) is a boolean algebra. Some f u n d a m e n t a l facts about boolean algebras come from the above axioms. We will find t h e m useful in later discussions. ( U n i q u e I d e n t i t i e s ) The zero element and the unit element of algebra B are unique.
a
boolean
808
Chapte," 12 Boolean Algebra and Combinatorial Circuits
PROOF: T h e zero e l e m e n t in B will be p r o v e d u n i q u e ; t h e o t h e r h a l f will be left as a r o u t i n e exercise. S u p p o s e t h e r e a r e two zero e l e m e n t s , 01 a n d 02, in B. Since 02 is a zero e l e m e n t , 01 + 02 : 01. Likewise, since 01 is a zero e l e m e n t , 02 § 01  02. B u t 02 + 01 = 01 + 02, by t h e c o m m u t a t i v e law. T h e r e f o r e , 01 = 01 + 02 02 § 01  02, SO t h e zero e l e m e n t is u n i q u e . I
•
(Unique Complement) a l g e b r a is u n i q u e .
T h e c o m p l e m e n t of e v e r y e l e m e n t in a b o o l e a n
PROOF: Let x be an a r b i t r a r y e l e m e n t in a b o o l e a n algebra. T h e n by t h e c o m p l e m e n t laws, x + x' = 1 a n d x x ' = 0. If x h a s a second c o m p l e m e n t y, x + y = 1 a n d
xy =O. T o s h o w t h a t y = x' (the r e a s o n for each step is given on t h e RHS):
y  yl
= y(x + x') = yx + yx' = xy + x'y
i d e n t i t y law c o m p l e m e n t law d i s t r i b u t i v e law c o m m u t a t i v e law c o m p l e m e n t law c o m p l e m e n t law c o m m u t a t i v e law d i s t r i b u t i v e law c o m p l e m e n t law i d e n t i t y law of every element in a boolean algebra is I
= 0 + x'y
 xx' + x'y  x ' x + x ' y
= x' (x + y)  x'l = x' Thus the unique. complement
T h e n e x t e x a m p l e uses T h e o r e m 12.2. ~ W i t h t h e b o o l e a n a l g e b r a D30 in E x a m p l e 12.2, verify each. (1) 6 @ 6 = 6 (4) 3 @ (3 Q 5)  3 SOLUTION: (1) 6 @ 6  lcm{6,6}  6 3O (3) 5 ' =   = 6 5 So (5')'  6' 3O 6 = 5 (2) 5  30  lcm {5, 30 }  30 (4) 3 G 5  gcd{3,5}  1 So3@(3Q5)30 1 (2) 5 @ 30 = 30 (5) (3 @ 5)' = 3' Q 5' (3) (5')' = 5 (6) (5 Q 6)'  5' @ 6'
= lcm {3, 1}  1
12.1
Boolean Algebra
809
(5) 3 @ 5  lcm {3, 5}  15 So (3 @ 5)'  15' = =2 3' ( 9 5 ' 10(96 3O 15
(6) 5 (9 6 = gcd {5, 6} = 1 So (5 (9 6)'  1' = =30 =5@6 = lcm {5, 6} = 5' @ 6'
I
3O 1
= gcd {10, 6}  2 = (3 @ 5)'
We now establish a few m o r e p r o p e r t i e s of boolean algebras. [ ~ ~ ~ ~ ~ ~ Let x a n d y be a r b i t r a r y e l e m e n t s in a boolean algebra (B, + , . , Then:
Idempotent
9 XffX
', 0, 1).
laws
X 9 XXX
Boundedness
9 x+11
laws
9 x0=0
Involution laws
9 (X')l __ X
9 0'=1
9 1'0
Absorption laws
9 x +xyx 9 x(x + y)  x
De Morgan's
laws
9 ( x y ) '  x' + y '
9 (x + y ) ' = x ' y '
PROOF:
9 T o p r o v e t h a t x + x  x:
x +x
(x + x ) l
= (x + x ) ( x + x')
 xx + x x t Jr x x q XX t
identity law c o m p l e m e n t law d i s t r i b u t i v e law i d e m p o t e n t law i d e n t i t y law identity law
1"
 X ~ XX I
=x+0
~X
9 To prove that x + 1 x + 1 x
+ (x + x ' )
c o m p l e m e n t law
810
Chapter 12 Boolean Algebra and Combinatorial Circuits
= (x + x) + x' = x + x' = 1
9 T o p r o v e t h a t (x')' = x:
associative law i d e m p o t e n t law c o m p l e m e n t law
Since x' is the c o m p l e m e n t of x, x + x' = 1 and x x ' = 0. U s i n g the c o m m u t a t i v e laws, these e q u a t i o n s can be r e w r i t t e n as x'+x=l and
x'x=O
(12.1)
Since x' is also an e l e m e n t of B, it has a c o m p l e m e n t (x')'. Therefore, x' + (x')' = 1 and
x'(x')' = 0
(12.2)
E q u a t i o n s (12.1) imply x is a c o m p l e m e n t of x'; by E q u a t i o n s (12.2), (x')' is also a c o m p l e m e n t of x'. But, by T h e o r e m 12.2, the c o m p l e m e n t of x' is unique; therefore, (x')'  x.
9 T o p r o v e t h a t x + x y  x" x+xy=xl +xy
=x(1 +y) = x ( y + 1) =xl ~x
9 T o p r o v e t h a t (x + y)'  x ' y "
identity law distributive law c o m m u t a t i v e law b o u n d e d n e s s law
By the c o m p l e m e n t laws, we m u s t show t h a t (x + y ) + x ' y ' (x + y ) ( x ' y ' )  O.
1 and
(1) (x + y ) + x'y'  x + (y + x ' y ' )
= x + (yl + x'y') = x + [y(x + x') + x 'y 'i
= x + [(yx + yx') + x'y'!
= x + [xy + (x'y + x ' y ' ) l = (x + x y ) + (x'y + x 'y ')
= x(1 + y ) + x'(y + y ' ) = x l + x'(y + y') = xl + x'l
x+x f
=1 (2) (x + y ) ( x ' y ' )  ( x ' y ' ) ( x + y )
= ( x 'y ') x + (x'y')y = x ( x ' y ' ) + ( x 'y ') y
associative law identity law c o m p l e m e n t law distributive law c o m m u t a t i v e law associative law distributive law b o u n d e d n e s s law c o m p l e m e n t law identity law c o m p l e m e n t law c o m m u t a t i v e law distributive law c o m m u t a t i v e law associative law
= (xx')y' + x'(y'y)
12.1 BooleanAlgebra
= (xx')y' + x'(yy')
811 commutative law complement law commutative law boundedness law identity law
= Oy' + x'O = y'O + x'O
=0+0 =0
x'y'.
Thus, by parts i and 2, x'y' is the complement of x + y; that is, (x + y)' = I
A close relationship exists between the boolean algebras D30 and P(U), where U  {a,b,c}. They are displayed by the Hasse diagrams in Figures 12.1 and 12.2, respectively, which have the same structure. This is not surprising because the algebras are i s o m o r p h i c . [Find an isomorphism f : D 3 0 ~ P ( U ) that preserves the operations; that is, f ( x G y) = f ( x ) U f ( y ) , f ( x Q y) = f ( x ) n f(y), and f( x ' ) = (f(x))'.] F i g u r e 12.1 30
6
15
2
5
1 F i g u r e 12.2
U
{a,b}
{b,c}
{a}
{c}
0 Any boolean algebra, then, has at least two elements, as well as two binary operators and a unary one that adhere to various laws.
812
Chapter 12 BooleanAlgebra and Combinatorial Circuits
Exercises 12.1
Using Example 12.2, evaluate each. 1. 6 @ 1 0 5. 3  1 7 4
9. (5 Q 10)'
2. 6 Q 1 0 6. ( 3 Q 5 ) Q 6
10. 5'@ 10'
3. 2 @ ( 3 @ 5 ) 7. (3@6)' 11. 2 Q 3 @ 5
4. ( 2 @ 3 ) @ 5 8. 3 '  12. 2 @ 3 G 5
The set D70  {1, 2, 5, 7, 10, 14, 35, 70} of positive factors of 70 is a boolean algebra under the operations @, Q, and ' defined by x @ y = lcm {x, y}, x  y = gcd {x, y}, and x' = 70/x. Compute each.
13. 7 @ 5 17. ( 7  14. 2 G 7 18. 7'@2' 15. (5@7)' 19. 1 0 @ 1 0 16. 5 ' Q 7 ' 20. 7 Q 7
Using the boolean algebra D70, verify each.
21. ( 5 ' ) ' = 5 22. 7 @ ( 7 Q 5 ) = 7
23. 5 Q ( 5 @ 7 ) = 5
24. ( 5 @ 7 ) ' = 5 ' G 7 '
25. With the boolean algebras in Examples 12.1, 12.2, and 12.3, and D70, predict the n u m b e r of elements in a finite boolean algebra. 26. Define the operations + , . , and ' on B = {0, 1} as follows: x § y = m a x { x , y } , x . y = min{x,y},0'  1, and 1' = 0. Is ( B , § ',0,1) a boolean algebra? Determine if (S, +,, ', 0, 1) is a boolean algebra for each subset S of the boolean algebra D30. 27. {1,2,15,30}
U = {a,b,c}?
28. {1,6,10,30}
29. {1,3,5,30}
30. {1,5,6,30}
Is (S, u, N,', O, U) a boolean algebra for each subset S of P ( U ) , where
31. {0, {a}, {b,c}, U}
32. {0, {a}, {b}, {a,b}, U}
34. {0, {c}, {a,b}, U}
33. {0, {b}, {a,c}, U}
35. Define the operations + , . , a n d ' on B = {a, b, 0, 1} so t h a t {B, +,., ', 0, 1) would be a boolean algebra.
Find the dual of each boolean property.
36. 0 '   1 37. x ( x + y )  x
38. (x + y ) z = xz + y z Prove algebraically.
39. (xy)'  x' + y'
40. The unit element in a boolean algebra is unique.
12.2 Boolean functions 41. x x = x 44. 1' = 0 47. x + y = (x'y')' 42. x 0 = 0 45. x(x + y) = x 48. x y = (x' + y')' 43. 0 '  1 46. (xy)' = x' + y'
813
49. (x + y ) z = x z + yz
50. xy' + x'y = (x + y)(xy)'
51. (xy) + z = (x + z)(y + z)
*52. Let U = {a,b,c}. Define a suitable function f : D 3 0 ~ P ( U ) t h a t preserves the operations; t h a t is, f (x @ y) = f (x) u f (y), f (x Q y) = f (x) n f (y), and f (x') = if(x))'.
This section introduces the concept of a boolean function and illustrates the possible equality of boolean expressions. We also see t h a t t h r o u g h the laws of boolean algebra, boolean expressions can be r e w r i t t e n in a s t a n d a r d form. Since the design and analysis of electronic devices rely on the twoelement boolean algebra
V
8"
x y
9,
x
z
~
10. x
:i
i~~] ~
~____
12.4
C o m b i n a t o r i a l Circuits
839
11. x
rt ~~
j v
v v
12.
x
I
:[
I,
I
13.
x
:D
))
,
I
"[
r
:E)
1417. Devise a logic table for each circuit in Exercises 69. 1821. Construct a combinatorial circuit for each boolean expression in Exercises 9, 10, 14, and 15 in Section 12.2. Using only NAND gates, design a combinatorial circuit that receives x and y as input signals and outputs: 22. x' 23. x + y 24. xy
2527. Redo Exercises 2224, using only NOR gates.
840
(~hapter12 BooleanAlgebra and Combinatorial Circuits 28. Make a combinatorial circuit for a hallway light fixture controlled by two switches x and y. Assume the light is off when both switches are. By means of the circuit in Figure 12.33, add each pair of binary numbers. 29. 101, 100 30. 110, 011 31. 110, 101 32. 101, 111
Using the fulladder in Figure 12.32, verify each algebraically. 33. si  xiYiCi + xiylci + xiYiC i + iYiCi
3 4 . Ci+l  xiYiCi + xiYiC I + xiylci + xlyici.
I f I X 1 I
Design a halfadder with: 35. NAND gates. 36. NOR gates.
With the set of gates {AND, OR, NOT} functionally complete, every combinatorial circuit can be represented by a boolean expression. So simplifying such expressions amounts to simplifying, or minimizing, circuits. Boolean algebraic laws can reduce boolean expressions, as two examples with three variables will demonstrate. However, Karnaugh maps give us an easier method. The DNFs of expressions of up to fbur variables will shrink through this graphic procedure. Section 12.2 explored boolean expressions that yield the same value no matter how we combine the values of variables. Consequently, the corresponding combinatorial circuits yield the same output for the same set of input values. Accordingly, we make the following definition. Equivalent Combinatorial Circuits Two combinatorial circuits are e q u i v a l e n t if the corresponding boolean expressions are equal.
~
F i g u r e 12.34
Are the combinatorial circuits C1 and C2 in Figure 12.34 equivalent?
x
:[}
y .________~ Combinatorial circuit C2

Combinatorial circuit C1
SOLUTION: First, find the boolean expressions representing the two circuits: E1 x t Y ' and E2 = (x + y)', respectively. Since E1 = E2 by De Morgan's law, the two combinatorial circuits are equivalent, m
12.5 Minimization of Combinatorial Circuits
841
Circuit C1 contains three gates; C2, only two. Accordingly, C2 is s i m p l e r than C1. Through two methods, algebraic and graphical, we can find a circuit C2 equivalent to but simpler than a given circuit C1. Since a gate in a combinatorial circuit corresponds to a boolean operator, the key to both procedures lies in locating a boolean expression with fewer boolean operators and literals. We can simplify a boolean expression through the laws of boolean algebra, as the next two examples demonstrate. However, we can use this method successfully only if we know the laws of boolean algebra well, so review them as often as needed. ~
F i g u r e 12.35
Find a simpler combinatorial circuit equivalent to the one in Figure 12.35.
SOLUTION: The boolean expression represented by the circuit is f ( x , y , z )  xyz t xyz' + xy'z + x'y'z. Simplify it as much as possible, justifying every step:
f (x, y, z)  xyz + xyz' + xy'z + x'y'z = x y ( z + z') + y ' z ( x + x') = xyl + y'zl = xy + y'z
Consequently, we can replace the given circuit by the much simpler version in Figure 12.36.
F i g u r e 12.36
I
842
Chapter 12 Boolean Algebra and Combinatorial Circuits
~
Find a simpler combinatorial circuit equivalent to the one represented by the boolean function f ( x , y , z ) = xyz + xyz' + xy'z + x'yz + xy'z' (see Example 12.13 also). SOLUTION:
f ( x , y , z ) = xyz + xyz' + xy'z + x'yz + xy'z' = xyz + xyz' + xy'z + x'yz + (xy'z' + xy'z') = xyz + (xy'z + xy'z') + (xyz' + xy'z') + x'yz = xyz + xy'(z + z') + xz'(y + y') + x'yz = xyz + x y ' l + x z ' l + x'yz = xyz + xy' + xz' + x'yz = xyz + x(y' + z') + x'yz + xyz = xyz + x(yz)' + yz(x + x') = x[yz + (yz)']+ y z l
=xl +yz
=x+yz
Therefore, the simpler twogate circuit in Figure 12.37 can replace the original ninegate one, saving seven gates.
F i g u r e 12.37
z
m
Simplifying a boolean expression can be arduous and frustrating. To a large extent, success depends on grouping the various terms properly and then ingeniously applying the laws of boolean algebra. But there is another way. Maurice Karnaugh developed the graphical method of the K a r n a u g h m a p at Bell Laboratories in 1953 to simplify the DNFs of boolean expressions. The essence of the Karnaugh map lies in grouping minterms that differ by exactly one literal. For instance, the minterms xy'z and xy'z' differ in exactly one literal and their sum can shrink: xy'z + xy'z' = xy'(z + z') = x y ' l = xy' (see also Example 12.29). The Karnaugh map has a rectangular grid of squares. Each square stands for a possible minterm in the DNF of the boolean expression that represents
i2.5 Minimization of Combinatorial Circuits
843
M a u r i c e K a r n a u g h (1924), a physicist, was born in New York City. After graduating from the City College of New York in 1948, he received his Ph.D. in physics from Yale in 1952. Karnaugh was a member of the research staff at Bell Telephone Labs from 1952 to 1966, and a research and development manager at the Federal Systems Division of A T & T during the next 4 years. In 1970, he joined the research staff at IBM. His research interests include knowledgebased computer systems.
the circuit. Each contains a 1 if the corresponding m i n t e r m exists in the expression. A 2 • 2 K a r n a u g h map can help simplify the D N F of a boolean expression in two variables x and y. Since four possible m i n t e r m s are feasible m x y , xy', x'y', and x ' y m the map consists of four squares. Label the squares so the m i n t e r m s in any two adjacent squares in each row and column differ by exactly one literal; two such squares are a d j a c e n t . For instance, the squares xy and xy' are adjacent, whereas the squares xy and x'y' are not. The resulting a r r a n g e m e n t appears in Table 12.16.
T a b l e 12.16
X X p
y
xy
y'
xy'
x'y
x'y'
Find the K a r n a u g h map for each boolean expression. (1) xy + x'y + x'y' (2) xy' + x'y
SOLUTION: Place a 1 in the square c o r r e s p o , d i n g to each m i n t e r m and leave the other squares blank, as shown in Tables 12.17 and 12.18, respectively. Table 12.17
y'
X X p
K a r n a u g h map for xy + x'y + x'y'.
844 Table 12.18
Chapter 12 Boolean Algebra and Combinatorial Circuits
y x x' 1 1
y'
Karnaugh map for xy' + xty.
1
m
The sum of minterms in adjacent squares can be simplified. By drawing a loop around such blocks containing l's, always beginning with the largest block, we can always reduce the minterms corresponding to the blocks, as Example 12.31 suggests. ~ Using a Karnaugh map, simplify the boolean expressions in Example 12.30, if possible. SOLUTION: (1) First, loop off adjacent squares containing l's. There are two such blocks in Table 12.19. Therefore,
xy + x'y + x'y' = (xy + x'y ) + (x'y + x'y')
= (x + x')y + x'(y + y')
= ly + x'l
=y+x' =x+y
(With a little practice, we can read this answer directly from the map.)
Table 12.19
X X p
y
y'
(2) The squares corresponding to the minterms xy' and x'y are not adjacent, so each forms a block by itself in Table 12.20. Consequently, the expression xy' + x'y cannot be simplified.
Table 12.20
TP
x p
m
A Karnaugh map for three variables x, y, and z expands to a 2 • 4 rectangular grid. Each of the eight squares in Table 12.21 corresponds to a possible minterm in x, y, and z. Again, two squares are a d j a c e n t if the
12.5 Minimization of Combinatorial Circuits
845
corresponding m i n t e r m s differ by exactly one literal. For instance, the s q u a r e s x y ' z a n d x y z are adjacent. To see this geometrically, cut out the grid, bend, and glue the two shorter edges to form the cylinder in Figure 12.38. Two adjacent squares share a boundary.
T a b l e 12.21
X Xp
yz xyz
yz' xyz t
y' z' xy tz ~
y' z xff z
x'yz
x'yz'
x'y'z'
r
I
x'y'z
F i g u r e 12.38
xyz_.~_..... ~
xy'z
,, x'y'z
To simplify a sum of m i n t e r m s in x, y, and z, identify blocks of m i n t e r m s that can be combined by drawing loops around them. Always begin with the largest block and use the m i n i m u m n u m b e r of blocks. This time a block may be a I x 1, 1 x 2, 1 x 4, 2 x 2, or a 2 x 4 rectangle. An example of each possibility appears in Figure 12.39.
F i g u r e 12.39
yz
X
yz"
y' z'
y'z
X
yz
yz'
y' z'
y'z
A few possible blocks in a 2 • 4 K a r n a u g h map.
X'
X'
(a)
yz yz" y'z ' y "z
(b)
X'
(c)
yz yz' y'z" y'z
X
Xp
yz
yz'
y'z"
y'z
Xp
(d)
(e)
846
C, hapter 12 Boolean Algebra and Combinatorial Circuits
Notice that the sum of the minterms in the block in Figure 12.39d can be simplified as follows:
xyz' + xy'z' + x'yz' + x'y'z'  yz'(x + x') + y'z'(x + x') = y z ' l + y'z' 1 = yz' + y'z' = (y+y')z' = lz'
Z/
The next example provides two additional exercises in simplifying boolean expressions. ~ Simplify each boolean expression with a Karnaugh map. (1) (2)
SOLUTION:
E 1 E2 
xyz + xyz' +xy'z + x~y'z xyz +xyz' + xy'z + x~yz + xy'z'
Figures 12.40 and 12.41 show the Karnaugh maps for the two expressions.
F i g u r e 12.40
x
p
yz
yz"
y'Z'
y'Z
X
0
yz yz' ~1'~~ y'z' y'z
F i g u r e 12.41
~1
(1) From Figure 12.40,
E1  (xyz + xyz') + (xy'z + x~y'z) = xy(z + z') + y'z(x + x') = xyl + y'zl = xy + y'z
(also see Example 12.28)
12.5 Minimization of Combinatorial Circuits
847
(2) Since a 1 occurs in both loops, count the corresponding m i n t e r m xyz twice using the idempotent law:
E2  (xyz + xyz' + xy'z' + xy'z) + (xyz + x'yz) = x[y(z + z') + y'(z' + z)] + (x + x')yz
= x(yl + y ' l ) + lyz
 x ( y + yt) + yz = x + yz
(also see Example 12.29)
Again, with a little experience and patience, we can read these expressions directly from the maps. I Finally, a K a r n a u g h map for four variables w, x, y, and z encompasses the 4 x 4 grid of Table 12.22. Each of the 16 squares r e p r e s e n t s a possible m i n t e r m in w, x, y, and z. Geometrically, the grid can be cut, bent, and glued to form the doughnutshaped surface called a t o r u s . Two squares are adjacent if they share a border on the torus. The cells wxy'z and wxyz are adjacent; wxyz and wx'yz' are not.
Table 12.22
yz
yz'
yP z p
yP z
K a r n a u g h map for xy + x'y + x'y'.
wx wx'
.
wxyz wx 'yz
WlxIyz
wxyz'
wxy'z'
wxy'z
wx'yz' w'x'yz'
wl xyz t
wx'yz' w'x'y'z'
wf xyf z p
wx'y' z w'x'y'z
wl xyl z
WPX p
w'x
w'xyz
As was done with two and threevariable expressions, place a 1 in the square corresponding to each m i n t e r m in the boolean expression and loop off the l ' s into blocks of m i n t e r m s t h a t can be combined, always looking for the largest block first and using as few blocks as possible. The blocks useful for minimization are 1 x 1, 1 x 2, I x 4, 2 x 2, 2 x 4, and 4 x 4. F o u r such blocks are highlighted in Figure 12.42. For example, the sum of the m i n t e r m s in Figure 12.42a can be simplified:
wxyz + wxy'z + w'xyz + w'xy'z = wxz(y + y') + w'xz(y + y')
 WXZ + W lXZ
= (w + w')xz
 XZ
We conclude this section with an example t h a t illustrates how to simplify fourvariable boolean expressions using a K a r n a u g h map.
848
Chapter 12 Boolean Algebra and Combinatorial Circuits
F i g u r e 12.42
WX
yz
yz'
y'z'
y'z
wx
yz
yz'
y'z p
y'z
WX ~
W PX p
WX p W PXp
WPX
w'x
(a)
(b)
yz
WX
yz'
y'z'
y'z
wx
yz
yz'
y'z"
y'z
WX p
W PX p
WX p W PX~
W'X
w'x
(c) (d)
~
Using a Karnaugh map, simplify each boolean expression. (1) E1  wxyz + wxyz' + wxy'z' + wxy'z + w'xyz + w'xy'z (2) E2  wxy'z' + wx'y'z' + wx'y'z + w'x'yz + w'x'yz' + w'xyz + w'xyz'
SOLUTION:
(1) Place a 1 in the square for each minterm in the expression. Loop off each block, beginning with the largest. The Karnaugh map is shown in Figure 12.43. Accordingly, the expression E1 can be pruned:
F i g u r e 12.43
wx
p
yz ~ 1
yz'
y'z' 1 ~
yPz
wx WX p WPX p
E1  (wxyz + wxyz' + wxy'z' + wxy'z) + (w'xyz + w'xy'z) = wxD,(z + z') + y'(z' + z)] + w'xz(y + y') = wx(y + y') + w'xz
 WX ~ W lXZ
12.5 Minimization of Combinatorial Circuits
849
Figure 12.44
WX WX t W'X'
yz
yz"
y'z'
y'z
w'x
(2) Using the Karnaugh map in Figure 12.44, we have:
E2 = (wxy'z' + wx'y'z') + (wx'y'z I + wx'y'z) + (w'x'yz + w'x'yz' + w'xyz + w'xyz') = wy'z'(x + x') + wx'y'(z + z') + [w'x'y(z + z') + w'xy(z + z')] = wy'z' + wx'y' + (w'x'y + w'xy) = wy'z' + wx'y' + w'y(x + x') = wy z + wx'y' + w'y
! t
m
These cases suggest that Karnaugh maps simplify boolean expressions more easily than algebraic laws, especially when variables are few.
Exercises 12.5
Simplify each boolean expression using the laws of boolean algebra.
1. xy + xy' 2. x(x + y) + xy' 3. (x + y )xy' 4. xy + xy' + x'y'
5. x'yz + x'y'z' + x'yz' + x'y'z 7. (x + y)(x + y + z)xy 9. (x + y)(y + z)(z + x) 11. (x + y)(x' + y)(x + y')
6. xy'z' + x'y'z' + xy'z + x'y'z 8. (x + y + z)xyz 10. (xy + yz + zx)xyz
12. (x + y' + z)(x + y + z')xy'z' 14. (x + yz)(y + zx)(z + xy) 16. wx' yz + wx' yz' + w 'x' yz' + w'xyz' y
1 1
13. (x + y)(y + z)(z + x)xyz 15. wxyz + w'xy'z' + wxyz' + w'xy'z
Find the boolean expression represented by each Karnaugh map.
1"I. y y' x[111]
18.
x
y y~ 1[
19.
y
y'
20.
y'
x
Display each sum of minterms in a Karnaugh map.
21. xy + x'y' 22. x'y + xy'
850
Chapter 12 Boolean Algebra and Combinatorial Circuits
Using a Karnaugh map, simplify each sum of minterms. 23. xy + xy' 24. xy + xy' + x'y'
Find the boolean expression represented by each Karnaugh map. 25. yz yz' y'z' y'z 26. yz yz' y'z' y'z
x x'
.
1
1
1 1
x x'
1 1 1 1
27.
x x'
yz yz' y'z ~ y~z 1 1 1 1
!
28.
x x'
yz yz' y'z' y'z 1 1 1
]
Using a Karnaugh map, simplify each boolean expression. 29. xy' z + xy' z'
31. xy'z' + xy'z + x'y'z' + x'y'z
30. xyz + xy' z + x' yz + x' y' z 32. xyz + xyz' + x'y'z' + x'y'z
333{}. Using a Karnaugh map, simplify the boolean expressions in Exercises 2528. Find the boolean expression represented by each Karnaugh map.
37.
wx wx' w'x' w'x
yz yz' y'z' y'z 1 1 1 1 1 1 1 1 yz yz' y'z' y'z 1 1 1 1 1 1
38.
yz yz' y'z' y'z wx 1 1 wx' w'x' w'x i 1 wx wx' w'x' w'x yz yz' y'z' y'z 1 1 1 1 1 1
39.
wx wx' w'x' w'x
40.
Represent each sum of minterms in a Karnaugh map. 41. wxy'z + w'xyz 42. wxyz + wxy'z + w'xyz + w'xy'z
43. wxy'z + wx'y'z + w'xy'z + w'x'y'z 44. wx'yz'+wx'y'z' +w'x'yz'+w'x'y'z' 4548. Using a Karnaugh map, simplify the boolean expressions in Exercises 3740. Using a Karnaugh map, simplify each boolean expression. 49. wxyz + wx'yz + w'x'yz + w'xyz 50. wx'yz' + wx'y'z' + w'x'yz' + w'x'y'z'
51. wx'yz + wx'yz' + wx'y'z' + w'x'y'z' + w'xy'z' + w'xy'z 52. wxyz + wxyz' + wxy'z' + wxy'z + wx'y'z + w'x'y'z + w'xy'z
12.6 Don'tCare Conditions
851
Occasionally, combinatorial circuits may occur that do not accept certain combinations of input signals. In other words, a boolean function f can be generated for which some value combinations never occur as input. Consequently, we d o n ' t care about corresponding o u t p u t values of f in such cases; they remain u n s p e c i f i e d . Such functions are i n c o m p l e t e l y s p e c i f i e d , and such input values are d o n ' t c a r e c o n d i t i o n s . Consequently, to simplify the sum of m i n t e r m s of such a combinatorial circuit with a K a r n a u g h map, place a d in each cell corresponding to a don't care condition. This simply means that the corresponding value of the function can be arbitrarily assigned: the m i n t e r m may or may not enter the simplification process. Whenever the t e r m can help minimization, its d counts as a 1 in the otherwise standard procedure. Three examples will illustrate this technique. Simplify the boolean expression E represented by the K a r n a u g h map in Figure 12.45.
F i g u r e 12.45
wx wx" w'x" yz d d 1 yz" d d 1 y'z" d y'z d
w'x
, .
1
1
SOLUTION: First, draw a loop around each block as usual. Since the top row contains no l's, exclude it in the minimization process; t h i n k of its cells as blanks. The resulting blocks are shown in Figure 12.46. You may verify that: F i g u r e 12.46
WX
yz d
yz' d
y 'z" d
y' z d
WX p
w'x' w'x
C
E = (wx'yz + wx'yz' + w'x'yz + w'x'yz') + (w'xyz + w'xy'z) = x'y + w'xz
N o t e : If the top row counts as a block, E  w x + x ' y + w ' x z . Thus, w x + x ' y + w ' x z  x ' y + w ' x z for all combined values of the variables for which the
852
Chapter 12 Boolean Algebra and Combinatorial Circuits
expression E is specified. As a result, the solution depends on a judicious choice of d's. m The next two examples provide interesting applications of don't care conditions. A system to represent decimal numbers is to replace each decimal digit with its binary version. Four bits are needed to encode the 10 digits, as Table 12.23 shows. With Table 12.23, 345 can be encoded as 001101000101. This binary representation is the b i n a r y c o d e d d e c i m a l ( B C D ) e x p a n s i o n . Although there are 16 possible arrangements of four bits, only 10 of them appear in the table. The others (see Table 12.24) are never used, meaning those BCD expansions manifest don't care conditions.
T a b l e 12.23 Decimal digit Binary representation
0000 0001 0010 0011 0100 0101 0110 0111 1000 1001
T a b l e 12.24
Decimal number
Binary representation
1010 1011 1100 1101 1110 1111
10 11 12 13 14 15
We would like to design a combinatorial circuit that will accept four input signals w, x, y, and z, and output 1 if they represent the BCD expansion of a decimal prime number < 10 (see Figure 12.47). That is, we would like to develop a boolean function f such that f(w,x,y,z) equals 1 if (wxyz)two is the BCD expansion of a prime number and equals 0 otherwise. Combining Tables 12.23 and 12.24, Table 12.25 reveals the logic table of the function f. Since the numbers 10 through 15 do not apply, they generate
12.6 Don'tCare Conditions
853
F i g u r e 12.47
W X
y
z
>
prime number detector
f(w, x, y, z)
T a b l e 12.25
Decimal value
0
BCD input
W x y z
Output
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15
0 0 0 0 1 1 1 1 0 0
0 0 1 1 1 1
0 0 1 1 0 0 1 1 0 0
1 1 0 0 1 1
0 1 0 1 0 1 0 1 0 1
0 1 0 1 0 1
don't care conditions
d o n ' t care c o n d i t i o n s , s i g n a l e d by t h e d ' s in t h e last c o l u m n . (Recall t h a t dcells crop u p in t h e m a p only if h e l p f u l in m i n i m i z a t i o n . ) T h e blocks a r e s h o w n in F i g u r e 12.48. It follows t h a t :
f ( w , x , y , z ) = (wxyz + wxy'z) + (wx'yz + wx'yz' + w'x'yz + w'x'yz') + (w'xyz + w'xy'z) = wxz(y + y') + wx'y(z + z') + w'x'y(z + z') + w'xz(y + y') = wxz + wx'y + w'x'y + w'xz yz wx
wx
W~X p
F i g u r e 12.48
yz" d
y"z" d
y'z d
d
w,x
C
854
Chapter 12 Boolean Algebra and Combinatorial Circuits
= xz(w + w') + x'y(w + w') = x z +x'y
Figure 12.49 gives the circuit.
Figure 12.49 x ~ ~>~
~ ~
r
f(w, x, y, z,)=x'y+xz
m
The next example presents a familiar application of don't care conditions to everyday life. ( D i g i t a l D i s p l a y s ) E l e c t r o n i c devices such as modern calculators, microwave ovens, and video cassette recorders display digits by lighting up a maximum of seven line segments, labeled a through g in Figure 12.50. Figure 12.51 provides the display strategy: the digit 0 lights up the segments a, b, c, d, e, and f; 6 lights up the segments c, d, e, f, and g; and
SO o n .
F i g u r e 12.50
a
rl lb el Ic
d
F i g u r e 12.51
@ lq P,
0 1 5 6
2
3
4
7
89
8
9
Design a combinatorial circuit to accept the BCD expansion (wxyz)two of a decimal digit and light up a segment s. In other words, develop a boolean expression to define the boolean function f ( w , x , y , z ) = s. As in Example 12.35, 16 arrangements are available. However, since there are only 10 digits, six arrangements are don't care conditions; they are listed in Table 12.26. (Verify them.) For instance, suppose we wish to find a minimal boolean expression for the boolean function f(w,x,y,z) = a. Table 12.26 gives the corresponding
12.6 Don'tCare Conditions
855
Table 12.26 D e c i m a l digit
w 0 0 0 0 0 0 0 0 1 1
BCD input
x 0 0 0 0 1 1 1 1 0 0 y 0 0 1 1 0 0 1 1 0 0 z 0 1 0 1 0 1 0 1 0 1
Inputs a
1 0 1 1 0 1 0 1 1 1
b
1 1 1 1 1 0 0 1 1 1
c
1 1 0 1 1 1 1 1 1 1
d
1 0 1 1 0 1 1 0 1 0
e
1 0 1 0 0 0 1 0 1 0
f
1 0 0 0 1 1 1 0 1 1
g
0 0 1 1 1 1 1 0 1 1
DNF, a  w'x'y'z' + w'x'yz' + w'x'yz + w'xy'z + w'xyz + wx'y'z' + wx'y'z, which leads to t h e K a r n a u g h m a p in F i g u r e 12.52. T h e c a r t o g r a p h i c a l c o n c l u s i o n , a = w + x'y + x'y'z' + w'xz, yields t h e d e s i r e d c i r c u i t in F i g u r e 12.53.
F i g u r e 12.52
yz
yz '
y' z'
y'z
wx
d
Figure 12.53
X
)
f(w, x, y, z)=a
z
)
~>o
I
In a c c o m m o d a t i n g t h r o u g h dcells, t h e d o n ' t c a r e c o n d i t i o n s of i n c o m p l e t e l y specified f u n c t i o n s in K a r n a u g h m a p s e m e r g e as s t r o n g w o r k h o r s e s in t h e s e a r c h for c o m b i n a t o r i a l c i r c u i t s t h a t m i m i c b o o l e a n e x p r e s s i o n s .
856
Chapter I~ Boolean Algebra and Combinatorial Circuits
Exercises 12.6
Simplify the boolean expression represented by each K a r n a u g h map.
1. x
Xf
yz 1
yz ~ y'z' d d 1 yz yz' y'z ~
1
y'z d
2. x
XI
yz d
yz'
y'z' 1
y'z d 1 y'z
d yz yz' d 1
1 y'z'
0
y'z
1
4.
WX WX ~ U2 P f X W~X
WX WX ~ WlX t WtX
1 d
d d
1
d d
d
d
1
d
d
1
yz yZ f
1
0
yz
WX WX t WlX I
yz' 1
y'z' 1
y'z 1
6.
WX WX l
y'z'
y'z
d
d
1
d
d
w~x ~
w 'x
1
1
d
w'x
Simplify the boolean expression defined by each table.
11,
x 0 0 0 0
1 1 1 1
y 0 0 1 1
0 0 1 1
z 0 1 0 1
0 1 0 1
f 1 1 0 0
d d 1 1
0
x 0 0 0 0
1 1 1 1
y 0 0 1 1
0 0 1 1
z 0 1 0 1
0 1 0 1
Using the BCD expansions of the decimal n u m b e r s 0 t h r o u g h 9, develop a boolean function f ( w , x , y , z ) such t h a t the decimal value of (wxyzltwo is divisible by: 9. Two 11. Two and three 10. T h r e e 12. Two or three
13. Seven or more 14. Not more t h a n five, but not one 1522. Find the D N F s of the boolean functions in Exercises 714. 2328. Generate a minimal boolean expression for each function b t h r o u g h g in Example 12.36.
Chapter Summary
857
Boolean operators and the laws of boolean algebra serve well in designing and simplifying electronic circuits.
Boolean Algebra
9 A boolean algebra H R 1/8 f p
z
+ 5 ? I S ] g q
{
6 @ J T A h r
I
7 A K U i s
}
. 8 B L V " j t
"
/ 9 C M W a k u
0 9 D N X b 1 v
1 ; E O Y c m w
Codes 0031 and 127 are nonprintable control characters. The symbol [] denotes the
blank character.
Determinants were invented by the Japanese mathematician Seki Kowa (16421708). However, German mathematician Gottfried Wilhelm Leibniz ( 1 6 4 6  1 7 1 6 ) is c r e d i t e d w i t h t h e i r c r e a t i o n , a l t h o u g h h i s c o n t r i b u t i o n (in 1 6 9 3 ) c a m e 10 y e a r s l a t e r t h a n K o w a ' s . L e i b n i z d e v e l o p e d d e t e r m i n a n t s while solving systems of linear equations. *Based on T. Koshy, College Algebra and Trigonometry with Applications, McGrawHill, New York, 1986, pp. 423432. 867
868 Determinant
Appendix A
A d e t e r m i n a n t is a function from t h e set of s q u a r e m a t r i c e s to R, t h e set of real n u m b e r s . The d e t e r m i n a n t of a s q u a r e m a t r i x A is d e n o t e d by det A or IAI. If A is an n • n m a t r i x , d e t A is called a d e t e r m i n a n t of o r d e r n. ( D e t e r m i n a n t s of n o n s q u a r e m a t r i c e s are not defined.) For example, the d e t e r m i n a n t of the m a t r i x A = (aij)2• is w r i t t e n as detAu s i n g vertical bars and is defined by d e t A = alia22  a21a12 For instance,
I Ia l l la21
a12 a22
14
2
3
= 2. (5)  4 . 3 =  2 2
If we k n o w how to e v a l u a t e 2 • 2 d e t e r m i n a n t s , we can t h e n e v a l u a t e higherorder determinants.
Minor and Cofactor
Let (aij) be a s q u a r e m a t r i x of order n. T h e d e t e r m i n a n t of t h e m a t r i x obtained by deleting row i and column j is the m i n o r of t h e e l e m e n t aij, denoted by Mij. The c o f a c t o r Aij ofaij is defined by Aij  (1)i+JMij. Aij = +Mij if i + j is even, and Mij if i + j is odd. T h e signs associated with each cofactor Aij can be displayed in a c h e c k e r b o a r d p a t t e r n like t h i s 3 • 3 matrix:
I!iil
The signs a l t e r n a t e for each row and column, b e g i n n i n g w i t h + for t h e first row and first column. For t h e m a t r i x A
i1 2 31
45 7 8
c o m p u t e M13,A13,M32, and A32.
A.2
Determinants
869
SOLUTION: M13 7
4 5[ = 4 . 8 8
 (7). (5) = 3
A13  (  1 ) 1+3M13 :  3 M32 1 4 3  1.64.3=6 6
i
A32  (  1) 3+2M32  + 6
E x p a n s i o n of d e t A Let Aij denote the cofactor of the e l e m e n t aij of the m a t r i x A T h e n the d e t e r m i n a n t of A is defined by d e t A  a i l A i l ~ ai2Ai2 ~ ... Jr ainAin  a l j A l j + a 2 j A 2 j + ... + a n j A n j (expansion by row i) (expansion by c o l u m n j ) (aij)n•
(det A can be e x p a n d e d by any row or a ny column.) ~ E v a l u a t e d e t A for the m a t r i x A in E x a m p l e A. 1. SOLUTION: E x p a n d i n g det A by t h e first row,
8

(2)
I 74
4
5

+3
7
8
= + 1(45  48) + 2 (  3 6 + 42) + 3(32  35) =0 We can e x p a n d d e t A by a n y o t h e r row or a ny c o l u m n to verify this. i
Suppose t h a t every e l e m e n t in a row or c o l u m n of a s q u a r e m a t r i x A is zero. E x p a n d i n g det A by t h a t row or c o l u m n shows t h a t det A = 0 (Verify this.). T h e o r e m A. 1 confirms this. If every e l e m e n t in a row or c o l u m n of a s q u a r e m a t r i x A is zero, t h e n d e t A = 0. i E v a l u a t i n g a d e t e r m i n a n t of a large orde r is not easy unless it c o n t a i n s m a n y zeros. However, t h e following t h e o r e m s can m a k e e v a l u a t i o n easier. ~ Let B be the m a t r i x o b t a i n e d by i n t e r c h a n g i n g a ny two rows or c o l u m n s of a s q u a r e m a t r i x A. T h e n det B =  det A.
870
Appendix A
PROOF: To prove this t h e o r e m for t h e m a t r i x A rows of A: B __ [a21 all and a22] a12

(aij)2•
interchange the
d e t A  a21a12  a l i a 2 2 =  ( a l i a 2 2  a21a12) = detA m
If a n y two rows or c o l u m n s of a s q u a r e m a t r i x A are identical, det A  0. PROOF: Let B be t h e m a t r i x o b t a i n e d by i n t e r c h a n g i n g two identical rows or c o l u m n s of A. By T h e o r e m A.2, det B   d e t A . B u t since A  B, det B = det A. T h e r e f o r e , det A   d e t A a n d h e n c e det A  0. m T h e following e x a m p l e i l l u s t r a t e s this. E v a l u a t e d e t A if
n
I
2 8 2
3 4 3
5 1 7 5
SOLUTION: Since rows 1 a n d 3 are the same, d e t A  0, by T h e o r e m A.3. E x p a n d i n g d e t A by c o l u m n 1 confirms this" detA2 4 3 7 5 (8) 3 3 5 3 5 +24 5 7
= 2(20  21) + 8(15  15) + 2(21  20) = 2 +0+ =0 as expected, m 2
T h e n e x t t h e o r e m shows t h a t we can factor out c o m m o n factors from a row or column.
A.2 Determinants
871
Let B be the m a t r i x produced by m u l t i p l y i n g one row or c o l u m n of a s q u a r e m a t r i x A by a real n u m b e r k. T h e n det B  k 9det A. PROOF" Let A = (aij)n x n and B be the m a t r i x derived from m u l t i p l y i n g its i t h r o w by k. E x p a n d i n g det B by row i, d e t B = (kail)Aii + (kai2)Ai2 + . . . + (kain)Ain
= k(ailAil ~ ai2Ai2 +.'.+ainAin)
= k. detA Using T h e o r e m A.4, evaluate detA, w h e r e m
A=
4/5 7
6
8 9
SOLUTION: Factor out 3 from row 112 detA = 3 45 7 Now factor out 3 from column 3: 12 detA  9 45 7 E x p a n d by row 1" detA  9 1 1 2 8 3 3 6 8 9
L
5 8
2 +2 3
4 7
2 +1 3
4 7
5 8
= 911(15  16) + 2 (  1 2 + 14) + 1 ( 3 2  35)1 =0 m
Let B be the m a t r i x obtained by adding k t i m e s a row or column to a n o t h e r row or column of a square m a t r i x A. T h e n det B  detA. PROOF: We shall prove the t h e o r e m for an a r b i t r a r y m a t r i x A  (aij)2x2. Let B be the m a t r i x o b t a i n e d by adding k times row 2 to row 1" B  Ia l l + ka21 I a21 a12 + ka22 a22
872
Appendix A
d e t A  (all H k a 2 1 ) a 2 2  a21 (a12 F ka22) = (alia22  a21a12) H k ( a 2 1 a 2 2  a21a22) = detA + k. 0 = det A Similarly, i f B is t h e m a t r i x obtained by adding k t i m e s row I to row 2, t h e n det B  detA. m ~ E v a l u a t e d e t A with T h e o r e m A.5, w h e r e
A
45 7
6 8 9
SOLUTION: Keep the 1 in c o l u m n 1 to m a k e the o t h e r two e n t r i e s in c o l u m n 1 (4 a n d  7 ) O's. To this end, add  4 times row 1 to row 2 and 7 t i m e s row 1 to row 3: 1 detA2 3 18 0 13 0
22  3 0
E x p a n d i n g d e t A by column 1, we see t h a t detA  1[(13). (30)  22.18]  6 m
The above properties used s i m u l t a n e o u s l y can speed up e v a l u a t i n g d e t e r m i n a n t s , as the next example shows. ~ E v a l u a t e d e t A if
3
n ~
1
21
3 2 1
05 4 0 2 0
i
SOLUTION: Use T h e o r e m s A.4 and A.5 to evaluate detA. F a c t o r out 2 from row 3: 3 3 detA  2 1 1 2 2 0 0 4 3
1
21 6
05
A.2
Determinants
873
N o w u s e t h e 1 in c o l u m n 1 to m a k e t h e r e m a i n i n g e n t r i e s in t h e c o l u m n zeros. To t h i s end, a d d 3 t i m e s r o w 3 to r o w 1 a n d  3 t i m e s r o w 3 to r o w 2; a d d r o w 3 to r o w 4: 0 0 det A  2 1 0 7 6 2 4 2 5 0 0 11 6
E x p a n d t h i s d e t e r m i n a n t by c o l u m n 1: 11 detA = 2 6 5 6
E x p a n d t h e n e w d e t e r m i n a n t by r o w 3: d e t A  2 [4 2 5 11 6 0+7 7 2
= 214(12= 858
55)  0 + 7 (  3 5 + 12)!
m
E x e r c i s e s A.2
Evaluate each determinant.
0 3
1
4. 0
" 0 0 3
5

3.
a
b
a
0
0 2 0
5.
I! il
7
4
0
0
l+x 1 1 a b
1 l+y 1 0 0
1 1 l+z
~ 91!
12. 1 0 1 2
b
3 4 9 b c a
5 7 15 c a b 0 1 2 4 1 1 3 6 1 1 4 8
10.
I
l
l
a2
l b b2
c c2
11.
c
d 0 0 O e f 0 g h
P r o v e each. 13. L e t A a n d B be t w o 2 x 2 m a t r i c e s . T h e n IABI  IAI. IBI. 14. Let A be a n y s q u a r e m a t r i x of o r d e r n. T h e n IkAI  knlAI.
874
AppendixA
15. Let A be an n • n m a t r i x such t h a t
An T :
In. T h e n IAI = +1.
16. Let A = (aij)nxn where aij = 0 f o r j > i. (Such a m a t r i x is called an u p p e r t r i a n g u l a r m a t r i x . ) T h e n IAI = a l i a 2 2 . . , ann. Solve each equation for f(x).
f(x) 1
x
a =0
If(x)
1
x
x2
17. f(a)
f(b)
1
1
b
If(a) 18. if(b )
If(c)
1
1
a
c
a2
b2 c2
1 b
=0
Exponential functions and their closest relatives, the logarithmic functions, form one of the most i m p o r t a n t classes of functions. They have a wide range of fine applications to computer science, life sciences, m a n a g e m e n t science, physical sciences, and social sciences. For example, an exponential function occurs in the n a t u r a l sciences: E. coli are microscopic unicellular bacteria t h a t multiply by halving into two bacteria about every 20 m i n u t e s u n d e r ideal conditions. Suppose initially there is one bacterium and 20 minutes is unit time. T h e n t h e r e will be two bacteria in one unit time, four in two units, eight in three units, and so on. In general, there will be f ( x ) = 2x bacteria in x units of time, a s s u m i n g none dies during the period. This function belongs to the large class of exponential functions, defined below.
Exponential Function Let a be a positive real n u m b e r different from 1 and x an a r b i t r a r y real number. Then the function f ( x ) = a x is called an e x p o n e n t i a l f u n c t i o n w i t h b a s e a. If a = 1 in this definition, the function becomes the constant function f( x ) = 1, which holds little practical interest. As a result, it is excluded from the definition. Figure A. 1 shows the graphs of several exponential functions with different bases a, where a > 1. They grow faster and more steeply as a increases. We can use such exponential functions to describe growths of populations; consequently, their graphs are called g r o w t h c u r v e s . Figure A.2 shows some exponential graphs y = a x, where 0 < a < 1. Each can be obtained by flipping the corresponding graph y = a x, where a > 1, about the yaxis.
*Based on T. Koshy, College Algebra New York, pp. 204241.
and Trigonometry with Applications,
McGrawHill,
A.3 Exponentialand Logarithmic Functions F i g u r e A.1
G r o w t h curves y  a x, a>l.
875
l
~l
Y ~ ~
y = lO x y=7 x y=3 x
~X
F i g u r e A.2
Decay curves y = a x, 0 x2 t h e n f ( x l ) > f(x2). If 0 < a < 1, t h e n f is a d e c r e a s i n g f u n c t i o n ; t h a t is, i f x l > x2 t h e n f ( x l ) < f(x2). 9 The xaxis is a h o r i z o n t a l a s y m p t o t e for t h e graph. 9 The g r a p h s y = a x a n d y = a x are reflections of each o t h e r a b o u t t h e yaxis. The next two examples explore some applications of the e x p o n e n t i a l function.
876
~
Appendix A
Consider the world population (in billions) given by Table A.1. We can use these to draw a smooth curve as in Figure A.3. The graph shows t h a t the world population is growing exponentially.
T a b l e A.1
Yeart
Population P(t)
II
1
1000 0.3
1650 0.5
1750 0.7
1800 0.9
1850 1.1
1900 1.6
1950 2.5
1970 3.6
1990 5.2
2000 6.1
0.25?
F i g u r e A.3
The exponential growth of the world population.
1 1750 1850 Years 1950 2000
1650
Suppose a bowl of n u t r i e n t broth contains 1,000,000 viable bacteria per millimeter. Table A.2 shows the n u m b e r of viable bacteria per millimeter in the broth after an agent has been added to kill them. The population at time t is only onetenth of t h a t at time t  10. Let N(t) denote the n u m b e r of viable bacteria at time t. Then N(t)  1060"it. If t can take on all possible values (at least theoretically), then the graph of N(t) looks like the decay curve in Figure A.4.
T a b l e A.2
Time t (in minutes)
II
0 106
10 105
20 104
30 103
40 102
50 10
60 1
N u m b e r of bacteria N(t) II
m
An extremely useful exponential function is f(x) = ex, where e is the irrational n u m b e r 2.718281828 . . . . (The letter e for the base was chosen in
*Based on R. M. Thrall et al. (ed.), "Extermination of Bacteria," Some Mathematical Models in Biology, The University of Michigan Press, Ann Arbor, MI, 1967, pp. PE3.1PE3.2.
A.3 Exponential and Logarithmic Functions
877
Figure A.4
Nqt)
106 1
O
Time
t
honor of Euler.) It is established in calculus t h a t eThe graphs of the functions lim
n* oo
( 1)n
1+ n and
g(x) 
f(x)
= ex
e x
appear in Figure A.5.
F i g u r e A.5
y
=e
x
 e x
,11 0
~X
Logarithms
Logarithms were invented by the Scottish m a t h e m a t i c i a n J o h n Napier (15501617), to facilitate numeric computations. He published a s u m m a r y of his results in 1614.
Logarithmic Function
In the exponential function y = a x , where a > 0 and a #= 1, the exponent x is the l o g a r i t h m o f y t o t h e b a s e a; it is denoted by loga (y) or
878
Appendix A
simply loga y. Thus,
logay = x if a n d only i f y = a x
The function g ( x ) = log a x is t h e l o g a r i t h m i c f u n c t i o n w i t h base a. W i t h the base a always positive, y = a x is positive for all values of x. As a result, x = log a y is defined only if y is positive. In o t h e r words, only l o g a r i t h m s of positive n u m b e r s are defined. T h u s the d o m a i n of t h e logarithmic function is (0,~z), a n d the r a n g e is (  ~ , ~ ) . Since the exponential function is bijective w i t h d o m a i n (  ~ , ~ ) a n d r a n g e (0, ~ ) , it has an inverse function. The next t h e o r e m shows t h e inverse is indeed the logarithmic function. T h e logarithmic function g ( x ) function f (x) = a x.

log a x is the inverse of t h e e x p o n e n t i a l
PROOF:
By definition a l~ Then x __ x for every x > 0, and t h a t l o g a ( a x) = x for every x.
(f og)(x)
 f(g(x))
 f ( log a x)  a l~
 x
and
(g o f ) ( x ) = g ( f ( x ) ) = g ( a x) = l o g a ( a x) = x
So the logarithmic function fi
function
g
is the
inverse
of t h e
exponential
II
According to this t h e o r e m , we can obtain the g r a p h of y = logax by reflecting t h a t o f y = a x a b o u t the line y = x, as shown in F i g u r e A.6.
F i g u r e A.6
Y
x
Y =a x
IY
2ey =x
/ / // /
~

'
~
/
a>l
~X
1/
f
O 1, g is a n i n c r e a s i n g function; if 0 < a < 1, g is a d e c r e a s i n g function. 9 T h e yaxis is a vertical a s y m p t o t e . 9 T h e f u n c t i o n s f ( x ) = a x a n d g ( x ) = loga x are i n v e r s e s of each o t h e r ; t h e i r g r a p h s are m i r r o r i m a g e s of each o t h e r a b o u t t h e line y  x. 9 T h e g r a p h s o f y  loga x a n d y  lOgl/a x are reflections of each o t h e r a b o u t t h e xaxis.
Frequently Used B a s e s
T h r e e bases of l o g a r i t h m s , 10, e, a n d 2, are f r e q u e n t l y used. L o g a r i t h m s to t h e base 10 are called c o m m o n l o g a r i t h m s or B r i g g s i a n l o g a r i t h m s , after t h e E n g l i s h m a t h e m a t i c i a n H e n r y Briggs (15611631). T h e i r base is u s u a l l y omitted: log x  log 10 x L o g a r i t h m s to t h e base e are n a t u r a l l o g a r i t h m s a n d a b b r e v i a t e d as ln" In x = log e x L o g a r i t h m s to t h e base 2, which we often use in c o m p u t e r science, a p p e a r as lg: lg x  log 2 x T h e f u n d a m e n t a l p r o p e r t i e s of l o g a r i t h m s are given in the next t h e o r e m . We can prove each u s i n g t h e definition of a l o g a r i t h m . Let a, b, x, a n d y be a n y positive real n u m b e r s such t h a t a r 1, b r 1, a n d n a n y real n u m b e r . Then" (1)
al~
(3) loga  = loga X Y (5) loga 1 = 0 (7) loga (a x) = x logax (9) logb x = log a b
x = X
X 
loga Y
(2) (4) (6) (8)
l o g a ( x y ) = loga x + loga y log a ( x n ) ~ n 1Oga X
1Ogaa = 1 1OgaX  1Ogay if a n d only if
Xy
m
880
Appendix A
The next t h r e e examples i l l u s t r a t e some useful e x p o n e n t i a l a n d logarithmic functions. T h e pH of a solution, which m e a s u r e s its acidity or alkalinity, is defined as t h e negative l o g a r i t h m of t h e h y d r o g e n c o n c e n t r a t i o n x in moles per liter of the solution; t h a t is, pH =  l o g x. At 25~ the p H of p u r e w a t e r is 7; if the pH of a solution falls below 7, t h e solution is acidic, a n d above 7 it is alkaline. C o m p u t e the pH of a solution w h e r e the c o n c e n t r a t i o n of h y d r o g e n ions is 6.7 x 10 4. SOLUTION: pH of the solution   log (6.7 x 10 4) =  log 6.7  log (10 4) = log6.7 + 4
0.8261 + 4 = 3.1739 The world population (in billions) at any t i m e t (in years) is given by m
p(t)

6.1e ~176 w h e r e p(0) denotes the population in 2000. At t h e given rate, w h e n will the population double the 2000 level? SOLUTION: P o p u l a t i o n in 2000  p ( 0 )  6.1 billion. We need the value of t w h e n 12.2 (Figure A.7). T h e n 6. le 00167t = 12.2 Divide by 6.1" e 0"0167t  2 Then In e 00167t  In 2 0 . 0 1 6 7 t  In 2 ln2
t m
p(t)

0.0167 42 y e a r s
m
F i g u r e A.7
12.2 t=? 2000 6.1
A.3 Exponential and Logarithmic Functions
881
The h a l f  l i f e of a radioactive substance is the time required for onehalf of t h a t substance to decay. The a m o u n t of 11C, an isotope of carbon, present at a future time t (in months) is given by A ( t )  100e 00338t. Find the halflife of the material. SOLUTION: Original a m o u n t = A(0)  100 grams. The halflife is the value of t when A ( t )  89  50 (Figure A.8). T h e n
100e 0"0338t  50
Divide by 100"
e 0.0338t 1
2
Then In e 00338t  In 1 2  0 . 0 3 3 8 t   In 2
t __
ln2 0.0338 20.5 m o n t h s
F i g u r e A.8
t=?
II
Exercises A.3 Prove each, where f denotes the exponential function.
1. f ( x + y ) = f ( x ) . f ( y ) 2. f ( x  y ) f(x)
3. f is injective,
f(y)
4. The logarithmic function g ( x ) = loga x is injective.
882
Appendix A
Prove, by contradiction, that each is an irrational number.
5. log 2
6. log 3
7. The concentration of hydrogen ions in a solution is 3.76 • 10 s. Compute the pH of the solution. 8. The pH of a solution concentration. is 5.3575. Compute its hydrogen ion
The intensity of sound the h u m a n ear can hear is measured in decibels, named after Alexander Graham Bell (18471922), the American scientist who invented the telephone. The n u m b e r of decibels in a sound of intensity I is given by I B  10 log I0 where I0 is the standard intensity of 1012 watts per square meter (W/m 2). Find the number of decibels in: 9. Threshold of hearing (10 12 W/m 2) 10. Normal conversation (10 6 W/m 2) 11. Street traffic (10 5 W/m 2) 12. Two sounds of intensity 11 and I2 have decibels B 1 and B2, respectively. I1 Show that B1  B2  10 log I2 13. The magnitude of a star is a measure of its brightness: the brighter the star, the smaller the magnitude. If b l and b2 denote the magnitudes of two stars, B1 and B2 their respective brightnesses, B1 bl  b2 = 2.512 log B~. The brightest stars, Aldebaran and Sirius, have magnitudes 1 and  1.6, respectively. How much brighter is Sirius than Aldebaran? The absolute magnitude M of a star, its apparent magnitude m, and the stellar distance d (in parsecs) from the earth are related by the formula rn = M § 5 log 0.1d. 14. Compute the stellar distance of Aldebaran if its absolute magnitude is  0 . 2 and its apparent magnitude 0.86. 15. The difference m  M is the d i s t a n c e m o d u l u s of the star. The distance modulus of Sirius is 2.86. Find its stellar distance. 16. With R u b i k ' s C u b e , a popular game, the goal is to turn its faces to one particular configuration. The minimum number of turns required from 4.3 • 1019 some position to the desired state is 1 + log15 ( 18 ). ~ Compute it. (C. Kluepfel, 1982)
A.4 GeneratingPermutations and Combinations
883
Sections 6.2 t h r o u g h 6.5 presented various types of p e r m u t a t i o n s and combinations, as well as formulas for computing the n u m b e r of rpermutations and rcombinations of a finite set S. In this section, we will study algorithms which e n u m e r a t e them.
Generating Permutations
For convenience, we choose S = {1, 2 , . . . , n }. We shall present an algorithm for generating the various p e r m u t a t i o n s of elements of the set. The algor i t h m is based on l e x i c o g r a p h i c ordering, the same type used to arrange words in a dictionary.
Lexicographic Order precedes) a p e r m u t a t i o n
9
In lexicographic ordering, a p e r m u t a t i o n A = a l a 2 . . , B  b i b 2 . . , b,~ if:
an
is less than (or
al < b l o r
for 1 _< i < k  1, and ak < bk.
9 ai  bi
This is precisely the order we use to alphabetize words (of the same length) in the dictionary. For instance, the word c o m p u t e precedes the word p e r m u t e and the word e s t a t e precedes the word e s t e e m . Consider the p e r m u t a t i o n a l a ' z a 3 a 4  2134 and b l b 2 b 3 b 4  2143 of the set {1,2,3,4}. They agree in the first two positions: al = bl and a2 = b2. But they differ in the third positions: a3 r b3. Since a3 < b3, the p e r m u t a t i o n 2134 precedes the p e r m u t a t i o n 2143 in lexicographic ordering, m Recall that there are 4! = 24 p e r m u t a t i o n s of the set {1, 2, 3, 4}. They are listed in Figure A.9 in lexicographic order columnwise, beginning with 1234. (Can you find a p a t t e r n for e n u m e r a t i n g them?)
Figure A.9
1234 1243 1324 1342
1423 1432 2134 2143
2314 2341 2413 2431
3124 3142 3214 3241
3412 3421 4123 4132
4213 4231 4312 4321
bib2..,
Suppose we have an algorithm to generate the next larger p e r m u t a t i o n bn from a given p e r m u t a t i o n a l a 2 . . , a n . Then, beginning with the p e r m u t a t i o n 123... n, we can invoke it to generate the remaining n  1 permutations. The procedure to generate the next larger p e r m u t a t i o n from a given p e r m u t a t i o n in lexicographic order is described below.
884
Appendix A
If a n  1 < an, switch them to obtain the new p e r m u t a t i o n a l a 2 . . , a n  2 a n a n  1 . Then the p e r m u t a t i o n a l a 2 . . , a n  2 a n  l a n precedes the p e r m u t a tion a l a 2 . . , a n  2 a n a n  1 in lexicographic ordering. For example, consider the p e r m u t a t i o n 2143; it succeeds the p e r m u t a tion 2134 in the listing in Figure A.9. On the other hand, suppose a n  1 > an. (Then by switching t h e m we cannot obtain a larger permutation.) So we look at a n  2 . If a n  2 < a n  l , find the smaller of the elements a n  1 and an t h a t is larger t h a n a n  2 ; place it in position n  2; and now arrange a n  2 and the remaining elements in increasing order. For instance, consider the p e r m u t a t i o n a l a 2 a 3 a 4  2341 in Figure A.9. Here a3 > a4, but a2 < a3; the smaller of a3 and a4 t h a t is larger t h a n a2 = 3 is a3  4; so hold 4 in position 2 to get 2 4 ; now arrange the r e m a i n i n g elements a2  3 and a4 = 1 in ascending order to yield the next larger p e r m u t a t i o n 2413 (see Figure A.9). If an2 > a n  l , we need to look at the elements a n  3 t h r o u g h an. More generally, all we need to do is the following. From right to left, find the first pair of elements ai, ai+l such t h a t ai < a i + l , where ai+l > ai+2 > . . . > an. Then find the smallest of the elements ai+l, a i + 2 , . . . , a n , say, aj, that is larger t h a n ai. Place aj in position i. Now arrange the elements ai, a i + l , . . . , a j  1 , aj, a j + l , . . . , a n in increasing order in positions i § 1 t h r o u g h n. The following example illustrates this procedure. Find the next p e r m u t a t i o n larger t h a n a l a 2 a 3 a 4  3421. SOLUTION: The first pair ai, ai+l from right to left for which ai < ai+l is the pair al, a2. So find the smallest of the elements a2  4, a3  2, and a4 = 1 t h a t is larger t h a n a l  3. Clearly, it is a2  4. So we place 4 in position 1. Now we arrange the elements 3, 2, and 1 in increasing order: 123. Consequently, the next larger p e r m u t a t i o n is 4123. (See Figure A.9.) 9 The above discussion can be translated into an algorithm for finding the next larger p e r m u t a t i o n t h a t follows a given p e r m u t a t i o n a l a 2 . . , an. It is presented in Algorithm A.1.
Algorithm nextpermutation (ala2...an)
(* This algorithm finds the permutation larger than the given permutation ala2...an. Assume i t is not the largest permutation n(n  1)...321. *) O. Begin (* algorithm *) (* From right to l e f t , find the f i r s t pair ai, ai+l for which ai < ai+l. *) 1. i K   n  1
A.4 GeneratingPermutations and Combinations
885
2. While ai > a i + l do (* continue searching to the l e f t *) 3. i ~ii (* When we e x i t the loop, we w i l l have found an i for which ai < a i + l . Find the smallest of the elements a i + l through an, say, ak, that is larger than ai. Since ai + 1 > ai +2 > "'" > an, scan from r i g h t to l e f t to find the element ak that is larger than ai. *) 4. k ~  n 5. While ak < ai do (* continue scanning to the l e f t . *) 6. k+k1 (* ak is the smallest of the elements ai+1 through an that is larger than ai. Swap ai and ak. *) 7. swap ai and ak 8. sort the elements in positions i + 1 through n into ascending order. 9. End (* algorithm *)
Algorithm A.I
Using Algorithm A. 1, e n u m e r a t e in lexicographic order all p e r m u t a t i o n s of elements of the set {1, 2, 3}. SOLUTION: We begin with the basic p e r m u t a t i o n ala2a3 : 123. Since a2 < a3 and the least of the elements(s) a3 t h a t is g r e a t e r t h a n a2 is a3, switch 2 and 3 to get the next larger p e r m u t a t i o n 132. Since 1 < 3 and the smaller of 3 a n d 2 t h a t is larger t h a n 1 is 2, swap 1 and 2, and sort the elements 3 and 1 to yield the p e r m u t a t i o n 213. C o n t i n u i n g like this, we get the r e m a i n i n g p e r m u t a t i o n s in lexicographic order: 123, 132, 213, 231,312, 321. (Also see the tree diagram in Figure A. 10.) F i g u r e A.10 3 2 3 1 2 1 ~ ~ ~ permutations > > > > > > 123 132 213 231 312 321 1
Generating Combinations
How can we g e n e r a t e the rcombinations of set {1,2,... ,n}, where 0 < r < n? First we list the various combinations in increasing numerical order.
886
Appendix A
F o r i n s t a n c e , t h e r e a r e t e n 3  c o m b i n a t i o n s of t h e set S = {1, 2, 3, 4, 5}; in lexicographic order, t h e y are: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345 H o w can we find t h e n e x t l a r g e r 3  c o m b i n a t i o n f r o m a given 3  c o m b i n a t i o n ? F o r i n s t a n c e , c o n s i d e r t h e 3  c o m b i n a t i o n 134 of t h e set S. T h e last e l e m e n t (from right) is less t h a n 5 (= n). So we i n c r e m e n t it by 1 to get t h e n e x t l a r g e r c o m b i n a t i o n 135. U s i n g t h e set S  {1, 2, 3, 4, 5}, find t h e c o m b i n a t i o n following a l a 2 a 3 135 in lexicographic order.

SOLUTION:
Notice t h a t a3 = 5, t h e l a r g e s t i n S . So look at a2; a2 < 5; so i n c r e m e n t a2 by 1 to get 4. Replace a3 by 1 m o r e t h a n t h e c u r r e n t a 2 : a 3 ~ a2 + 1 = 4 + 1 = 5. T h u s t h e n e x t l a r g e r c o m b i n a t i o n is 145. m F i n d t h e c o m b i n a t i o n t h a t follows in lexicographic o r d e r t h e 3  c o m b i n a t i o n 145 of t h e set {1, 2, 3, 4, 5}.
SOLUTION:
Let a l a 2 a 3 = 145. Clearly, a3 = 5 a n d a2 = 4. Since a2 < 5, if we i n c r e m e n t it by 1 to get 4 + 1 = 5, t h e n e w a3 s h o u l d be 5 + 1; u n f o r t u n a t e l y , it does not exist in t h e set. So go to a l = 1, w h i c h is less t h a n 5. (Notice t h a t a3  5  3 + 3; a 2  4  5  3 + 2; b u t a l  1 r 5  3 + 1. T h e r e t b r e , a l is t h e first e l e m e n t f r o m t h e r i g h t such t h a t a i r n  r + i.) I n c r e a s e a l by 1 : a l * a l + 1; so a l * 2. Now assign a2 = al + 1 = 3 a n d a:3 = al + 2 = 4. T h e r e s u l t i n g c o m b i n a t i o n is 234. m
T h u s , to find t h e c o m b i n a t i o n t h a t follows t h e r  c o m b i n a t i o n a l a 2 . . , a r , we proceed as follows. F r o m r i g h t to left, we find t h e first e l e m e n t a i s u c h t h a t a i ~= n  r + i. I n c r e m e n t a i by 1 : a i ~ a i + 1. ( U s i n g t h e n e w a i ) a s s i g n t h e v a l u e s a i + 1 , a i + 2 , . . . to a i + l , a i + 2 , . . . , a r , respectively; t h a t is, a j ~ a i + j  i, w h e r e i + 1 < j _< r.
F i n d t h e c o m b i n a t i o n t h a t follows in lexicographic c o m b i n a t i o n 245 of t h e set {1, 2, 3, 4, 5}.
ordering
the
3
SOLUTION:
H e r e n  5, r = 3, a n d n  r = 2. Let a l a 2 a 3  245. F r o m r i g h t to left, find t h e first a i s u c h t h a t a i ~= n  r + i = 2 + i. Clearly, a3  5  2 + 3, a2 = 4 = 2 + 2; b u t a l = 2 r 2 + 1. T h e r e f o r e , t h e first such a i is a l . U p d a t e a l as a l + l : a l ~ 2 + 1  3 . N o w assign t h e v a l u e a l + j  1 to a j for t h e r e m a i n i n g positions j , n a m e l y , 2 a n d 3: W h e n j = 2, a2 = a l +21 =3+ 1 =4
A4
Generating Permutations and Combinations
887 1 =3+2=5 m
W h e n j = 3 , a3   a l + 3 The r e s u l t i n g c o m b i n a t i o n is 345.
These discussions lead us to A l g o r i t h m A.2. Algorithm nextcombination (ala2...ar)
(* This a l g o r i t h m f i n d s the combination t h a t f o l l o w s the r  c o m b i n a t i o n a l a 2 . . . a r of the set {1,2 . . . . . n}. Assume the given combination is not the l a r g e s t combination (n r + Z)...(nl ) n . *) O. Begin (* n e x t  c o m b i n a t i o n *) (* Find the f i r s t ai from r i g h t to l e f t f o r which ai ~ n  r + i . *) I . i _ 0 and il + i2 + ... + ik  n. The il elements of the subset $1 can be selected in ( n ) il ways. This leaves
n  il elements in S  $1. Therefore, the i2 elements of $2 can be selected in ( n  i l ) y S ' i 2 wa
(nil\
Similarly, the i3 elements of S3 can be selected in
13
t2] ways, and so on. The ik elements of Sk can be ch ;en in
/
( n  i l  i2  ' "
 ik1) ways
C~
by the multiplicati~
ciple, the n u m b e r of ways of choosing the mutually disjoint subsets $1, $ 2 , . . . , Sk is
(i )(nil
) ( ni i i .....
892
Appendix A
n! i l ! ( n  i l ) ! n! ( n  i l ) ! i 2 ! ( n  i l  i2)! ( n  i l . . . . i k ! ( n  i l . . . . . + ik) ik1) ik)!
( N o t e 9 n = i l + i2 + . . .
(
il!i2!...ik!
n
)
m
i1,i2,... , i k This concludes the proof.
Find the number of ways of dividing a set of size five into three mutually disjoint ordered subsets of sizes 2, 1, and 2. SOLUTION: By Theorem A.10, the number of ways of dividing the set in the desired way is given by the multinomial coefficient 5 ) 5! 2, 1,2 = 2!1!2! = 30 m
Find the number of w a y s of dividing the set S  {a, b, c } into three mutually disjoint ordered subsets $1, $2, and $3 of sizes 1, 1, and 1, respectively, and list them. SOLUTION: Number of possibilities = They are listed in Table A.4. 1, 1, 1 1!1!1!  6
Table A.4
$1 {a} {a} {b} {b} {c} {c}
$2 {b} {c} {a} {c} {a} {b}
S:~ {c} {b} {c} {a} {b} {a}
A n i m p o r t a n t o b s e r v a t i o n " The division { {a}, {b}, {c} } is different from the division { {a}, {c}, {b} }. However, they form the same partition, m
Exercises A.5
Evaluate each multinomial coefficient. 1.
(\2, 5 1 ) 2,
2. ( 1 2 ,
3. (2, 3,
72)
4 (085) 3,
A.5 The Multinomial Theorem
893
5.
(8)
2, 1, 3,2
6.
(10)(10)(lo)
1, 2, 3,4 7. 3, 1, 4,2 8.
1, 4, 4,1
Find the coefficient of each.
9. x2yz 2 in the expansion of (x + y + z) 5
10. xy2z 3 in the expansion of (x + y + z) 6
11. yz in the expansion of (x + y  z) 2
12. xy 2 in the expansion of (x  y  z) 3 13. xy2z 5 in the expansion of (x + y + z) s
14. xy3z 3 in the expansion of (x  y + 2z) 8
Expand each.
15. ( x + y  z ) 18. ( x  y 
2
z) 3
16. ( x  y  z )
2
17. ( x  y + z )
20.
3
19. (x + 2y + Z) 3
( x  2y + 2z) 3
Consider the various ways of dividing the set S = {a, b, c, d} into three mutually disjoint ordered subsets $1, $2, and $3 with sizes 1, 1, and 2, respectively. 21. Find the number of possibilities. 22. List the possibilities. Find the number of ways of dividing a set of size n into k mutually disjoint ordered subsets of sizes il, i 2 , . . . , ik in each case. 23. n = 10, three subsets with sizes 2, 3, and 5. 24. n = 10, four subsets with sizes 2, 2, 3, and 3. 25. n = 12, six subsets with sizes 1, 1, 2, 2, 3, and 3. 26. n = 15, four subsets with sizes 1, 1, 5, and 8. 27. Find the sum of the multinomial coefficients
( il, i 2 ,n. . . , ik ) "
It follows by Theorem A.10 that the n u m b e r of permutations of n items of which il are alike, i2 are alike, ..., and ik are alike is given by the . multinomial coefficient ( i l ni 2 ,). . . , ik , Using this fact, compute each.
28. The n u m b e r of ways of scrambling the letters of the word ABRACADABRA. 29. The number of ways of scrambling TINTINNABULATION. the letters of the word
30. The n u m b e r of binary words of length 10 and containing exactly three l's and seven 0's.
894
Appendix A
31. The n u m b e r of bytes c o n t a i n i n g exactly five 0's. 32. The n u m b e r of t e r n a r y words of length 12 and c o n t a i n i n g t h r e e 0's, four l's, and five 2's. Find the n u m b e r of t e r m s in the expansion of each. 33. (x + y + z) a 35. (w + x + y + Z) 12 Find the coefficient of each.
37.
34. (x + y + z) 10 36. (2w  3x + 4y  4 z ) 15
x2y2z 3 in xy3z in
the expansion of (x + y + z) 7
38.
the expansion of (x + 2y  z) 5
Using the m u l t i n o m i a l t h e o r e m , expand each.
39. ( x  2 y + z )
3
40. ( x + y  z )
4
Find the n u m b e r of t e r m s in the expansion of each.
41. ( x + y  z )
6
9
42. ( x  y  z )
44. ( w + x  y  z )
s 1~
43. ( 2 x + 3 y  4 z )
Find the n u m b e r of ways of dividing a set with size 15 into each. 45. F o u r m u t u a l l y disjoint subsets with sizes 7, 2, 3, and 3. 46. Five m u t u a l l y disjoint subsets with sizes 3, 6, 2, 1, and 3.
A B F A E Z H () I K A M
a /~ y 5 e r ~ O r K )~ tt
alpha beta gamma delta epsilon zeta eta theta iota kappa lambda mu
N
P
nu
E 0 FI P T T X s2
~ o rr p a r v r x ~ w
xi omicron pi rho sigma tau upsilon phi chi psi omega
A.7
Web Sites
895
The following Web sites provide valuable information for further exploration and enrichment. The status of a Web site could change with time, so it may not exist when you look for it; if it does not, use a search engine to locate a similar Web site. 1. Wilhelm Ackermann wwwgap.dcs.stand, ac.uk/~history/Mathematicians/ Ackermann.html 2. John Backus www.digitalcentury.com/encyclo/update/backus.html wwwgap.dcs.stand.ac.uk/~history/Mathematicians/Backus.html www.acm, org/awards/turing_citations/backus.html www.cs.nyu, edu/cs/faculty/shashaoutofmind/backus.html 3. George Boole wwwgap.dcs.stand.ac.uk/~history/Mathematicians/Boole.html www.digitalcentury.com/encyclo/update/boole.html homepages.enterprise.net/rogerp/george/boole.html 4. Georg Cantor wwwgroups.dcs.stand.ac.uk/~ history/Mathematicians/Cantor, html www.treasuretroves.com/bios/CantorGeorg.html www.aug.edu/dvskel/JohnsonSU97.html 5. Arthur Cayley wwwgap.dcs.stand.ac.uk/~history/Mathematicians/Cayley.html scienceworld.wolfram.com/biography/Cayley.html www.stetson.edu/~efriedma/periodictable/html/C.html www.optisyn.com/presentations/Cayley.html www.math.ukans.edu/~engheta/bio/cayley.html www.geometry.net/Biographer/Cayley.html 6. Edsger Dijkstra www.digidome.nl/edgser_wybe_dijkstra.html www.acm.org/classics/oct95/ www.cs.utexas/users/EWD/obituary.html news.com.com/21001001949023, html 7. Leonhard Euler
www.maths.tcd.ie/pub/HistMath/People/Euler/RouseBall/
RB Euler.html
www.shu.edu/html/teaching/math/reals/history/euler.html
8. Pierre de Fermat www.maths.tcd.ie/pub/HistMath/People/Fermat/RouseBall/ RB Fermat.html 9. Karl F. Gauss wwwgap.dcs.stand.ac.uk/~history/Mathematicians/Gauss.html www.english.upenn.edu/~jlynch/FrankenDemo/People/gauss.html scienceworld.wolfram.com/biography/Gauss.html www.indiana.edu/~intell/gauss.html www.brown.edu/Students/OHJC/hm4/gauss.html
896
Appendix A 10. Christian Goldbach
11.
12.
13.
14.
15.
16.
17.
18.
19.
wwwgap.dcs.stand.ac.uk/~history/Mathematicians/Goldbach.html www.geocities.com/Heartland/Hills/7972/mathgoldbach.html www.wikipedia.org/wiki/Christian_Goldbach plus. maths, org/issue 11/news/Goldbach Sir William R. Hamilton www.chembio.uoguelph.ca/educmat/chm386/rudiment/tourclas/ hamilton.html scienceworld.wolfram.com/biography/HamiltonWilliamRowan.html www.iaste.com/hall of fame/hamilton.html Maurice Karnaugh www.informatik.unitrier.de/~ley/db/indices/atree/k/Karnaugh" Maurice.html www.maxmon.com/library.html wwwcse, st an ford. edu/classes/cs 103 a/h 9BooleanAlgebra. pdf Alfred B. Kempe wwwgap.dcs.stand.ac.uk/~history/Mathematicians/Kempe.html mappa.mundi.net/locus_014 www.uwinnipeg.ca/~ooellerm/guthrie/FourColor.html www.mathsyear2000.org/explorer/morphing/13usedownload.shtml Stephen C. Kleene www.library.wisc.edu/libraries/Math/kleen.html wwwgroups.dcs.stand.ac.uk/~history/Mathematicians/Kleene.html www.dcs.ed.ac.uk/homes/als/lics/newsletters/19.html www.student.math.uwaterloo.ca/~cs462/Hall/kleene.html Donald E. Knuth sunburn.stanford.edu/~knuth wwwcsstaff.stanford.edu/~knuth/index.html www.digitalcentury.com/encyclo/update/knuth.html laurel.actlab.utexas.edu/~cynbe/muq/muf3_20.html Kazimierz Kuratowski wwwgroups.dcs.stand.ac.uk/~history/Mathematiticains/ Kuratowski.html www.stetson.edu/~efriedma/periodictable/html/Kr.html Gabriel Lame wwwgroups.dcs.stand.ac.uk/~history/Mathematicians/Lame.html www.bath.ac.uk/~ma0dmp/Lamelife.html Edmund Landau www.ma.huji.ac.il/~landau/landuniv.html wwwgroups, dcs. stand, ac. uk/~ hi st ory/M athem atician s/Landau, ht m 1 www. ard. huj i. ac.il/publications/25years/chap 13. html PierreSimon Laplace wwwgroups.dcs.stand.ac.uk/~history/Mathematicians/Laplace.html www.maths.tcd.ie/pub/HistMath/People/Laplace/RouseBall/ RB_Laplace.html www. stetson.edu/~efriedma/periodictable/html/La.html
A.7 Web Sites
897
20. Gottfried W. Leibniz
www.maths.tcd.ie/pub/HistMath/People/Leibniz/RouseBall/
RB Leibniz.html
21. J a n Lukasiewicz
22.
23.
24.
25.
26.
27.
28.
29.
30.
wwwgroup s. dcs. s tand, ac. u k / ~ history/Mat hematiticain s/ Lukawiewicz.html www. h p m u s e u m . o r g / r p n . h t m l www.wikipedia.com/wiki/Jan_Lukawiewicz George H. Mealy www.informatik.unitrier.de/~ley/db/indices/atree/m/Mealy: George_H = .html Marin Mersenne wwwgroups, dcs. stand.ac.uk/~history/Mathematiticains/ Mersenne.html www2.andrews.edu/~calkins/math/biograph/biomerse.html Blaise Pascal members.aol.com/KatherenaE/private/Philo/Pascal/pascal.html www. m at hs. tcd. ie/pub/HistM at h/P eo ple/P ascal/Rouse Ball/ RB Pascal.html www. cs. was h i ngt o n. edu/h o m e s/j b ae r/clas se s/b laise/b lai se. h tml www.math, sfu.ca/histmath/Europe/17thCenturyAD/Balise.html wwwgroups.dcs.stand.ac.uk/~history/Mathematiticains/ Pascal.html Bertrand Russell www. m cm aster, c a/ru s sd o c s/ru sse ll. h tml plato.stanford.edu/entries/russell desktop 12.cis.macmaster.cal/~bertrand Claude E. S h a n n o n wwwgroups.dcs.stand.ac.uk/~history/Mathematiticains/ Shannon.html www.belllabs.com/news/2001/february/26/1.html www.digitalcentury.com/encyclo/update/shannon.html Alan M. Turing www.turing.org.uk/turing e i. cs. vt. edu/~ histo ry/Tu ring. h tml wwwgroups.dcs.stand.ac.uk/~history/Mathematiticains/ Turing.html AlexandreTheophile Vandermonde wwwgroups.dcs.stand.ac.uk/~history/Mathematiticains/ Vandermonde.html J o h n Venn wwwgroups.dcs, stand, ac.uk/~history/Mathematicians/Venn.html userwww.sfsu.edu/~rsauzier/Venn.html Boolean Algebra educ.queensu.ca/~compsci/units/BoolLogic/titlepage.html www.bit.umkc.edu/vu/course/cs281/lectures/booleanalgebra/ booleanalgebra.html
898
Appendix A
32.
33.
34.
35.
36.
37.
ever.phys.ualberta, ca/~ gingrich/phys395/notes/node 121.html www. maxmon, com/1847ad, h t m l www.yale, edu/ynhti/curri culu m/u nits/ 1989/7/89.07.07.x.h tml FiniteState Machines www.cs.brown.edu/people/jes/book/BOOK/nodel0.html classwww.gsfc.nasa.gov/CAGESite/pages/cage_gpf_fsm.html www. cs. ariz ona. edu/classes/cs352/summer 02/fsa. ht m 1 www.c3.1anl.gov/megamath/workbk/machine/mabkgd.html www.beigarath.demon, co.uk/j ava/fsme.html members.aol.com/asakharov/fsm.html Fuzzy Sets www.answermath.com/fuz z y m a t h . h t m l sun 16. cecs. missouri, edu/index, html news:comp.ai.fuzzy G e n e r a t i n g Functions www.cs.wpi.edu/~cs504/s00m/notes/ln/1999/class06/class06.html www. cs. wp i. edu/~ cs504/s00m/classes/clas s05/Class 05. h tml msl. cs.uiuc, edu/~ lavalle/cs576/proj ects/wmchan www. eco. rug. nl/gau ss/GAUSS00/mho narc. db www. m a t h s. surrey, ac. u k/per sonal/st/d, fi sher/M S 103/MS 1055. mws Graph Theory www. u t m. e du/de p a r t m e n t s/m at h/gr ap h www.c3.1anl.gov/megamath/workbk/graph/graph, html www groups, dcs. st and. ac. u k / ~ histo ry/M athe mat ician s/E rdo s. h tml www.nada.kth.se/~viggo/problemlist/compendium.html www. shodor.org/interactive/lessions/frac 1.html www.scism.sbu.ac.uk/law/Section3/chapter3/s3c3int.html www. cs. cm u. edu/~ cbu rch/survey/recur se www.nd.edu/~cholak/computability/computability, html Hilbert's Paradoxes www.wordsmith.demon.co.uk/paradoxes eluzions.com/Puzzles/Logic/Paradoxes.shtml www.c3.1anl.gov/megamath/workbk/infinity/infinity.html www. cs. tpu. ee/~j aagu p/uk/fm m/m at h/1.2.5.3.h tml The FourColor Problem www.math.gatech.edu/~thomas/FC/fourcolor.html www.cs.uidaho.edu/~casey931/megamath/gloss/math/4ct.html wwwgroups.dcs, stand.ac.uk/~ history/HistTopics/The_four_colour theorem.html
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904
References
110. G.L. Ritter et al., "An Aid to the Superstitious," Mathematics Teacher, Vol. 70 (May 1977), pp. 456457. 111. S. Sahni, Concepts in Discrete Mathematics, 2nd ed., Camelot, Fridley, MN, 1985. 112. B. J. Schwartz, Solution to Problem 1014, J. Recreational Mathematics, Vol. 14:4 (19811982), p. 309. 113. J. Sedlacek, "On the Skeletons of a Graph or Digraph," Proceedings of the Calgary International Conference of Combinatorial Structures and their Applications, Gordon & Breach, New York, pp. 387391. 114. D. E. Shasha, Out of their Minds: The Lives and Discoveries of 15 Great Computer Scientists, Copernicus, New York, 1995, pp. 89101. 115. D.R. Sherbert, "Difference Equations with Applications," UMAP Module 322, Arlington, MA, 1980. 116. R. M. Smullyan, What is the name of this book?, PrenticeHall, Englewood Cliffs, NJ, 1978. 117. R. M. Smullyan, Alice in PuzzleLand: A Carrollian Tale for Children Under Eighty, Penguin Books, New York, 1982. 118. R. M. Smullyan, "Leaps of Logic," Discover (March 1993), p. 96. 119. S.K. Stein, "The Mathematician as an Explorer,"Scientific American, Vol. 204 (May 1961), pp. 149158. 120. S. K. Stein, Mathematics: The ManMade Universe, W. H. Freeman, San Francisco, CA, 1969. 121. A. Sterrett, "Gambling Doesn't Pay," Mathematics Teacher, Vol. 60 (March 1967), pp. 210214. 122. P. Stevens, Patterns in Nature, Little Brown, Boston, 1974. 123. D. R. Stone, "A Different Prime Proof," Mathematics Teacher, Vol. 83 (Jan. 1990), p. 63. 124. A. S. Tanenbaum, Structured Computer Organization, PrenticeHall, Englewood Cliffs, NJ, 1976, pp. 420423. 125. The Official LSAT PrepBook, Law Services, Newtown, PA, 1991. 126. R. M. Thrall, "Insulin Requirements as a Linear Process in Time," Some Mathematical Models in Biology (R. F. Baum, ed.), The University of Michigan Press, Ann Arbor, MI, 1967, pp. 0L2.10L2.4. 127. P. M. Tuchinsky, "International Standard Book Numbers," The UMAP Journal, Vol. 6:1 (1985), pp. 4153. 128. A. Tucker, Applied Combinatorics, Wiley, New York, NY, 1980. 129. T. Tymoczko, "Computers, Proofs, and Mathematics: A Philosophical Investigation of the FourColor Proof," Mathematics Magazine, Vol. 53 (May 1980), pp. 131138. 130. J. H. van Lint and R. M. Wilson, A Course in Combinatorics, Cambridge University Press, New York, 1992. 131. M. vos Savant, Ask Marilyn, St. Martin Press, New York, p. 228. 132. S. Warshall, "A Theorem on Boolean Matrices," J. of the Association of Computing Machinery, Vol. 9 (1962), pp. 1112. 133. J. Williams, "Graph Coloring Used to Model Traffic Lights," Mathematics Teacher, Vol. 85 (March 1992), pp. 212214.
References
905
134. R. J. Wilson and J. J. Watkins, Graphs: An Introductory Approach, Wiley, New York, 1990. 135. D. Wood, "Towers of Brahma and Hanoi Revisited," J. Recreational Mathematics, Vol. 14:1 (19811982), pp. 1724. 136. R. V. Young (ed.), Notable Mathematicians, Gale Research, Detroit, MI, 1997.
Solutions
to OddNumbered Exercises
Exercises 1.1 (p. 17)
1. yes 3. no 11. ~ p v q 5. F 13. ~ p A ( q v r ) 7. T 15. T
9. 1 + 1 ~ = 0
17. F
19. T
p T F F q F T F ~p F F T T ~q F T F T ~p v ~q F T T T
21. F 25.
p q TT TF FT FF pvq T T T F ~q ( p v q ) v ~ q F T F T T T T T
23.
TT
27. F
29. T
31.
p T T F F
q T F T F
p XOR q F T T F
33. If two lines are p e r p e n d i c u l a r to the same line, t h e n they are parallel. 35. If x = 1, t h e n
X 2 ~
1.
907
908
Solutions to OddNumbered Exercises
37. converse: If Paris is in England, t h e n London is in France. inverse: If London is not in France, t h e n Paris is not in England. contrapositive: If Paris is not in England, t h e n London is not in France.
39. ~q vp ~ r 41. ~p A ~ q o ~r 43. no
45. yes
51. T 55.
p q pAq ~p
47. T
53. T
pAq* ~'p
49. F
T T F F
57.
P
T F T F
T F F F
F F T T
F T T T
q
pvq
pAq
pVq~pAq
TT
T F F
59. yes 67. (p A q) ~
v
F T F
T T T F
T F F F
T F F T
61. yes
( ( ~ p) v ( ~ q))
63. yes
69. (p ~ q) o
65. no
((~p) v q)
71. (A' A C)
73.
(B'
v
C')
v
(A A B)
75.


i
i
@ @
@ 9
77. Ellen m u s t use computer 1.
Exercises 1.2 (p. 29)
1. F
3.
P
,,~p F T
~ (~p) T F
0
P
pvp
T F
T F
T F
L identical J"
t
~j
J~
=r
~J b~ < ><
Q~
< < > b~
Q~ m
i~~
r
~D D.io
L~ >
IL
ii~
e~
L~
L~
>
910
Solutions to OddNumbered Exercises
27.
P
q
p NOR q
29. T 37. T
31. T 39. T 43. ~ ( u p v q )
33. T 41. F u(up)
 pAuq
35. F
T T F F
T F T F
F F F T
 ~pv~(~q) _= ~ p v q
A~q
45. ~ ( p A u q ) 49. p Aq 55. (A
A
47. p  +
~q upv~q ~'(pAq) 53. pA u q
51. p v q B')
v
[(A
A
B)
v
CI  A
v
C
@
57. 59.
P
q
P ^ q
Plq
(P[q)l(Plq) T F F F t
T F
F T
F T
1"
t
T T F F
T F T F
T F F F 1"
F T T T identical
qlq
61.
P
T T F F
q
T F T F
P * q T F T T t
PIP F F T T
(Pho)l(PlP) T T F F identical
((P[P)I(P[P))I(qlq)
F T F T
T F T T t
63. p X O R q  (((P[P)l(qlq))l(((Plq)l(Plq))l((Plq)l(Plq))))l (((p~)l(qlq))l(((plq)l(plq))l((plq)l(plq)))) 65. 77. 1 67. 1 69. 0.3 71. 0.7 73. 0 75. 0.5
/ 1x x
/
Exercises 1.3 (p. 36)
if 0_ 0) 23. (Vx)(3y)(xy = x)
25. F
27. T
29. The s q u a r e of every i n t e g e r is nonnegative. 31. T h e r e are i n t e g e r s x a n d y such t h a t x § y  7. 33. T h e r e is an i n t e g e r x such t h a t y  x  y for every i n t e g e r y.
35. T 49. T 63. T
37. T 51. T 65. F
39. T 53. T 67. T
41. T 55. F
43. F 57. F
45. T 59. T
47. T 61. T
Exercises 1.4 (p. 45)
1. p ~q ~q
o
p T F F
q F T F
p vq T T T F
p   . (p v q ) T T T T 19. Carol is a baby.
5. invalid 7. valid 9. valid 13. q is false.
T T
11. valid 15. r is true. 17. The p r o g r a m is r u n n i n g . 21. B e n j a m i n to Cindy a n d Aaron to Daphne; t h e y are 34, 27, 28, and 29 y e a r s old, respectively.
23. blue
25. B
27. A is a k n i g h t and B a knave. 31. k n i g h t 35. Kitty is guilty. 29. yes 33. "I a m red." 37. At least one cowgirl will escape injury.
Exercises 1.5 (p. 54)
1. yes
3. yes
912
Solutions to OddNumbered Exercises
5. L e t x a n d y b e a n y t w o e v e n i n t e gers. Thenx  2m andy 2 n f o r s o m e i n t e g e r s m a n d n. Then x + y  2m + 2n = 2 ( m + n) w h i c h is a l s o a n e v e n i n t e g e r . 11. L e t x b e a n y e v e n i n t e g e r a n d y any odd integer. Then x 2m a n d y  2n + 1 for s o m e i n t e g e r s m a n d n. T h e n x y = ( 2 m ) ( 2 n + 1)  2 ( 2 r a n + m ) , a n even integer. 15. L e t x b e a n y i n t e g e r . A s s u m e it is n o t e v e n ; it is o d d a n d is o f t h e f o r m 2 m + 1. T h e n x 2  ( 2 m + 1) 2  4 m 2 + 4 m + 1 = 2(2m 2 + 2m) + 1 w h i c h is a n o d d i n t e g e r . So t h e g i v e n h y p o t h e s i s is f a l s e a n d t h e r e s u l t follows. 19. S u p p o s e ~/2 is n o t a n i r r a t i o n a l n u m b e r ; t h a t is, , / 2 is a r a t i o n a l n u m b e r . L e t ~/2  a / b , w h e r e a and b have no positive common f a c t o r s e x c e p t 1. T h e n ( a / b ) 2 = 2 o r a 2  2b 2. .'. 2 is a fact o r o f a 2 a n d h e n c e o f a. T h e n a  2 m for s o m e i n t e g e r m. .'. ( 2 m ) 2 _ 2b 2 o r b 2  2 m 2 . .'. 2 is a f a c t o r o f b 2 a n d h e n c e o f b. C o n s e q u e n t l y , 2 is a f a c t o r o f b o t h a a n d b, w h i c h c o n t r a d i c t s the assumption.
7. L e t x  2 m b e a n y e v e n i n t e g e r . T h e n X 2   ( 2 m ) 2  2 ( 2 m 2) is also an even integer. 9. L e t x b e a n y o d d i n t e g e r . T h e n x 2m + 1 for some integer m. .. x 2  ( 2 m + 1)2 = 2(2m 2 + 2m) + 1 w h i c h is a n o d d i n t e g e r . 13. L e t x = 4k + 1. T h e n x 2 = (4k + 1) 2  16k 2 + 8k + 1 = 4 ( 4 k 2 + 2 k ) + 1  4 m + 1, w h e r e m  4k 2 + 2k is a n e v e n integer. .'. x 2 is a l s o a n i n t e g e r o f t h e same form. Let x and y be any two integers. Assume that the given conclus i o n is false; t h a t is, a s s u m e that both x and y are odd integers. Then x  2m + 1 and y  2n + 1 for s o m e i n t e g e r s m andn.." xy = (2m + l)(2n + l) = 2(2ran+re+n)+ 1, a n o d d integer. This negates the given h y p o t h e s i s a n d so t h e r e s u l t follows. A s s u m e ~/p is a r a t i o n a l n u m b e r a/b, w h e r e a a n d b h a v e no positive common factors except 1. T h e n v / p  a / b , ( a / b ) 2  p o r a 2 = p b 2. C o n s e q u e n t l y , p is a f a c t o r o f a 2 a n d h e n c e o f a. So let a  mp. Then (mp) 2 pb 2 o r b 2  p m 2. A s b e f o r e , t h i s s h o w s p is a f a c t o r o f b. T h u s p is a c o m m o n f a c t o r o f a a n d b, a contradiction.
17.
21.
25. p r o o f : E v e r y i n t e g e r n is o f t h e
f o r m 3k, 3k + 1, o r 3k + 2. c a s e 1 L e t n  3k. T h e n n 3  n  (3k) 3  (3k) = 27k33k  3(9k3_k) c a s e 2 L e t n  3k + 1. T h e n n 3  n  (3k + 1) 3  (3k + 1) = (27k 3 + 2 7 k 2 + 9 k + 1)  ( 3 k + 1) = 2 7 k 3 + 2 7 k 2 + 6k = 3 ( 9 k 3 + 9k 2 + 2k)
23.
proof:
c a s e I n is a n e v e n i n t e g e r , s a y , 2m. Thenn2+n = (2m)2+2m 2(2m 2 + m), an even integer. c a s e 2 n is a n o d d i n t e g e r , s a y , 2m+1.
Chapter 1
The Language of Logic
913
c a s e 3 L e t n = 3k + 2. T h e n n 3  n  (3k + 2) 3  (3k + 2) = (27k 3 + 5 4 k 2 + 3 6 k + 8)  ( 3 k + 2) = 27k 3 + 54k 2 + 33k + 6 = 3 ( 9 k 3 + 18k 2 + l l k + 2) T h u s , i n e v e r y c a s e , n 3  n is d i v i s i b l e b y 3. 29.
proof: Choose x x 2  x. 1. C l e a r l y ,
Then n 2 + n  ( 2 m + 1) 2 + ( 2 m + 1) = ( 4 m 2 + 4 m + 1) + ( 2 m + 1) = 2 ( 2 m 2 + 3 m + 1), again, an even integer.
27.
proof: c a s e 1 L e t x , y > O. T h e n I x . y l = x y = Ixl. [yl c a s e 2 L e t x < 0, y < 0. T h e n I x [   x a n d [y[ =  y . Ix.y[  x y  (  x ) . (  y )  Ix 9lYl c a s e 3 L e t x >__ 0, y < 0. T h e n ]x . y [   ( x y ) = x. (y)  JxJ . [y] c a s e 4 L e t x < 0, y >_ 0. T h i s c a s e is s i m i l a r t o c a s e 3.
31.
proof: Let a be any nonzero integer. T h e n 1729a 3  a 3 + (12a)3 = (9a) 3 + ( 1 0 a ) 3 S i n c e a is a r b i t r a r y , 1 7 2 9 a 3 h a s infinitely many choices. 41.
33.
0
35.
x 2 
2
4.
3 7 . x  2, P(x)" 39.
C a n c e l i n g a  b is i n v a l i d s i n c e ab. p r o o f : L e t a 9 b  0 a n d a r 0. Sincea.0  0, a . b  a.0. Canceling a from both sides, we g e t b  0. 45.
p r o o f : S i n c e a < b, t h e r e is a positive real number x such t h a t a + x  b. .. (a + x ) + c = b + c T h a t is, (a + c) + x  b + c ..a+c
(w+y
_ y, t h e n x + z _>_y + z. c o n t r a p o s i t i v e : I f x + z >_ y + z, t h e n x >__y. 33. I f x _< 3 a n d x >_  3 , t h e n Ixl _< 3. 37. no 39. yes
35. yes
41. y e s
43.
T
45. T
47. F
49. ( A ' A B ) v ( A ' A B ' ) v ( A A B ' ) = A ' v B '
i
51. invalid
@
53. F 55. T 57. F
59. F
61. F
63. p r o o f : L e t x a n d x + I b e
6 5 . p r o o f : n 4  n 2  l ( n  1 ) n ( n + 1)In contains a product of three two consecutive integers. consecutive integers. Therefore, c a s e 1 L e t x be e v e n . T h e n b y c a s e s , it is d i v i s i b l e b y 3. x  2 m for s o m e i n t e g e r m. T h e n x ( x + 1) = 2 m ( 2 m + 1) 67. i n d i r e c t p r o o f : A s s u m e t h a t t h e = 2 1 m ( m + 1)1, g i v e n c o n c l u s i o n is false, t h a t is, an even integer. a < 6 a n d b < 6. T h e n a + b < c a s e 2 L e t x b e odd. T h e n 6 + 6, t h a t is, a + b < 12, w h i c h x = 2 m + 1 for s o m e i n t e g e r m . contradicts the hypothesis. .'. x ( x + 1)  ( 2 m + 1 ) ( 2 m + 2) 69. p r o o f : A s s u m e t h a t a 2  b 2 a n d = 2 [ ( m + 1 ) ( 2 m + 1)i, a ~: b. S i n c e a 2  b 2 = (a  b) again an even integer. (a + b ) = 0 a n d a  b r O, a + b = O. .. a =  b .
71. 0.5
73. 0.5
75. 0
77. 0.3
79. S i n c e t ( p )  x, t ( p ' ) = 1  x. T h e n t ( p v p ' ) = m a x { x , 1  x}.
S u p p o s e t ( p v p ' ) = 1. T h e n m a x { x , 1  x}  1. S u p p o s e 0 < x < 1/2. T h e n max{x, 1  x } = 1  x  1, so x  0; t h a t is, t ( p )  O. O n t h e o t h e r h a n d , let 1/2 < x < 1. T h e n m a x I x , 1  x } = x  1; so t ( p ) = 1. I n e i t h e r case, t ( p ) = 0 or 1. C o n v e r s e l y , let t ( p )  0 o r 1. T h e n t ( p v p ' ) = m a x { t ( p ) , t ( p ' ) } m a x { t ( p ) , 1  t ( p ) } . I f t ( p ) = O, t h e n t ( p v p ' ) = m a x { 0 , 1} = 1; o n t h e o t h e r h a n d , if t ( p )  1, t h e n a l s o t ( p v p ' ) = m a x { l , 0}  1. I n b o t h cases, t ( p v p ' ) = 1. T h u s t ( p v p ' )  1 if a n d o n l y if t ( p )  0 o r t ( p )  1.
Chapter I
The Language of Logic
915
81. proof:
t ( ( p v q ) ' )  1  t ( p v q) = 1  max{t(p),t(q)} = 1  max{x,y}
1
/y
i f x __a o r x _<  a . c o n t r a p o s i t i v e " I f (x _<  a ) o r (x >_ a), t h e n Ixl >__a.
3. ~ q
9. ~ [(Vx)(Vy)(xyyx)
5.
~pv~q
7. y e s
=_ ( R x ) ( 3 y ) ( x y # y x )
11.
~ [ ( V x ) ( 3 y ) ( 3 z ) ( x + y  z) =_ ( 3 x ) ( V y ) ( V z ) ( x + y r z)
13. proof: L e t K = (n 2 9 n ) ( n 2 9 n + 1)(n 2 + n + 2)
and L  (n 2  n ) ( n 2  n + 1)(n 2  n + 2). 2 n ( 3 n 4 + 7n 2 + 2)  6 n 5 + 1 4 n 3 + 4 n = (4n 3 + 6 n ) n 2 + (2n 4 + 8 n 2 + 4 ) n = (n 2 + n ) ( n 4 + 2 n 3 + 4 n 2 + 3 n + 2)  (n 2  n ) ( n 4  2 n 3 + 4 n 2  3 n + 2)
=KL S i n c e K is t h e p r o d u c t o f t h r e e c o n s e c u t i v e i n t e g e r s , it is d i v i s i b l e b y 3. B u t n 2 + n  n ( n + 1) is d i v i s i b l e b y 2. So n 2 + n + 2 is d i v i s i b l e b y 4. T h u s K is d i v i s i b l e b y 24. L i k e w i s e , L is a l s o d i v i s i b l e b y 24. T h e r e f o r e , K  L is d i v i s i b l e b y 24. T h u s n ( 3 n 4 + 7n 2 + 2) is d i v i s i b l e b y 12.
15. 40
17. t ( p v p ' ) =
max{t(p),t(p')} = m a x { t ( p ) , 1  t(p)} o n l y if t ( p )  0 o r t ( p )  1.
1
19. N o t a t h r e e  v a l u e d t a u t o l o g y . 21. Is a t h r e e  v a l u e d t a u t o l o g y .
916
Solutions to OddNumbered Exercises
23.
p v q 0 u 1 u
u
(p v q)' 1 u 0 u
u
p' 1 1 1 u
u
q' 1 u 0 1
u
p ' ^ q' 1 u 0 u
u
1 1 1 1
0 0 0 0
u 0 0 0
0 1 u 0
0 0 0 0
1"
identical
t
Exercises 2.1 (p. 76) 1. {April, August} 5. { x E Z l 0 < x < 5 } 9. yes
15. T 17. F
3. {Jan, March, May, July, Aug, Oct, Dec}
7. {xlx is a member of the United Nations}
11. yes
19. F 21. T 23. T
13. F 25. T 27. T
29. F 35. {3,6,9}
31. {O,A} 37. {2,3,4,6,8,9} 41. b, b2 bab, b3,ba2b 49. 00,01,10,11
33. 2 n
39. {0,{1},{2},{3},{1,3}} 45. 3 47. 5
43. ~, b, a 2, a2b, aba 51. 0, 1, 2
53. Consider the implication x e 0 ~ x e A. Since the hypothesis is false, this is a true implication. ". 0 _c A. Exercises 2.2 (p. 93) 1. {a,b,c,g, j,k} 7. { a , c , d , e , f , h , i , j,k} 13. {w,y, z} 3. {d, h}
9. {a}
55. Let x be an arbitrary element in A. Since A _ B, x e B. Since B C, x e C. Thus, every element in A is also in C..'. A c C.
m
5. { a , b , c , f , g , i , j,k} 11. {a, b, g} 17. {x,y}
15. {a, b}
Chapter 2
The Language of Sets
917
19. {(b, x), (c, x) }
21. 0
23. {(b, x), (b, z), (c, x), (c, z) }
27. T 35. F 41. A 45. no 49. A N (A u B)  (A N A) u (A N B)
=Au(ANB)
25. {(b,x,x), (b,x,z), (c,x,x), (c,x,z)} 33. T
39. A = { a } , B  O , C  { a }
29. F 37. F
31. F
{a},B = {b}, C  {a,b}
43. no 47. Let x e (A')'. Then x r A', so x e A. Thus (A')' _c A. Conversely, let x e A. Then x r A', so x e (A')'. .'. A c_ (A')'. Thus (A')' = A.
51. A o A = ( A  A ) u ( A  A ) =OuO=O
 A, since ANB _c A.
53. A ~ B = (A  B ) u (B  A ) = (B  A) u (A  B)
=B~A
55. (A u B u C)' = = =
[(A u B) u C]' (A u B)' N C' (A' n B') N C' A' n B' n C'
57. A N ( A  B )  A  B 59. ( A  B')  ( B  A') = 0 63. (A N B)' U (A u B')  U 67. A u
61. A u B  ( A N B ) '  A N B 65. (A' u B')' u (A' N B) = B
(N
ieI ieI
Bi) =
N(A u Bi)
ieI ieI
n N (U Bi) = U (n n Bi)
69. 0
71. {Angelo 0.4, Bart 0.7, Cathy 0.6, Dan 0.7, Elsie 0.2, F r a n k 0.6} 75. {Angelo 0.4, Bart 0.7, Cathy 0.6}
73. 0
77. {(Angelo, Dan) 0.3, (Angelo, Elsie) 0.4, (Angelo, Frank) 0.4, (Bart, Dan) 0.3, (Bart, Elsie) 0.7, (Bart, Frank) 0.4, (Cathy, Dan) 0.3, (Cathy, Elsie) 0.6, (Cathy, Frank) 0.4} 79. {(Angelo, Angelo) 0.4, (Angelo, Bart) 0.4, (Angelo, Cathy) 0.4, (Bart, Angelo) 0.4, (Bart, Bart) 0.7, (Bart, Cathy) 0.6, (Cathy, Angelo) 0.4, (Cathy, Bart) 0.6, (Cathy, Cathy) 0.6} 81. p r o o f : By De Morgan's law in ordinary sets, it suffices to show t h a t d(AnB)'(x)  dA'uB'(X) for every element x.
918
Solutions to OddNumbered Exercises
Let x e (A N B)'. Then d(AnB),(x)  1  d(AnB)(X) = 1  min {dA(x), dB(x)} I 1  dA(x) if clA (x) ~
~j~
+
~ ~ "
~c~ ~.~ 0"~ I I I
~
~
9
+
+
+
+
I
I
~ ~
~ ~
%~" C
+
,, ~ =.al
,
II
~
~
i
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I
~.~.:
~
~.
.~
"~:
II I
' I ~:~. L'~
I ~I~ ~i
'~c~
I
I
L~ ' ~
+
+
II L'~
,
I
u
i
~
~
i
'
' I
I
I
I
p..,
I
I
+
+
9
n
Ib~
930
Solutions to OddNumbered Exercises
57. p r o o f : Since B is an inverse of A , A B  I  B A . Likewise, A C  I = CA. Then B  BI  B(AC)  (BA)C  IC  C 59. p r o o f : (AB)(B1A 1)  A [ B ( B  1 A  1 ) ]
= A[(BB1)A = A(IA 1) 1]
=AA1 / Similarly, ( B  1 A  1 ) ( A B ) .. (AB) 1  B  1 A  1
 I.
Review Exercises (p. 177)
1. 558 9. W e d n e s d a y 17. no 3. 761 11. M o n d a y 19. yes 27. 0 A 33. CBS1AA M NCR4SK 35. AQZ5CD C B S I A A C N N I T K 6 N C R 4 S K ABC5ZZ 37. Use m = 19, n = 6. 39.  2 7 V P S 3 S L N B C 4 G K CBA3BA CBA3BA N CNNITK AQX5CD O C D 5. T u e s d a y 13. April 1 7. J u l y 15. M a r c h 28
21. no
29. 5540 K
23. 8
31. 5
25. 0
.
o
.
NBC4GK X Y
VPS3SL
ABC5ZZ
41.  3 9 47. F 53. 3.4, 4.5 59. 1, 1,2, 3
43. 1 49. F 55. 2, 3.5 61. 25
45. 36 51. 3.4, 4.6 57. 1, 1, 2, 3 63. 72
65. p r o o f : Let bl, b2 ~ B such t h a t g ( b l )  g(b2). T h e n f ( g ( b l ) )  f(g(b2)); t h a t is, (f o g)(bl)  (f o g)(b2). T h u s 1B(bl)  1B(b2); t h a t is, bl  b2. .'. g is injective.
Chapter4 Supplementary Exercises (p. 179)
Inductionand Algorithms
931
1. (A + B) T  (aij + bij) T
 (aji + bji)  (aft) + ( b j i ) = A T + B T
3. (ANT) T  (AT)TAT
: AA T, since (AT) T  A. So A n w is symmetric.
7. 63.
5. p r o o f (by contradiction)Suppose each of t h e n u m b e r s is less t h a n t h e i r a v e r a g e A. Then n
~ai l
Sn
ak
 a +
(k 
1)d,
i=1 ( 1 ~_~
n i=1 a contradiction.
= n ai
)
n
= ~
i=1
ai,
15. L e t S n  a + a r + . . . + a r
rSn  ar + ar 2 +... a r n  1 _jr_a r n .'. r S n  S n = a r n  a a ( r n  1) Sn = (r # 1) +
n1
n Z ak k=l n = ~[a+(k1)d k=l n n k=l k=l ( n  1)n
2
= ~a+d~(k1) =na+d. n
= ~ [ 2 a + (n  1)d]
r1
17. Let X = { a , b , c , d , e , g } , A = { a , b , c } , B = {b,c,d,e}, a n d Y = {0,3,5}. Let f : X ~ Y defined by f ( a ) = 0, f(b) = f(c) = 3, f(d) = f(e) = f(g) = 5. T h e n A n B = {b, c}, f ( A ) = {0, 3}, f ( B ) = {3, 5}, f ( A n B ) = {3} = f ( A ) ~ f ( B ) . N o n e t h e l e s s , f is not injective.
19. proof: Let a l . . . . .
T h e n ~ ai  n i=1
an. n + 1
OG
21.
an 
+ 1 = 2n
i=n+l
~
ai
oo
=n+l So, by Exercise 18, if n + 1 pigeons occupy n pigeonholes, one pigeonhole m u s t c o n t a i n at least two pigeonholes.
= an+l +
~
ai
i=n+2 = an+l + an+l = 2an+l 1 an+l = ~an
Exercises 4.1 (p. 188)
1. yes 3. 1 2 ; 5 5.  3 ; 5
932
Solutions to OddNumbered Exercises
7. {0,1} 11. p r o o f (by c o n t r a d i c t i o n ) : A s s u m e t h e r e is a n i n t e g e r n s u c h t h a t 0 < n < 1. L e t S {ai0 < a < 1}. Since n ~ S, S r D. T h e r e f o r e , by t h e W O P , S h a s a least e l e m e n t e, w h e r e 0 < e < 1..'. 0 < ~2 < e a n d h e n c e 0 < ~2 < 1. So e 2 E S , w h e r e e 2 < ~, a c o n t r a d i c t i o n . 9. { 0 , 1 , 2 , 3 , 4 , 5 , 6 } 13. p r o o f (by c o n t r a d i c t i o n ) : A s s u m e t h e r e is a n i n t e g e r n ~ ~ S such t h a t n ~ < no + t~*  1. T h e n no + t~*  1 < t~*. Since n '  n o + 1 ~ S* a n d n ~n0+l 6, which is impossible. Exercises 4.3 (p. 205) 1. 13 7. 3360eight 13. 1Dsixteen 3. 1022 9. 15eight 15. 75sixtee n 21. ll010two 5. 10000110100two 11. 72eight 17. 110110two
43. Let n a n d n + 1 be a n y two consecutive integers. Clearly, one of t h e m m u s t be even. Suppose it is n. T h e n 2in. B u t n is a prime, so n  2 a n d h e n c e n + 1 = 3. T h e case n § 1 is even yields n  1, which is a contradiction.
19. 1000110111two 25. lOlOOtwo 29. 110,
101010 1011,
23. 87ECsixteen
27. 3071730two
10110,
11011,
31. 0, 1 35. 5 39. 1) 1 ~ S .
2) If x ~ S, t h e n x0, lx, x i ~ S. 45. 10EFsixteen
33. 2
37. 1) 0 e S. 2) If x ~ S, t h e n Ox, lx ~ S. 41. 165 Exercises 4.4 (p. 221) 1. 666 43. 101
12 3. E i i=1
12.13 2
= 78
(Note: In Exercises 513, P(n) d e n o t e s the given s t a t e m e n t . )
934
Solutions to OddNumberedExercises 5. B a s i s step: W h e n n  1, L H S = 1  12  R H S . . ' . P(1) is t r u e .
k
I n d u c t i o n step: A s s u m e P(k) is true" ~ (2i  1)  k 2
k i=1
Then ~(2i1)+(2k+l)k
i=l
2+(2k+l)(k+l)
2
.'. P(k + 1) is t r u e . T h u s , t h e given r e s u l t follows by P M I .
7. B a s i s step: W h e n n  1, L H S  13  1  R H S . . ' . P(1) is t r u e . I n d u c t i o n step: A s s u m e P(k) is t r u e : ~ i 3 i=l
k
[ k ( k + 1)] 2 ~ ~
2
+ (k + 1)2
Then
k
Ei3+(k+l)3[k(k+l)]
i=1
3 2
(k § 1)2(k § 2) 2
 I ( k + l ) k + 2 ) ] 2 ( 2
.. P(k + 1) is t r u e . T h u s t h e r e s u l t is t r u e by i n d u c t i o n .
9. B a s i s step: W h e n n  1, n 2 + n  1 + 1  2 is divisible by 2..'. P ( 1 ) is
true.
I n d u c t i o n step: A s s u m e P(k) is true" k 2 + k is divisible by 2. L e t k 2 + k  2m for s o m e positive i n t e g e r m. T h e n
(k + 1)2 § (k + 1)  k 2 § 3k + 2 = (k 2 § k) + 2(k + 1) = 2m + 2(k § 1) w h i c h is divisible by 2 . . . P(k + PMI. 11. B a s i s step" W h e n n = 2 P(2) I n d u c t i o n step: A s s u m e P(k) j o i n i n g Q to e a c h of t h e o r i g i n a l 1) is t r u e . T h u s t h e r e s u l t follows by 1  ~c21/ . P(2) is t r u e . is t r u e . C o n s i d e r a n e w p o i n t Q. By k p o i n t s , we get k n e w lines.
2 " 9 9
.. T o t a l no. of p o i n t s w i t h k + 1 p o i n t s 
k(k
1)
2 (k + 1)[(k + 1 ) 2
+ k =
k2 § k 2 1]
.. P(k + 1) is t r u e . T h u s t h e r e s u l t follows by P M I .
13. B a s i s step: Clearly, P(1) is t r u e . I n d u c t i o n step: A s s u m e P(k) is t r u e : I f p l a k, t h e n pla. S u p p o s e p la k+ 1. Since a k +1 _ a k . a,pl(a k 9a). T h e n p la k or p [a, by E x e r c i s e 33
Chapter 4
Induction and Algorithms
935
in Section 4.2. Since p la k by t h e IH, P(k + 1) is t r u e . T h u s t h e r e s u l t follows by induction. 15. 28,335
17.  n2/4
/ (n 2  1)/4
if n is even otherwise
n
19. x 
~(2i1)n
2
21. n(n + 1)/2
i=l
23. n(n + 1)(n + 2)/6
25. n(n + 1)(2n + 1)/6
29. 2 n2(n+l)
27. (n!) 2
31" s u m =
~  ~ P s  p i + l  1  p 1 =0
s
33. s u m = (s~_O ) ( ? pS =0
qS ) ( s ~ r s )
=0 1
pi+l p1
35.
1
9
q]+l
rk+l 1 r1
q1 6
n Eii=l
n(n + 1) 2
37.
~Ej=
n
i
n(n + 1)(n + 2)
i=lj=l
39. Recall t h a t t h e kth t e r m of the 41. c a s e 1 Let n be even. T h e n geometric sequence an = 0 + 1 + 1 + . . . a, ar, ar 2 is ak  ar k 1 + (n/2  1) + (n/2  1) + n/2 .'. no. of g r a i n s on t h e last = 211 + 2 + . . . + (n/2  1)1+ n/2 s q u a r e = n2th t e r m in the 2(n/2  1)(n/2 + 1) n n2 sequence 1, 2, 22, 23, .... H e r e 2 2 4 a = 1, r  2 , a n d k  n 2. c a s e 2 Let n be odd. T h e n .'. no. of g r a i n s on the last an = 0 + 1 + 1 + . . . + ( n  1)/2 s q u a r e = 1.(2 n21)  2 n21 . + ( n  1)/2 = 2 1 1 + 2 + . . . + (n  1)/2] (We could prove this P M I also.) 2 [ ( n  1)/21[(n + 1)/2] 43. 93 2 n2_1 45. 249 4 47. 8tn + 1 = 8n(n + 1)/2 + 1 49. L H S = 4n(n + 1) + 1 = [(n  1)n/2] 2 + [n(n + 1)/2] 2 = (2n + 1) 2  n2[(n  1) 2 + (n + 1)2]/4 =n2(n 2+1)/2t RHS
n2
51. b a s i s s t e p : W h e n n = 0, L H S  A u B1 = RHS. So P(1) is true. k k i n d u c t i o n s t e p : A s s u m e P(k) is true: A u ( N Bi)  N (A U Bi). T h e n
AU \i=lBi
(k l)
=AU
(
i=1~B i n B k + l
)
i=l
 [ n u
( )
i=1~ Bi
i=l
] n ( n U B k + 1)
936
Solutions to OddNumbered Exercises
k k+l i=1
 ~ ( A U B i ) A ( A U B k + I ) i=1
A (AUBi)
.'. P(k) ~ P(k + 1), so t h e r e s u l t follows by PMI.
53. b a s i s s t e p : W h e n n  1, L H S  P l  R H S . . " P(1) is t r u e . i n d u c t i o n s t e p : A s s u m e P(k) is t r u e . T h e n
(p A.. Apk Apk+ 1)   ~ [(Pl A... Apk) Apk+ 1
~ ( P l A  . . A p k ) V ~ P k + l
( ~ P l v . . . . pk)V ~Pk+l
.. P(k + 1) is t r u e . T h u s t h e r e s u l t follows by PMI.
55. b a s i s s t e p : Clearly, P(2) is t r u e . i n d u c t i o n s t e p : A s s u m e P(k) is t r u e . By t h e division a l g o r i t h m ,
k  3q + r, w h e r e 0 __no. Since q(no) is t r u e , p(no) is t r u e . A s s u m e p(k) is t r u e , w h e r e k >__n0. By c o n d i t i o n (2), q ( k + l ) is t r u e . T h e r e f o r e , p(k+ 1)q(no)A...Aq(k)Aq(k+ 1) is t r u e . So p ( k )  . p(k + 1). So by t h e first principle, p(n) is t r u e Vn >_no. T h u s q(n) is t r u e Vn > _no. _ 59. n ( 3 n 4 + 7n 2 + 2)/12
Exercises 4.5 (p. 234)
1. W h e n n = 0, a0  1 = x ~ ". P(0) is t r u e . A s s u m e P(k) is t r u e : ak = x k. Then
ak+l = ak * x,
by step 4
= x k .x, by t h e i n d u c t i v e h y p o t h e s i s
_~ X k + l
.'. P(k + 1) is t r u e . T h u s P(n) is a loop i n v a r i a n t by PMI.
Chapter4
Induction and Algorithms
937
3. Let gcd{a,b}  d. W h e n n  0, g c d { x n , Y n } = gcd{x0#0} = gcd{x,y}. .'. P(0) is t r u e . A s s u m e P ( k ) is t r u e : g c d { x k O ' k }  gcd{x,y}  d. T h e n Xk  Y k  1, Yk = r k  1, rk  Xk m o d Yk, a n d x k = qkYk + rk. (1) Since d l x k a n d d[Yk, dlrk. T o s h o w t h a t P ( k + 1) is true: gcd{xk+l, Yk+l}  d. L e t gcd{Xk+l, Yk+l}  d'. (2) Xk+l Yk , Yk+l rk. .'. dlXk+l a n d d[Yk+l. So d i d ' . (3) F r o m (1), d ' l x k + l a n d d'L)'k+l..', d'L)'k a n d d ' l r k . . ' , d ' l x k by (1). T h u s d'lxk a n d d'[Yk. .'. d ' l d . (4) T h u s by (3) a n d (4), d  d'. T h u s P ( n ) is a loop i n v a r i a n t .
= =
5. W h e n n = 0, so  x  x + 0 . . ' . P(0) is t r u e . A s s u m e P ( k ) is t r u e : Sk = x + k. T h e n Sk+l  Sk + 1  (x + k) + 1  x + (k + 1)..'. P ( k + 1) is t r u e . T h u s P ( n ) is a loop i n v a r i a n t . 7.3 11. 7, 18, 19, 23, 53
9. 4 13. 7, 18, 19, 23, 53
17.
15.
Algorithm factorial (n) Begin (* algorithm *)
fact K I i ~i while i < n do fact ~ fact * i End (* algorithm *)
Algorithm product (A, B, C)
Begin (* algorithm *)
for i = 1 to n do for j = I to n do n cij = ~ a i t b t j t=l End (* algorithm *)
19.
Algorithmsum(X) Begin (* algorithm *)
sum ~ 0 for i (* : I to n do *) sum , sum + x i
21.
Algorithm max(X) Begin (* algorithm *)
max ~ x 1 for i = 2 to n do i f xi > max then
End
23.
algorithm
End
Algorithm print(X) Begin (* algorithm *) i ~1 flag ~ true while (i < n) and (flag) do i f si = Sn+l_ i then i ~i+1 else flag ~ false End (* algorithm *) 25.
max K x i (* algorithm
*)
Algorithm palindrome (S) Begin (* algorithm *)
for i = I to n do w r i t e (x i) End (* algorithm *)
(In E x e r c i s e s 2636, P ( n ) d e n o t e s t h e s t a t e m e n t t h a t t h e a l g o r i t h m w o r k s correctly.) 27. Initially, flag = t r u e , a n d i = j = 1. b a s i s s t e p : W h e n n = 1, if azz = b11, flag is still t r u e , so A = B; o t h e r w i s e , flag = false (line 10), so A r B. In b o t h cases, P(1) is t r u e .
938
Solutions to OddNumberedExercises i n d u c t i o n step: A s s u m e P(k) is true. To show t h a t P ( k + 1) is true,
it suffices to show t h a t the a l g o r i t h m works correctly for the (k + 1)st rows and columns of A and B. W h e n j  k + 1, t h e n inner loop (lines 610) checks w h e t h e r or not ai(k+l)  bi(k+l). So as i varies from 1 to k + 1, the o u t e r loop (lines 412) checks if the (k + 1)st columns ofA a n d B are equal. W h e n i  k + 1, the i n n e r loop checks if the (k + 1)st rows of A and B are equal. Thus, P(k) > P(k + 1). So, by induction, the a l g o r i t h m works correctly u > 1. 29. Initially, flag = true, and i = j  1. b a s i s step: W h e n n = 1, if a l l _< b11, flag is still true, so A  B; otherwise, flag  false (line 10), so A < B. In both cases, P(1) is true. i n d u c t i o n step: A s s u m e P(k) is true. To show t h a t P ( k + 1) is true, it suffices to show t h a t the a l g o r i t h m works correctly for the (k + 1)st rows and columns of A and B. W h e n j  k + 1, t h e n i n n e r loop (lines 610) checks w h e t h e r or not ai(k+l) 1.
31. b a s i s step: W h e n n  1, A = !all I and B  Ibll I. So
Cll  allbll by line 3.." P(1) is true. i n d u c t i o n step: Assume P(k) is true, t h a t is, the a l g o r i t h m works correctly for any two k x k matrices. In o t h e r words, line 3 gives the correct value of c(i for 1 _< i,j _1. _ 33. b a s i s step: W h e n n = 1, lines 24, are skipped. So the a l g o r i t h m r e t u r n s the correct value of m i n form line 1..'. P(1) is true. i n d u c t i o n step: Assume P(k) is true. Suppose the algorithm is invoked with n = k + 1 elements. As i varies from 2 t h r o u g h k, the for loop finds the m i n i m u m of the first k n u m b e r s . W h e n n = k + 1, this value is c o m p a r e d to xk+ 1 and the m i n i m u m of x 1 t h r o u g h Xk+ 1 is r e t u r n e d in line 4. .'. P(k) ~ P(k + 1). So, by induction, the a l g o r i t h m works correctly Vn>l.
35. b a s i s step: W h e n n  1, Xn_ i+l
true.

Xl_ 1+1

Xl is p r i n t e d . . ' . P(1) is
i n d u c t i o n step: Assume P(k) is true. Suppose the algorithm is
invoked with n = k + 1 elements. W h e n i  1, Xk+l is printed. T h e n as
Chapter4
Induction and Algorithms
939
i varies from 2 t h r o u g h k, Xk, X k  l , . .  , X l a r e p r i n t e d , by t h e i n d u c t i v e h y p o t h e s i s . . ' . P(k) ~ P(k + 1). So, by i n d u c t i o n , t h e a l g o r i t h m w o r k s correctly Vn >_ 1. 37. 2, 3, 5, 6, 8, 13
39. b a s i s step: W h e n n  1, t h e f o r loop is skipped. So P(1) is t r u e by default. i n d u c t i o n step: A s s u m e P(k) is true. S u p p o s e t h e a l g o r i t h m is invoked w i t h n  k + 1 e l e m e n t s . W h e n i = 2, t h e a l g o r i t h m places t h e l a r g e s t of t h e k + 1 e l e m e n t s in t h e correct position. This leaves a sublist w i t h k e l e m e n t s . By t h e i n d u c t i v e hypothesis, t h e a l g o r i t h m correctly sorts it.." P(k) ~ P(k + 1). So, by induction, t h e a l g o r i t h m works correctly Vn > 1. Exercises 4.6 (p. 246)
1. O(n) 7. O ( n l g n ) 3. O(n 3)
9. O(n a) 15.
tt
5. O ( l g n )
11. O(n 2)
~ i k 0. .'. If(n)] < A B l h ( n ) ] = Clh(n)], w h e r e C = A B .
< Clh(n)] +Clh(n)l < 2Clh(n)] =O(h(n))
where C = max{A, B }
=O(h(n))
51. Suppose f(n) =  T h e n f ( n ) = O(g(n)), f ( n )  g2(g(n)). So ]f(n)l < Big(n)] a n d If(n)] >_ Alg(n)] for s o m e c o n s t a n t s A a n d B. T h u s Alg(n)] < If(n)] _< Blg(n)l. C o n v e r s e l y , letAlg(n)] < ]f(n)l < Blg(n)]. SinceAlg(n)] < ] f ( n ) l , f ( n ) = g2(g(n)). L i k e w i s e f ( n )  O(g(n)). So f ( n ) = 
Exercises 4.7 (p. 25 I)
1. No. o f e n t r i e s in e a c h m a t r i x  n 2 .'. no. o f a d d i t i o n s r e q u i r e d = n 2  O ( n 2)
9.
Cn 
3.
Sn
•
n(n

1)
5. No. o f a d d i t i o n s = n = O(n)
7. an = n = O(n)
11. 37, 2 13.
n 
1 = O(n)
O(n)
Review Exercises (p. 254) 1. 2 9. 11110101two 3. 3 11. 2305two 19. 463eight 5. 1024 13. 21. 100001two
B5sixtee n
7. 2 9 8 9 15. D033sixteen 23. 133sixteen
17. 265eight 25. n2
2 7 . an 
n(n + 2)/4 (n + 1)2/4
if n is e v e n otherwise
2 9 . p r o o f : L e t d = gcd{a  b,a + b}. T h e n dl(a  b) a n d dl(a + b); so dl[(a  b) + (a + b)]. T h a t is, dl2a. S i m i l a r l y , dl2b. B u t a a n d b a r e r e l a t i v e l y p r i m e . So dl2.." d  1 or d  2. 31. Clearly, P ( 1 ) is t r u e . A s s u m e P(k) is t r u e ; t h a t is, 3](k 3  k). (k + 1) 3  (k + 1)  (k 3 + 3k 2 + 3k + 1)  (k + 1) S i n c e 3](k 3  k), t h e R H S is divisible by 3.." P(k + 1) is t r u e . T h u s P(n) is t r u e Vn > 1. 3 3 . W h e n n = 1, L H S
= (21)(2+1)
1

1 3
_ 
R H S . 9 P ( 1 ) is t r u e .
Chapter 4
Induction and Algorithms
1 k §   . 2k+1
941
k A s s u m e P ( k ) is true" ~
/=1
(2i

1)(2i + 1)
Then
k+l
i=l
k 1 V~ + A_~ ( 2 i  1)(2i + 1) = Z_, ( 2 i  1)(2i + 1) (2k + 1)(2k + 3)
i l
1
=
k
2k + 1
4
1
k ( 2 k + 3) + 1
(2k + 1)(2k + 3) k + 1 (k+l)+l
(2k + 1)(2k + 3)
(k + 1)(2k + 1) (2k + 1)(2k + 3)
.'. P ( k ) ~ P ( k + 1). So the r e s u l t holds by induction.
35. Let P(n): We m u s t select at least 2n + 1 socks to e n s u r e n m a t c h i n g pairs. By the P H P , P(1) is true. A s s u m e P ( k ) is true, t h a t is, we m u s t select at least 2k + 1 socks to e n s u r e k m a t c h i n g pairs. Add one m a t c h i n g pair. This e n s u r e s k + 1 m a t c h i n g pairs by selecting (2k + 1) + 2 = 2(k + 1) + 1 s o c k s . . . P ( k ) ~ P ( k + 1). T h u s P ( n ) is t r u e Vn > 1.
D
37. 45. 47.
5(n + 1)n 2 2
m
39. 2 n + 2  n 
4 41. 3 n(n+l)(n+2)/6 43. (n + 1 ) !  1
~ [5i/nj, w h e r e m is the largest i n t e g e r such t h a t 5 m _< n.
i=1
t19 + t20 + t21 + t22 + t23 + t24  t25 + t26 + t27 + t28 t29 + t30 + t31 + t32 + t33 + t34 + t35  t36 + t37 + t38 + t39 + t40
49. tn + t n  l t n + l  n ( n + 1)/2 + (n  1)n/2. (n + 1)(n + 2)/2 = [n(n + 1)/2][1 + (n 2 + n  2)/2] = ln(n + 1)/211n(n + 1)/2] = t2n Supplementary Exercises (p. 256)
1. (m 2  n2) 2 + (2mn) 2 = (m 4  2m2n 2 + n 4) + 4m2n 2 = m 4 + 2m2n 2 + n 4 _ (m 2 + n2) 2
3. tk = k ( k + 1)/2. Let n  m ( m + 1)/2. T h e n
(2k + 1)2n + tk 
(2k + 1) 2. m ( m + 1) 2 2
+
k ( k + 1)
2
[(2k + 1)2(m 2 + m ) + k ( k + 1)] [(2k + 1)2m 2 + (2k + 1 ) 2 m ) + k ( k + 1)] 2 [(2k + 1)m + k][(2k + 1)m + (k + 1)] 2
N ( N + 1)
942
Solutions to OddNumbered Exercises
where N number.

(2k + 1)m + k
". (2k + 1)2n + tk is a t r i a n g u l a r
5. p r o o f : S u p p o s e 111 is a perfect s q u a r e a 2 in some b a s e b, so a 2 b 2 4b + 1 < (b + 1) 2. T h e n (b + 1/2) 2   b 2 4b + 1/4 < b 2 4b + 1. T h a t is, (b + 1/2) 2 < a 2 < (b + 1) 2. This yields b + 1/2 < a < b4 1; t h a t is, a lies b e t w e e n b + 1/2 a n d b + 1, which is impossible. T h u s 111 c a n n o t be a s q u a r e in a n y base. 7. 25 a n d 36 are Duffinian, b u t 18 a n d 43 are not. 9. a ( n ) = ~ d + n  s + n..'. N o n e of t h e n ' s factors, except 1, divides s
dtn d#r~
if a n d only if n o n e divides c~(n). T h u s n is Duffinian if a n d only if n o n e of n's factors, except 1, divides z (n). 11. In Exercise 59 in Section 4.4, we e s t a b l i s h e d t h a t Sn  S n  1  n ( 3 n 4 + 7n 2 +2)/12. Since b o t h S n a n d S n  1 a r e positive integers, it follows t h a t n(3n 4 + 7n 2 + 2 ) / 1 2 is also an integer. C o n s e q u e n t l y , 121n(3n 4 + 7n 2 +2).
13. n 3
15. Georgia
Exercises 5.1 (p. 273)
1. 1 , 4 , 7 , 1 0 7. 1 , 3 , 4 , 7 , 1 1 , 1 8
3. 1 , 2 , 3 , 4 9. 3
5. 1 , 1 , 2 , 4
11. A ( 0 ) A(n)
1000
1.015A(n1),n > 1
13. $1024
15. a l 
1
n > 2
anan_l+3,
17. a o = O
an = 2 a n  l + 3, n > 1
19. 00000, 00001, 0 0 0 1 0 , 0 0 0 1 1 , 0 0 1 0 0 , 0 0 1 0 1 , 0 0 1 1 0 , 0 1 0 0 0 , 0 1 0 0 1 , 0 1 0 1 0 , 01011,01100,01101
21. S1 = A1
Sn  Sn1 U An,n
23. S o > 2
1
25. a l  a
an  r a n _ l , n >_ 2
S n  2 S , , _ 1, n > 1
Chapter 5
Recursion
943
27. 91
29. 91
31. f(99)  f ( f ( l l 0 ) )  f(100) = f ( f ( l l l ) ) = f(101) =91 35. 429
33. Let k be t h e smallest i n t e g e r such 90 < x + l l k < 100. T h e n f ( x )  f ( f ( x + 11)) = f(f(f(x + 11.2)))
37. 1, 2, 1.66666667, 1.5, 1.66666667, 1.6, 1.625, 1.61538462, 1.61904762, 1.61764706, 1.61818182, = fk+l(x +llk) 1.61797753 Since 90 < x + 11k < 100 a n d k > 1, 39. 2 F n  2 + Fn3 f ( x + 11k) = 91, by Exercise 43. .'. f k + l ( x + l l k ) = fk(91) = F n  2 + (Fn2 + F n  3 ) Since f(91)  91 by Exercise 32, = Fn2 + Fnx fk(91)  9 1 . . ' . f ( x )  91 for = Fn 0_ 2
+
n ( n + 2) an (n + 1)2/4
n(n +
if n is e v e n otherwise
31.
an 
n(n
1)/2,n > 1
33.
an 
1)(2n + 1)/6, n > 1
m
35. an  n ( n + 1)(n + 2)/6, n >_ 1 39.
tn n(n
37. an
[n(n + 1)/2] 2 n > 1
+ 1)/2
Chapter 5
Recursion 9
947 43. P n  n ( 3 n  1)/2
1 hn1 + 4 ( n if n  1
41.
45. hn 
1)+ 1
if n _> 2
47. P n + tn  n  n ( 3 n  1)/2 + n ( n + 1)/2  n
= 2n 2  n  (2n  1)n  h n
49. T 1  1
Tn T n  1 + n ( n + 1)/2, n _> 2
51. W e s h a l l p r o v e b y P M I t h a t T n  n ( n + 1)(n + 2)/6 Vn > 1. W h e n n  1, T1 = 1 2 . 3 / 6  1, w h i c h is t r u e . So t h e f o r m u l a w o r k s w h e n n  1. A s s u m e it w o r k s for a n a r b i t r a r y p o s i t i v e i n t e g e r k. U s i n g t h e r e c u r rence relation,
T k + l  T k + (k + 1)(k + 2)/2  k ( k + 1)(k + 2)/6 + (k + 1)(k + 2)/2
= (k + 1)(k + 2)(k + 3)/6 T h u s , b y P M I , t h e f o r m u l a h o l d s for e v e r y n > 1.
53. Sn = S n  1 + n 2
 Sn2 + (n  1) 2 + n 2 = S n  2 + (n  2) 2 + (n  1) 2 + n 2
= $1 + 22 + 32 + . . . = 12+22+32+...+n
= n ( n + 1)(2n + 1)/6
+ n2 2
So we c o n j e c t u r e t h a t S n  n ( n + 1)(2n + 1)/6, w h e r e n >_ 1.
55.
1
57. 2 61. a n + an2, n > 2
59. a 0  1 , a l  2
an  a n  1
l1
[ (1 + a n  1 ) / k
if n  0
if n > 1
63. p r o o f (by PMI)" L e t P ( n ) d e n o t e t h e g i v e n s t a t e m e n t . C l e a r l y P ( 0 ) is
t r u e . So a s s u m e P ( m ) is t r u e for a n y m >_ 0. T h e n
a m + l = (1 + a m ) / k km + k  2 km+l + k  2 km+l(k
=1+
km(k
1)k
1)
T h u s P ( k ) i m p l i e s P ( k + 1), so t h e r e s u l t follows b y P M I .
Exercises 5.3 (p. 296)
1. y e s
3. y e s
5. n o
948
Solutions to OddNumbered E x e r c i s e s
7. y e s 11.
15. 19.
21.
9.
an an an
= 2 (  1)n _~_ 2 n, n > 0

an Ln an 
3 (  2 ) n + 2 . 3 n, n > 0
a n ~2n 
13.
17.
Fn+2, n > 0
2.3 n n . 3 n, n > 0
~n
3n + n . 3n
+
,
=
n > 0
m
an
2 n + (2)n+l
2.3 n 
(4)n,
n
> 0
23. 25. 29.
a n = 3 . 2 n  n 2 n + n 2 2 n + 2 . 3 n, n > 0 a (n  a n + b p)
27.
a(np) 
an 2 + bn + c
a(np) ( a n + b ) 2 n if 2 is a c h a r a c t e r i s t i c n m ( a n + b ) 2 n.
root;
otherwise,
a(np)

31.
35.
a(np) = a n 2 2 n
an an  2 n  1, n > 0 = 1 + n(n +
33. 37.
a(np)  n 2 ( a n 2 + b n + c ) 2 n a n  11 9 4 n  1 1 9 3 n  n 9 3 n + l n > 0
39.
1)/2, n > 0
41.
S i n c e rn a n d Sn a r e s o l u t i o n s o f e q u a t i o n
Sn  a S n  1 + b s n  2 . ". rn + Sn  a ( r n  1 a n is a s o l u t i o n o f (9).
(9), rn  a r n  1
+ brn2
and
+ S n  1 ) 4 b ( r n  2
+ Sn2). Thus
43.
S i n c e a is a r o o t o f x 3  a x 2  b x  c = 0 w i t h m u l t i p l i c i t y
X 3  ax 2
3,
bx  c 
(x  or)3 _ x 3 _ 3 a x 2 + 3 a 2 x _ a 3
.'. 3 a = a , 3 a 2 
b, and a 3 = c
(1) (2)

Also, a 3 = aa2 + ba + c a) M u l t i p l y e q u a t i o n .'. a n is a s o l u t i o n . b) W h e n a n  n a n,
aan_l + ban2 + Can3
(2) b y a n  3 . a n
aa n1 + ba n2 + ca n3
(3)
= a(n 
1)a n1 + b(n  2)a n1 + c(n  3)a n3
+ ba n2 + ca n3) _ (aa n1 + 2ba n2 + 3ca n3)
_ n(aan1 
n a n  ( 3 a n  6 a n + 3 a n), b y e q u a t i o n s an = na n
(1)
.'. n a n is a s o l u t i o n .
c) W h e n a n  n 2 a n,
aan1 4 b a n  2 b C a n  3
= a(n 
1)2a n  1 + b ( n  2 ) 2 a n  1 + c ( n  3 ) 2 a n  3
Chapter 5
Recursion
949
2 n ( a a n  1 H 2bc~ n  2 H 3cc~n  3 )
 n 2 ( a o l n  1 Jrbot n  2 F cot n  3 )
 
H (aol n  1 F 4bol n  2 F 9c(z n  3 )
 n2o~n
 a n
=
_
2 n (3c~n  6c~n + 3c~n) + (3c~n  12c~ n + 9c~n), b y (1)
n2ol n
.'. n2ot n is a s o l u t i o n .
Exercises 5.4 (p. 306)
1. 7.
2 x1 1x x2 + 2 f
1 x+3 2x x2 + 3
l,n
n 
3. 9.
2 1+2x x1 x2+1
§
3 1  3x
2x + 1
5.
8 13(2+3x)
+
7x+4 13(x 2 + 1 )
11 9 an  2 n ,n>O
x2x+l
15. a n = 3 (   2 ) n  2 n, n _> 0
an 
13. a n   2 n 17. an 5.2
>_ 1
3 n, n > 0
19.
Fn+3 , n > 0
21.
25.
an 
(2n + 3)2 n, n > 0
2 3 . a n  3 (  2 ) n + 2 n  3 n, n > 0 2 7 . an  n 92 n + l  n 2 2 n , n > O
a n  2 n + 3 n . 2 n  3 n, n > 0
2 9 . a n  (3n 2  n + 2 ) (  1 ) n  2 n+l n > 0 E x e r c i s e s 5.5 (p. 314)
1.2
7. 7
3.13
9. an=Fn1
5.2
11. Algorithm Fibonacci (n) Begin
if (* a l g o r i t h m *) (n = I) or (n = 2) then answer _ 1.
19. 0
21. 3
23. l g n
25. Clearly, the a l g o r i t h m works when n  1 or n  2. So P(1) and P(2) are true. A s s u m e P(k) is true for any k > 2. Suppose the a l g o r i t h m is invoked with n  k + 1 (> 3). In line 9, the previous n u m b e r (Fk_ 1) and the c u r r e n t n u m b e r (Fk) are added to yield Fk+l. So P(k) ~ P(k + 1). T h u s the result follows by PMI.
Exercises 5.7 (p. 329)
1. O(n 2) 9.3
17. bn  n , n > 0
3. O(n 2) 11.3
5. O(n 2) 13. 5
7. O(n2 n) 15. 5
19. O(n 2)
25. O(n 4) if n if n > 2 1 29. O(n 2)
21. O(n 3)
23. O(n 2)
27. b n = 31. r
{o
bnl+(n1)
1, n >__0
33. Cn [n(n + 1 ) / 2  lib, n > 1 37. f ( n ) [lgnJ + 1
nb+
35. f ( n )  ( n  1 )
2+1,n>__1
(~hapter 5
Recursion
953
39. f ( n )  (no. of bits in t h e b i n a r y r e p r e s e n t a t i o n of n) + (no. of m u l t i p l i c a t i o n s ) < (no. of bits) + (no. of bits in n  1) = ([lg nJ + 1) + [lg nJ = 2[lg nJ + 1 = O(lg n) 41. O(n 3) 45. f ( n ) = af(n/b) + g(n) 43.
Cn '
(a + 1)n  1  O(n )
= a2f(n/b 2) + ag(n/b) + g(n) = a3f(n/b 3) + a2g(n/b 2) + ag(n/b) + g(n)
k1
= akf(n/bk) + E
i=O k1
aig(n/bi)
= akf(1) + E aig(n/bi)
i =0
47. Let n  2 h. By Exercise 45,
k1
49. By Exercise 48,
an  2 . 0 + ~ 2i(n/2 i)
0
f(n)
 da k + cn[ k1 (a/b)i] ~
0
= 0 + nk = n lg n
= O(nlg n) 51. C n  2 n  2 55. Since n is a p o w e r of 2,
= 0 +cn(k~~)1)   c k n
0
= O(nlg n) 53. 5a + 12b, 7a + 19b
t(n)
1
[a 2t(n/2) + bn
if n  1 otherwise
.'. By Exercise 45,
k1 bn t(n) = 2kt(1) + ~ 2 i i=O 2i = a2 k + kbn = an + kbn = Cn + Ckn, w h e r e C  max{a, b} < Ckn
 O(nlg n), since k = lg n
954
Solutions to OddNumbered Exercises
57. By Exercise 56,
f (n)  2cn 2 + ( d < C n 2 JrCn < 2Cn 2 2c)n
5 9 . hn = ~ 1 hnl+l/n
/
if n = 1 if n > 2
= O(n 2) w h e r e C  max{ 2c, d  2c}. 61. p r o o f (by induction) Let P ( m ) be the given s t a t e m e n t . Since h20 hi  1 > 1 + 0/2, P(0) is true. A s s u m e P ( k ) is t r u e  h 2 k > 1 + k/2. T h e n
h2k + 1 
2k+l 1 ~ i=1 l
2k 1 _/~1~__ ~v, > > 1+~
( k)
1+ ~
1 2k+l
1
1 ~_..._~ 2 k+l 2k+l
+2k+1+.+ + 1 2k+1 k+l
n 1 i /~1_<
1+~
1)
+~1~
~
.'. P ( k ) ~ P ( k + 1). So the result follows by PMI.
63. h n = .
~dx 1 x ~
n
= lnxl
1
=lnnIn I  l n n = I n n  (ln 2)lg n = O(lg n)
65. proof: case 1
0
Let n be a power of b, say, n  b k . T h e n
f (n)  a f (n/b) + c  a 2 f (n/b 2) + ac + c = a 3 f ( n / b 3) + a2c + ac +c
o
= a k f ( n / b k) + a k  l c + . . .
k1
+ a2c + ac + c
= akd + c E
i=O
ai'
(1) since f ( n / b k) = f(1)  d.
Chapter 5
Recursion
955
Since a = 1, f (n ) = d + ck = d + c log b n = O(lg n) (Note: log b n  log 2 n 9log b 2)
c a s e 2 Let n not be a p o w e r of b. S u p p o s e t h a t b h < n < b k+l for s o m e positive i n t e g e r k. Since f is n o n d e c r e a s i n g , this implies t h a t f ( n ) < f(b k+l). If a = 1, t h e n by subcase (1) above,
f ( b k + l ) = d + c(k + 1) = (c + d ) + ck
_< (c + d) + clogb n,
f (n) b
.'.
2,thena k >b2k;thatis,
f (n ) < A a k + B a k
a k > n 2.
= (A + B ) a k = (A + B ) a l~  (A + B ) n l~ O(n l~
n
956
Solutions to OddNumbered Exercises
Review exercises (p. 334)
1. 7
5. h ( n )  2 n ( n + 1), n > 1
9. an2 n1, n >_ 1
3. a l   0   a 2
an an1 4. a n  2 4.
1, n > 3
7. an  n ( n 4. 1)(n + 2)/3, n > 1
11. L e t P ( n ) : an P ( 1 ) is t r u e . k ( k + 1)(k + (k + 2)(k + by PMI.
 n ( n + 1)(n + 2)/3. W h e n n = 1, az  1 9 2 . 3 / 3 = 2. So A s s u m e P ( k ) is t r u e . T h e n a k + l = ak + (k + 1)(k + 2) 2)/3 + (k + 1)(k + 2)  (k + 1)(k + 2)[k/3 + 1] = (k + 1) 3)/3. T h u s P ( k ) ~ P ( k + 1). So t h e r e s u l t follows
13. L e t P(n)" an  2 n  1. S i n c e a l  21  1 = 1, P ( 1 ) is t r u e . A s s u m e P ( k ) is t r u e : ak  2 k  1. T h e n ak+l  ak + 2 k  (2 k  1) + 2 k = 2 k+l  1. So P ( k ) ~ P ( k + 1). T h u s t h e r e s u l t follows b y P M I .
15. an  F n + 2 6 ~ 17. an = ~ (  1 ) n 4. 2n  28 28 ~  ~ (  2 ) n 4 2~3n,n >_ 0
333 . 19. an . . . 144 
173 5 10 3 n 4 ~ 9 7 n + ~x~n + z,n _> 0 IZ
21. an = L(n + 1)/2J
m
23.
Cn   (n 
1)n/2
< (n + 1)/2
27. Cl = 0
Cn  ~  C n  1 4. 3n, n > 2
= O(n)
25. 8
29. Cn  3 n ( n
+ 1)/2
31. 9
33. bn  2 n
= O ( n 2) 35. a l   2 , a 2   4 , a 3   7
ana n  1 4 . a n  2 4. an_3, n >_ 4
37. an
 Fn
39. Algorithm sum (n) Begin (* sum *)
if n = 1 then sum _ 2
45.
Cn   O ( n )
4 1 . L e t P(n)" T h e a l g o r i t h m w o r k s c o r r e c t l y V n > 1. W h e n n = 1, s u m = 1, b y line 2. So P ( 1 ) is true. A s s u m e P ( k ) is t r u e . S u p p o s e n = k + 1. T h e n , b y line 4, s u m  s u m (k) + (2k + 1), w h i c h is t h e s u m of t h e first k o d d p o s i t i v e i n t e g e r s , b y t h e IH. .. P ( k ) ~ P ( k + 1). T h u s t h e r e s u l t follows b y P M I .
Chapter 5
Recursion
957
47. p r o o f (by P M I ) : W h e n n = 1, L H S = 1 = R H S ; so t h e r e s u l t is t r u e w h e n n = 1. N o w a s s u m e it is t r u e for a n for a n a r b i t r a r y p o s i t i v e i n t e g e r k. T h e n
k+l k F2i 1  ~ F2i 1 ~ F2k + 1 i1 i=1
= F2k + F 2 k + l  F2(k+l)
49. p r o o f (by P M I ) : W h e n n  1, L H S = L1 = 1  L3; so t h e r e s u l t is t r u e w h e n n  1. N o w a s s u m e it is t r u e for a n a r b i t r a r y p o s i t i v e i n t e g e r k. Then
k+l
Li i=1
k ~ Li JrLk+l i=1
So, by P M I , t h e r e s u l t is t r u e V n > 1.
m
 (Lk+2  3) fLk+ 1
= Lk+3  3
T h u s , t h e r e s u l t is t r u e V n > 1. 51. p r o o f (by P M I ) : W h e n n  1, L H S  3 = R H S ; so t h e r e s u l t is t r u e w h e n n = 1. N o w a s s u m e it is t r u e for a n a r b i t r a r y p o s i t i v e i n t e g e r k. Then
k+l k+l L2i = ~ L2i 'F L2k+2   ( L 2 k + l i=1 i=1 = L2k+3  1
1)+ L2k+2
T h u s , by P M I , t h e r e s u l t is t r u e Vn > 1.
5 3 . g 2 + l  g 2 _ l  ( F n + l f g n  1 ) ( g n + l  g n  1 ) = LnFn = F2n
55. L e t P ( n ) be t h e g i v e n s t a t e m e n t . W h e n n = 2, R H S = F 2 x + F1  x + 1 = x 2  L H S . . " P ( 2 ) is t r u e . A s s u m e P ( k ) is t r u e :
x k  x F k + F k _ 1. T h e n
x k+l = x2Fk + x F k _ l
= (x + 1)Fk + x F k _ l
= x(Fk + Fk_ ~) + Fk = xFk+l + gk
T h u s P ( k ) ~ P ( k + 1). So t h e r e s u l t follows by P M I .
2n1 5 9 . A ( n )  (n + 1) +  ~ A ( i ) n i=2 n1 n A ( n ) = n ( n + 1) + 2 ~ A ( i ) i=2 n2 .'. (n  1)A(n  1) = (n  1)n + 2 ~ A ( i ) i=2 n A ( n )  (n  1)A(n  1) = 2n + 2 A ( n  1)
n A ( n ) = 2 n + (n + 1 ) A ( n 
1)
958
Solutions to OddNumbered Exercises
A(n) n+l
A(nn
1) 4
2 n+l 2 2 +  + ~ n n+l 2 2
A ( n  2) n1 A(1) 2 2
0 =~+~+...+~ n+l
+ + g + 4 + ' " + n~ l 2 2
1
=2~S u p p l e m e n t a r y E x e r c i s e s (p. 338)
n+l
i=3 t
1. t l  1 tn  t n  1
+ n ( n + 1)/2, n >__ 2
3. f 2 _ 2fn + 2  (22'2 + 1) 2  2(22" + 1) + 2 _ 22''+1 + 22+1 '' + 1 22+1 ''  2 + 2 __ 2 2''+1

+ 1 =fn+l
5. ( s o l u t i o n by V. G. F e s e r ) a) C o n s i d e r t h e s e q u e n c e {a,}, w h e r e an+2  a n + l + a,,. T h e n a,,+3 an+2 + a n + l  an+2 + (an+2  an). T h u s a , + an+3 = 2an+2 .'. T h e s e q u e n c e s a t i s f i e s t h e g i v e n r e c u r r e n c e r e l a t i o n . b) U s i n g t h e g i v e n c o n d i t i o n s , a4  2a3  a l  2a3  1 a5  5 a6  2a5  a3 = 1 0  a3 (1)
a12 = 2 9 0  70a3  144 .'. a3 = 2 a n d a4 = 3. S i n c e a5 = 2 a 4  a 2  5, a2 = 1. T h u s t h e s e q u e n c e {an} s a t i s f i e s t h e g i v e n c o n d i t i o n s a n d t h e r e c u r r e n c e r e l a t i o n , so it m u s t b e t h e F i b o n a c c i s e q u e n c e ; t h a t is, an = F,~. 7.
Sn 
(3n 2 + 5)n3/8
9. an  an+5, n E Z
11. 8, 8 is e v e n for n > 2 and
13. f(n) = r ~(2) + r
Since r  1 + 1 = 2 is e v e n , f ( n ) is e v e n .
15. f~  2, f2  3 fnfn1 +fn2, n > 3
17. f ( n , 1) = 1 1 f (n, n ) f(n,k) 0 ifnOorl otherwise
= f ( n  2 , k  1) + f ( n  1,k)
Chapter 6
Combinatorics
959
Exercises 6.1 (p. 349)
1. 1,317 9. 200 17. 0 25. 1,757,600 33. 36,504
3. 1,449
5. 312
7. 18
11. 36[n/2]
19. 260 27. 0 35. 32 43. 24 51. 1,053
1
13. 3 n
21. 100,000 29. 1,572,120,576 37. 128 45. 2 n2
15. 1000
23. 0 31. 186,624 39. 192 47. 364 53. 1 + m + m2 59.
41.
16
49. 9
m n+l
55.
57. m!
n!
(nm)!
m1
61. Let m  plal . al p~,, bl
b1 bm p~" and n = . q l 9 . q.m Since gcd { m , n }  1, m n . . . b,,, P l "'" ql ...qm is the prime factorization of m n . ". r ( m n ) = (al + 1)... (am + 1)(bl + 1)... (bin + 1 )  r ( m ) r ( n )
Exercises 6.2 (p. 358)
1.5 7. T 3.60 9. F 5. F 11. F
13. F
19. 362,880 25. 1 31. 103,680 37. 120 45. P ( n 
15. F
21. 40,320 27. 562,432 33. 1,693,440
17. 20
23. 358,800 29. 208,827,062,548 35. 60,480
39. 6
1, r ) + r P ( n 1, r 1)
41. 6
43. 7
(nr
( n  1)! r ( n  1)! + 1)! (nr)!
(nr)! n!  P ( n , r)
(n  1)!(n  r + r)
(nr)!
960
Solutions to OddNumbered Exercises
47.
(n + l)!  n!  (n + l)n!  n! = (n + 1  1)n! = n ( n ! )
49.
Algorithm permutations (n,r,answer) (* This algorithm computes P(n,r) where 0 < r __ n and the result < is returned in the variable answer. The recursive factorial subalgorithm is invoked by the algorithm.*) Begin (* permutations *)
answer _ 2
1 is e v e n .
15. 8 19. 14
23. 4
21. 25. 29.
66
27.
8 10 31.
co  0
Cn  Cn1
~ 2, n > 1
16
33. Cn  2 n = O ( n ) 37. 0 . 3 6 8 8 7 9 39. 0.367879
Dn Dn1
35. a n  ( n  3)(n  2) = O ( n 2)
4 1 . P n  P n  1 =
n!
(n
1)!
Dn  nDn 1
n~ (1) n ~ , by e q u a t i o n (4) n! 43. D n = ( n Dn
1)(Dn1 + D n  2 )
( n  1)  ~ ( D n  1 t D n  2 ) n! n! n1 1 Pn = Pn1 +Pn2 n n
Chapter 6
Combinatorics
961
45.
Notice thatpo E x e r c i s e 43,
= 1 andpl
 0. L e t g n
Pn Pn1. Thengl

1. By
gn (1)gn1
: (1)
( ~ n  1 gn2
1
1)
1
n_l)"'(~)g'
(1) n n!
47. Algorithm derangements (n)
(* This a l g o r i t h m computes the number of derangements of n items using the a l t e r n a t e r e c u r s i v e d e f i n i t i o n . *) Begin (* derangements *) i f n = 0 then derangements ~ I else i f n is odd then derangements + n . derangements(ni)  I else derangements . n . derangements(nl) + I End (* derangements *)
Exercises 6.4 (p. 372)
1. 36 3. 84 5. 1 6 , 1 7 0 11. y e s 7. 45
9. f n = f n _ l
+n
= f n  2 + (n  1) + n
13. 78
15. No. o f g i f t s s e n t o n t h e n t h d a y n i = n(n + 1 ) / 2  C(n + 1 , 2 ) i=1
=fo+(1+2+...
= 1 + n(n +
+n) 1)/2, n >_ 0
17. a n y n o n n e g a t i v e i n t e g e r . 19. 8
21. C(n,r) 31. 40
23. 33.
243
25.
260
27.
957
29. 42
37.
105
35. 5
n ( n  1)! RHS =. r ( r  1)!(n  r)! n ( n  1)!
39. R H S =
C ( n  1, r ) + C ( n  1, r  1)
(n1)!
l r ( r  1)!](n  r)!
= n!
r!(n  r  1)!
+
(n
1)!
(r  1)!(n  r)! (n  1)!(n  r + r)
r!(nr)!
= LHS
r!(nr)! n! = LHS r!(nr)!
962
Solutions to OddNumbered Exercises
41. L e t P ( n ) " g n  C ( n , O) + C ( n , 2 ) +
C ( n , 4). T h e n C(0, 0) + C(0, 2) + C(0, 4)
43. 2
= 1 +0+0=
1 =go.
.'. P(0) is t r u e . A s s u m e P ( k ) is true: gk = C(k, 0) + C ( k , 2) + C(k, 4)
g k + l  gk zt C ( k , 1) + C ( k , 3) = C ( k , O) + C ( k , 2) + C ( k , 4) + C ( k , 1) + C ( k , 3) = C ( k , 0) + [C(k, 1) + C ( k , 2)1 + [C(k, 3) + C ( k , 4)1 = C ( k + 1, 0 ) + C ( k + 1, 2 ) + C ( k + 1, 4) .'. P ( k ) ~ P ( k + 1). T h u s t h e r e s u l t follows by PMI.
0 45. A ( n , r) 0 A(nl,r) +A(nl,r1) + l
ifrO if r = n ifO2"3 l) '
AssumeP(k)'i~2(2)
Thenk~_]l ~ (~)_~_ ( k + 1 ) = (~)_
i=2 i=2 2
(k+l) + (k+l) 3 / 2
.. P(k +
1) is t r u e . T h u s , by P M I , t h e r e s u l t h o l d s Vn >_ 2.
/ n \ / A N
51. p r o o f : L e t P ( n ) d e n o t e t h e g i v e n s t a t e m e n t , w h e r e n >_ 3. W h e n n  3, L H S  ( 3 )  1 \, /
(4)\ /
RHS... "
P(3)is
true. AssumeP(k)
9
~ (3)i=3 Then
( k + 'lw h e r e k > 3 4 )
i=3
i=3
.'. P(k +
Exercises 6.7 (p. 408)
1. 74
1) is t r u e . T h u s , by P M I , t h e f o r m u l a h o l d s Vn >_ 3.
3. 533
5. 266
7. 21
9. 30
11. L e t S b e t h e s e t of s o l u t i o n s w i t h x < 3, y _< 4, a n d z < 4. T h e n IS]C(3+111,11)=78. Letp1"x>3, p2y>4, andp3z> 5. Then"
N(pl) N(p3)
= no. of s o l u t i o n s w i t h x >__ 4  C(3 + 7  1, 7)  36 = no. o f s o l u t i o n s w i t h z >_ 6  C(3 + 5  1, 5)  21
N ( p 2 ) = no. of s o l u t i o n s w i t h y >_ 5  C(3 + 6  1, 6)  28 N ( p l p 2 )  no. o f s o l u t i o n s w i t h x >_ 4 a n d y >_ 5  C(3 + 2  1, 2)  6 N ( p l P 3 )  no. o f s o l u t i o n s w i t h x >__4 a n d z >_ 6  C(3 + 1  1, 1) = 3 N ( p 2 P 3 ) = no. of s o l u t i o n s w i t h y >_ 5 a n d z >_ 6  C(3 + 0  1, 0) N ( p l p 2 P 3 )  no. o f s o l u t i o n s w i t h x >__4, y >_ 5, a n d z >__6 = C(3  4  1 ,  4 )  0 .. answer 1
N(p'lp'~' 3) 
78  (36 + 28 + 21) + (6 + 1 + 3)  0  3
13. W h e n n  5, L H S  600 = 5(5!)  R H S ; w h e n n  6, L H S  3 6 0 0 5(6~) RHS.
Chapter 6
Combinatorlcs
967
Exercises 6.8 (p. 416)
1 13 1890 11. 1000000
1.  21. 0
2 5. 13 98901 13. 15. 1000000 2 23. 25. 5
3. ~
1 36 1 w 17
0
1 12 13 17. 102 39 27. 95
7.
9 1000000 2 19. 429 9 29. 19
9.
Exercises 6.9 (p. 425)
1.
11
2
3.
13
5
5. 1
4
7. 1
4 7
9.0
11
"
1 13. 2
27.
15. ~1
29. $1.75
17. 161
31.
19. 0
21.
23. 0 37. 425
15 25. 91
20 273 624 625
125 5 92.11r 33. 3~ 35. ~
n
256 39. 6~ 41.
n
43. E 
~ k C ( n , k ) p k q n  k = ~ kC(n,k)pkq nh
k=0 k=l
n
By the binomial theorem, n(1 + x) n 1 = ~ kC(n, k)x h 1. Let x = p/q.
k=l
Then n(p + q)n1 n pk1 n pk1 qn1 = ~ kC(n,k) qk1 = ~ kC(n,k) qk1
k=l k=l
Since p + q = 1 in a Bernoulli trial, this yields
n
np  y~ kC(n, k )pkqnk = E
k=l
Review Exercises (p. 429)
1. 1,038
3. 1,454
5. 1,572,120,576 13. 80
21. 29.
7. 436,700,160 15. 1,296 23. 59,049 31. 34,650
9. 1,542,636,576 11. 0 17. 2,401 25. 176 33. 20 19. 103,680 27. 2,598,960
39,916,800 3,744
968
Solutions to OddNumbered Exercises
35. h ( n )  h ( n  1)+ n u m b e r of h a n d s h a k e s m a d e by t h e n t h couple  h ( n  1) + 4n, w h e r e h(0)  0. 37. 84 39. 26 41. 165 43.  3 3 0
45. x 6 + 6 x 5 y + 15x4y 2 + 20x3y 3 + 15x2y 4 + 6 x y 5 + y 6 47. 243x 5  4 0 5 x 4 y + 270x3y 2  90x2y 3 + 15xy 4  y 5 49. p r o o f (by s t r o n g induction)" Let P ( n ) d e n o t e t h e given s t a t e m e n t . W h e n n  0, r  0. T h e n C ( n , r )  C(0,0)  1 is an integer. So P(0) is true. A s s u m e P ( i ) is t r u e Vi _< k, w h e r e k > 0. By T h e o r e m 6.12, C ( k + 1, r) C ( k , r  1) + C ( k , r ) . Both n u m b e r s on t h e R H S are integers, by t h e inductive hypothesis. So t h e i r s u m is also an integer. ". P ( k + 1) is true. Thus, by PMI, t h e result holds u > 0.
n
51. No. of gifts s e n t on t h e n t h day = ~ i  n ( n + 1)/2 = C ( n + 1, 2)
i=1
53. p r o o f : Consider a p e r m u t a t i o n of a set of n e l e m e n t s . Suppose exactly k of t h e m occupy t h e i r n a t u r a l positions. N o n e of t h e r e m a i n i n g n  k e l e m e n t s occupy t h e i r n a t u r a l positions; the n u m b e r of such a r r a n g e m e n t s is D n  k . B u t the original k e l e m e n t s can be chosen in C ( n , k ) different ways. Therefore, the n u m b e r of p e r m u t a t i o n s with exactly k e l e m e n t s in t h e i r n a t u r a l positions is C ( n , k ) D n _ k .
55. 2 n 57. A(n, 1 )  n  1
59. p r o o f ( b y PMI)" L e t A ( n , r )  C ( n , r )  l . W h e n n  1, R H S  C ( n , 1 ) 1  n  1  A ( n , 1)  LHS. So P(1) is true. A s s u m e P ( k ) is true: A ( k , r)  C ( k , r)  1. T h e n
A ( k + 1, r)  A ( k , r = IC(k,r.'. 1)+A(k,r)+ 1 IC(k,r
1) ll+lC(k,r)
11+ 1
1) + C ( k , r ) l  1  C ( k + 1, r)  1  A ( k + 1, r)
P ( k )  ~ P ( k + 1). T h u s the r e s u l t follows by PMI.
61. $375,000
Supplementary Exercises (p. 432)
63. 0.40951
1. Let k be the least of the r consecutive positive integers. T h e i r p r o d u c t is k ( k + 1 ) . . . (k + r  1). Since
k ( k + 1 ) . . . (k + r 1)
rt is an integer, k ( k + 1 ) . . . (k + r  1) is divisible by rt.
Cn  C ( 2 n , n ) / ( n + 1) is an integer; so n + 1 I C ( 2 n , n ) . C ( 2 n + 1, n) 2n + 1 =
 C ( k + r  1, r)
3.
5. R H S 
(2n + 1)!
n t ( n + 1)t(2n + 1)
=
(2n)t
n t ( n + 1)!
= Cn  L H S
(~hapter 6 7. 13; 80
Combinatorics
969
9. We shall use the following identities:
1. n
(;_~)
=k
(;)
n (;) 2. ~ k=O
2 n
n (;) 3. ~~k k=l k  n Z k=O
 n 2 n1
( k  1 + 1)
k=O
k2  ~ k=O
=n
~ k =0
n (~ ~)
k.k  ~ n k=O
(k1)+n
~(~ ~)
k=0
nk~o(n1) =n(n1)
(k 2) §
( k~)
n2 ( ~ _ 2 ) ni (;_ ~ +n k=2 k=l
+ n n ) ( 1  1 ~ n k k=O = n(n  1). 2 ''2 + n . 2 ''1  n(n + 1)2 ''2 ) = n(n  1) k=o'~2( n _ k2 11. d=lj=0 k=0
=F~
d=lj=0
J + 1
1E(~11) +
_
1]
d=l
= ~ n+l ,=o r
'~'()
()(
n+l 0
_
n+l 1
)
(:~11) ,n 1, +
=2 ''~11(n+1)1(n1) = 2 ''+1 13. no, no
2n  2
n
15.
f(n,k) 0
if k  1
ifkn
f(n2,
k1)+f(nl,k)
if 1 < k < n
17. By Exercise 16, f ( n , k )  C(n  k + 1, k). .'. Total no. of subsets t h a t do not contain consecutive integers is given by
n t/ n
f ( n , k )  ~ C(n  k + 1, k)  ~ C((n + 2)  k  1,k)  F,,+2 k=o k=o k =o
19. Fn+2
L'~
m
~
~
r
o
II
II
.e.,,. , ~~. ('D ~ ~ L. I
II
II II ' '' ' '
II
II
II
II
''
'
m.
z
0
'"
e@
~. II ~' ~ :
~,:
..1 O0
II "'
3
~
~
~'11
3x
~.
m
'~" 1'
II
,.., m 
'..
,'4
>
o
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r._,.
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0
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~5I'D
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I" I I I i i
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II
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t~ 0 ...% "~~
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r ~
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IV
~
Chapter 7
Relations
977
41. R  {(a,a), (a,b), (b,a), (b,b), (c,c)} a n d S  {(a,a), (a,c), (b,b), (c,a), (c,c)} are equivalence r e l a t i o n s on { a , b , c } , b u t R u S  {(a,a), (a, b), (a, c), (b, a), (b, b), (c, a), (c, c)} is not. 43. M o n d a y 45. T u e s d a y
47. L e t a  b ( m o d m ) andcd(modm).Thenab=lmandcdk m for s o m e l , k E Z. T h e n ( a  b ) + ( c  d ) = lm+km, (a + c)  (b + d)  (l + k)m. ". a + c  (b + d) (mod m).
t h a t is,
49. By the division a l g o r i t h m , a = b m + r for some b ~ Z..'. a _= r (mod m). 51. By the division a l g o r i t h m , a  d i m + rl a n d b  d 2 m + r2, w h e r e 0 < rl, r2 < m. T h e n a  b = (dl  d 2 ) m 4 (rl  r2)..', a = b (mod m) if a n d only if rl ~ r2 (mod m).
n
53. Let N

~ a i l O ni i=O
be
the
decimal
expansion
of N .
Since
10  1 (mod 9), 10 k  1 (mod 9) by Exercise 48. T h u s a i l O "  i  aiVi by Exercise 48.
1l
.'. N =  ~ ai (mod 9). C o n s e q u e n t l y , N is divisible by 9 if a n d only if =
i0
/t
~ ai is divisible by 9.
i=0
55. 0
57. 8
59. 4
61. 1 5 , 6 2 1
Exercises 7.9 (p. 503)
1. no 9. no 17. yes 3. yes 11. no 19. (2, 6) < (3, 5) 5. yes 13. yes 7. yes 15. yes
21. (2, 3, 2), (2, 3, 3), (2, 2, 4)
23. luxury, m a x i m u m , m o m e n t , neglect, neighbor, n e o p h y t e 25. discount, discourse, discreet, discrete, discretion, d i s k e t t e
27.
6 2 1
18
9 3
29. c
b
I
978
Solutions to OddNumbered Exercises
31. aaa
aab
aba
abb
baa
bab
bba
bbb
Vaa
a
Vab
Vba
b ).
Vbb
33. c , d ; a 41. 18, 1
35. none; none
37. 18; 1
39. none; n o n e
43. F
51. (Z, G3 be i s o m o r p h i s m s . T h e n g o f : G1 ~ G3 is also a n i s o m o r p h i s m . . ' . T h e r e l a t i o n is t r a n s i t i v e . T h u s t h e r e l a t i o n is an equivalence relation.
Exercises 8.4 (p. 554)
1.2 3.4 5.7 7.1
Chapter 8
Graphs
983
9. badb, bacb, bdcb 17. 5 25. 10 33. 0
35. s t a r t   ab   a aA A w AB ~
11. n o
13. n o
15. y e s
19. 10 27. 0
21. 45 29. 10
23. 0 31. 2
aA
AC ~
A
aA w C ~ ~A~aA
cC
CcC
37. I n i t i a l l y , t h e f a r m e r t a k e s a 2 , a 4 , . . . ,a2n a c r o s s t h e r i v e r a n d t h e n r e t u r n s . S h e t a k e s a3, a 5 , . . . , a2n1 a c r o s s a n d b r i n g s a2, a 4 , . . . , a2n b a c k to t h e o r i g i n a l side. S h e t a k e s a l a n d a2n+l to t h e o t h e r side, a n d t h e n r e t u r n s . F i n a l l y , s h e t a k e s a2, a 4 , . . . , a2n to t h e o t h e r side.
39. Algorithm graph (G,A,connected)
(* This a l g o r i t h m determines i f the graph G with n v e r t i c e s is connected using Theorems 8.4 and 8.5, and i t s adjacency m a t r i x A. The boolean v a r i a b l e connected denotes whether or not G is connected. *) Begin (* a l g o r i t h m *) (* f i n d S = E A i *) B ~A (* i n i t i a l i z e *) S +  A (* i n i t i a l i z e *) f o r i = 2 to n  1 do begin (* f o r *) B ~ AB S ~S+B
endfor
(* check i f any e n t r y of S is nonzero. *) i ~i j ~i done _ 2.
~176
65.
T,, is E u l e r i a n Vn >_ 1.
59. a  c  b  d  e  a , 53
61. n o n e
63.
none
1151245111697214133610178
67. Algorithm Eulerian path (G,A)
(* G is a connected graph w i t h n v e r t i c e s and w i t h adjacency m a t r i x A = (aij)n• Using Theorem 8.8, t h i s a l g o r i t h m determines i f G contains an Eulerian path. Odd is a counter t h a t keeps t r a c k of the number of odd v e r t i c e s in G. *) Begin (* a l g o r i t h m *) odd ,  0 (* i n i t i a l i z e the counter *) f l a g , f a l s e (* boolean f l a g to e x i s t when odd > 2 *) i , 1 (* i n i t i a l i z e row index *) w h i l e (i _< n) and (not f l a g ) do (* compute each row sum *) begin (* w h i l e *) sum 3 5. yes 7. yes
9. Label the vertex of degree 11. ~ vi, ~ v i v j n with 1 and the others i~:j 2 , 3 , . . . , ( n + 1)" 13. Let G = (V1,E1) and H = (V2,E2) be two simple graphs. A s s u m e f: 2 3 n n+l G ~ H is an isomorphism. Let ! ! G !  (V1,E 1) and H ' (V2,E 2) be their complements. Let g: G ! H ' defined by g ( u ) = f ( u ) . Since 1 I E l l  IE21, l E V I  lEVI. To show t h a t g preserves the adjacency rela15. C5 tionship, a s s u m e {u, v} c E~. T h e n 17. Suppose the graph contains e {u,v} ~ E l . So { f ( u ) , f ( v ) } ~ E2; edges. T h e n e  C(n, 2)/2 t h a t is, {g(u),g(v)} ~ E'2. T h u s if n ( n  1)/4..'. 4In or 4 1 ( n  1), {u,v} E E'1, t h e n {g(u),g(v)} ~ E'2. .'. g is an isomorphism. t h a t is, n  0 (mod 4) or Conversely, let G' and H ! be n = 1 (mod 4). isomorphic. T h e n (G')! = G and 19. ( n  1)!/2 (H') ! = H are isomorphic, by the 21. (n!)2/2 first part. 23. f
e
d
a
b
c
Delete ad. The ensuing subgraph is planar: d
Exercises 9.1 (p. 613)
1. yes
9. n  8 ,
3. no
e7n1
5. no
7. no; contains cycles. e11n1
11. n  1 2 ,
988
Solutions to OddNumbered Exercises
n
13. By T h e o r e m 8.1, ~ d e g
i=l
(vi) n
2(no. o f e d g e s ) . B u t , b y T h e o r e m 9.2,
n u m b e r of e d g e s  n  1.." ~ d e g i=1 15. y e s 17. n o
(vi) 
2(n  1)  2 n  2.
19. m = l o r n =
1
21.
25. 4, 6, 3, 5, 4, 4, 4, 5, 5, 5, 6
2 7 . Algorithm tree
(A,n)
(* Using Theorem 9.2, t h i s a l g o r i t h m determines i f a graph w i t h adjacency m a t r i x A = ( a i j ) n x n is a t r e e . Flog is a boolean v a r i a b l e and s t o r e s the value t r u e i f the graph is a t r e e and f a l s e o t h e r w i s e . *) O. Begin (* t r e e *) 1. f l a g , f a l s e (* i n i t i a l l y assume t h a t the graph is not a t r e e *) (* compute the sum of the degrees of the v e r t i c e s *) 2. sum
~@
47.
%
Cg: ".
2
Exercises 11.4 (p. 777)
1. 81 3. S1
5. 0
9.
7. 1
a/1
(~ a/O
11.
..~
(~a~O (~
.....
a~(~
,~
b/O
3o "
g
a
80 Sl 82 SO S2 S2
b
81 S2 80
1008
Solutions to OddNumbered Exercises
15.
a
f
b a
g
b
so Sl s2 83
SO 81 S2 S3
Sl S2 83 Sl
0 0 0 0
1 1 1 1
17. 0101 25.
0o
19. 0 0 1 0 1 1
21.
0110
23. 0000
f
g
01 CO C1
lO CO C1
11 C1 C1
oo 0 1
01 1 0
lO 1 0
11 0 1
CO C1
CO CO
27.
11001
29.
110000
31.
(~ a/O
b/1
a/O 33" _~
35. no 39. Words e n d i n g in aa. 43. g(s,x) = 0
(~a/O ,~
(~a/O ~
(~ a/1 ~. s 3 ( ~ ~ b/O
37. yes 41. (~b/O
(~b/1 ~. ~ , a / O ~ ~ ~,s2~~b/O
~a/1
Exercises 11.5 (p. 78 I)
1. no
3. no
5. so ~ as1, sO > bso, Sl ~ as1, 81 > bs2, s2 ~ as1, 82 > bs3, 83 >as4, 83 > bso, 83 >a, s4 >a, s 4 ~ b
Chapter 11
Formal Languages and FiniteState Machines
1009
7. so + a s o , so + b s l , Sl + a s o , Sl + bs2, s2 + a s o , s2 + bs2, S l + b, s2 '~ b 9. so ~ a s o , so ~ s3 > bs3, s2 ~ b s l , 81 >a s l , Sl ~ b, s3 >a, s3 ~ b bs2, s2 >a s 2 , s2 > bs3, s3 > a s 3 ,
11. so ~ a s l , s o > b s o , s l ~ a s 1 , Sl ~ bs2, s2 ~ a s 1 , s2 ~ bso, Sl > b
13. so ~ a s 1 , so ~
s3 > bs3, Sl ~ bs2, Sl > a s l , 81 > bs3, s2 ~ b, s3 > b a s 2 , s2 > b82, s3 ~ as1,
15. so ~
a s l , so ~ s2 > b
bso, Sl > a s 2 , Sl > b s l , s2 > a s 2 , s2 > bs2, S l > a,
17. so ~ a s 2 , so > b s l , Sl > a s l , Sl > b S l , s2 >a s 3 , 82 > b s l , s3 + a s 3 ,
83 ~ bs3, s2 ~ a, 83 + a, s3 > b
19. so ~ a s l , so ~ bso, Sl ~ a s l , Sl ~ bs2, s2 > a s 3 , s2 > bso, S3 > a s 3 ,
s3 ~ bs3, s2 ~ a, s3 ~ a, s3 > b bs3, s2 >a s 3 , s2 ~ a, s4 ~ b bs4, s3 ~
21.
so ~ a s l , so ~ 83 > bs3, s4 ~
bs2, Sl ~ a s 4 , Sl ~ a s 4 , s4 ~ bs4, s4 ~
as3,
Exercises 11.6 (p. 785)
1. _+
a ~ ~ a b ~Q b
3.
~
a
~
a
,
b
~
a
b
b
5.
)
a
(~ a,b
~ a ]~Q
7.
a
so 81 s2
b
{so,s1} {si} {s2}
{s2} {s2} {s2}
a,b
1010
Solutions to OddNumbered Exercises
11. yes; a  A  B  B  F
sO 81 s2 83 s4 s5
{so, s1} {sl,s2} {s2} {s3} {s4} {ss}
{ss} {ss} {s3} {s4} {s4} {ss}
. . . . . .
13. yes; aAAAF 15. yes; sosl8182 17. no
19. no
21. yes; 8 0  8 1  8 2  8 3  8 4
23.
25.
b
29.
~
a ~ a
31.
J b 33.
a,b
Chapter11
35.
__.> @
Formal Languagesand FiniteState Machines
a a ~. a ~ b ~.
1011
Exercises 11.7(p. 792)
0
b __~ a
a
Q
__+(~
a ~. @ , ,
b r. 
b ~
a
0
0
~
a,b
a,b
a
b
1012
Solutions to OddNumbered Exercises
0
b
11.
b
a
13"~ b< a ~a,b(___ 15.~ a ~a..~b/7"~,[ a ~a~
b
17. Words beginning w i t h a a .
19. Words b e g i n n i n g w i t h bba.
21. Words with at least one a. 23. Words c o n t a i n i n g a b a as a substring. 25.
b
a
.__)
27.
a,b
a
a
b
Chapter 11
29.
FormalLanguages and FiniteState Machines
1013
a
31.
a a
Review Exercises (p. 794)
1. { a, aa, ab, aab, bca, bcab }
3. {)~,a, aa, bc, abc, bca, bcbc }, {)~,a, aa, bc, abc, bca, bcbc, an,abca, bca 2, bcbca, a2bc, abcbc, bcabc, bcbcbc } 5. a, b, ac 11. (np) (art) the 13. no 17. b a a A B b
21. o ~ b o , o ~b 23. o ~ b o b , o ~b
7. ab, abab, ababab (sentence) (verb) (noun) (eats) wolf (art) the 15. yes
19.
q3
9. yes
(np) (noun) cabbage
(adj) discrete
I
I
I
I
a b
A G a A~
I
1014
Solutions to OddNumbered Exercises
25. while
do
I
(vble>
x 9= y 27. ~ a b a b _~~ b + z
x
~
y
I
I
I
I
/a
9e
~"
so Sl
81 82
S2
S2
80 80 s2
31. W o r d s b e g i n n i n g w i t h abb. 35.
33. W o r d s c o n t a i n i n g aa.
a, o
37.
(~a/1 b/O a/1
(~ a/1
Chapter 11
Formal Languages
and FiniteState Machines
1015
39. a
s0 81 s2 SO 82 S2
b
Sl 81 S2
a
b
0 0 1
1 1 0
41. 0010111
43. 010110
45.
a o
f~
a/1 b/1 r ~ ~ b/1
a/1
___+~b/O~..b/ ~/~a/O (~b/O __> a/Or ~ ( ~ ~ ~ ~ a/O D ~ aD,D ~ bD,D+
47.
49. ~ . aA, a + bC, A+ bB, A+ aC, B ~ aC, C + aC, C + bC, B + bD, a,D+ b
51. ~ ( ~ b
53.
s0 Sl s2 s3
{81} {80,81} {Sl}
{82} {Sl}
(s3}
{s3} {s2,s2}
57.
no
55. yes 59. yes
61. yes
1016
Solutions to OddNumbered Exercises
63.
(~b
~b
a,b{
.~b
~a
b_
67. W o r d s c o n t a i n i n g exactly one a. 69. By E x e r c i s e 68 in Section 11.1, (A u B*)* = (A u B)* = (A* u B)*.
Supplementary Exercises (p. 798)
1. b 3. b
b
b
Q
c ~:d,.
ccd, E
~ ,8
7. a ~ aA, a ~ bC, A o aA, A ~ bB, B ~ aA, B ~ b B, C ~ a C, C o bC, B> b
9. (wfnp> ::= ~[(wfnp)
Chapter 12
Boolean Algebra and Combinatorial Circuits
1017
11. 0 1 1 ~ 0 1 ~ 1 0
13. f input
s
g 1
0 so 82 S2
sO Sl 82
sl 80 81
1 0 1
17. 101110 15. 1101
19. S u p p o s e x = x l x 2 . . . X n E L. T h e n t h e r e is a D F S A A t h a t e x c e p t s x, so t h e r e is a p a t h t h a t b e g i n s at so a n d e n d s at a n a c c e p t i n g s t a t e Sn. R e v e r s i n g t h e p a t h yields t h e s t r i n g X n X n  l . . . X l . So m a k i n g Sn t h e initial s t a t e a n d so t h e a c c e p t i n g s t a t e yields a N D F S A A' t h a t a c c e p t s L R. (IfA' c o n t a i n s m o r e t h a n one a c c e p t i n g state, i n t r o d u c e a n e w s t a t e s; t h e n c o r r e s p o n d i n g to every i n c o m i n g edge to an a c c e p t i n g state, add an edge t e r m i n a t i n g at s.) N o w t h e r e s u l t follows by T h e o r e m s 11.3 a n d 11.4.
Exercises 12.1 (p. 812) 1. 30 11. 5 3. 30 13. 35 5. 1 15. 2 7. 5 17. 70 23. 5 E ) ( 5 G 7 ) = 5 Q 3 5 = 5 29. no 31. yes 33. yes 9. 6 19. 10
21. 5 ' = 70/5 = 14 .'. ( 5 ' ) ' = 1 4 ' = 70/14 = 5 25. 2" 35. 27. yes
+ 0 1 a b
0 0 1 a b
1 1 1 1 1
a a 1 a 1
b b 1 1 b
0 0 0 0 0
1 0 1 a b
a O a a 0
b O b 0 b
0' = 1, 1' = 0 , a ' = b,b' = a. 37. x + x y = x 39. (x + y)' = x ' y '
1018
Solutions to OddNumbered Exercises
4 1 . xx = xx + 0  xx + xx' = x ( x + x')  x l = x
45. x ( x + y ) = x x + x y = x + x y  x
49. 51.
(x + y ) z  z(x + y) = zx + zy = x z + y z (x + z)( = xy + = xy + = xy + = xy + =xy+z
43. By c o m p l e m e n t law, 0 + O' = 1. B u t by i d e n t i t y law, 0 + O'  0'. .'. O'  1. 47. (x + y ) ' = x'y'. ". [(x + y ) ' ] ' = (x'y')' i.e. x + y = (x'y')'
y + z) = x( y + z) + z( y + z) x z + zy + zz x z + zy + z (z + zx) + zy (z + zy)
Exercises 12.2 (p. 821) 1. 1, 0
o
3. 2
5. yes
7. yes
x
y
x' 1 1 0 0
y' 1 0 1 0
x + y' 1 0 1 1
x' + y 1 1 0 1
(x + y ' ) ( x ' + y) 1 0 0 1
0 0 1 1 11.
x
y 0 0 1 1 0 0 1 1
z 0 1 0 1 0 1 0 1
xy 0 0 0 0 0 0 0 1
y' 1 1 0 0 1 1 0 0
y'z 0 1 0 0 0 1 0 0
z' 1 0 1 0 1 0 1 0
yz' 0 0 1 0 0 0 1 0
xy +y'z +yz' 0 1 1 0 0 1 1 1
0 0 0 0
1
1 1 1
13.
x
y 0 0 1 1 0 0 1 1
z 0 1 0 1 0 1 0 1
y'
1 1 0 0 1 1 0 0
x + y' + z
1 1 0 1 1 1 1 1
xy'z
0 0 0 0 0 1 0 0
(x + y' + z) ( x y ' z )
0 0 0 0 0 1 0 0
0 0 0 0 1 1 1 1
Chapter 12
Boolean Algebra and Combinatorial Circuits
1019
15.
x 0 0 0 0 1 1 1 1
y 0 0 1 1 0 0 1 1
z 0 1 0 1 0 1 0 1
yz 0 0 0 1 0 0 0 1
xyz 0 0 0 0 0 0 0 1
(yz)'
. _ .
x(yz)' 0 0 0 0 1 1 1 0
xyz + x(yz)' 0 0 0 0 1 1 1 1
1 1 1 0 1 1 1 0
17. x y
xy
0 0 0 1 identical
x + xy
0 0 1 1
0 0 1 1
0 1 0 1
19. X y oo 0 1 1 0 1 1
x+y 0 1 1 1
(x + y)' 1 0 0 0
1"
x' 1 1 0 0
y' 1 0 1 0
x'y' 1 0 0 0
identical
21.
x 0 0 1 1
y 0 1 0 1
x+y 0 1 1 1
(x + y)' 1 0 0 0 1'
x' 1 1 0 0
y' 1 0 1 0
x' + y' 1 1 1 0 1"
not identical
23. yes 27. xy
25. xyz, xyz', xy' z, xy' z', x' yz, x' yz', x' y' z, x' y' z' 29. xy + x'y' 33. xy' z' + x' yz' + x' y' z 37. xy'
31. xyz + xy' z'
35. xy + x'y + xy'
1020
Solutions to OddNumbered Exercises
39. x y z + x y z ' + x ' y z 43. (x + y ) x y ' = = = xxy' + yxy' xy' + x(yy') xy' + xO xy'
41. x + y = = =
x l + l y  x ( y + y ' ) + (x + x ' ) y xy + xy' + xy + x'y (xy + x y ) + x y ' + x ' y xy + xy' + x'y y x + y z = xy + y z x y ( z + z') + (x + x ' ) y z xyz + xyz' + xyz + x'yz xyz + xyz' + x'yz
47. 0 49. 1
45. y ( x + z) = = =
51. 0 53. 55.
x$y
1 1 1 0
(x$y)$(x$y) 0 0 0 1
x 0 1
y 0 0
x$y 1 1
(x$y)$(xSy) 0 0
1 1
y $y 1 0 1 0
0 1
1 0
0 1
57.
x $x 0 0 1 1 0 1 0 1 1 1 0 0 (x $ x ) $ (y $ y) 0 0 0 1
59. xy' + x'y + x'y' 61.
x 0 0 1 1 y 0 1 0 1 x +y 0 1 1 1 1" x $x 1 1 0 0 y $y 1 0 1 0 identical (x $ x ) $ (y $ y) 0 1 1 1 1'
63.
x +y 0 1 1 1 1". . . . . . x Sy 1 0 0 0 identical (x Sy) $ (x Sy) 0 1 1 1 1"
65. (0,1,1)
Chapter 12
Boolean Algebra and Combinatorial Circuits
1021
67. Since { 1"} is f u n c t i o n a l l y c o m p l e t e , x'  x 1" x a n d x + y = (x 1" x) 1' (y 1" Y), {+, '} is also f u n c t i o n a l l y c o m p l e t e .
69.
1
71. 0 75. no
79. x@x 0 0 x~y 0 1 1 0 y~x 0 1 1 0
73. 0
77.
81.
x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 y~z 0 1 1 0 0 1 1 0 x ~ (y ~ z) 0 1 1 0 1 0 0 1 1" x ~y 0 0 1 1 1 1 0 0 identical (x ~ y ) ~ z 0 1 1 0 1 0 0 1
83. (x + y)(x + y') 85. (x + y + z)(x + y + z')(x + y' + z')
87. (x + y + z)(x + y + z')(x +y' + z)(x +y' + z')(x' + y + z)(x' + y + z')(x' +y' + z)
Exercises 12.3 (p. 829)
1.0 3.1 Cy 1 1 1 0 5.1 7.1 9.0
11.
x y
13. xy' + x'y + x'y'
15. xyz + x'yz + x'y'z + x'yz' + x'y'z'
0 0 1 1
0 1 0 1
1022
Solutions to OddNumbered Exercises
17. x y z NAND 000 001 010 011 100 101 110 111
19. t r u e
21. false
Exercises 12.4 (p. 837)
1. x  O 3. x = y  1
5. x'y I
13. (xy + y z + zx)xy'z (x + y ) ' y 0 0 0 0
7. (x + y)'y
9. x + ( yz)' 15.
X
11. (x + y)z' + x ' z
x +y 0 1 1 1 (x + y)' 1 0 0 0
0 0 1 1
17.
Z
yz
0 0 0 1 0 0 0 1
(yz )' 1 1 1 0 1 1 1 0
x + (yz )' 1 1 1 0 1 1 1 1
0 1 0 1 0 1 0 1
19. x
Y
)
I
x(y'z+yz')
Chapter12 21. x
Boolean Algebra and CombinatorialCircuits
1023
23. x
~
~
x+y
2o x_ _ ~
29. 1001 31. 1011
27.
33. s  (x + y)(xy)' = xy' + x'y
.'. 8i   8Ci + 8'Ci  
(xy' + x'y)c I + (xy' + x ' y ) ' c i = xy'c I + x'yc'i + (x' + y ) ( x +y')ci = xy'c'i + x'yc'i + (xx' + x'y' + xy + yy')ci
= xy'c I + x'yc I + (o + x'y' + xy + O)c~ = xyci + xy'c I + x'yc I + x'y'ci
35.
[5 } [} D
Exercises 12.5 (p. 849)
1. X 3. xy t
)s = (x +y) (xy)'
c=xy
5.
X!
7. xy
13. xyz
9. xy + yz + zx + xyz
15. x(wy + w'y')
ll.
xy
17. x
1024
Solutions to OddNumbered Exercises
19. xy' + x'y 23. x 21. x
yt
25. xyz + xyz' + xy'z' + x'y'z' 27. xyz' + xy'z' + xtyz ' + x'y'z'
33. xy + y ' z '
x'
29. xy'
35. z' 31. y'
37. wxyz' + wxy'z' + wx'yz' + wx'y'z' + w'x'yz' + w'x'y'z' + w'xyz' + w'xy'z t 39. wxyz + wxtyz + w'x'y'z' + w'x'y'z + w'xy'z' + w'xy'z
yz
WX
yz ~ y'z'
y'z
WX
yz
yz ~ y~z'
y' z 1
41.
wx'
WlX I
43.
wx'
WlX f
1 1
w'x
45. z'
49. yz
1
w'x
47. wyz + w' y'
51. wx + y'z
Exercises 12.6 (p. 856)
1. x + y' z'
7. x + y '
13. wx' + xyz
3. y' + w' y
9. z'
5. wx + xyz + w 'xy'
11. x y z ' + w'x'y'z'
15. x' y' z' + x' y' z + xyz' + xyz
17. w'x'y'z' + w'x'yz' + w'xy'z' + w'xyz' + wx'y'z' + wx'yz' 19. w'x'y'z' + w'xyz'
21. w'xyz + wx'y'z' + wx'y'z + wx'yz'
23. yz + y'z' + wx' + w'x'
25. yz' + w'x'y + x'y'z' + w'xy'z 27. xy'z + xyz' + w'x + y'z'
Review Exercises (p. 859)
1. yes 7.2
3.
no
5. W h e n n is a product of distinct primes.
9. 30
11. (x + y)(x' + y) = y
Chapter 12
Boolean Algebra and Combinatorial Circuits
1025
13.
x
y
z
x +y 0 0 1 1 1 1 1 1
x +y +z 0 1 1 1 1 1 1 1
x' 1 1 1 1 0 0 0 0
y' 1 1 0 0 1 1 0 0
x'y' 1 1 0 0 0 0 0 0
z' 1 0 1 0 1 0 1 0
x' y' z'
0 0 0 0 1 1 1 1
15.
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
x 0 0 0 0 1 1 1 1
y 0 0 1 1 0 0 1 1
z 0 1 0 1 0 1 0 1
yz 0 0 0 1 0 0 0 1
x' 1 1 1 1 0 0 0 0
x'y'z 0 0 0 1 0 0 0 0
x + x'yz 0 0 0 1 1 1 1 1
x + yz 0 0 0 1 1 1 1 1
t identical ._1,
17. x' yz' + xy' z + xyz' 21. xyz + xyz' + xy' z + x' yz
25.
19. xyz + xyz' + xy'z + x'yz 23. 0
x 0 0 1 1
y 0 1 0 1
xlx 1 1 0 0
YlY 1 0 1 0
(xt x)$ (yl y)
27. xy + x y ' + x ' y + x ' y '
31. x
29. xy' z' + x'yz' + x' y' z 33. 1110
z
~
)
xyz + x'y'z'
1026
Solutions to OddNumbered Exercises yz
wx
35.
yz ~ y' z ~ y~z 1 1 1 1 1 1
37. x ~ + y '
39. w y + w ' y '
41. y~
1
43. wx + wyz' + x' y' z' + y' z
wx'
WfXI W~X
45. x' + z'
47. wy + xy'z + w'x~y 'z'
49. wy ~ + yz + xyz' + x'y~z '
1
51. w'x'y'z' + w'xy'z + wx'yz'
53. wx'y'z' + w'xyz + w'x'y'z + w'x'y'z' + w'xyz' + w'x'yz
Supplementary Exercises (p. 862)
1. yes 3. p r o o f (by PMI) Clearly, x (1) = x'. So a s s u m e x (k) equals x if k is even a n d x' if k is odd, w h e r e k >_ 1. Then" x(k+l) = (x(k)),  {x' (x')'
g
if k is even otherwise ifkiseven otherwise
_ ix' x
/
T h u s t h e r e s u l t follows by induction. 5. p r o o f (by s t r o n g induction): T h e r e s u l t is clearly t r u e w h e n n  1. So a s s u m e it is t r u e for a n y boolean e x p r e s s i o n w i t h 2, 3, 4 , . . . , or k variables. T h e n ( X l X 2 . . . XkXk+ 1 )' : [ ( X l X 2 . . . Xk )Xk+ 1 ]'  [(X l X 2 . . . Xk )'+ ' ' ' ' Xk+l] (XPl +X 2 + ' ' ' + x k) + Xk+l Xl' +x2+''" +xk' + x'k+l" T h u s t h e r e s u l t follows by PMI.
7. xy + (x + y)' + xy'
11. {x'y' + l ( y ' ) ' l ' ] + Ix' + (0')']}
9. [ x'y' + (x' + y')' + x'( y') ' I '
13. yes 15. no
Exercises A.2 (p. 873)
1. 6 3. a 2 + b 2 5. 0
Appendix A
1027
c 2)
7. a b + b c + c a  a 2  b 2  c 11. (ad  bc)(eh  fg)
2
9.
(a + b + c ) ( a b + bc + c a  a 2  b 2

13. IABI = ( a d  b c ) ( e h  f g ) = IAI. IBI
=
15.
IAI.
[AIx b ab
IAI.
[AT[ = [A. AT[   I I n l
xa ba f(b)
1, so [AI = :El.
17. f ( x ) = ~
f(a) +
Exercises A.3 (p. 881)
1. f (x + y ) = a x +Y = a x . aY = f (x) . f (y)
3. Let f ( x ) = f ( y ) . T h e n a x = ay, so x = y. T h u s f ( x ) = f ( y ) implies x = y; so f is injective. 5. Suppose log 2 = a / b , w h e r e a a n d b are relatively p r i m e i n t e g e r s a n d a < b. T h e n 2 = l Oa/b; t h a t is, 2 b = 10 a. So 5 a = 2 b  a . This implies 215, which is a contradiction. T h u s log 2 is an i r r a t i o n a l n u m b e r . 7. pH =  log(3.76 • 10 8) ~ 7.4248 9. 0 decibals 13. 10.84006 11. 70 decibals 15. 2.67917 parsecs
Exercises A.4 (p. 887)
1. 2134 3. 4213 5. 21354 7. 23145
9. 2 3 1 , 3 1 2 , 321
13. 3421, 4123, 4132 17. 12,21 19. 134
11. 2341, 2413, 2431 15. 23541, 24135, 24153 21. 1345 23. 3456
25. 12, 13, 23
29. Algorithm next b y t e ( a z a z  . 
27. 12, 13, 14, 23, 24, 34
a8)
(* This a l g o r i t h m finds byte t h a t follows byte ala2 ... a8 in l e x i c o g r a p h i c order. *) Begin (* a l g o r i t h m *) i ,8 while bi  I do (* continue u n t i l bi  0 *) begin (* while *) bi ,   0 i ,i I
endwhi 1e
bi ~ 0 (* update bi *)
End (* algorithm *)
31. D , { 1 } 33. D , { 1 } , { 2 } , { 3 } , { 1 , 2 } , { 1 , 3 } , { 2 , 3 } , { 1 , 2 , 3 }
1028
Solutions to OddNumbered Exercises
Exercises A.5 (p. 892)
1. 30
9. 30
3. 210
11.  2
5. 1680
13.
7. 12,600
168
15.
X2
+ y2 +
Z 2 + 2xy
2yz
2zx
17. x 3  y 3 + z 3 _ 3x2y + 3x2z + 3xy2 + 3y2z + 3xz 2 _ 3yz 2 _ 3xyz 19. x 3 + 8y 3 + z 3 + 6x2y + 3x2z + 12xy 2 + 12y2z + 3xz 2 + 6yz 2 + 6xyz
21.
12
23. 2520
25. 3,326,400 31. 56 37. 210
27. k n 33. 10
29. 12,080,096,000 35. 455 39.
X 3 + 8 2 3 + Z 3 
6x2y + 3x2z + 12xy 2  12y2z + 3xz 2 43. 165

6yz 2 
6xyz
41. 10
45. 1,801,800
Credits
Cover image (sunflower head) courtesy of Runk, Schoenberger/Grant Heilman Photography, Inc. Text from The Divine Proportion by H. E. Huntley (Mineola, New York: Dover Publications, 1970) reprinted courtesy of the publisher. Coffee pot illustration (Figure 1.7) by Michael Crawford, New York. All biographical portrait illustrations by Nancy Crumpton.
1029
Index
A ABRACADABRA, 378379, 387388 Absolute value, 33, 126 Absorption laws, 83 Accepted string, 764 Accepting state, automaton, 760, 761, 783 Ackermann, Wilhelm, 278 function, 278 Acyclic graphs, 610611 See also Dags Adder full, 836837 half, 835836 Addition in base b, 199201 of functions, 123 of generating functions, 300301 of matrices, 166 rule (principle), 43, 99, 344345, 415 Adjacency edges and, 520 graphs and, 520 list representation, 452453, 538 matrix, 444445, 520522, 528, 716 Adjacent squares, in Karnaugh maps, 843 Algebra See also Boolean algebra origin of word, 96 ALGOL (ALGOrithmic Language), 748 Algorithms addition, 199200 binary multiplication, 203 binary search, 228230, 250251,311312, 327328 binary search tree, 665
breathfirst search (BFS), 621622 bubble sort, 230233, 248249, 317318 complexities of, 247252, 319333, 420422 connectivity, 471475, 481 correctness of, 224237 correctness of recursive, 316319 defined, 96 depthfirst search (DFS), 618621 Dijkstra, 717723 disjunctive normal form (DNF), 819 divide and conquer, 228, 327329 division, 185191 equivalence relation, 490 Euclidean, 191197, 322323 Eulerian circuit, 561 Eulerian graph, 559560 exponentiation, 318319 factorial, 226 Fibonacci, 310 findmax, 251 Floyd, 725 Gilbert's, 728 greatest common divisor (gcd), 311 greedy, 633 handshake, 308309 Horner's, 336 Huffman, 671675 identifier, 767768 insertion sort, 236237 Kruskal's, 616617, 626628 legally paired sequence, 106107 linear search, 227228, 249250, 316317, 319320, 421422 merge, 312313
merge sort, 313314, 328329 minmax, 252 multiplication, 225 nextcombination, 887 nextpermutation, 884885 nextsubset, 9697 nondecimal bases (baseb expansion), 197207 origin of word, 96 Prim's, 628633 product, 171 recursive, 307333 selection sort, 233234 space complexity, 247 spanning tree, 616617 subsets, 102 time complexity, 247248 topological sort, 501 tower, 309 Warshall's, 477481 Alphabet, 75 Greek, 894 input, 772 Alphanumeric character, 348 Ambiguous grammar, 753755 Ancestors, 636 And Boolean, 438 conjunctions and use of, 5 AND gate, 825826 Angle brackets, 744 Antisymmetric relations, 456458 Appel, Kenneth, 589 Arguments, 38 valid and invalid, 3949 Aristotle, 1, 2 Arithmetic, fundamental theorem of, 218221 Arithmetic mean, 54 Arithmetic sequence, 180 Around the World puzzle, 564565 Array, predecessor, 720
1031
1032
Index
ASCII, 67O character set, 867 Assignment operator, 7 Assignment statement, 7 Associated linear homogeneous recurrence relations with constant coefficients (ALHRRWCCs), 287, 293296 Associative law Boolean algebra, 806 of logic, 22 of sets, 83 Associative properties, 470, 804, 805 Automata. See Finitestate automata (FSA) Averagecase time complexity, 248, 420422 Axioms defined, 185 dual, 807
B
Bachmann, Paul Gustav Heinrich, 237, 238 Backtracking, 618 Backus, John W., 748 BackusNaur Form, 748 BackusNormal Form (BNF), 748751 Balanced trees, 640643 Barber paradox, 4243 Baseb expansion, 197207 Basis clause, 105, 262, 814 Basis step, 209 Begin, 18 Bell, Eric T., 397 Bell numbers, 397, 492 Bernoulli, Jacob I., 212, 422 inequality, 212 trials, 422425 Bestcase time complexity, 247 Biconditional statements, 1415 Bigoh estimate, 618 Bigoh notation, 237239 Bigomega notation, 243245 Bigtheta notation, 245 Bijection function, 137138 Binary alphabet, 75 Binary coded decimal (BCD) expansion, 852854 Binary digits, 95 Binary expression trees, 655 Binary multiplication, 201203
Binary numbers, 394395 Binary operators, 12, 31, 806 Binary predicate, 3334 Binary relation, 443444 Binary search algorithm, 228230, 250251, 311312, 327328 Binary search tree algorithm, 665 Binary search trees, 664668 Binary subtraction, 203205 Binary trees, 639, 646663 Catalan numbers and, 660 parenthesized triangulations and, 659660 Binet, Jacques Phillipe Marie, 277 Binet form of Fibonacci number, 276277, 290, 304305 Binomial coefficient, 366, 386, 889 Binomial expansion, 388 Binomial probability, 423 Binomial theorem, 386399, 888889 Bipartite graphs, 526 Birthday paradox, 182, 412 Bits operations, 438 string, 95 Boole, George, 1, 4, 803 Boolean algebra applications, 803 complement, 442, 806 defined, 806811 examples, 804806 expressions, 7, 814817 expression, selfdual, 863 matrices, 438443 operators, 12, 806 power, 441 precedence rules, 807 product, 439440, 806 sum, 806 unit element, 806 variables, 2, 813 zero element, 806 Boolean functions, 134 conjunctive normal form (CNF), 824 defined, 813814 disjunctive normal form (DNF), 817820 equality, 815817 functional completeness, 820 minterm, 817820 NAND and NOR, 820821 recursive definition, 814
Bound variables, 33 Breathfirst search (BFS) algorithm, 621622 Bridge, 686 Briggs, Henry, 879 Briggsian logarithms, 879 Bubble sort algorithm, 230233, 248249, 317318 Bus topology, 529 Bytes, 348349 C Cabbagegoatwolf puzzle, 552554 Cantor, Georg, 67, 68 diagonalization process, 141 Card dealing, 132133 Cardinality of sets, 98102, 140 Carroll, Lewis, 44, 45 Carry, 199 Cartesian plane, 88 Cartesian products, 8789 Catalan, Eugene Charles, 108 numbers, 108, 388390,393, 395396, 660 parenthesization problem, 394 Cayley, Arthur, 164, 519, 609 formula, 626 Ceiling function, 126130 Center, tree, 614 Characteristic equation, 287288 Characteristic function, 131132 Characteristic roots, 288 Chessboard puzzle eight queen puzzle, 620 four queen puzzle, 619620 knight's tour problem, 558560, 567571 two queen puzzle, 133134 Child, 636 left and right, 646 Chomsky, (Avram) Noam, 751, 752 hierarchy of grammar, 751 Chromatic number, 588 Chuckaluck game, 423424 Circle problem, 370372 Circuit, combinatorial, 830840 minimization of, 840850 Circuits, graph, 548 Eulerian, 556 Circuits, sequential, 830 Class, equivalence, 486490 Closed interval, 73 Closedopen interval, 73
Index
1033
Closed path, 548 Closure Kleene, 739741 reflexive, 482 symmetric, 482 transitive, 475482 Coconuts and monkey problem, 437, 493 Codeabar system, 493 Codes, gray, 446447 Coding scheme, Huffman, 670 Codomain of functions, 118 Coefficients binomial, 366, 386, 889 multinomial, 891 Cofactor, 868 Collision, 138139 Coloring, graph, 586598 Column vector, matrix, 165 Combinations, 365375 generating, 383384, 885887 with repetitions, 379381 Combinatorial circuits, 830840 minimization of, 840850 Combinatorics binomial theorem, 386399 combinations, 365375 defined, 343 derangements, 360365 thndamental counting principles, 344351 generalized inclusion/exclusion principle, 399409 permutations, 351360 repetitions, 375386 Common difference, 180 Common factor, 191 Common ratio, 180 Communication link, 2stage, 550 Commutative law Boolean algebra, 806 of logic, 22 of sets, 83 Commutative properties, 804, 805 Comparable elements, 495 Complements Boolean, 442, 806 of graphs, 536 properties, 805, 806 relation, 469 of sets, 81 Complete bipartite graphs, 527 Complete graphs, 524526 Completeness, functional, 820
Complete npartite graph, 605 Complexities of algorithms, 247252 Complexities of graphs, 622623 Complexities of recursive algorithms, 319333 Components, 5 Composite numbers, 189190 Composition function, 150153 of relations, 463464 Compound propositions, 5 Computer operations, with sets, 9498 Computer representations of graphs, 538539 of relations, 449454 Concatenation, 76 of languages, 736739 Conclusion, 9 Concurrent lines, 266 Conditional probability, 417419 Conditional statements, defined, 9 Congruence to b modulo m, 484 relation, 484486 Conjunction rule, 43 Conjunctions, 57 symbol for, 5 Conjunctive normal form (CNF), 824 Connected digraphs strongly, 698700 weakly, 705 Connected graphs, 549554 Connectives, 5 Connectivity relation, 471475, 481 Constructive existence proof, 5253 Contextfree grammar, 751 Contextfree language, 751753 Contextsensitive grammar, 751 Contextsensitive language, 751753 Contingencies, 16 Contradictions, 16 proof by, 51 Contrapositive, 1112 proof of, 51 Contrapositive law, 22, 43 Converse, 1112 Correctness of algorithms, 224237
Correctness of recursive algorithms, 316319 Correct program, 224225 Correspondence, onetoone, 137138 Countable sets, 140141 Counterexample, 53 Counting principles, 344351 Cycle digraphs, 698 graphs, 526 Hamiltonian, 565566 of a path, 446, 548 Cyclic permutations, 355356
D
Dags (directed acyclic graphs), 707715 Databases, 464466 Data field, 451 de Bruijn, Nicolaas G., 702 graph, 703 sequence, 702704 Decay curves, 875 Decimal expansion, 197 Decision trees, 676680 Degree of a region, 581582 Degree of vertex, 520, 694698 De Moivre, Abraham, 298, 299 De Morgan, Augustus, 23, 207 De Morgan's law of logic description of, 2224 for negating quantifiers, 3536 De Morgan's law of sets, 83 Dependent events, 419420 Depthfirst search (DFS) algorithm, 618621 Derangements, 360365 counting, 404408 Derivation, language and, 746747 Derivation trees, 743744 Descartes, Ren~, 87 Detachment, law of, 43 Determinants, 867874 Deterministic finitestate automata (DFSA), 779782 Diagonal, matrix, 166 Diagrams Hasse, 496499 transition, 760, 772 tree, 6 Venn, 7273
1034
Index
Difference between sets, 80 symmetric, 82 Differential and Integral Calculus, The (De Morgan), 23 Digraphs (directed graphs) closed, 698 dags, 707715 de Bruijn sequence, 702704 degree of vertex, 694698 features of, 691692 open, 698 relations and, 443447 Fibonacci numbers and, 692694 nstage win, 698 paths, reachability, and cycles, 698 strongly connected, 698700 teleprinter's problem, 700702 weakly connected, 705 weighted, 715725 Dijkstra, Edsger Wybe, 717 algorithm, 717723 Dirac, Gabriel Andrew, 571 theorem, 570 Directed acyclic graphs. See Dags Directed edges, 692 Directed graphs. See Digraphs Directly derivable, 746 Direct proofs, 50 Dirichlet, Peter Gustav Lejeune, 144, 145 Dirichlet box principle. See Pigeonhole principle Disconnected graphs, 549 Discrete probability, 409427 Disjoint sets, 71 Disjunctions defined, 78 exclusive, 8, 18 inclusive, 8 symbol for, 7 Disjunctive normal form (DNF), 817820 Distributive law Boolean algebra, 806 of logic, 22 of sets, 83 Distributive properties, 804, 805 Div function, 132134 Divide and conquer algorithm, 228, 327329 Dividend, 186
Divisibility properties, 189197 Division algorithm, 185191 Divisor, 186 greatest common divisor (gcd), 191 Doctrine of Chances, The (De Moivre), 299 Domain of functions, 118 of relations, 443 Domination law of logic, 21 of sets, 83 Don't care conditions, 851856 Dot product, 493 Double complementation law, 83 Double negation law, 22 Dual axioms, 807 Duality, principle of, 807 Dudeney, Henry Ernest, 4041 Duffinian, 257 Dump state, 767
E
Easter Sunday, 135, 195 Eccentricity, tree, 614 Edges, graph, 445, 517 adjacent, 520 directed, 692 parallel, 518 Eight coins puzzle, 678 Eight Queens puzzle, 620 Elementary operations, graph, 583 Elements comparable, 495 greatest and least, 500 matrix, 165 maximal, 499 minimal, 499 noncomparable, 495 of sets, 68 Elements (Euclid), 192 Ellipsis, 73 Empty language, 735 Empty relation, 455 Empty sets, 70 Empty word, 75 Endpoints, path, 546 Endwhile, 18 Equality of Boolean expressions, 815817 of functions, 123
of generating functions, 299300 of matrices, 166 of relations, 455 of sets, 70 of words, 734735 Equivalence class, 486490 Equivalence relations, 482484 Equivalence statement, 483 Equivalent combinatorial circuits, 840850 Equivalent finitestate automata, 764768 Equivalent nondeterministic finitestate automata, 784785 Equivalent switching networks 2526 ErdSs, Paul, 147 theorem, 146147 Euclid, 191 Euclidean algorithm, 191195, 322323 Euler, Leonhard, 53, 72,393, 516517 circuit, 556 formula, 578581 graphs, 556564 path, 556 phifunction, 196 theorem, 512 Even parity, 765 Event(s) defined, 410 dependent and independent, 419420 mutually exclusive, 414415 probability of, 410414 Eves, Howard, 67, 391 Excluded middle, law of, 28 Exclusive disjunction, 8, 18 Existence proof, 5253 Existential quantifiers, 32 Expansion baseb, 197207 binary coded decimal (BCD), 852854 binomial, 388 decimal, 197 Expected value, 419420 Experiment, 409 Exponential function, 125, 874877 Expressions, Boolean, 7, 814817, 863
Index
1035
F
Factorial, 108, 226, 308 Fallacy, 39 Family tree, 611 Fermat, PierreSimon de, 4, 5 Last Theorem, 4, 145, 511 number, 338 primes, 53 Fibonacci, Leonardo, 268, 269 Fibonacci numbers, 269271 algorithm, 310 Binet form of, 276277, 290, 305305 complexity of iterative version, 320321 digraphs and, 692694 paraffins and, 523524 permutations and, 356358 recursive, 310 Fibonacci trees, 646647 Final state, automaton, 761, 783 Finite sequences, 158 Finite sets, 7374 number of partitions of, 490 Finitestate automata (FSA) applications, 759761 defined, 761764 deterministic, 779782 equivalent, 764768, 784785 languages accepted, 787792 nondeterministic, 782787 Finitestate machines (FSMs), 771 applications, 733 characteristics, 772778 Mealy, 772 Moore, 799800 simply minimal, 800 unit delay, 774 Fixedlength code, 670 Floor function, 126130 Floyd, Robert W., 725, 726 algorithm, 725 Football pools, 426427 Forest, 611 For loop, 233, 381382 Formal languages. See Languages Formal power series, 299 Formulas Cayley's, 626 Euler's, 578581 midpoint, 145 molecular, 519 recursion, 262 structural, 519520
wellformed, 107 Zeller's, 492 FORTRAN, 748 Fourcolor problem, 589595 Four Queens puzzle, 619620 Free variables, 33 Fridaythe13th, 485486 Friedberg, S. H., 325 Fulladder, 836837 Full tree, 639 Functional completeness, 820 Functions absolute value, 126 Ackermann, 278 applications, 117118 bijection, 137,38 Boolean, 134, 813824 cardinality, 140 ceiling, 126130 characteristic, 131132 codomain of, 118 composition, 150153 defined, 118120, 448 div, 132134 domain of, 118 equality of, 123 Euler's phi, 196 exponential, 125, 874877 floor, 126130 generating, 298307, 383384 graphing, 122123 greatest integer, 127 growth of, 237247 hashing, 138139 identity, 136 injection, 136137 inverse, 153155 least integer, 127 linear, 125 logarithmic, 125126, 877879 McCarthy's 91,275 mod, 132134 nextstate, 761,783 onetoone, 136137 onetoone correspondence, 137138 onto, 137 order, most common, 239241 output, 772 piecewise definition, 121123 pigeonhole principle and, 144150 polynomial, 125 product ot, 123
properties of, 136143 quadratic, 125 range of, 120121 recursive definition of, 262278 sum of, 123 surjection, 137 transition, 761,772, 783 Fundamental theorem of arithmetic, 218221 Fuzzy decisions, 2829 Fuzzy logic, 2628 Fuzzy sets, 9192 Fuzzy subsets, 9192 G Game tree, 637 Gates AND, 825826 defined, 824 logic, 824829 NAND, 828 NOR, 828829 NOT, 827 OR, 826 XOR, 18 Gauss, Karl Friedrich, 145, 222, 484 Generalized inclt~sion/exclusi~n principle (GIEP), 399409 Generalized pi~.~eonhole principle, 147149 Generating combinati~ns, 383384,885887 Generating functions, 298307, 383384 Generating permutations, 352358, 883 Geometric mean, 54 Geometric sequence, 180 Gilbert, E. N., 728 algorithm, 728 Goldbach, Christian, 4, 5 conjecture, 4 Golden ratio, 179, 276 Graceful graph, 604 Gracelhl tree, 686 Grammars, 743 759 ambiguous, 753755 Chomsky hierarchy, 751 contextsensitiw,, contextl'rt'e, and regular, 751 Type 0, 751 Type 1, 751 3'ype 2, 751 Type 3, 751
1036
Index
Graphing functions, 122123 Graphs See also Digraphs (directed graphs) acyclic, 610611 adjacency and incidence, 520 bipartite, 526 coloring, 586598 complement of, 536 complete, 524526 complete bipartite, 527 complete npartite, 605 complexity, 622623 computer representations of, 538539 connected, 549554 cycle, 526 de Bruijn, 703 degree of vertex, 520 disconnected, 549 edges, 445, 517 Eulerian, 556564 graceful, 604 Hamiltonian, 565567 homeomorphic, 583 isomorphic, 541544 nodes, 445, 517 nonplanar, 582 paths, 446447, 546547 Petersen, 530531 planar, 576586 planar representation, 576 ret~mlar, 536 selfcomplementary, 605 simple, 518520 square, 575 sub, 522523 underlying, 702, 705 undirected, 517 union of, 532533 vertex, 445, 517 weighted, 527528 wheel, 526, 623 Gray, Frank, 447 Gray codes, 446447,566 Greatest common divisor (gcd), 191 recursive, 311 Greatest element, 500 Greatest integer function, 127 Greedy algorithm, 633 Growth curves, 874 Guthrie, Francis, 589
H
Haken, Wolfang, 589 Halfadder, 835836 Halflife, 881 HAM (Hampton Court) game, 572573 Hamilton, William Rowan, 564, 565 Around the World puzzle, 564565 graph, 565567 Hamming, Richard W., 180 distance, 179, 446 Handshake problem, 264265, 280, 308309, 431,524526, 533 Harmonic mean, 351 Harmonic number, 332 Hash function, 138139 Hasse, Helmut, 496, 497 diagrams, 496499 Heap, 669 Heapsort, 669 Heawood, Percy John, 589 Height, tree, 637 Heterogeneous trees, 655 Hexagonal number, 284 Hilbert, David, 74 hotel paradoxes, 7475 Hoare, C. Anthony R., 316 Homeomorphic graphs, 583 Homogeneous trees, 664 Horner, William G., 207 algorithm, 336 method, 207 Huffman, David Albert, 671 algorithm, 671675 coding scheme, 670 trees, 670675 Huygens, Christian, 419 Hypothesis, 9 inductive, 209
I
Ibn Ezra, Rabbi, 366 Icosian calculus, 565 Idempotent law of logic, 21 of sets, 83 Identity function, 136 Identity law Boolean algebra, 806 of logic, 21 of sets, 83 Identity matrices, 166
Identity properties, 805 Ifandonlyif statement, 1415 If statements, Boolean expressions in, 7 Ifthenelse statements composition, 152 piecewise definition, 122 Ifthen statements, 753754 Image, function, 119 Immediate predecessor, 692 Immediate successor, 692 Implication conversion law, 22 Implications, 911 converse, inverse, and contrapositive of, 1112 Incidence, graph, 520 Inclusion/exclusion principle, 99101,345, 414415 generalized, 399409 Inclusive disjunction, 8 Indegree, vertex, 694 Independent events, 419420 Index, summation, 158, 160161 Indexed sets, 86 Indirect prool's, 51 Induction. See Mathematical induction Induction step, 209 Inductive definitions, 263 Inductive hypothesis, 209 Inequality, mathematical induction and, 212 Inference rules, 43 Inferential form, 38 Infinite sequences, 158 Infinite sets, 7374 Infinity symbol, 74 Infix notation, 653 Initial state finitestate automaton, 760, 761,783 finitestate machine, 772 Initial vertex, 692 Injection function, 136137 Inorder traversal, tree, 647648 Input alphabet, 772 Input symbols, 761,783 Insertion sort algorithm, 236237 Interior points, 477 Internal vertex, 636 International Standard Book Number (ISBN), 509
Index
1037
Intersections relations, 462463 of sets, 7880 Inter