Ralph p grimaldi discrete and combinatorial mathematics ,5e (instructors solution manual) pearson addison wesley (2004)

INSTRUCTOR'S SOLUTIONS MANUAL DISCRETE AND COMBINAtORIAL MATHEMATICS FIFTH EDITION Ralph P Gritnaldi Rose-Hulman Institute o/Technology Boston San Fra.'1cisco New York London Toronto Sydney Tokyo Mexico City Munich Paris Town Madrid Reproduced by Pearson Addison-Wesley from electronic files supplied by the author Copyright © 2004 Pearson Education, Inc Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116 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 Printed in the United States of America written permission of the ISBN 0-201-72660-2 CRS 06 05 04 03 PEARSON Addison Wesley Dedicated to the memory of Nellie and Glen /Fuzzy/ Shidler CONTENTS PART FUNDAMENTALS OF DISCRETE Chapter Fundamental Principles Counting Chapter :Fundamentals of Logic 26 Chapter Set Theory 59 Chapter Properties of the Integel's: Mathematical Induction 95 Chapter Relations and Functions 134 Chapter Languages: Finite State Machines 167 Chapter Relations: The Second Time Around 179 PART FURTHER TOPICS IN ENUMERATION 207 Chapter The Principle of Inclusion and Exclusion 209 Chapter Generating Functions 229 Chapter 10 Recurrence Relations 243 GRAPH 1.'HEORY AND APPLICATIONS 281 4: Chapter 14 Riugs and Modular Arithmetic 369 Chapter 15 Boolean Algebra and Switching Functions 396 Chapter 16 Groups, Coding Theory, and Polya's Method Enumeration 413 Chapter Fini te Fields and Comhinatorial Designs 440 THE APPENDICES 459 Appendix Exponential and Logarithmic Functions 461 Appendix Properties of Matrices 464 Appendix Countable and Uncountable Sets 468 PART FUNDAMENTALS OF DISCRETE MATHEMATICS CHAPTER Sections and ( a) By the rule of sum, there are = 13 for the eventual winner (b) Sim~e there are eight Republica,ns and Democrats, the rule of product we have X ::;:;; 40 possible pairs of opposing candidates (c) The rule of sum in part (a); the rule of product in part (b) By the rule of product there are x x x x x = 56 license plates where the first two symbols are vowels and the last four are even digits By the rule of product there are (a) X 12 X x = 288 distinct Buicks that can be manufactured Of these, (b) x x x = 24 are blue ( a) From the rule of product there are 10 X X X = P( 10, 4) = 5040 possible slates (b) (i) There are x x x = 1512 slates where a physician is nominated for president (ii) The number of slates with exactly one physician appearing is x [3 x x x 5] = 2520 (iii) There are x x x 840 slates where no physician is nominated for any of the four offices Consequently, 5040 - 840 = 4200 slates include at least one physician = Based on the evidence supplied by Jennifer and Tiffany, from the rule of product we find that there are x x x 10 x 10 x = 800 different license plates (a) Here we are dealing with the permutations of 30 objects (the runners) taken (the first eight finishing positions) at a trophies can be awarded in P(SO,S) = 30!/22! ways (b) n.!.Hl>en;(:!, and runners in ways each these ways, there are P(28,6) ways for the other finishers (in the top 8) to finish the race the are 6· to with two runners a.U:lon,~ the top are (a.) 12! re31~nCl"lOl,LI); (b) (4!)( 8!) ways so that the four ,." ."' and (c) (4!)(51)(3!) where the top , """,., tare PT(:JCe:8SE: 6/AV( v-I) = 21" ==> A(v - IS even AV(v -1) = vr(k -1) = bk(k -1) = (b) Here ),.=1 By part (a) 6Iv(v-l)=>3Iv(v-1)==>3Iv or 31(v-l),since3is prime Also, by part (a) A(v - 1) = (v -1) is even, so v is odd (i) 31v ==> v = 3t, todd => v = 3(209 1) = 6s and v (mod 6) (ii) 31(v-l)==>v-l=3t,t even =}v-l=6x=>v=6x+l and v=l (mod 6) = = Here v = 9, k = 3, b 12, l' = and A(v -1) so the design is a Steiner triple system k = 3, ) 10 (a) v= = b= => v = 9, r = P2) P3) b= 21; r=7 = r(k -1) = 4(2) => 8A = ==> ) = 1, 14 Here = n, b = p, k r = bk/v = fJ = m (a) (b) >.(v -·-1) = r(k 15 ::::; >.(n -1) = (pm/n)(m 1) ::::; > = => n = 5, so (b) n + n + = 57 ::::; n = (a) n are = [pm(m -l)J/[n(n -1») +n = 31 points in this projective plane are n + = points on each line of this plane 16 The lines y = x and y = x z, for example, would intersect at the two distinct points (0,0,0) and (1,1,0) This contradicts conditions (PI) and (P2) of Definition 17 17 (a) v (b) (c) 18 = k = 6; > = v = b = 57; r = k = 8; > = v = b = 73; 'f' = k = 9; > = =b= 31; r (a) There are rune (0,0), (1,0), (0,1), (1,1), (0,2), (1,2), and 12 lines: points: (2,0) (2,1) (2,2) ° = x=l x=2 x y=o y=l y=2 y=x y=x y=x+2 y = 2x = 2x y = 2x +2 y Here there are four parallel classes, and the parameters for the associated balau.ced incomplete block design are v = 9, b = 12, r = 4, k = 3, ) = (b) From the :nine points in part (a) we get (0,0,1), (1,0,1), (2,0,1) (0,1,1), (1,1,1), (2,1,1) (0,2,1), (1,2,1), (2,2,1) To these nine we adjoin the line z Co:nsequently, = and the additional poi:nts (1,0,0) and ,0), (1,1,0), (2,1,0) for "''''+'nT'''' plane has 1= + = 13 points x = 0: y= x = z: y = z; x = 2z: y = 2z: y=x: y = x +z: y = x 2z: y= 2x: y = 2x + z: y = 2x +2z: z = (Roo): ° {(O,O,1 ),(O,l,l ),(0,2,1 ),(0,1,0)} {(0,0,1 ),(1,0,1 ),(2,0,1 ),(1,0,0)} {(I,O,l ),(1,1,1 ),(1,2,1 ),(O,l,On {(0,1, 1),(1,1,1 ),(2,1,1 ),( ,O,O)} {(2,0,1 ),(2,1,1 ),(2,2,1 ),(O,l,O)} {(O,2,1 ),(1,2,1 ),(2,2,1 ),(1,0,0)} {(O,O,l ),(1,1,1 ),(2,2,1 ),(l,l,O)} {(1,2,1 ),(2,0,1 ),(0,1,1 ),(1,1,0)} {(1,O,1 ),(0,2,1 ),(2,1,1 ),(1,1,0)} {(0,0,1 ),(1,2,1 ),(2,1,1 ),(2,1,0)} {(O,l,l ),(1,0,1 ),(2,2,1 ),(2,1,0)} {(0,2,1 ),(1,1,1 ),(2,0,1 ),(2,1,0)} {(I ,0,0 ),(0,1,0),(1,1,0 ),(2,l,0)} Since there a.re four points on £00 there are four parallel classes Finally, the parameters for the associated balanced incomplete block design are v:::;.: b :::;.: 13, " = k = 4, A = Supplementary Exercises n=9 (a) = fer /13) = an(r l.s)1/, a n-l (1' /s )71-1 + + al(,-js) + ao =? = anrn + an_lrn-ls + '" + a]rs n - + atJ!ll Since /3 divides aud s is a factor of all summands except the first, it follows that s divides a n 7,n.With gcd(r,.9) = 1, sian In similar fashion, rlao (b) (i) f(x) = 2x 3x - 2x - From part (a) the possible rational roots are ±1, ±3, ±1/2, +3/2 f(l) = 2(1 ) 3(12) - 2(1) - = 0, so is a root of f(x) and x - I is a factor By long division of polynomials (or synthetic division) f(x) = (x - 1)(2x2 ox (x - 1)(2x 3)(x 1), so the other roots of f(x) are -3/2 (ii) f (x) == X4 + x - x - 2x - The possible rat:iOU,W -2=?a so c=l, = -1, x-n~ are 31 (b) b)=x +2x-n:=.:::?b-a=2 we find that n = ab S; 960, so are 30 = n When a S; 30 values of n this case -a)(x b=a (c) In this case there are 29 , alues of n Each n has form a(a 5) for S; a < If (x - a)(x b) = g(x), b- a = k = n When k = 1000, b = a 1000 and ab = a 1000a > 1000 For k = 999, with a = and b = 1000 we have n = ab = 1000 and x +999x-lOOO = 1000)(x-l) In fact, for each S; k :s; 999, a = b = k+l andn=ab=k 1, and it follows thatx kx-n=x kx-(k l)=[x (k 1)](x-1) "', "''"' the smallest positive integer k for whicll g( x) cannot be so factored is k = 1000 If F = Z:'h then j(l) = 1 + + = 0, so a root of and (x = (x + 1) a factor If Ff:.Z2 then -lEF and -1 Here 1(-1)=1-1-1+1=0,80 is a root ( x l ) is a factor 18 For all a E ZPl aP = a (See part (a) Exercise 13 at the end Section 16.3), 80 a is a P root of x - x and x - a is a factor of xl' - x Since (Zp, +, ) is a field, the polynomial xl' - x can have at most p roots Therefore XV - x = nl.EZp(x - a) lex) = + a'll._lXn-1 + Xn alX (a) The coefficient of xn - comparing coefficients we have + aO = (x - (x - f'l)(X - an-l = -1'1 - (b) The constant term in (x - rd(x comparison of coefficients we find that ao r'l)(X - r2)'" (x - 1':d" (x 1'2 - 1'2)'" = (-1)101"17'2'" ••• - '1'n) 1'n (x - Tn) r'H or , is or IS e-l)"ao = (_1)2n r1r2 ·•• T'r., = 1'17'2'" r'n)' -1'1 - ,,'), - (-1)"1'11'2'" ••• Tn, SO Tn' by Again by 1'10' {1,2,4}, {2,3,5}, {4,5,1} Ii: = \ = ==> r = 31 , b = v= n 1= ==> n = 10 :=.:::? n = ==> n = :=.:::? »2 = 9, + n + ::;;::: 91, to k r (b) A Jt a v X b whose (i,j) entry is k, since there are k 1'8 Hence J'I}' A = each column of A and every entry in J11 (c) The (i, j) entry in A· is obtained from the componentwise multiplication of rows j i = j this results the 1'8 in row i, which is r i j, the number of 1'8 is the number times Xi and Xj appear in the same block by) Hellce A· Atr = (r - >')111 )'Jv' this is 't (d) ) ;\ > > > I Ar r ) ) ) r > > r ) ) A ) ) ) 'f' > > \ ) ) ) r r fr + (v -1);\](r - \)V-l r-A 7' -) o o o o o r- \ 0 o > = (r _ 0 0 r -) '1'-> \ \ > )11-1 o o o ) I ) - r r-A o o o (v - 1)>" A-r A-r ) -'1' o 0 r - o o o > r-A r(k - 1)1 = rk(r _ >"),,-1 (1) Multiply column by -1 and add it to the other v - columns (2) Add rows through v to row L 12 (a) Here V = {I, 2, ,9} and the 12 blocks are 2 {) {) 4: (} 4: 4: r,t = k' ::::; 1 1 - we ;;;;;;:b- ; ; ; : v-k, 3 4: () 4: 8 124: 236 123 234: THE APPENDICES LOGARITHMIC FUNCTIONS (b) (a.) 125- 4/ = 1/(125)4/3 = 1/[(125)1/3)4 = 1/54 = 1/625 = 0.09 (4/3)(1/8)-2/3 = (4/3)[1/(1/8)2/3] = (4/3)[1/[(1/8)1/3]2] = (b) 0.027 / = [(0.027)1/3P~ = (0.3)2 (c) (4/3)[1/(1/4)J = (4/3)(4) = 16/3 (a.) (53/4)(513/4) = 5[{3/4)+(13/4)} = 516/ = 54 = (a) 5&:2 or x = = 7-3 = 1/73 = 1/343 = (51 / )(4.5)1/2 = (51/2)(41/2)(51/2) = = 553)+2 ::} 3x = 5x 2=} 3x (b) = (1/2)'~X-l ::} 2{x-l} = fix = ::} x = 1/2 - 5x - 2-(4;:;-1) ::} = 625 (b) (73/5)/(718/5) = 7[(3/5)-(18/5)J = 7{3-18)/5 = 7- 15/ (c) (51 / )(201 /2) (4/3)[1/(1/2)2) 2(51 /2)2 = 2(5) = 10 = O::} (3x 2(x - l)(x - 2) = -(4x -1) ::} 2x - 11 o x= = ::} x = -1/3 = -4x ::} Proof: Let x = 10gb rand 'tI - 10gb s y = logbs ~ bY = s, we have Then, because x - 10gb r ~ bX - rand = b:& /bY = b:&-'!J, part (2) of Theorem ALL = ba:- y ~ 10&,('1'/8) = X - 'til it follows that '1'- (a) Proof (by Mathematical Induction): For n = the statement is 10gb '1'1 = 10gb r, so the result is true for this first case Assuming the result for n = k C;;::: 1) we have: 10gb rk = k 10gb r Now for the case where n = k + we find that 10gb rk+ = log",(r rio) lo&, '1' + 1ogb'1'k (by part (a) of Theorem A 1.2) = 10gb r + k 10gb r (by the induction hypothesis) = (1 + k) 10gb r = (k + 1) 10gb r Therefore the result follows for all n E Z+ by the Principle of Mathematical Induction = (b) For all n E Z+, lo&, r- n = logb(l/'1' n ) = 10gb 1-1o~ '1'n (by part (b) of Theorem A1.2) = - nlogb r (by part (a) above) = (-n) lo~ r 10 (a) log21O = log2(2· 5) = log22 + loga = + 2.3219 = 3.3219 (b) log2100 = log2102 = 21og2 10 = 2(3.3219) = 6.6438 (c) logz(7/5) = log27 -loga5 = 2.8074 - 2.3219 = 0.4855 (d) log2175 = log2 (7 ·25) = logz 11 (a) Let x = loga Then X 21og2 = 2.8074 + 2(2.3219) = and x(ln 2) = 1.0986/0.6931 : U5851 (b) logs 2:= 5= (&) :r, ;: 2/1n5 = 0.6931/1.6094 : 0.4307 : 1.4650 3= loglO(2 5) =:;} x = 10 = 7.4512 2:1: = In 3, so loga = x = 3/ln2 = (c) = log3(x2 4x + - 10g3(2x - 5) = loga[(x 4x (X 4x 4)/(2x 5) ::=> 9(2x - 5) = x 4x 4::::.=;} x -14x 49 = O::=> (x - 7)2 = O::::.=;} x = 14 log2 x = (1/3){10g2 - log:;! 5] + (2/3) log2 + 10g2 17 10g2[17(108/5)1!3} ::=> x = 17(108/5)1/3 4)/(2x - 5)J ==> 32 = = - 45 = x ::=> = log2(3/5)1/3 log'}, 6'2/3 Proof: Let x = aiogb C and y = b Then x = alor;b C==> 10gb X = 10gb [alD~ C] = {10gb C )(logb a) and y= b = } lo&y = 10gb [dog" a] = (lo&a)(logb c), Consequently, we find that 10& x = 10gb Y from which it follows that x = y j (a) (c) A +0 3251 =[0 J (b) [~ ~ ~] (d) = (e) 2A = [4-2 02 8] (g) 20 30 = (i) 2B - 4C = [ _ (k) 2(3B) 3a = [~ = 2, - 12 = [2~ ~8 12 ~i51 20 :1 ;~] 2~] 3d - = , 1" [ :] -11 if A is an n X n matrix with two identical rows or two identical columns, then det(A) = O 15 (a) (i) ! -1 -1 11+ 3(_1)3+2 ! =2(_1)3+ 1 -1 -1 = 2( -1) = 1 i = 2( - (ii) (b) (i) 16 ( -1 » - 3( -1) (iii) 25 (U) 306 (iii) 510 There are n entries in the matrix product AB For each entry we perform n multiplications and n - additions Therefore, we perform n multiplications and n 2(n - 1) = - f!i additions APPENDIX COUNTABLE UNCOUNTABLE (Ii},) True (b) False (c) True (d) True (e) True (f) True (g) False: Let = Z+ U (0,1] and B = (0,1] Then A, B are both uncountable, but A - B = {2, 3, 4, } is countable (a) The function I: z+ ~ A defined by I(n) = n is a one-to-one correspondence (b) Let y: z+ ~ {2,6, 10, 14, } be defined yen) = (n-l)4+2 Then is a one-to-one correspondence If B were countable, then by Theorem A3.3 it would follow that A is countable This leads us to a contradiction since we are given that A is uncountable The set [ of irrational numbers is uncountable If not, then R - by virtue of Theorems A3.8 and A3.9 (or, A3.7) Since S, T are countably infinite, we know from Theorem A3.2 that we can write S = {ShSa,S31"'} and T = {tb t", } - two (jnfinit-e) sequences of distinct terms Define the function f: S X T ~ z+ = Q U[ would be countable = by f(Si,tj) 2i ai, for all i,j E Z+ If i,j,k,l E Z+ with !(Si,t;) = f(Sk,tl), then j'(Si' tj) = j'(81:;, tt) ::::} = 2k3£ ::::} i = k, j = I (By the Fundamental Theorem of :::¢ 8;, = 13k tj = t.e ::::} (Si' tj) = (Sk' f is a functioll and S x "" f(8 x T) C Z+ from Theorem A3.3 we know that S X T is x x = f(a21 b2 , C2), - ::::} - f(Z+ X Z+ = ~,Cl X f(o" b, c) = 18 nnf"-T.n,-or~p (Z - {O}) X Z X Z is 468 ::.'::: C2 COUIU,RD.l.e for all (a, b, c) E (Z - {O}) x Z x Z there are at most two (distinct) real solutions for the quadratic ax ox + c = it set bx + c = where a, OJ c E Z a all real solutions of the quadratic is countable (a) f(x)=3x, (e) O