CONTENTS GRAPH THEORY

Transcription

1 CONTENTS i GRAPH THEORY

2 GRAPH THEORY By Udit Agarwal M.Sc. (Maths), M.C.A. Sr. Lecturer, Rakshpal Bahadur Management Institute, Bareilly Umeshpal Singh (MCA) Director, Rotary Institute of Management and Technology, Moradabad UNIVERSITY SCIENCE PRESS (An Imprint of Laxmi Publications Pvt. Ltd.) BANGALORE CHENNAI COCHIN GUWAHATI HYDERABAD JALANDHAR KOLKATA LUCKNOW MUMBAI RANCHI NEW DELHI

3 Published by : UNIVERSITY SCIENCE PRESS (An Imprint of Laxmi Publications Pvt. Ltd.) 113, Golden House, Daryaganj, New Delhi Phone : Fax : info@laxmipublications.com Copyright 2009 by Laxmi Publications Pvt. Ltd. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher. Price : Rs Only. First Edition : 2009 OFFICES India Bangalore Jalandhar Chennai Kolkata Cochin Lucknow Guwahati , Mumbai , Hyderabad Ranchi UGT GRAPH THEORY C Typeset at : Sukuvisa Enterprises, New Delhi. Printed at :

4 CONTENTS Preface (xi) 1. Elementary Combinatorics Introduction Basic Counting Principles The Inclusion Exclusion Principle Permutations (Arrangement of Objects) Combinations Probability 22 Exercises Discrete Numeric Functions and Generating Functions Discrete Numeric Functions Manipulation of Numeric Functions Definitions of S i a and S i a for a Numeric Function Generating Function Operations on Generating Functions Finding a Generating Function of the Fibonacci Sequence Solution of Combinatorial Problems using Generating Functions 52 Exercises Recurrence Relations Introduction Order of Recurrence Relation Degree of the Recurrence Relation Solution of Linear Recurrence Relation with Constant Co-efficients Solution of Recurrence Relation by the Method of Generating Function Backtracking Method Forward Chaining Method Summation Method 88 Exercises 88 v

5 vi CONTENTS 4. Graphs Introduction What is a Graph? Basic Terminologies Regular Graph Complete Graph Null Graph Bipartite Graph Degree Sequence Graphs Isomorphism Complement of a Simple Graph Homeomorphic Graphs Subgraphs Operations of Graphs Connected Graphs and Disconnected Graphs Walk, Paths and Circuits 115 Exercises Euler and Hamiltonian Graphs Euler Graphs Unicursal Graph Constructing Eulerian Paths and Cycles Hamiltonian Graphs Introduction to Travelling Salesman Problem Chinese Postman Problem 137 Exercises Trees Introduction Trivial and Non-trivial Tree Properties of Trees Rooted Trees Pendant Vertex Distance and Centers in a Tree Binary Trees Counting Tree Spanning Tree Shortest Path Problems Fundamental Circuits 162 Exercises 163

6 CONTENTS vii 7. Cut-sets and Network Flows Cut-sets Fundamental Cut-set Connectivity Separable Graph Network-flow Problems (Transport Network) Ford and Fulkerson Algorithm Cut and Its Capacity Max-flow Min-cut Theorem 181 Exercises Planar and Dual Graphs Introduction Kuratowski s Graphs Region Euler s Formula Some Planar Graphs Detection of Planarity of a Graph Geometric Dual Combinatorial Dual Thickness and Crossing 204 Exercises Vector Spaces of a Graph Introduction Sets with One Operation Modular Arithmetic Galois Fields Vectors Vector Space Vector Space Associated with a Graph Basis Vectors of a Graph Circuit and Cut-set Subspaces Orthogonal Vectors and Spaces 218 Exercises Matrix-representation of Graphs Introduction Incidence Matrix Circuit Matrix Cut-set Matrix 226

7 viii CONTENTS 10.5 Relationship between A f, B f and C f Path Matrix Adjacency Matrix 231 Exercises Coloring of Graphs Graph Coloring Chromatic Number Chromatic Partitioning and Independent Sets Chromatic Polynomial Matching Covering Four Color Problem The Five Color Theorem The Six Color Theorem 262 Exercises Directed Graphs Introduction Directed Graph Types of Digraphs Directed Paths and Connectedness Binary Relations and Directed Graphs Euler Directed Graphs Trees with Directed Edges Arborescence Acyclic Digraphs Fundamental Circuit in Directed Graphs Incidence Matrix of a Digraph Circuit Matrix of a Digraph Adjacency Matrix of a Digraph Orientations and Tournaments 281 Exercises Enumeration of Graphs Types of Enumeration Cayley s Theorem Counting Unlabeled Trees Rooted Unlabeled Trees Free Unlabeled Trees Permutation 289

8 CONTENTS ix 13.7 Polya s Counting Theorem Graph Enumeration with Polya s Theorem 293 Exercises Applications of Graph Theory Introduction Graphs in Computer Science Miscellaneous Applications 300 Exercises 300 Index

9 PREFACE The textbook Introduction to Graph Theory has been designed primarily to meet the requirements of B.E./B.Tech students of all technical colleges affiliated to U.P. Technical University. The book will also be fruitful to the candidates appearing in UGC, NET, GATE and other competitive examinations. All the principles and fundamental concepts have been explained very clearly leaving no scope of illusion and confusion. The language used in this book is simple and lucid in style. No previous knowledge of Graph Theory is required to follow this book. The book has been made as much self-contained as could be possible. Many people have contributed directly or indirectly to the completion of this book. We gratefully acknowledge our indebtedness to various authors and publishers whose books have been freely consulted during the preparation of this book. We specially thank to Mrs. Veena Mathur, Er. Navin P. Mathur, Dr. Manish Sharma, Dr. Ravinder Sing and Dr. Neeraj Saxena, who always encouraged us and gave us constructive suggestions. We are thankful to all of them. We wish to thank our colleagues of the departments of Mathematics and Computer Science for their many helpful corrections, comments and suggestions. Any suggestions for improving the contents would be warmly appreciated. AUTHORS xi

10 1 Chapter ELEMENTARY COMBINATORICS 1.1 INTRODUCTION In our daily life, we often come across the problems of finding the number of ways in which a set of objects may be arranged or selected under certain given condition or conditions. This process is also known as the sophisticated counting or counting without counting. In this process of arranging a set of objects, we are generally concerned with the problems some of which are given below : (i) Seating arrangements (linear or circular). (ii) Arrangements of books on a shelf. (iii) Formation of numbers with given digits. (iv) Arrangements of the letters of a given word, etc. In the process of selection some of the problems are of the following nature : (i) Selection of a team or a committee from a group of persons. (ii) Finding the number of lines passing through a given number of points in the plane, etc. Combinatorics is the study of arrangements of objects and counting of objects with certain properties. Combinatorics is concerned with the study of how to combine, select and partition elements of sets in certain way. Thus, aspects of combinatorics include counting the objects satisfying certain criteria (enumerative combinatorics), deciding when the criteria can be met, and constructing and analyzing objects meeting the criteria, finding largest, smallest, or optimal objects (extremal combinatorics), and finding algebraic structures these objects may have (algebraic combinatorics). 1

11 2 GRAPH THEORY 1.2 BASIC COUNTING PRINCIPLES In this section, we will discuss two basic counting principles and illustrate the use of these counting principles in solving different counting problems Sum Rule (The Principle of Disjunctive Counting) If one action can be performed independently in n 1 different ways and another disjoint action can be performed independently in n 2 different ways, and yet another disjoint action can be performed independently in n 3 different ways, and so on. (the number of actions being finite), then the total number of ways in which either of these actions can be performed is n 1 + n 2 + n In this, it is presumed that only one action can be performed at a time. EXAMPLE 1. How many ways can we get a sum of 8 when two indistinguishable dice are rolled? SOLUTION. We should obtain a sum of 8 by the outcomes (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2) but the dice are similar so the outcomes {(2, 6) and (6, 2)}, and {(3, 5) and (5, 3)} cannot be differentiated. Thus, we obtain the sum of 8 in only 3 ways. EXAMPLE 2. If there are 36 boys and 24 girls in a class, find the number of ways of selecting one student as class representative. SOLUTION. Using sum rule, there are = 60 ways of selecting one student (either a boy or girl) as a class representative. EXAMPLE 3. In how many ways can we draw a heart or a spade from an ordinary deck of cards? SOLUTION. In a deck of cards, there are 13 spades and 13 hearts. So, we may draw a heart in 13 ways and a spade also in 13 ways. Thus, by sum rule, a spade or a heart may be drawn in = 26 ways. EXAMPLE 4. In how many ways can we choose a prime number or an even number between 10 and 20 (excluding both numbers)? SOLUTION. A prime number between 10 and 20 can be chosen in 4 ways i.e., 11, 13, 17 and 19 and an even number between 10 and 20 can be chosen in 4 ways i.e., 12, 14, 16, 18. Thus, by sum rule, a prime or an even number between 10 and 20 can be chosen in = 8 ways. EXAMPLE 5. What is the value of k after the following code has been executed? k 0 for i 1 to n 1 k k + 1 for j 1 to n 2 k k + 1 for m 1 to n 3 k k + 1

12 ELEMENTARY COMBINATORICS 3 SOLUTION. The initial value of k is 0. This block of code is made up of 3 different loops and no. two of these loops can be done at the same time. Each time a loop is traversed, 1 is added to k. The sum rule shows that the final value of k is n 1 + n 2 + n Product rule (The Principle of Sequential Counting) If one action can be performed independently in n 1 different ways, and after it has been done, if a second action can be performed independently in n 2 different ways, and after both have been done, if a third action can be performed independently in n 3 different ways and so on (the number of actions being finite), then the total number of ways in which all the actions can be performed together is : n 1 n 2 n 3... Here we have assumed that each way of performing the first action may be followed by any way of performing the second action, and is then followed by any way of performing the third action and so on. Sum rule is in contrast to the product rule where we consider the sequence of actions occurring together and thereby the number of ways get multiplied. EXAMPLE 6. A coin is tossed four times and the result of each toss is recorded. How many different sequences of heads and tails are possible? SOLUTION. Each tossing of a coin results in either head or tail i.e., there are two possible outcomes for each toss. Thus, by product rule, there are = 2 4 different sequence of heads and tails. EXAMPLE 7. How many different bit strings are there of length 7? SOLUTION. Each of the seven bits can be chosen in two ways since each bit is either 0 or 1. Therefore, the product rule shows there are a total of 2 7 = 128 different bit strings of length seven. EXAMPLE 8. An office building contains 27 floors and has 40 offices on each floor. How many offices are there in the building? SOLUTION. There are 27 ways to choose the floor and 40 ways to choose an office for each of these floor. Thus, by product rule, there are = 1080 offices. EXAMPLE 9. How many 4-digit numbers can be formed using the digits 1, 3, 4, 6, 7 and 8? How many 4-digit numbers can be formed if no digit can be repeated? SOLUTION. The number of digits that can be used to form a 4-digit number is 6. So each of the four digits can be chosen in 6 ways. Therefore, by the product rule, = 6 4 = 1296 four digit numbers can be formed. If the repetition of digits is not allowed, then thousand s place can be filled in 6 ways, hundred s place can be filled in 5 ways, ten s place can be filled in 4 ways and unit s place in 3 ways. Thus, by product rule, = 360 different 4-digit numbers can be formed.

13 4 GRAPH THEORY EXAMPLE 10. How many two-digit or three-digit numbers can be formed using digits 1, 2, 3,5, 7, 9, if no repetition is allowed? SOLUTION. The number of digits which can be used to form the numbers are six. Number of ways in which two-digit numbers can be formed = 6 5 = 30 and the number of ways in which three-digit numbers can be formed = = 120. By sum rule, the number of two-digit or three-digit numbers that can be formed = = 150. EXAMPLE 11. Each user on a computer system has a password, which is 6 to 8 characters long, where each character is an upper case letter or a digit. Each password must contain at least one digit. How many possible passwords are there? SOLUTION. Let P be the total number of passwords. P 1 be the passwords of length 6. P 2 be the passwords of length 7. P 3 be the passwords of length 8. By sum rule P = P 1 + P 2 + P 3 To find P 1, it is easier to find the number of strings of upper case letters and digits that are six characters long and subtract from this the number of strings with no digits. By the product rule, the number of strings of six characters is ( ) 6 = 36 6 and the number of strings with no digits is (26) 6. (26 English letters and 10 digits) Hence P 1 = (36) 6 (26) 6 = 1,867,866,560 Similarly, P 2 = (36) 7 (26) 7 = 70,332,353,920 and P 3 = (36) 8 (26) 8 = 2,612,282,842,880 So, P = P 1 + P 2 + P 3 = 2,684,483,063,360. EXAMPLE 12. How many different license plates are available if each contains a sequence of four letters followed by three of the digits 0 9 (not necessarily distinct)? SOLUTION. There are 26 choices for each of the four letters. So by the product rule, the number of ways in which the four letters can be chosen in = (26) 4 ways and the three digits can be chosen in = (10) 3 ways. Hence, by the product rule, there are total (26) 4 (10) 3 = 456,976,000 different license plates. EXAMPLE 13. What is the value of m after the following code has been executed? m 0 for i 1 1 to n 1

14 ELEMENTARY COMBINATORICS 5 for i 2 1 to n 2 for i 3 1 to n for i m 1 to n m m m + 1 SOLUTION. By product rule, it follows that the rested loop is traversed n 1 n 2... n m times. Hence the final value of m is n 1 n 2... n m. EXAMPLE 14. How many numbers in the range to do not have any repeated digits? SOLUTION. There are nine choices for the first digit from 1 to 9. Once this has been chosen, there remains still 9 choices for the second because now 0 be used. There are now 8 choices for the third digit, seven for the fourth digit and six for the fifth. So there are = possible numbers. EXAMPLE 15. How many even numbers in the range have no repeated digits? SOLUTION. This event can be partitioned into two mutually exclusive cases. Case 1 : The number ends in 0. In this case, there are 9 choices for the first digit (1 9) and then 8 choices for the second (since 0 and the first digit must be excluded). So there are 9 8 = 72 numbers of this type. Case 2 : The number does not end in 0. Now, there are four choices for the final digit (2, 4, 6 and 8) and 8 choices for the first digit (0 and last digit are excluded) and 8 choices for the second digit (the first and last digits are excluded). There are = 256 numbers. By addition rule, there are = 328 even numbers. EXAMPLE 16. Three persons enter into a room, where there are 5 chairs. In how many ways can they take up their chairs? SOLUTION. The first person has a choice of 5 chairs and can sit in any one of those 5 seats. So there are 5 ways. Second person has a choice of 4 seats and the third person has a choice of 3 chairs. Hence the required number of ways = = 60. EXAMPLE 17. Suppose that we draw a card from a deck of 52 cards and replace it before the next draw. In how many ways can 10 cards be drawn so that the tenth card is a repetition of a previous draw? SOLUTION. First we count the number of ways, we can draw 10 cards so that the 10 th card is not a repetition. First choose what the 10 th card will be. This can be done in 52 ways. If the first 9 draws are different from this, then each of the 9 draws can be chosen from 51 cards. Thus, there are (51) 9 ways to draw the first 9 cards different from the 10 th card.

15 6 GRAPH THEORY Hence there are (51) 9 52 ways to choose 10 cards with the 10 th card different from any of the previous 9 draws. There are (52) 10 ways to draw 10 cards with replacements. So, there are (52) 10 [(51) 9 52] ways to draw 10 cards where the 10 th is a repetition. EXAMPLE 18. In how many ways can 10 people be seated in a row so that a certain pair of them are not next to each other? SOLUTION. There are i.e., 10! ways of seating all 10 people. The number of ways of seating the 10 people where the certain pair of people (Say x and y) are seated next to each other. If we treat the pair xy as one entity, then there are 9 total entities. These are arrange in 9! ways. But, x and y can be seated next to each other in 2 different orders i.e., xy and yx. Thus, there are (2) (9!) ways of seating all 10 people where x and y are next to each other. So, the number of ways in which 10 people be seated in a row and certain pair of them are not next to each other is 10! (2)(9!). EXAMPLE 19. Find the number of three digit natural numbers having digits in increasing order from left to right. SOLUTION. First place can be filled by numbers 1, 2, 3,... 7, second place by 2, 3, 4,... 8 and third by 3, 4, 5,... 9 i.e., the number of ways of filling each place is seven but total number of ways is not = 343 because number of ways of filling each place is not independent. Reason behind this is that corresponding to 1 at first place, second place can filled up by anyone of seven digits i.e., 2 to 8 but when we put 2 at first place, the number of ways of filling second place is only six. The rule of product is used only when the number of ways of doing each part is independent. So, the right approach for this problem is that first select three distinct non-zero digits which can be done in 9 C 3 ways then arrange them in increasing order which can be done in only one way. Therefore, the required number of natural numbers are 9 C 3 1 = = 84 9! ! = 3!(9-3)! ! For each positive integer we define n! = n.(n 1)(n 2) = the product of all integers from 1 to n. Also define 0! = 1 Note that 1! = 1 Thus 7! = (We read n! as n factorial ) The relation n! = n[(n 1)!] enables us to compute the values of n! for small n fairly quickly.

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