MTH202 Discrete Mathematics LECTURE 31 K-COMBINATIONS K-SELECTIONS

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1 LECTURE 31 K-COMBINATIONS K-SELECTIONS K-COMBINATIONS: With k-omintion the order in whih the elements re seleted does not mtter nd the elements nnot repet. DEFINITION: A k-omintion of set of n elements is hoie of k elements tken from the set of n elements suh tht the order of the elements does not mtter nd elements n t e repeted. The symol C(n, k) denotes the numer of k-omintions tht n e hosen from set of n elements. NOTE: k-omintions re lso written n s or n C k k REMARK: With k-omintions of set of n elements, repetition of elements is not llowed, therefore, k must e less thn or equl to n, i.e., k n. Let X = {,, }. Then 2-omintions of the 3 elements of the set X re: {, }, {, }, nd {, }. Hene C(3,2) = 3. Let X = {,,, d, e}. List ll 3-omintions of the 5 elements of the set X, nd hene find the vlue of C(5,3). Then 3-omintions of the 5 elements of the set X re: {,, }, {,, d}, {,, e}, {,, d}, {,, e}, {, d, e}, {,, d}, {,, e}, {, d, e}, {, d, e} Hene C(5, 3) = 10 PERMUTATIONS AND COMBINATIONS: Let X = {A, B, C, D}. The 3-omintions of X re: {A, B, C}, {A, B, D}, {A, C, D}, {B, C, D} Hene C(4, 3) = 4 The 3-permuttions of X n e otined from 3-omintions of X s follows. ABC, ACB, BAC, BCA, CAB, CBA ABD, ADB, BAD, BDA, DAB, DBA ACD, ADC, CAD, CDA, DAC, DCA BCD, BDC, CBD, CDB, DBC, DCB So tht P(4, 3) = 24 = 4 6 = 4 3! Clerly P(4, 3) = C(4, 3) 3! In generl we hve, P(n, k) = C(n, k) k! In generl we hve, P(n, k) = C(n, k) k! or Cnk (, ) = Pnk (, ) k! Virtul University of Pkistn Pge 202

2 But we know tht Hene, n! Pnk (, ) = ( n k)! n! Cnk (, ) = ( n k)! k! Compute C(9, 6). 9! C (9,6) = (9 6)!6! 9876! = 3! 6! 987 = 32 = 84 COMPUTING C(n, k) SOME IMPORTANT RESULTS () C(n, 0) = 1 () C(n, n) = 1 () C(n, 1) = n (d) C(n, 2) = n(n-1)/2 (e) C(n, k) = C(n, n k) (f) C(n, k) + C(n, k + 1) = C(n + 1, k + 1) A student is to nswer eight out of ten questions on n exm. () Find the numer m of wys tht the student n hoose the eight questions () Find the numer m of wys tht the student n hoose the eight questions, if the first three questions re ompulsory. () The eight questions n e nswered in m = C(10, 8) = 45 wys. () The eight questions n e nswered in m = C(7, 5) = 21 wys. An exmintion pper onsists of 5 questions in setion A nd 5 questions in setion B. A totl of 8 questions must e nswered. In how mny wys n student selet the questions if he is to nswer t lest 4 questions from setion A. There re two possiilities: () The student nswers 4 questions from setion A nd 4 questions from setion B.The numer of wys 8 questions n e nswered tking 4 questions from setion A nd 4 questions from setion B re C(5, 4) C(5, 4) =5 5 = 25. () The student nswers 5 questions from setion A nd 3 questions from setion B.The numer of wys 8 questions n e nswered tking 5 questions from setion A nd 3 questions from setion B re C(5, 5) C(5, 3) =1 10 = 10. Thus there will e totl of = 35 hoies. A omputer progrmming tem hs 14 memers. () How mny wys n group of seven e hosen to work on projet? Virtul University of Pkistn Pge 203

3 () Suppose eight tem memers re women nd six re men (i) How mny groups of seven n e hosen tht ontin four women nd three men (ii) How mny groups of seven n e hosen tht ontin t lest one mn? (iii)how mny groups of seven n e hosen tht ontin t most three women? () Suppose two tem memers refuse to work together on projets. How mny groups of seven n e hosen to work on projet? (d) Suppose two tem memers insist on either working together or not t ll on projets. How mny groups of seven n e hosen to work on projet? () How mny wys group of 7 e hosen to work on projet? Numer of ommittees of 7 C (14, 7) = 14! (14 7)! 7! = = 3432 () Suppose eight tem memers re women nd six re men (i) How mny groups of seven n e hosen tht ontin four women nd three men? Numer of groups of seven tht ontin four women nd three men 8! 6! C(8,4) C(6,3) = (8 4)! 4! (6 3)! 3! = 4! 3! = = = 1400 () Suppose eight tem memers re women nd six re men (ii) How mny groups of seven n e hosen tht ontin t lest one mn? Totl numer of groups of seven 14! C (14,7) = (14 7)! 7! = = 3432 Numer of groups of seven tht ontin no men 8! C(8,7) = (8 7)!7! = 8 Hene, the numer of groups of seven tht ontin t lest one mn C(14,7) C(8, 7) = =3424 ()Suppose eight tem memers re women nd six re men Virtul University of Pkistn Pge 204

4 (iii) How mny groups of seven n e hosen tht ontin t most three women? Numer of groups of seven tht ontin no women = 0 Numer of groups of seven tht ontin one womn = C(8,1) C(6,6) = 8 1 = 8 Numer of groups of seven tht ontin two women = C(8,2) C(6,5) = 28 6 = 168 Numer of groups of seven tht ontin three women = C(8,3) C(6,4) = = 840 Hene the numer of groups of seven tht ontin t most three women = = 1016 () Suppose two tem memers refuse to work together on projets. How mny groups of seven n e hosen to work on projet? Cll the memers who refuse to work together A nd B. Numer of groups of seven tht ontin neither A nor B 12! C (12, 7) = (12 7)! 7! = 792 Numer of groups of seven tht ontin A ut not B C(12, 6) = 924 Numer of groups of seven tht ontin B ut not A C(12,6) = 924 Hene the required numer of groups re C(12,7) + C(12,6) + C(12, 6) = = 2640 (d)suppose two tem memers insist on either working together or projets. How mny groups of seven n e hosen to work on projet? Cll the memers who insist on working together C nd D. Numer of groups of seven ontining neither C nor D C(12, 7) = 792 Numer of groups of seven tht ontin oth C nd D C(12, 5) = 792 Hene the required numer = C(12, 7) + C(12, 5) = = 1584 () How mny 16-it strings ontin extly 9 1 s? ()How mny 16-it strings ontin t lest one 1? 16! () 16-it strings tht ontin extly 9 1 s= C (16,9) = = (16 9)!9! () Totl no. of 16-it strings = 2 16 Hene numer of 16-it strings tht ontin t lest one = not t ll on Virtul University of Pkistn Pge 205

5 = K-SELECTIONS: k-seletions re similr to k-omintions in tht the order in whih the elements re seleted does not mtter, ut in this se repetitions n our. DEFINITION: A k-seletion of set of n elements is hoie of k elements tken from set of n elements suh tht the order of elements does not mtter nd elements n e repeted. REMARK: 1. k-seletions re lso lled k-omintions with repetition llowed or multisets of size k. 2. With k-seletions of set of n elements repetition of elements is llowed. Therefore k need not to e less thn or equl to n. THEOREM: The numer of k-seletions tht n e seleted from set of n elements is C(k+n 1, k) or k+n -1 k A mer shop stoks ten different types of tteries. () How mny wys n totl inventory of 30 tteries e distriuted mong the ten different types? () Assuming tht one of the types of tteries is A76, how mny wys n totl inventory of 30 tteries e distriuted mong the 10 different types if the inventory must inlude t lest four A76 tteries? () k = 30 n = 10 The required numer is C( , 30) = C(39, 30) = 39! ( 39 30)!30! = () k = 26 n = 10 The required numer is C( , 26) = C(35, 26) 35! = (35 26)!26! = WHICH FORMULA TO USE? ORDER ORDER DOES MATTERS NOT MATTER REPETITION ALLOWED REPETITION NOT ALLOWED k-smple n k k-permuttion P(n, k) k-seletion C(n+k-1, k) k-omintion C(n, k) Virtul University of Pkistn Pge 206

6 LECTURE 32 ORDERED AND UNORDERED PARTITIONS PERMUTATIONS WITH REPETITIONS K-SELECTIONS: k-seletions re similr to k-omintions in tht the order in whih the elements re seleted does not mtter, ut in this se repetitions n our. DEFINITION: A k-seletion of set of n elements is hoie of k elements tken from set of n elements suh tht the order of elements does not mtter nd elements n e repeted. REMARK: 1. k-seletions re lso lled k-omintions with repetition llowed or multisets of size k. 2. With k-seletions of set of n elements repetition of elements is llowed. Therefore k need not to e less thn or equl to n. THEOREM: The numer of k-seletions tht n e seleted from set of n elements is C(k+n 1, k) or k+n-1 k A mer shop stoks ten different types of tteries. () How mny wys n totl inventory of 30 tteries e distriuted mong the ten different types? () Assuming tht one of the types of tteries is A76, how mny wys n totl inventory of 30 tteries e distriuted mong the 10 different types if the inventory must inlude t lest four A76 tteries? () k = 30 n = 10 The required numer is C( , 30)= C(39, 30) 39! = ( 39 30)!30! = () k = 26 n = 10 The required numer is C( , 26) = C(35, 26) = 35! (35 26)!26! = WHICH FORMULA TO USE? Virtul University of Pkistn Pge 207

7 REPETITION ALLOWED REPETITION NOT ALLOWED ORDER MATTERS k-smple n k k-permuttion P(n, k) ORDER DOES NOT MATTER k-seletion C(n+k-1, k) k-omintion C(n, k) ORDERED AND UNORDERED PARTITIONS: An unordered prtition of finite set S is olletion [A 1, A 2,, A k ] of disjoint (nonempty) susets of S (lled ells) whose union is S. The prtition is ordered if the order of the ells in the list ounts. Let S = {1, 2, 3,, 7} The olletions P = [{1,2}, {3,4,5}, {6,7}] 1 And P = [{6,7}, {3,4,5}, {1,2}] 2 determine the sme prtition of S ut re distint ordered prtitions. Suppose ox B ontins seven mrles numered 1 through 7. Find the numer m of wys of drwing from B firstly two mrles, then three mrles nd lstly the remining two mrles. The numer of wys of drwing 2 mrles from 7 = C(7, 2) Following this, there re five mrles left in B. The numer of wys of drwing 3 mrles from 5 = C(5, 3) Finlly, there re two mrles left in B. The numer of wy of drwing 2 mrles from 2 = C(2, 2) Thus m = ! 5! 2! = 2!5! 2!3! 2!0! 7! = = 210 2!3!2! THEOREM: Let S ontin n elements nd let n, n,, n e positive integers with 1 2 k n +n + +n = n. 1 2 k Then there exist n! n1! n2! n3! nk! different ordered prtitions of S of the form [A, A,, A ], where 1 2 k A ontins n elements 1 1 A ontins n elements 2 2 A ontins n elements Virtul University of Pkistn Pge 208

8 A k ontins n k elements REMARK: To find the numer of unordered prtitions, we hve to ount the ordered prtitions nd then divide it y suitle numer to erse the order in prtitions. Find the numer m of wys tht nine toys n e divided mong four hildren if the youngest hild is to reeive three toys nd eh of the others two toys. We find the numer m of ordered prtitions of the nine toys into four ells ontining 3, 2, 2 nd 2 toys respetively. Hene 9! m = 3!2!2!2! = 2520 How mny wys n 12 students e divided into 3 groups with 4 students in eh group so tht (i) one group studies English, one History nd one Mthemtis. (ii) ll the groups study Mthemtis. (i) Sine eh group studies different sujet, so we seek the numer of ordered prtitions of the 12 students into ells ontining 4 students eh. Hene there re 12! = 34, 650 4!4!4! suh prtitions (ii) When ll the groups study the sme sujet, then order doesn t mtter. Now eh prtition {G 1, G 2, G 3 } of the students n e rrnged in 3! wys s n ordered prtition, hene there re 12! 1 4!4!4! 3! unordered prtitions. How mny wys n 8 students e divided into two tems ontining (i) five nd three students respetively. (ii) four students eh. (i) The two tems (ells) ontin different numer of students; so the numer of unordered prtitions equls the numer of ordered prtitions, whih is 8! = 56 5!3! (ii) Sine the tems re not leled, so we hve to find the numer of unordered prtitions of 8 students in groups of 4. Firstly, note, there re 8! = 70 ordered prtitions into two ells ontining four students eh. 4!4! Sine eh unordered prtition determine 2! = 2 ordered prtitions, there re 70 = 35 2 unordered prtitions Virtul University of Pkistn Pge 209

9 Find the numer m of wys tht lss X with ten students n e prtitions into four tems A, A, B nd B where A nd A ontin two students eh nd B nd B ontin three students eh. There re 10! = 25,200 ordered prtitions of X into four ells 2!2!3!3! ontining 2, 2, 3 nd 3 students respetively. However, eh unordered prtition [A, A, B, B ] of X determines ! 2! = 4 ordered prtitions of X. Thus, 25, 200 m = = Suppose 20 people re divided in 6 (numered) ommittees so tht 3 people eh serve on ommittee C nd C, 4 people eh on ommittees C nd C, 2 people on ommittee C nd 4 people on ommittee C. How mny possile rrngements re there? 5 6 We re sked to ount leled group - the ommittee numers leled the group.so this is prolem of ordered prtition. Now, the numer of ordered prtitions of 20 people into the speified ommittees is 20! 3!3!4!4!2!4! = If 20 people re divided into tems of size 3, 3, 4, 4, 2, 4, find the numer of possile rrngements. Here, we re sked to ount unleled groups. Aordingly, this is the se of ordered prtitions. Now numer of ordered prtitions 20! 1 = 3!3!4!4!2!4! 3!2! = GENERALIZED PERMUTATION or PERMUTATIONS WITH REPETITIONS: The numer of permuttions of n elements of whih n re like, n re like,, n re like 1 2 k is n! n! n! n! 1 2 k REMARK: n! The numer is often lled multinomil oeffiient, nd is denoted y the n symol. 1! n2! nk! n n1, n2,, n k Find the numer of distint permuttions tht n e formed using the letters of the word BENZENE. Virtul University of Pkistn Pge 210

10 The word BENZENE ontins seven letters of whih three re like (the 3 E s) nd two re like (the 2 N s) Hene, the numer of distint permuttions re: 7! 3!2! = 420 How mny different signls eh onsisting of six flgs hung in vertil line, n e formed from four identil red flgs nd two identil lue flgs? We seek the numer of permuttions of 6 elements of whih 4 re like nd 2 re like. There re 6! = 15 4!2! different signls. (i)find the numer of words tht n e formed of the letters of the word ELEVEN. (ii)find, if the words re to egin with L. (iii)find, if the words re to egin nd end in E. (iv)find, if the words re to egin with E nd end in N. (i)there re six letters of whih three re E; hene required numer of words re 6! = 120 3! (ii)if the first letter is L, then there re five positions left to fill where three re E; hene required numer of words re 5! 3! = 20 (iii)if the words re to egin nd end in E, then there re only four positions to fill with four distint letters. Hene required numer of words = 4! = 24 (iv)if the words re to egin with E nd end in N, then there re four positions left to fill where two re E. 4! Hene required numer of words = 2! = 12 (i)find the numer of permuttions tht n e formed from ll the letters of the word BASEBALL (ii)find, if the two B s re to e next to eh other. (iii)find, if the words re to egin nd end in vowel. (i)there re eight letter of whih two re B, two re A, nd two re L. Thus, 8! Numer of permuttions = 2!2!2! = 5040 (ii)consider the two B s s one letter. Then there re seven nd two re L. Hene, 7! Numer of permuttions = 2!2! = 1260 letters of whih two re A Virtul University of Pkistn Pge 211

11 (iii)there re three possiilities, the words egin nd end in A, the words egin in A nd end in E, or the words egin in E nd end in A. In eh se there re six positions left to fill where two re B nd two re L. Hene, Numer of permuttions 6! = 3 = 540 2!2! Virtul University of Pkistn Pge 212

12 LECTURE 33 TREE DIAGRAM INCLUSION - EXCLUSION PRINCIPLE TREE DIAGRAM: A tree digrm is useful tool to list ll the logil possiilities of sequene of events where eh event n our in finite numer of wys. A tree onsists of root, numer of rnhes leving the root, nd possile dditionl rnhes leving the end points of other rnhes. To use trees in ounting prolems, we use rnh to represent eh possile hoie. The possile outomes re represented y the leves (end points of rnhes). A tree is normlly onstruted from left to right. Leve Root Brnh Leve Brnh Leve A TREE STRUCTURE Find the permuttions of {,, } The numer of permuttions of 3 elements is 3! P (3,3) = = 3! = 6 (3 3)! We find the six permuttions y onstruting the pproprite tree digrm. The six permuttions re listed on the right of the digrm. Find the produt set A B C, where A = {1,2}, B = {,,}, nd C = {3,4} y onstruting the pproprite tree digrm. Virtul University of Pkistn Pge 213

13 The required digrm is shown next. Eh pth from the eginning of the tree to the end point designtes n element of A B C whih is listed to the right of the tree. 3 (1,,3) (1,,4) (1,,3) (1,,4) (1,,3) (1,,4) (2,,3) (2,,4) (2,,3) (2,,4) (2,,3) 4 (2,,4) Tems A nd B ply in tournment. The tem tht first wins two gmes wins the tournment. Find the numer of possile wys in whih the tournment n our. We onstrut the pproprite tree digrm. A B A B A B A B A B The tournment n our in 6 wys: AA, ABA, ABB, BAA, BAB, BB How mny it strings of length four do not hve two onseutive 1 s? Virtul University of Pkistn Pge 214

14 The following tree digrms displys ll it strings of length four without two onseutive 1 s. Clerly, there re 8 it strings st it nd it rd it 0 4th it Three offiers, president, tresurer, nd seretry re to e hosen from mong four possile: A, B, C nd D. Suppose tht A nnot e president nd either C or D must e seretry. How mny wys n the offiers e hosen? We onstrut the possiility tree to see ll the possile hoies. C A D B C D D C C strt A D D President seleted B A B Tresurer seleted Seretry seleted D C C Virtul University of Pkistn Pge 215

15 From the tree, see tht there re only eight wys possile to hoose the offies under given onditions. THE INCLUSION-EXCLUSION PRINCIPLE: 1.If A nd B re disjoint finite sets, then n(a B) = n(a) + n(b) 2.If A nd B re finite sets, then n(a B) = n(a) + n(b) - n(a B) REMARK: Sttement 1 follows from the sum rule Sttement 2 follows from the digrm A B A B In ounting the elements of A B, we ount the elements in A nd ount the elements in B. There re n(a) in A nd n(b) in B. However, the elements in A B were ounted twie. Thus we sutrt n(a B) from n(a) + n(b) to get n(a B). Hene, n(a B) = n(a) + n(b) - n(a B) There re 15 girls students nd 25 oys students in lss. How mny students re there in totl? Let G e the set of girl students nd B e the set of oy students. Then n(g) = 15; n(b) = 25 nd n(g B) =? Sine, the sets of oy nd girl students re disjoint; here totl numer of students re n(g B) = n(g) + n(b) = = 40 Among 200 people, 150 either swim or jog or oth. If 85 swim nd 60 swim nd jog, how mny jog? Let U e the set of people onsidered. Suppose S e the set of people who swim nd J e the set of people who jog. Then given n(u) = 200; n(s J) = 150 n(s) = 85; n(s J) = 60 nd n(j) =? By inlusion - exlusion priniple, n(s J) = n(s) + n(j) - n(s J) 150 = 85 + n(j) - 60 n(j) = = 125 Hene 125 people jog. Virtul University of Pkistn Pge 216

16 Suppose A nd B re finite sets. Show tht n(a\b) = n(a) - n(a B) Set A my e written s the union of two disjoint sets A\B nd A B. A B A\B A B B\A i.e., A = (A\B) (A B) Hene, y inlusion exlusion priniple (for disjoint sets) n(a) = n(a\b) + n(a B) n(a\b) = n(a) - n(a B) REMARK: n(a ) = n(u\a) where U is the universl set = n(u) - n(u A) = n(u) - n(a) Let A nd B e susets of U with n(a) = 10, n(b) = 15, n(a )=12, nd n(a B) = 8. Find n(a B ). A B U From the digrm A B = U \ (B \ A) Hene n(a B ) = n(u \ (B\A)) = n(u) - n(b\a) (1) Now U = A A where A & A re disjoint sets n (U) = n(a) + n(a ) = = 22 Also n(b\a) = n(b) - n(a B) Virtul University of Pkistn Pge 217

17 = 15-8 = 7 Sustituting vlues in (1) we get n (A B ) = n(u) - n(b\a) = 22-7 = 15 Ans. Let A nd B re suset of U with n(u) = 100, n(a) = 50, n(b)= 60, nd n((a B) ) = 20. Find n(a B) Sine (A B) = U \ (A B) n ((A B) ) = n (U) - n (A B) 20 = n (A B) n (A B) = = 80 Now, y inlusion - exlusion priniple n (A B) = n (A) + n (B) - n (A B) 80 = n (A B) n (A B) = = 30 Suppose 18 people red English newspper (E) or Urdu newspper (U) or oth. Given 5 people red only English newspper nd 7 red oth, find the numer r of people who red only Urdu newspper. Given n (E U) = 18 n (E\U) = 5, n (E U) = 7 r = n (U \ E) =? From thedigrm E U E\U E U U\E E U = (E\U) (E U) (U\E) nd the union is disjoint. Therefore, n(e U) = n(e\u) + n(e U) + n(u\e) 18 = r r = r = 6 Ans. Fifty people re interviewed out their food preferenes. 20 of them like Chinese food, 32 like fst food, nd 12 like neither Chinese nor fst food.how mny like Chinese ut not fst food. Let U denote the set of people interviewed nd C nd F denotes the sets of people who like Chinese food nd fst food respetively. Now given n (U) = 50, n (C) = 20 Virtul University of Pkistn Pge 218

18 n (F) = 32, n ((C F )) = 12 To find n (C F ) = n (C\F) Sine n ((CUF) )= n (U) - n (C F) 12 = 50 - n (C F) n (C F)= = 38 Next n (C F)= n (C\F) + n (F) 38 = n (C\F) + 32 n (C\F) = = 6 C F C\F Virtul University of Pkistn Pge 219

19 LECTURE 34 INCLUSION-EXCLUSION PRINCIPLE PIGEONHOLE PRINCIPLE.How mny integers from 1 through 1000 re multiples of 3 or multiples of 5?.How mny integers from 1 through 1000 re neither multiples of 3 nor multiples of 5? Let A nd B denotes the set of integers from 1 through 1000 tht re multiples of 3 nd 5 respetively. Then A B ontins integers tht re multiples of 3 nd 5 oth i.e., multiples of 15. Now 1000 na ( ) 333 nd 1000 = = nb ( ) = = nd 1000 na ( B) = = Hene y the inlusion - exlusion priniple n(a B) = n(a) + n(b) - n(a B) = = 467 ()The set (A B) ontins those integers tht re either multiples of 3 or multiples of 5. Now n((a B) ) = n(u) - n(a B) = = 533 where the universl set U ontin integers 1 through INCLUSION-EXCLUSION PRINCIPLE FOR 3 AND 4 SETS: If A, B, C nd D re finite sets, then 1. n(a B C)= n(a) + n(b) + n(c)- n(a B) - n(b C) - n(a C) + n(a B C) 2.n(A B C D) = n(a) + n(b) + n(c) + n(d)- n(a B) - n(a C) - n(a D) - n(b C) - n(b D) - n(c D)+ n(a B C) + n(a B D) + n(a C D) + n(b C D) - n(a B C D) A survey of 100 ollege students gve the following dt: 8 owned r (C) 20 owned motoryle (M) 48 owned iyle (B) 38 owned neither r nor motoryle nor iyle No student who owned r, owned motoryle How mny students owned iyle nd either r or motoryle? Virtul University of Pkistn Pge 220

20 C M U B No. of elements in the shded region to e determined Let U represents the universl set of 100 ollege students. Now given tht n (U) = 100; n (C) = 8 n (M) = 20; n (B) = 48 n((c M B) ) = 38; n (C M) = 0 nd n (B C) + n (B M) =? Firstly note n((c M B) ) = n (U) - n (C M B) 38 = n (C M B) n (C M B) = = 62 Now y inlusion - exlusion priniple n (C M B) = n(c) + n(m) + n (B) - n (C M) - n(c B) - n (M B) + n (C M B) 62 = n (C B) - n (M B) -0 ( n (C B) = 0) n (C B) + n (M B) = = = 14 Hene, there re 14 students, who owned iyle nd either r or motoryle. REVISION OF FUNCTIONS: Clerly the ove reltion is not funtion euse 2 does not hve ny imge under this reltion. Note tht if wnt to mde it reltion we hve to must mp the 2 into some element of B whih is lso the imge of some element of A. Now The ove reltion is funtion euse it stisfy the onditions of funtions(s eh element of 1 st set hve the imges in 2 nd set). The following is funtion. Virtul University of Pkistn Pge 221

21 The ove reltion is funtion euse it stisfy the onditions of funtions(s eh element of 1 st set hve the imges in 2 nd set). Therefore the ove is lso funtion. PIGEONHOLE PRINCIPLE: A funtion from set of k + 1 or more elements to set of k elements must hve t lest two elements in the domin tht hve the sme imge in the o-domin. If k + 1 or more pigeons fly into k pigeonholes then t lest one pigeonhole must ontin two or more pigeons. EXAMPLES: 1. Among ny group of 367 people, there must e t lest two with the sme irthdy, euse there re only 366 possile irthdys. 2.In ny set of 27 English words, there must e t lest two tht egin with the sme letter, sine there re 26 letters in the English lphet. Wht is the minimum numer of students in lss to e sure tht two of them re orn in the sme month? There re 12 (= n) months in yer. The pigeonhole priniple shows tht mong ny 13 (= n + 1) or more students there must e t lest two students who re orn in the sme month. Given ny set of seven integers, must there e two tht hve the sme reminder when divided y 6? The set of possile reminders tht n e otined when n integer is divided y six is {0, 1, 2, 3, 4, 5}. This set hs 6 elements. Thus y the pigeonhole priniple if 7 = integers re eh divided y six, then t lest two of them must hve the sme reminder. How mny integers from 1 through 100 must you pik in order to e sure of getting one tht is divisile y 5? There re 20 integers from 1 through 100 tht re divisile y 5. Hene there re eighty integers from 1 through 100 tht re not divisile y 5. Thus y the pigeonhole priniple 81 = integers from 1 though 100 must e piked in order to e sure of getting one tht is divisile y 5. Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose six integers re hosen from A. Must there e two integers whose sum is 11. The set A n e prtitioned into five susets: {1, 10}, {2, 9}, {3, 8}, {4, 7}, nd {5, 6} Virtul University of Pkistn Pge 222

22 eh onsisting of two integers whose sum is 11. These 5 susets n e onsidered s 5 pigeonholes. If 6 = (5 + 1) integers re seleted from A, then y the pigeonhole priniple t lest two must e from one of the five susets. But then the sum of these two integers is 11. GENERALIZED PIGEONHOLE PRINCIPLE: A funtion from set of n k + 1 or more elements to set of n elements must hve t lest k + 1 elements in the domin tht hve the sme imge in the o-domin. If n k + 1 or more pigeons fly into n pigeonholes then t lest one pigeonhole must ontin k + 1 or more pigeons. Suppose lundry g ontins mny red, white, nd lue soks. Find the minimum numer of soks tht one needs to hoose in order to get two pirs (four soks) of the sme olour. Here there re n = 3 olours (pigeonholes) nd k + 1 = 4 or k = 3. Thus mong ny n k + 1 = = 10 soks (pigeons), t lest four hve the sme olour. DEFINITION: 1. Given ny rel numer x, the floor of x, denoted x, is the lrgest integer smller thn or equl to x. 2. Given ny rel numer x, the eiling of x, denoted x, is the smllest integer greter thn or equl to x. Compute x nd x for eh of the following vlues of x.. 25/ /4 = 6 + ¼ = 6 25/4 = 6 + ¼ = = = = = = = = = = = = 2 Wht is the smllest integer N suh tht. N/7 = 5. N/9 = 6. N = 7 (5 1) + 1 = = 29. N = 9 (6 1) + 1 = = 46 PIGEONHOLE PRINCIPLE: If N pigeons fly into k pigeonholes then t lest one pigeonhole must ontin N /k or more pigeons. Among 100 people there re t lest 100/12 = 8 + 1/3 = 9 who were orn in the sme month. Wht is the minimum numer of students required in Disrete Mthemtis lss to e sure tht t lest six will reeive the sme grde, if there re five possile grdes, A, B, C, D, nd F. Virtul University of Pkistn Pge 223

23 The minimum numer of students needed to gurntee tht t lest six students reeive the sme grde is the smllest integer N suh tht N/5 = 6. The smllest suh integer is N = 5(6-1)+1= = 26. Thus 26 is the minimum numer of students needed to e sure tht t lest 6 students will reeive the sme grdes. Virtul University of Pkistn Pge 224

24 LECTURE 35 INTRODUCTION TO PROBABILITY INTRODUCTION: Comintoris nd proility theory shre ommon origins. The theory of proility ws first developed in the seventeenth entury when ertin gmling gmes were nlyzed y the Frenh mthemtiin Blise Psl. It ws in these studies tht Psl disovered vrious properties of the inomil oeffiients. In the eighteenth entury, the Frenh mthemtiin Lple, who lso studied gmling, gve definition of the proility s the numer of suessful outomes divided y the numer of totl outomes. DEFINITIONS: An experiment is proedure tht yields given set of possile outomes. The smple spe of the experiment is the set of possile outomes. An event is suset of the smple spe. When die is tossed the smple spe S of the experiment hve the following six outomes. S = {1, 2, 3, 4, 5, 6} Let E1 e the event tht 6 ours, E2 e the event tht n even numer ours, E3 e the event tht n odd numer ours, E4 e the event tht prime numer ours, E5 e the event tht numer less thn 5 ours, nd E6 e the event tht numer greter thn 6 ours. Then E1 = {6} E2 = {2, 4, 6} E3 = {1, 3, 5} E4 = {2, 3, 5} E5 = {1, 2, 3, 4} E6 = Ф When pir of die is tossed, the smple spe S of the experiment hs the following thirty-six outomes S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} or more omptly, {11, 12, 13, 14, 15, 16, 21, 22, 23, 24, 25, 26, 31, 32, 33, 34, 35, 36, 41, 42, 43, 44, 45, 46, 51, 52, 53, 54, 55, 56, 61, 62, 63, 64, 65, 66} Let E e the event in whih the sum of the numers is ten. Then E = {(4, 6), (5, 5), (6, 4)} DEFINITION: Virtul University of Pkistn Pge 225

25 Let S e finite smple spe suh tht ll the outomes re eqully likely to our. The proility of n event E, whih is suset of smple spe S, is the numer of outomes in E n( E) P ( E) = = the numr of totl outomes in S n( S) REMARK: Sine Ф E S therefore, 0 n(e) n(s). It follows tht the proility of n event is lwys etween 0 nd 1. Wht is the proility of getting numer greter thn 4 when die is tossed? When die is rolled its smple spe is S={1,2,3,4,5,6} Let E e the event tht numer greter thn 4 ours. Then E = {5, 6} Hene, n( E) 2 1 P( E) = = = n( S) 6 3 Wht is the proility of getting totl of eight or nine when pir of die is tossed? When pir of die is tossed,its smple spe S hs the 36 outomes whih re s fellows: S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} Let E e the event tht the sum of the numers is eight or nine. Then E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3)} Hene, n( E) 9 1 P( E) = = = n( S) 36 4 An urn ontins four red nd five lue lls. Wht is the proility tht ll hosen from the urn is lue? Sine there re four red lls nd five lue lls so if we tke out one ll from the urn then there is possiility tht it my e one of from four red nd one of from five lue lls hene there re totl of nine possiilities. Thus we hve The totl numer of possile outomes = = 9 Now our fvourle event is tht we get the lue ll when we hoose ll from the urn. So we hve The totl numer of fvorle outomes = 5 Now we hve Fvorle outomes 5 nd our smple spe hs totl outomes 9.Thus we hve Virtul University of Pkistn Pge 226

26 The proility tht ll hosen = 5/9 Two rds re drwn t rndom from n ordinry pk of 52 rds. Find the proility p tht (i) oth re spdes, (ii) one is spde nd one is hert. There re 52 = wys to drw 2 rds from 52 rd 13 (i) There re = 78 wys to drw 2 spdes from 13 spdes (s spdes re 13 in 52 2 rds); hene numer of wys 2 spdes n e drwn p = numer of wys 2 rds n e drwn = = 1 17 ii) Sine there re 13 spdes nd 13 herts, there re spde nd hert; hene p = = = = wys to drw In lottery, plyers win the first prize when they pik three digits tht mth, in the orret order, three digits kept seret. A seond prize is won if only two digits mth. Wht is the proility of winning () the first prize, () the seond prize? Using the produt rule, there re 10 3 = 1000 wys to hoose three digits. () There is only one wy to hoose ll three digits orretly. Hene the proility tht plyer wins the first prize is 1/1000 = () There re three possile ses: (i)the first digit is inorret nd the other two digits re orret (ii)the seond digit is inorret nd the other two digits re orret (iii)the thirds digit is inorret nd the other digits re orret To ount the numer of suesses with the first digit inorret, note tht there re nine hoies for the first digit to e inorret, nd one eh for the other two digits to e orret. Hene, there re nine wys to hoose three digits where the first digit is inorret, ut the other two re orret. Similrly, there re nine wys for the other two ses. Hene, there re = 27 wys to hoose three digits with two of the three digits orret. It follows tht the proility tht plyer wins the seond prize is 27/1000 = Wht is the proility tht hnd of five rds ontins four rds of one kind? (i) For determining the fvorle outomes we note tht Virtul University of Pkistn Pge 227

27 The numer of wys to pik one kind = C(13, 1) The numer of wys to pik the four of this kind out of the four of this kind in the dek = C(4, 4) The numer of wys to pik the fifth rd from the remining 48 rds = C(48, 1) Hene, using the produt rule the numer of hnds of five rds with four rds of one kind = C(13, 1) C(4, 4) C(48, 1) (ii) The totl numer of different hnds of five rds = C(52, 5). From (i) nd (ii) it follows tht the proility tht hnd of five rds ontins four rds of one kind is C(13,1) C(4,4) C(48,1) = C(52,5) 2,598,960 Find the proility tht hnd of five rds ontins three rds of one kind nd two of nother kind. (i) For determining the fvorle outomes we note tht The numer of wys to pik two kinds = C(13, 2) The numer of wys to pik three out four of the first kind = C(4, 3) The numer of wys to pik two out four of the seond kind = C(4, 2) Hene, using the produt rule the numer of hnds of five rds with three rds of one kind nd two of nother kind = C(13, 2) C(4, 3) C(4, 2) (ii) The totl numer of different hnds of five rds = C(52, 5). From (i) nd (ii) it follows tht the proility tht hnd of five rds ontins three rds of one kind nd two of nother kind is C(13,2) C(4,3) C(4,2) = C(52,5) ,598, Wht is the proility tht rndomly hosen positive two-digit numer is multiple of 6? 1.There re 99/6 = 16 + ½ = 16 positive integers from 1 to 99 tht re divisile y 6. Out of these 16 1 = 15 re two-digit numers(s 6 is multiple of 6 ut not two-digit numer). 2.There re 99 9 = 90 positive two-digit numers in ll. Hene, the proility tht rndomly hosen positive two-digit numer is multiple of 6 = 15/90 = 1/ DEFINITION: Let E e n event in smple spe S, the omplement of E is the event tht ours if E does not our. It is denoted y E. Note tht E = S\E Let E e the event tht n even numer ours when die is tossed. Then E is the event tht n odd numer ours. Virtul University of Pkistn Pge 228

28 THEOREM: Let E e n event in smple spe S. The proility of the omplementry event E of E is given y P(E ) = 1 P(E). Let 2 items e hosen t rndom from lot ontining 12 items of whih 4 re defetive. Wht is the proility tht (i) none of the items hosen re defetive, (ii) t lest one item is defetive? The numer of wys 2 items n e hosen from 12 items = C(12, 2) = 66. (i) Let A e the event tht none of the items hosen re defetive. The numer of fvorle outomes for A = The numer of wys 2 items n e hosen from 8 non-defetive items = C(8, 2) = 28. Hene, P(A) = 28/66 = 14/33. (ii)let B e the event tht t lest one item hosen is defetive. Then lerly, B = A It follows tht P(B) = P(A) =1 P(A) = 1 14/33 = 19/33. Three light uls re hosen t rndom from 15 uls of whih 5 re defetive. Find the proility p tht (i) none is defetive, (ii) extly one is defetive, (iii) t lest one is defetive. 15 There re = 455 wys to hoose 3 uls from the 15 uls (i)sine there re 15-5 = 10 non-defetive uls, there re = 120 wys to hoose 3 3 non-defetive uls. Thus p = = (ii) There re 5 defetive uls nd = 45 different pirs of non-defetive uls; hene there re = 5.45 = 225 wys to hoose 3 uls of whih one is defetive. 1 2 Thus P = = Virtul University of Pkistn Pge 229

29 (iii)the event tht t lest one is defetive is the omplement of the event tht none re defetive whih hs y (i), proility Hene p( tlest oneis defetive) = 1 p( noneis defetive) = 1 = Virtul University of Pkistn Pge 230

30 LECTURE 36 ADDITION LAW OF PROBABILITY THEOREM: If A nd B re two disjoint (mutully exlusive) events of smple spe S, then P (A B) = P(A) + P(B) In words, the proility of the hppening of n event A or n event B or oth is equl to the sum of the proilities of event A nd event B provided the events hve nothing in ommon. PROOF: By inlusion - exlusion priniple for mutully disjoint sets, n (A B) = n(a) + n (B) Dividing oth sides y n(s), we get na ( B) na ( ) + nb ( ) = ns ( ) ns ( ) na ( ) nb ( ) = + ns ( ) ns ( ) PA ( B) = pa ( ) + PB ( ) Suppose die is rolled. Let A e the event tht 1 ppers & B e the event tht some even numer ppers on the die. Then S = {1, 2, 3, 4, 5, 6}, A = {1} & B = {2, 4, 6} Clerly A & B re disjoint events nd 1 3 PA ( ) =, PB ( ) = 6 6 Hene the proility tht 1 ppers or some even numer ppers is given y PA ( B) = PA ( ) + PB ( ) 1 3 = = = Ans. 6 3 A g ontins 6 white, 5 lk nd 4 red lls. Find the proility of getting white or lk ll in single drw. Let A e the event of getting white ll nd B e the event of getting lk ll. Totl numer of lls = = PA ( ) =, PB ( ) = Virtul University of Pkistn Pge 231

31 Sine the two events re disjoint (mutully exlusive),therefore PA ( B) = PA ( ) + PB ( ) 6 5 = = 15 Ans A pir of die is thrown. Find the proility of getting totl of 5 or 11. When two die re thrown, the smple spe hs 6 * 6 = 36 outomes. Let A e the event tht totl of 5 ours nd B e the event tht totl of 11 ours. Then A = {(1,4), (2,3), (3,2), (4,1)} nd B = {(5,6), (6,5)} Clerly, the events A nd B re disjoint (mutully exlusive) with proilities given y Now y using the sum Rule for Mutully Exlusive events we get PA ( B) = PA ( ) + PB ( ) = + = = Ans For ny two event A nd B of smple spe S. Prove tht P(A\B) = P(A B ) = P(A) - P(A B) The event A n e written s the union of two disjoint events A\B nd A B. i.e. A = (A\B) (A B) Hene, y ddition lw of proility P(A) = P(A\B) + P(A B) P (A\B) = P(A) - P (A B) A B S A\B A B GENERAL ADDITION LAW OF PROBABILITY THEOREM If A nd B re ny two events of smple spe S, then P(A B) = P(A) + P(B) - P(A B) PROOF: The event A B my e written s the union of two disjoint events A\B nd B. i.e., A B = (A\B) B Hene, y ddition lw of proility (for disjoint events) P(A B)= P(A\B) + P(B) = [P(A) - P(A B)] + P(B) = P(A) + P(B) - P(A B) (proved) Virtul University of Pkistn Pge 232

32 A B A\B REMARK: By inlusion - exlusion priniple n (A B) = n (A) + n(b) - n (A B) (where A nd B re finite) Dividing oth sides y n(s) nd denoting the rtios s respetive proilities we get the Generlized Addition Lw of proility. i.e P(A B) = P(A) + P(B) - P(A B) Let A nd B e events in smple spe S, nd let P(A) = 0.65, P(B) = 0.30 nd P(A B) = 0.15 Determine the proility of the events ()A B () A B ()A B ()As we know tht P(A B ) = P(A) - P(A B) (s A-B=A B ) = = 0.50 ()By ddition Lw of proility P(A B)= P(A) + P(B) - P(A B) (s A B φ ) = = 0.80 () By DeMorgn s Lw A B = (A B) P(A B )= P(A B) = 1 - P (A B) = = 0.20 Ans. Let A, B, C nd D e events whih form prtition of smple spe S. If P(A) = P(B), P(C) = 2 P(A) nd P(D) = 2 P(C). Determine eh of the following proilities. ()P(A) ()P(A B) ()P(A C D) () Sine A, B, C nd D form prtition of S, therefore S = A B C D nd A, B, C, D re pir wise disjoint. Hene, y ddition lw of proility. P(S) = P(A) + P(B) + P(C) + P(D) 1 = P(A) + P(A) + 2P(A) +2 P(C) 1 = 4P(A) + 2 (2P(A)) 1 = 8 P(A) 1 PA= ( ) 8 () P(A B) = P(A) + P(B) Virtul University of Pkistn Pge 233

33 = P(A) + P(A) [ P(B) = P(A)] = 2 P(A) = = 8 4 () P(A C D) = P(A) + P(C) + P(D) = P(A) + 2 P(A) + 2 (2P(A)) [ P(C) = 2 P(A) & P(D) = 2 P(C)] = 7 P(A) = = Ans. 8 8 A rd is drwn from well-shuffled pk of plying rd. Wht is the proility tht it is either spde or n e? Let A e the event of drwing spde nd B e the event of drwing n e. Now A nd B re not disjoint events. A B represents the event of drwing n e of spdes. Now 13 4 PA ( ) = ; PB ( ) = PA ( B) 1 = 52 PA ( B) = PA ( ) + PB ( ) PA ( B) = = = A lss ontins 10 oys nd 20 girls of whih hlf the oys nd hlf the girls hve rown eyes. Find the proility tht student hosen t rndom is oy or hs rown eyes. Let A e the event tht oy is hosen nd B e the event tht student with rown eyes is hosen.then A nd B re not disjoint events. Inft, A B represents the event tht oy with rown eyes is hosen. PA ( ) = = PB ( ) = = nd 5 5 P( A B) = = ( s someoys lso hverown eyes) PA ( B) = PA ( ) + PB ( ) PA ( B) = = = 30 3 Ans. Virtul University of Pkistn Pge 234

34 An integer is hosen t rndom from the first 100 positive integers. Wht is the proility tht the integer hosen is divisile y 6 or y 8? Let A e the event tht the integer hosen is divisile y 6, nd B e the event tht the integer hosen is divisile y 8. A B is the event tht the integer is divisile y oth 6 nd 8 (i.e. s their L.C.M. is 24) Now na ( ) = = 16; nb ( ) = = nd 100 na ( B) = = 4 24 Hene PA ( B) = PA ( ) + PB ( ) PA ( B) = = = Ans OR Let A denote the event tht the integer hosen is divisile y 6, nd B denote the event tht the integer hosen is divisile y 8 i.e 16 A={6,12,18,24,,90,96} n(a)=16 PA ( ) = 100 B={8,16,24,40,,88,96} n(b)=12 A B={24,48,72,96} n(a B)=12 12 PB ( ) = PA ( B) = 100 A student ttends mthemtis lss with proility 0.7 skips ounting lss with proility 0.4, nd ttends oth with proility 0.5. Find the proility tht (1)he ttends t lest one lss (2)he ttends extly one lss (1)Let A e the event tht the student ttends mthemtis lss nd B e the event tht the student ttends ounting lss. Then given P(A) = 0.7; P(B) = = 0.6 And P(A B) = 0.5, P(A B) =? By ddition lw of proility P(A B) = P(A) + P(B) - P(A B) = = 0.8 Virtul University of Pkistn Pge 235

35 (2) Students n ttend extly one lss in two wys ()He ttends mthemtis lss ut not ounting i.e., event A B or ()He does not ttend mthemtis lss nd ttends ounting lss i.e., event A B Sine the two event A B nd A B re disjoint, hene required proility is P(A B ) + P(A B) Now P(A B )= P(A\B) = P(A) - P(A B) = = 0.2 nd P(A B)= P(B\A) = P(B) - P(A B) = = 0.1 Hene required proility is P(A B ) + P(A B) = = 0.3 PROBABILITY OF SUB EVENT THEOREM: If A nd B re two events suh tht A B, then P(A) P(B) PROOF: Suppose A B. The event B my e written s the union of disjoint events B A nd B A i.e., B = (B A) ( B A ) But B A = A ( s A B ) So B = A ( B A ) P(B) = P(A) + P( B A ) But P( B A ) 0 Hene P(B) P(A) Or P(A) P(B) A B S B A B A Let A nd B e susets of smple spe S with P(A) = 0.7 nd P(B) = 0.5. Wht re the mximum nd minimum possile vlues of P(A B). Virtul University of Pkistn Pge 236

36 By ddition lw of proilities P(A B) = P(A) + P(B) - P(A B) = P(A B) = P(A B) Sine proility of ny event is lwys less thn or equl to 1, therefore mx P(A B) = 1, for whih P(A B) = 0.2 Next to find the minimum vlue, we note A B B P(A B) P(B) = 0.5 Thus for min P(A B) we tke mximum possile vlue of P(A B) whih is 0.5. Hene min P(A B) = mx P(A B) = = 0.7 is the required minimum vlue. ADDITION LAW OF PROBABILITY FOR THREE EVENTS: If A, B nd C re ny three events, then P(A B C) = P(A) + P(B) + P(C) - P(A B) - P (A C) - P(B C) + P(A B C) REMARK: If A, B, C re mutully disjoint events, then P(A B C)= P(A) + P(B) + P(C) Three newsppers A, B, C re pulished in ity nd survey of reders indites the following: 20% red A, 16% red B, 14% red C 8% red oth A nd B, 5% red oth A nd C 4% red oth B nd C, 2% red ll the three For person hosen t rndom, find the proility tht he reds none of the ppers Given PA ( ) = 20% = = 0.2; PB ( ) = 16% = = PC ( ) = 14% = = 0.14; PA ( B) = 8% = = PA ( C) = 5% = = 0.05; PB ( C) = 4% = = nd 2 PA ( B C) = 2% = = Now the proility tht person reds A or B or C = P(A B C) = P(A) + P(B) + P(C) - P(A B) - P(A C) - P(B C)+ P(A B C) = = 100 Virtul University of Pkistn Pge 237

37 Hene, the proility tht he reds none of the ppers = P(( A B C) ) = 1 PA ( B C) 35 = = 100 = 65% Let A, B nd C e events in smple spe S, with A B C=S, A (B C) =φ, P(A) = 0.2, P(B) = 0.5 nd P(C) = 0.7. Find P(A C ), P(B C),P(B C). P(A ) = 1 - P(A) = P(A ) = 0.8 Next, given tht the events A nd B C re disjoint, sine A (B C)= φ, therefore P(A (B C)) = P(A) + P(B C) P(S) = P(B C) 1 = P(B C) P(B C) = = 0.8 Finlly, y ddition lw of proility P(B C)= P(B) + P(C) - P(B C) 0.8 = P(B C) P (B C)= 0.4 is the required proility. Virtul University of Pkistn Pge 238

38 LECTURE 37 CONDITIONAL PROBABILITY MULTIPLICATION THEOREM INDEPENDENT EVENTS. Wht is the proility of getting 2 when die is tossed?. An even numer ppers on tossing die. (i) Wht is the proility tht the numer is 2? (ii) Wht is the proility tht the numer is 3? When die is tossed,the smple spe is S={1,2,3,4,5,6} n(s)=6. Let A denote the event of getting 2 i.e A={ 2 } n(a)=1 P( 2 ppers when the die is tossed) = ( ) ( ) n A n S = 1 6. (i) Let S 1 denote the totl numer of even numers from smple spe S,when die is tossed (i.e S 1 S ) S 1 ={ 2,4,6 } n(s 1 )=3 Let B denote the event of getting 2 from totl numer of even numer i.e B={ 2 } n( B ) = 1 n( B) P(2 ppers; given tht the numer is even)=p(b)= n S = 3 ( ) 1 1 (ii) Let C denote the event of getting 3 in S 1 (mong the even numers)i.e C={ } n(c)=0 nc ( ) P(3 ppers; given tht the numer is even) =P(C ) = 0 n S = 3 = ( ) 1 0 Suppose tht n urn ontins 3 red lls, 2 lue lls, nd 4 white lls, nd tht ll is seleted t rndom. Let E e the event tht the ll seleted is red. Then P(E) = 3/9 (s there re 3 red lls out of totl 9 lls) Let F e the event tht the ll seleted is not white. Then the proility of E if it is lredy known tht the seleted ll is not white would e P(red ll seleted; given tht the seleted ll is not white) = 3/5 (s we ount no white ll so there re totl 9 lls(i.e 2 lue nd 3 red lls ) ) This is lled the onditionl proility of E given F nd is denoted y P(E F). DEFINITION: Let E nd F e two events in the smple spe of n experiment with P(F) 0. The onditionl proility of E given F, denoted y P(E F), is defined s P(E F) P(E F) = P(F) Let A nd B e events of n experiment suh tht P(B) = 1/4 nd P(A B) = 1/6. Virtul University of Pkistn Pge 239

39 Wht is the onditionl proility P(A B)? P(A B) 1/ P(A B) = = = = P(B) 1/ Let A nd B e events with PA ( ) =, PB ( ) = nd PA ( B) = Find () i P( A B) ( ii) P( B A) ( iii) P( A B) ( iv) P( A B ) Using the formul of the onditionl Proility we n write () i P( A B) PA ( B) = PB ( ) 1/4 3 = = 1/3 4 ( ii) P( B A) PB ( A) = PA ( ) 1/4 2 1 = = = 1/2 4 2 (As PB ( A) = PA ( B) = 1/ 4 ) ( iii) P( A B) = P( A) + P( B) P( A B) = = 12 ( iv) P( A B ) PA ( B) = PB ( ) P(( A B) ) = (By using DeMorgn's Lw) PB ( ) 1 PA ( B) = [ PE ( ) = 1 PE ( )] 1 PB ( ) 1 7/12 5/ = = = = 1 1/3 2/ Find P(B A) if (i) A is suset of B (ii) A nd B re mutully exlusive (i) When A B, then B A = A ( As A A B A Α B A..(i) Virtul University of Pkistn Pge 240

40 lso we know tht B A A..(ii), From (i) nd (ii) lerly B A = A ) PB ( A) PB ( A) = ( sb A= A PB ( A) = PA ( )) PA ( ) PA ( ) = = 1 PA ( ) (ii) When A nd B re mutully exlusive, then B A= PB ( A) PB ( A) = PA ( ) P( φ) = ( Sine B A = φ P( B A) = 0) PA ( ) 0 = = 0 ( s P( φ ) = 0 ) PA ( ) Suppose tht n urn ontins three red lls mrked 1, 2, 3, one lue ll mrked 4, nd four white lls mrked 5, 6, 7, 8. A ll is seleted t rndom nd its olor nd numer noted. (i) Wht is the proility tht it is red? (ii) Wht is the proility tht it hs n even numer mrked on it? (iii) Wht is the proility tht it is red, if it is known tht the ll seleted hs n even numer mrked on it? (iv) Wht is the proility tht it hs n even numer mrked on it, if it is known tht the ll seleted is red? Let E e the event tht the ll seleted is red. (i) P(E) = 3/8 let F e the event tht the ll seleted hs n even numer mrked on it. (ii) P(F) = 4/8 (s there re four even numers 2,4,6 & 8 out of totl eight numers). (iii) E F is the event tht the ll seleted is red nd hs n even numer mrked on it. Clerly P(E F) =1/8 (s there is only one ll whih is red nd mrked n even numer 2 out of totl eight lls). Hene, P(Seleted ll is red, given tht the ll seleted hs n even numer mrked on it.) = P(E F) P(E F) 1/8 = = = 1/4 P(F) 4 /8 (iv) P(Seleted ll hs n even numer mrked on it, given tht the ll seleted is red) = P(F E) P(E F) 1/8 1 = = = P(E) 3/8 3 Let pir of die e tossed. If the sum is 7, find the proility tht one of the die is 2. Virtul University of Pkistn Pge 241

41 Let E e the event tht 2 ppers on t lest one of the two die, nd F e the event tht the sum is 7. Then E = {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 2) (4, 2), (5, 2),,(6, 2)} F = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} E F = {(2, 5), (5, 2)} P(F) = 6/36 nd P(E F) = 2/36. Hene, P(Proility tht one of the die is 2, given tht the sum is 7) = P(E F) P(E F) 2 / 36 1 = = = P(E) 6 / 36 3 A mn visits fmily who hs two hildren. One of the hildren, oy, omes into the room. Find the proility tht the other hild is lso oy if (i) The other hild is known to e elder, (ii) Nothing is known out the other hild. The smple spe of the experiment is S = {, g, g, gg} (The outome g speifies tht younger is oy nd elder is girl, et.) Let A e the event tht oth the hildren re oys. Then, A = {}. (i) Let B e the event tht the younger is oy. Then, B = {, g},nd A B = {}. Hene, the required proility is P(Proility tht the other hild is lso oy, given tht the other hild is elder) = P(A B) 1/ 4 1 P(A B) = = = P(B) 2 / 4 2 (ii) Let C e the event tht one of the hildren is oy. Then C = {, g, g}, nd A C = {}. Hene, the required proility is P(Proility tht oth the hildren re oys, given tht one of the hildren is oy) = P(A C) P(A C) 1/ 4 1 = = = P(C) 3/ 4 3 MULTIPLICATION THEOREM Let E nd F e two events in the smple spe of n experiment, then P(E F) = P(F)P(E F) Or P(E F) = P(E)P(F E) Let E 1, E 2,..., E n e events in the smple spe of n experiment, then P(E 1 E... E ) = P(E )P(E E ) P(E 2 E n E 2 )... P(E E E... E ) n 1 2 n Virtul University of Pkistn Pge 242

42 A lot ontins 12 items of whih 4 re defetive. Three items re drwn t rndom from the lot one fter the other. Wht is the proility tht ll three re nondefetive? Let A 1 e the event tht the first item is not defetive. Let A 2 e the event tht the seond item is not defetive. Let A 3 e the event tht the third item is not defetive. Then P(A 1 ) = 8/12, P(A 2 A 1 ) = 7/11, nd P(A 3 A 1 A 2 ) = 6/10 Hene, y multiplition theorem, the proility tht ll three re non-defetive is P(A 1 A 2 A 3 ) = P(A 1 ) P(A 2 A 1 ) P(A 3 A 1 A 2 ) = = INDEPENDENCE: An event A is sid to e independent of n event B if the proility tht A ours is not influened y whether B hs or hs not ourred. Tht is, P(A B) = P(A). It follows then from the Multiplition Theorem tht, P(A B) = P(B)P(A B) = P(B)P(A) We lso know tht, P(A B) P(B A) = P(A) P(A)P(B) = Beuse P(A B) = P(A)P(B),due to independene P(A) = P(B) Let A e the event tht rndomly generted it string of length four egins with 1 nd let B e the event tht rndomly generted it string of length four ontins n even numer of 0s. Are A nd B independent events? A = {1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111} B = {0000, 0011, 0101, 0110, 1001, 1010, 1100, 1111} Sine there re 16 it strings of length four, we hve P(A) = 8/16 = 1/2, P(B) = 8/16 = 1/2 Also, A B = {1001, 1010, 1100, 1111} so tht P(A B) = 4/16 =1/4 We note tht, P(A B) = = = P(A)P(B) Hene A nd B re independent events. Let fir oin e tossed three times. Let A e the event tht first toss is heds, B e the even tht the seond toss is heds, nd C e the event tht extly two heds re tossed in row. Exmine pir wise independene of the three events. The smple spe of the experiment is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} nd the events re A = {HHH, HHT, HTH, HTT} Virtul University of Pkistn Pge 243

43 B = {HHH, HHT, THH, THT} C = {HHT, THH} A B = {HHH, HHT}, A C = {HHT}, B C = {HHT, THH} It follows tht P(A) = 4/8=1/2 P(B) = 4/8= 1/2 P(C) = 2/8=1/4 nd P(A B)=2/8=1/4 P(A C)= 1/8 P(B C) =2/8= 1/4 Note tht, P(A)P(B) = = = P(A B), so tht A nd B re independent P(A)P(C) = = = P(A C), so tht A nd C re independent P(B)P(C) = = P(B C), so tht B nd C re dependent The proility tht A hits trget is 1/3 nd the proility tht B hits the trget is 2/5. Wht is the proility tht trget will e hit if A nd B eh shoot t the trget? It is ler from the nture of the experiment tht the two events re independent. Hene, P(A B) = P(A)P(B) It follows tht, P(A B) = P(A) + P(B) P(A B) = P(A) + P(B) P(A)P(B) (due to independene ) = = 3 5 Virtul University of Pkistn Pge 244

44 LECTURE 38 RANDOM VARIABLE PROBABILITY DISTRIBUTION EXPECTATION AND VARIANCE INTRODUCTION: Suppose S is the smple spe of some experiment. The outomes of the experiment, or the points in S, need not e numers. For exmple in tossing oin,the outomes re H (heds) or T(tils), nd in tossing pir of die the outomes re pirs of integers. However, we frequently wish to ssign speifi numer to eh outome of the experiment. For exmple, in oin tossing, it my e onvenient to ssign 1 to H nd 0 to T; or in the tossing of pir of die, we my wnt to ssign the sum of the two integers to the outome. Suh n ssignment of numeril vlues is lled rndom vrile. RANDOM VARIABLE: A rndom vrile X is rule tht ssigns numeril vlue to eh outome in smple Spe S. OR It is funtion whih mps eh outome of the smple spe into the set of rel numers. We shll let X(S) denote the set of numers ssigned y rndom vrile X, nd refer to X(S) s the rnge spe. In forml terminology, X is funtion from S(smple spe) to the set of rel numers R, nd X(S) is the rnge of X. REMARK: 1. A rndom vrile is lso lled hne vrile, or stohsti vrile(not lled simply vrile, euse it is funtion). 2. Rndom vriles re usully denoted y pitl letters suh s X, Y, Z; nd the vlues tken y them re represented y the orresponding smll letters. A pir of fir die is tossed. The smple spe S onsists of the 36 ordered pirs i.e S = {(1,1), (1,2), (1,3),, (6,6)} Let X ssign to eh point in S the sum of the numers; then X is rndom vrile with rnge spe i.e X(S) = {2,3,4,5,6,7,8,9,10,11,12} Let Y ssign to eh point in S the mximum of the two numers in the outomes; then Y is rndom vrile with rnge spe. Y(S) = {1,2,3,4,5,6} PROBABILITY DISTRIBUTION OF A RANDOM VARIABLE: Let X(S) = {x 1, x 2,, x n } e the rnge spe of rndom vrile X defined on finite smple spe S. Define funtion f on X(S) s follows: f(x i ) = P (X = x i ) = sum of proilities of points in S whose imge is x i. This funtion f is lled the proility distriution or the proility funtion of X. The proility distriution f of X is usully given in the form of tle. x 1 x 2 x n f(x 1 ) f(x 2 ) f(x n ) The distriution f stisfies the onditions. Virtul University of Pkistn Pge 245

45 ( i) f( x ) 0 nd ( ii) f( x ) = 1 i A pir of fir die is tossed. Let X ssign to eh point (, ) in S = {(1,1), (1,2),, (6,6)}, the sum of its numer, i.e., X (,) = +. Compute the distriution f of X. X is lerly rndom vrile with rnge spe X(S) = {2,3,4,5,6,7,8,9,10, 11, 12} ( euse X (,) = + X (1,1) = 1+1=2, X (1,2) = 1+2=3, X (1,3) = 1+3=4 et). The distriution f of X my e omputed s: 1 f(2) = P(X=2)=P({(1,1)}) = 36 2 f(3) = P(X=3) = P({(1,2), (2,1)}) = 36 3 f(4) = P(X=4) = P({(1,3), (2,2),(3,1)}) = 36 4 f(5) = P(X=5) = P({(1,4), (2,3),(3,2),(4,1)}) = 36 n i= 1 i 5 f(6) = P( X = 6) = P({(1,5),(2, 4),(3,3),(4, 2),(5,1)}) = 3 f(7) = P( X = 7) = P({(1,6),(2,5),(3, 4),(4,3),(5, 2),(6, 2) 5 f(8) = P( X = 8) = P({(2,6),(3,5),(4,4),(5,3),(6,2)}) = 36 4 f(9) = P( X = 9) = P({(3,6),(4,5),(5,4),(6,3)}) = 36 3 f(10) = P( X = 10) = P({(4,6),(5,5),(6,4)}) = 36 2 f(11) = P( X = 11) = P({(5,6),(6,5)}) = 36 1 f(12) = P( X = 12) = P({(6,6)}) = 36 The distriution of X onsists of the points in X(S) with their respetive proilities. x i f(x i) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 A ox ontins 12 items of whih three re defetive. A smple of three items is seleted from the ox. Virtul University of Pkistn Pge 246

46 If X denotes the numer of defetive items in the smple; find the distriution of X. 12 The smple spe S onsists of = 220 tht is 220 different smples of size 3. 3 The rndom vrile X, denoting the numer of defetive items hs the rnge spe X(S) = {0,1,2,3} 3 9 There re = 84 smples of size 3 with no defetive items; 0 3 hene 84 p0 = P( X = 0) = There re = 108 smples of size 3 ontining one defetive item; hene p1 = P( X = 1) = 220 There re 3 9 = smples of size 3 ontining two defetive items; hene 27 p2 = P( X = 2) = Finlly, there is = defetive items; hene 1 p3 = P( X = 3) = 220, only one smple of size 3 ontining three The distriution of X follows: x i p i 84/ /220 27/220 1/220 EXPECTATION OF A RANDOM VARIABLE Let X e rndom vrile with proility distriution x i x 1 x 2 x 3 x n f(x i ) f(x 1 ) f(x 2 ) f(x 3 ) f(x n ) The men (denoted μ) or the expettion of X (written E(X)) is defined y μ = E(X) = x 1 f(x 1 ) + x 2 f(x 2 ) + + x n f(x n ) n = xi f( xi) i= 1 Virtul University of Pkistn Pge 247

47 Wht is the expettion of the numer of heds when three fir oins re tossed? The smple spe of the experiment is: S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let the rndom vrile X represents the numer of heds ( i.e 0,1,2,3 ) when three fir oins re tossed. Then X hs the proility distriution. x i x 0 =0 x 1 =1 x 2 =2 x 3 =3 f(x i ) 1/8 3/8 3/8 1/8 Hene, expettion of X is E( X ) = x 0 f(x 0) + x1f(x 1) + x 2f(x 2) + x3f(x 3) = = = A plyer tosses two fir oins. He wins Rs. 1 if one hed ppers, Rs.2 if two heds pper. On the other hnd, he loses Rs.5 if no heds pper. Determine the expeted vlue E of the gme nd if it is fvourle to e plyer. The smple spe of the experiment is S = {HH, HT, TH, TT} Now P(Two heds) = P( HH) = = P(One hed) = P( HT, TH) = + = P(No heds) = P( TT) = = Thus, the proility of winning Rs.2 is 1, of winning Rs 1 is 1 nd of losing Rs 5 is Hene, E = 2( ) + 1( ) 5( ) = = Sine, the expeted vlue of the gme is negtive, so it is unfvorle to the plyer. 3 1 A oin is weighted so tht nd PH ( ) =. nd PT ( ) = The oin is tossed three times. 4 4 Let X denotes the numer of heds tht pper. () Find the distriution of X () Find the expettion of E(X) () The smple spe is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Virtul University of Pkistn Pge 248

48 The proilities of the points in smple spe re p( HHH ) = = p( HHT ) = = p( HTH) = = p( THH) = = phtt ( ) = = ptht ( ) = = ptth ( ) = = pttt ( ) = = The rndom vrile X denoting the numer of heds ssumes the vlues 0,1,2,3 with the proilities: 1 P(0) = P( TTT) = P(1) = P( HTT, THT, TTH) = + + = P(2) = P( HHT, HTH, THH ) = + + = P (3) = P( HHH ) = 64 Hene, the distriution of X is x i P(x i ) 1/64 9/64 27/64 27/64 () The expeted vlue E(X) is otined y multiplying eh vlue of X y its proility nd tking the sum. The distriution of X is x i P(x i ) 1/64 9/64 27/64 27/64 Hene EX ( ) = = 64 = 225 Virtul University of Pkistn Pge 249

49 VARIANCE AND STANDARD DEVIATION OF A RANDOM VARIABLE: Let X e rndom vrile with men μ nd the proility distriution x 1 X 2 x 3 x n f(x 1 ) f(x 2 ) f(x 3 ) f(x n ) The vrine of X, mesures the spred or dispersion of X from the men μ nd is denoted nd defined s n 2 2 σ = Vr( X ) = ( x μ) f( x ) x i i i= 1 2 = E(( X μ) ) 2 2 = EX ( ) μ 2 2 = xi f( xi) μ The lst expression is more onvenient form for omputing Vr(X). The stndrd derivtion of X, denoted y, is the non-negtive squre root of Vr(X): σ X Where σ X = Vr( X ) Find the expettion μ, vrine σ 2 nd stndrd devition σ of the distriution given in the following tle. x i f(x i ) μ = E(X) = x i f(x i ) = 1(0.4) + 3(0.1) + 4(0.2) + 5(0.3) = = 3.0 Next E(X 2 ) = x 2 i f(x i ) = 1 2 (0.4) (0.1) (0.2) (0.3) = = 12.0 Hene σ 2 = Vr(X)= E(X 2 ) - μ 2 = (3.0) 2 = 3.0 nd σ = Vr( X ) = A pir of fir die is thrown. Let X denote the mximum of the two numers whih ppers. () Find the distriution of X () Find the μ, vrine σ 2 x = Vr(X), nd stndrd devition σx of X () The smple spe S onsist of the 36 pirs of integers (,) where nd rnge from 1 to 6; Virtul University of Pkistn Pge 250

50 tht is S = {(1,1), (1,2),, (6,6)} Sine X ssigns to eh pir in S the lrger of the two integers, the vlue of X re the integers from 1 to 6. Note tht: 1 f(1) = P( X = 1) = P({(1,1)}) = 36 3 f(2) = P( X = 2) = P({(2,1),(2, 2),(1, 2)}) = 36 5 f(3) = P( X = 3) = P({(3,1),(3, 2),(3,3),(2,3),(1,3)}) = 36 7 f(4) = P( X = 4) = P({(4,1),(4,2),(4,3),(4,4),(3,4),(2,4),(1,4)}) = 36 Similrly f (5) = PX ( = 5) = {(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5)} 9 = 36 nd f (6) = PX ( = 6) = {(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6, 2),(6,3),(6, 4),(6,5),(6,6)} 11 = 36 Hene, the proility distriution of x is: x i f(x i ) 1/36 3/36 5/36 7/36 9/36 11/36 ()We find the expettion (men) of X s μ = EX ( ) = xf( x) i i = = Virtul University of Pkistn Pge 251

51 Next Then nd EX 2 2 ( ) = xf i ( xi) = = σ = Vr( X) = E( X ) μ 2 2 x σ = x = 22.0 (4.5) = Virtul University of Pkistn Pge 252

52 LECTURE 39 INTRODUCTION TO GRAPHS INTRODUCTION: Grph theory plys n importnt role in severl res of omputer siene suh s: swithing theory nd logil design rtifiil intelligene forml lnguges omputer grphis operting systems ompiler writing informtion orgniztion nd retrievl. GRAPH: A grph is non-empty set of points lled verties nd set of line segments joining pirs of verties lled edges. Formlly, grph G onsists of two finite sets: (i) A set V=V(G) of verties (or points or nodes) (ii)a set E=E(G) of edges; where eh edge orresponds to pir of verties. e 1 v 1 v 2 e 4 e 3 e 2 v 4 e 5 v 3 e 6 v 5 The grph G with V(G) = {v 1, v 2, v 3, v 4, v 5 } nd E(G) = {e 1, e 2, e 3, e 4, e 5, e 6 } SOME TERMINOLOGY: e 1 v 1 v 2 e 3 e 2 v 4 e 4 e 5 v 3 e 6 v 5 1. An edge onnets either one or two verties lled its endpoints (edge e 1 onnets verties v 1 nd v 2 desried s {v 1, v 2 } i.e v 1 nd v 2 re the endpoints of n edge e 1 ). 2. An edge with just one endpoint is lled loop. Thus loop is n edge tht onnets vertex to itself (e.g., edge e 6 mkes loop s it hs only one endpoint v 3 ). 3. Two verties tht re onneted y n edge re lled djent; nd vertex tht is n endpoint of loop is sid to e djent to itself. 4. An edge is sid to e inident on eh of its endpoints(i.e. e 1 is inident on v 1 nd v 2 ). 5. A vertex on whih no edges re inident is lled isolted (e.g., v 5 ) 6. Two distint edges with the sme set of end points re sid to e prllel (i.e. e 2 & e 3 ). Define the following grph formlly y speifying its vertex set, its edge set, nd tle giving the edge endpoint funtion. Virtul University of Pkistn Pge 253

53 e 1 v 1 v 2 e 2 v 3 e3 v 4 Vertex Set = {v 1, v 2, v 3, v 4 } Edge Set = {e 1, e 2, e 3 } Edge - endpoint funtion is: Edge Endpoint e 1 {v 1, v 2 } e 2 {v 1, v 3 } e 3 {v 3 } For the grph shown elow (i) find ll edges tht re inident on v 1 ; (ii)find ll verties tht re djent to v 3 ; (iii)find ll loops; (iv)find ll prllel edges; (v)find ll isolted verties; e 1 e2 e 3 v 2 v 1 e 6 e 4 e e 7 5 v 3 v 5 v 4 (i) v 1 is inident with edges e 1, e 2 nd e 7 (ii) verties djent to v 3 re v 1 nd v 2 (iii) loops re e 1 nd e 3 (iv) only edges e 4 nd e 5 re prllel (v) The only isolted vertex is v 4 in this Grph. DRAWING PICTURE FOR A GRAPH: Drw piture of Grph H hving vertex set {v 1, v 2, v 3, v 4, v 5 } nd edge set {e 1, e 2, e 3, e 4 } with edge endpoint funtion Edge Endpoint e 1 {v 1 } e 2 {v 2,v 3 } e 3 {v 2,v 3 } e 4 {v 1,v 5 } Virtul University of Pkistn Pge 254

54 Given V(H) = {v 1, v 2, v 3, v 4, v 5 } nd E(H) = {e 1, e 2, e 3, e 4 } with edge endpoint funtion SIMPLE GRAPH A simple grph is grph tht does not hve ny loop or prllel edges. e 1 v 1 v 2 e 2 e 3 e4 v 5 v 4 It is simple grph H V(H) = {v 1, v 2, v 3, v 4, v 5 } & E(H) = {e 1, e 2, e 3, e 4 } Drw ll simple grphs with the four verties {u, v, w, x} nd two edges, one of whih is {u, v}. There re C(4,2) = 6 wys of hoosing two verties from 4 verties. These edges my e listed s: {u,v},{u,w},{u,x},{v,w}, {v,x},{w,x} One edge of the grph is speified to e {u,v}, so ny of the remining five from this list my e hosen to e the seond edge. This required grphs re: u v u v v 3 w x w u u v x v w x w x Virtul University of Pkistn Pge 255

55 5. u v w x DEGREE OF A VERTEX: Let G e grph nd v vertex of G. The degree of v, denoted deg(v), equls the numer of edges tht re inident on v, with n edge tht is loop ounted twie. Note:(i)The totl degree of G is the sum of the degrees of ll the verties of G. (ii) The degree of loop is ounted twie. For the grph shown v 2. v 1 e 1 e 2 v 3 deg (v 1 ) = 0, sine v 1 is isolted vertex. deg (v 2 ) = 2, sine v 2 is inident on e 1 nd e 2. deg (v 3 ) = 4, sine v 3 is inident on e 1,e 2 nd the loop e 3. Totl degree of G = deg(v 1 ) + deg(v 2 ) + deg(v 3 ) = = 6 e 3 REMARK: The totl degree of G, whih is 6, equls twie the numer of edges of G, whih is 3. THE HANDSHAKING THEOREM: If G is ny grph, then the sum of the degrees of ll the verties of G equls twie the numer of edges of G. Speifilly, if the verties of G re v 1, v 2,, v n, where n is positive integer, then the totl degree of G = deg(v 1 ) + deg(v 2 ) + + deg(v n ) = 2. (the numer of edges of G) PROOF: Eh edge e of G onnets its end points v i nd v j. This edge, therefore ontriutes 1 to the degree of v i nd 1 to the degree of v j. If e is loop, then it is ounted twie in omputing the degree of the vertex on whih it is inident. Aordingly, eh edge of G ontriutes 2 to the totl degree of G. Thus, the totl degree of G = 2. (the numer of edges of G) COROLLARY: The totl degree of G is n even numer Drw grph with the speified properties or explin why no suh grph exists. (i) Grph with four verties of degrees 1, 2, 3 nd 3 (ii) Grph with four verties of degrees 1, 2, 3 nd 4 Virtul University of Pkistn Pge 256

56 (iii)simple grph with four verties of degrees 1, 2, 3 nd 4 (i) Totl degree of grph = = 9 n odd integer Sine, the totl degree of grph is lwys even, hene no suh grph is possile. Note:As we know tht for ny grph,the sum of the degrees of ll the verties of G equls twie the numer of edges of G or the totl degree of G is n even numer. (ii) Two grphs with four verties of degrees 1, 2, 3 & 4 re d or d The verties,,, d hve degrees 1,2,3, nd 4 respetively(i.e grph exists). (iii) Suppose there ws simple grph with four verties of degrees 1, 2, 3, nd 4. Then the vertex of degree 4 would hve to e onneted y edges to four distint verties other thn itself euse of the ssumption tht the grph is simple (nd hene hs no loop or prllel edges.) This ontrdits the ssumption tht the grph hs four verties in totl. Hene there is no simple grph with four verties of degrees 1, 2, 3, nd 4, so simple grph is not possile in this se. Suppose grph hs verties of degrees 1, 1, 4, 4 nd 6. How mny edges does the grph hve? The totl degree of grph = = 16 Sine, the totl degree of grph = 2.(numer of edges of grph) Hndshking theorem ] 16 = 2.(numer of edges of grph) [y using Numer of edges of grph = 16 2 = 8 In group of 15 people, is it possile for eh person to hve extly 3 friends? Suppose tht in group of 15 people, eh person hd extly 3 friends. Then we ould drw grph representing eh person y vertex nd onneting two verties y n edge if the orresponding people were friends. But suh grph would hve 15 verties eh of degree 3, for totl degree of 45 (not even) whih is not possile. Hene, in group of 15 people it is not possile for eh to hve extly three friends. COMPLETE GRAPH: A omplete grph on n verties is simple grph in whih eh vertex is onneted to every other vertex nd is denoted y K n (K n mens tht there re n verties). The following re omplete grphs K 1, K 2,K 3, K 4 nd K 5. Virtul University of Pkistn Pge 257

57 v 1 v 1 K 1 v 1 v 2 K 2 v 2 v 3 v 2 v 1 v 4 v 1 v 5 v 2 v 3 K 4 v 3 v 4 K 5 For the omplete grph K n, find (i) the degree of eh vertex (ii)the totl degrees (iii)the numer of edges (i) Eh vertex v is onneted to the other (n-1) verties in K n ; hene deg (v) = n - 1 for every v in K n. (ii)eh of the n verties in K n hs degree n - 1; hene, the totl degree in K n = (n - 1) + (n - 1) + + (n - 1) n times = n (n - 1) (iii)eh pir of verties in K n determines n edge, nd there re C(n, 2) wys of seleting two verties out of n verties. Hene, Numer of edges in K n = C(n, 2) nn ( 1) = 2 Alterntively, The totl degrees in grph K n = 2 (numer of edges in K n ) n(n-1) = 2(numer of edges in K n ) nn ( 1) Numer of edges in K n = REGULAR GRAPH: 2 A grph G is regulr of degree k or k-regulr if every vertex of G hs degree k. In other words, grph is regulr if every vertex hs the sme degree. Following re some regulr grphs. (i) 0-regulr (ii) 1-regulr Virtul University of Pkistn Pge 258

58 (iii) REMARK: The omplete grph K n is (n-1) regulr. 2-regulr Drw two 3-regulr grphs with six verties. f f e d BIPARTITE GRAPH: A iprtite grph G is simple grph whose vertex set n e prtitioned into two mutully disjoint non empty susets A nd B suh tht the verties in A my e onneted to verties in B, ut no verties in A re onneted to verties in A nd no verties in B re onneted to verties in B. The following re iprtite grphs A v 1 v 2 v 3 A B v 1 e d v 4 B v 4 v 5 v 2 v 3 v 5 v 6 DETERMINING BIPARTITE GRAPHS: The following leling proedure determines whether grph is iprtite or not. 1. Lel ny vertex 2. Lel ll verties djent to with the lel. 3. Lel ll verties tht re djent to vertex just leled with lel. 4. Repet steps 2 nd 3 until ll verties got distint lel ( iprtite grph) or there is onflit i.e., vertex is leled with nd (not iprtite grph). Find whih of the following grphs re iprtite. Redrw the iprtite grph so tht its iprtite nture is evident. Virtul University of Pkistn Pge 259

59 (i), (onflit) The grph is not iprtite. (ii) By leling proedure, eh vertex gets distint lel. Hene the grph is iprtite. To redrw the grph we mrk lels s s 1, 2 nd s s 1, 2, Redrwing grph with iprtite nture evident COMPLETE BIPARTITE GRAPH: A omplete iprtite grph on (m+n) verties denoted K m,n is simple grph whose vertex set n e prtitioned into two mutully disjoint non empty susets A nd B ontining m nd n verties respetively, suh tht eh vertex in set A is onneted (djent) to every vertex in set B, ut the verties within set re not onneted. K 2,3 K 3,3 Virtul University of Pkistn Pge 260

60 LECTURE 40 PATHS AND CIRCUITS KONIGSBERG BRIDGES PROBLEM A B C D It is possile for person to tke wlk round town, strting nd ending t the sme lotion nd rossing eh of the seven ridges extly one? A A B C D C D B Is it possile to find route through the grph tht strts nd ends t some vertex A, B, C or D nd trverses eh edge extly one? Equivlently: Is it possile to tre this grph, strting nd ending t the sme point, without ever lifting your penil from the pper? DEFINITIONS: Let G e grph nd let v nd w e verties in grph G. 1. WALK A wlk from v to w is finite lternting sequene of djent verties nd edges of G. Thus wlk hs the form v 0 e 1 v 1 e 2 v n-1 e n v n where the v s represent verties, the e s represent edges v 0 =v, v n =w, nd for ll i = 1, 2 n, v i-1 nd v i re endpoints of e i. The trivil wlk from v to v onsists of the single vertex v. 2. CLOSED WALK A losed wlk is wlk tht strts nd ends t the sme vertex. 3. CIRCUIT A iruit is losed wlk tht does not ontin repeted edge. Thus iruit is wlk of the form v 0 e 1 v 1 e 2 v n-1 e n v n where v 0 = v n nd ll the e i,s re distint. 4. SIMPLE CIRCUIT Virtul University of Pkistn Pge 261

61 A simple iruit is iruit tht does not hve ny other repeted vertex exept the first nd lst. Thus simple iruit is wlk of the form v 0 e 1 v 1 e 2 v n-1 e n v n where ll the e i,s re distint nd ll the v j,s re distint exept tht v 0 = v n 5. PATH A pth from v to w is wlk from v to w tht does not ontin repeted edge. Thus pth from v to w is wlk of the form v = v 0 e 1 v 1 e 2 v n-1 e n v n = w where ll the e i,s re distint (tht is e i e k for ny i k). 6. SIMPLE PATH A simple pth from v to w is pth tht does not ontin repeted vertex. Thus simple pth is wlk of the form v = v 0 e 1 v 1 e 2 v n-1 e n v n = w where ll the e i,s re distint nd ll the v j,s re lso distint (tht is, v j v m for ny j m). SUMMARY Repeted Repeted Strts nd Ends t Sme Point Edge Vertex wlk llowed Allowed llowed losed wlk llowed Allowed yes(mens, where it strts lso ends t tht point) iruit no Allowed yes simple iruit no first nd lst only yes pth no Allowed llowed simple pth no no No In the grph elow, determine whether the following wlks re pths, simple pths, losed wlks, iruits, simple iruits, or re just wlks. v 1 v 3 e 10 v 0 e 1 e 2 e 3 e 8 e 9 v 2 e 7 e 5 e 4 v 5 e 6 v 4 () v 1 e 2 v 2 e 3 v 3 e 4 v 4 e 5 v 2 e 2 v 1 e 1 v 0 () v 1 v 2 v 3 v 4 v 5 v 2 () v 4 v 2 v 3 v 4 v 5 v 2 v 4 (d) v 2 v 1 v 5 v 2 v 3 v 4 v 2 (e) v 0 v 5 v 2 v 3 v 4 v 2 v 1 (f) v 5 v 4 v 2 v 1 () v 1 e 2 v 2 e 3 v 3 e 4 v 4 e 5 v 2 e 2 v 1 e 1 v 0 Virtul University of Pkistn Pge 262

62 v 0 () v 1 e 1 e 2 v 2 e 3 v 3 e e 4 5 v 5 This grph strts t vertex v 1,then goes to v 2 long edge e 2,nd moves ontinuously, t the end it goes from v 1 to v 0 long e 1.Note it tht the vertex v 2 nd the edge e 2 is repeted twie, nd strting nd ending, not t the sme points.hene The grph is just wlk. v 4 () v 1 v 2 v 3 v 4 v 5 v 2 v 1 v 3 v 0 v 2 v 5 In this grph vertex v 2 is repeted twie.as no edge is repeted so the grph is pth. v 4 () v 4 v 2 v 3 v 4 v 5 v 2 v 4 v 1 v 3 v 0 v 2 v 5 As verties v 2 & v 4 re repeted nd grph strts nd ends t the sme point v 4,lso the edge(i.e. e 5 )onneting v 2 & v 4 is repeted, so the grph is losed wlk. (d) v 2 v 1 v 5 v 2 v 3 v 4 v 2 v 1 v 4 v 3 v 0 v 2 v 5 In this grph, vertex v 2 is repeted nd the grph strts nd end t the sme vertex (i.e. t v 2 ) nd no edge is repeted, hene the ove grph is iruit. (e) v 0 v 5 v 2 v 3 v 4 v 2 v 1 v 4 v 1 v 3 v 0 v 2 Here vertex v 2 is repeted nd no edge is repeted so the grph is pth. v 5 v 4 (f) v 5 v 4 v 2 v 1 Virtul University of Pkistn Pge 263

63 v 1 v 3 v 0 v 2 v 5 Neither ny vertex nor ny edge is repeted so the grph is simple pth. CONNECTEDNESS: Let G e grph. Two verties v nd w of G re onneted if, nd only if, there is wlk from v to w. The grph G is onneted if, nd only if, given ny two verties v nd w in G, there is wlk from v to w. Symolilly: G is onneted verties v, w V (G), wlk from v to w: Whih of the following grphs hve onnetedness? v 4 () () v v 2 v 3 3 v 2 v 4 v 5 v 4 v 1 v 1 v 6 v 6 v 5 () v 2 v 5 v 6 (d) v 4 v 3 v 4 v 2 v 1 v 5 v 1 v 3 v 8 v 7 v 6 EULER CIRCUITS DEFINITION: Let G e grph. An Euler iruit for G is iruit tht ontins every vertex nd every edge of G. Tht is, n Euler iruit for G is sequene of djent verties nd edges in G tht strts nd ends t the sme vertex uses every vertex of G t lest one, nd used every edge of G extly one. THEOREM: A grph G hs n Euler iruit if, nd only if, G is onneted nd every vertex of G hs n even degree. KONIGSBERG BRIDGES PROBLEM A A B C D C D We try to solve Konigserg ridges prolem y Euler method. Here deg()=3,deg()=3,deg()=3 nd deg(d)=5 s the verties hve odd degree so there is no possiility of n Euler iruit. Determine whether the following grph hs n Euler iruit. B Virtul University of Pkistn Pge 264

64 v 3 v 4 v 6 v 2 v 1 v 5 v 0 v 7 v 9 v 8 As deg (v 1 ) =5, n odd degree so the following grph hs not n Euler iruit. Determine whether the following grph hs Euler iruit. i h From ove lerly deg()=2, deg()=4, deg()=4, deg(d)=4, deg(e)=2, deg(f)=4, deg(g)=4, deg(h)=4, deg(i)=4 Sine the degree of eh vertex is even, nd the grph hs Euler Ciruit. One suh iruit is: d e f g d f i h g h i EULER PATH DEFINITION: Let G e grph nd let v nd w e two verties of G. An Euler pth from v to w is sequene of djent edges nd verties tht strts t v, end nd w, psses through every vertex of G t lest one, nd trverses every edge of G extly one. COROLLARY Let G e grph nd let v nd w e two verties of G. There is n Euler pth from v to w if, nd only if, G is onneted, v nd w hve odd degree nd ll other verties of G hve even degree. HAMILTONIAN CIRCUITS DEFINITION: Given grph G, Hmiltonin iruit for G is simple iruit tht inludes every vertex of G. Tht is, Hmiltonin iruit for G is sequene of djent verties nd distint edges in whih every vertex of G ppers extly one. Find Hmiltonin Ciruit for the following grph. g f d d e h e The Hmiltonin Ciruit for the following grph is: Another Hmiltonin Ciruit for the following grph ould e: PROPOSITION: g f d e f g h d e f g h Virtul University of Pkistn Pge 265

65 If grph G hs Hmiltonin iruit then G hs su-grph H with the following properties: 1. H ontins every vertex G 2. H is onneted 3. H hs the sme numer of edges s verties 4. Every vertex of H hs degree 2 Show tht the following grph does not hve Hmiltonin iruit. d e g f Here deg()=5,if we remove 3 edges from vertex then deg()< 2, deg(g)< 2 or deg(f)< 2,deg(d)< 2. It mens tht this grph does not stisfy the desired properties s ove, so the grph does not hve Hmiltonin iruit. Virtul University of Pkistn Pge 266

66 LECTURE 41 MATRIX REPRESENTATIONS OF GRAPHS MATRIX: An m n mtrix A over set S is retngulr rry of elements of S rrnged into m rows nd n olumns: j 1 n j 2n ith row of A A = i1 i 2 ij in m 1 m2 mj mn jth olumn of A Briefly, it is written s: A = [ ij ] m n A = A is mtrix hving 3 rows nd 4 olumns. We ll it 3 4 mtrix, or mtrix of size 3 4(or we sy tht mtrix hving n order 3 4). Note it tht 11 = 4 (11 mens 1 st row nd 1 st olumn), 12 = -2 (12 mens 1 st row nd 2 nd olumn), 13 = 0, 14 = 6 21 = 2, 22 = -3, 23 = 1, 24 =9 et. SQUARE MATRIX: A mtrix for whih the numer of rows nd olumns re equl is lled squre mtrix. A squre mtrix A with m rows nd n olumns (size m n) ut m=n (i.e of order n n) hs the form: i 1 n i 2n A = Digonl entries i1 i2 ii in Note: n1 n2 ni nn The min digonl of A onsists of ll the entries 11, 22, 33,, ii,, n n TRANSPOSE OF A MATRIX: The trnspose of mtrix A of size m n, is the mtrix denoted y A t of size n m, otined y writing the rows of A, in order, s olumns.(or we n sy tht trnspose of mtrix mens write the rows insted of olums or write the olumns insted of rows. Thus if Virtul University of Pkistn Pge 267

67 Then n m n t m2 A, then A = = m1 m2 mn 1n 2n mn A = t A = SYMMETRIC MATRIX: A squre mtrix A = [ ij ] of size n n is lled symmetri if, nd only if, i.e., for ll i, j = 1, 2,, n, ij = ji Let A=, nd B = A t = A t t Then A = 3 2, nd B = Note tht B t = B, so tht B is symmetri mtrix. MATRIX MULTIPLICATION: Suppose A nd B re two mtries suh tht the numer of olumns of A is equl to the numer of rows of B, sy A is n m p mtrix nd B is p n mtrix. Then the produt of A nd B, written AB, is the m n mtrix whose ijth entry is otined y multiplying the elements of the ith row of A y the orresponding elements of the jth olumn of B nd then dding; p 11 1 j 1 n 11 1 j 1 n 21 2 j 2n i1 i2 ip = i 1 ij in p1 pj pn mi m2 mp m1 mj mn where p ij = i 11j + i 22 j + + ippj = ikkj k = 1 REMARK: If the numer of olumns of A is not equl to the numer of rows of B, then the Virtul University of Pkistn Pge 268

68 produt AB is not defined. Find the produt AB nd BA of the mtries A= nd B= Size of A is 2 2 nd of B is 2 3, the produt AB is defined s 2 3 mtrix. But BA is not defined, euse no. of olumns of B = 3 2 = no. of rows of A AB = (1)(2) + (3)(3) (1)(0) + (3)(6) (1)( 4) + (3)(6) = (2)(2) ( 1)(3) (2)(0) ( 1)( 2) (2)( 4) ( 2)(6) = Find AA t nd A t A, where A = A t is otined from A y rewriting the rows of A s olumns: 1 3 i.e t A = Now 1 3 t AA = 2-1 = = nd 1 3 t AA= = = ADJACENCY MATRIX OF A GRAPH: Let G e grph with ordered verties v 1, v 2,..., v n.the djeny mtrix of G is the mtrix A = [ ij ] over the set of non-negtive integers suh tht Virtul University of Pkistn Pge 269

69 ij = the numer of edges onneting v i nd v j for ll i, j = 1, 2,, n. OR The djny mtrix sy A= [ ij ] is lso defined s 1 if { vi, v j} is nedgeof G ij = 0 otherwise A grph with it s djeny mtrix is shown. v 1 e 1 v 4 e 3 e 2 e 4 v 2 v 3 e 5 v 1 v1 0 v 2 = 0 A v 3 1 v 4 1 Note tht the nonzero entries long the min digonl of A indite the presene of loops nd entries lrger thn 1 orrespond to prllel edges. Also note A is symmetri mtrix. Find grph tht hve the following djeny mtrix Let the three verties of the grph e nmed v 1, v 2 nd v 3. We lel the djeny mtrix ross the top nd down the left side with these verties nd drw the grph ordingly(s from v 1 to v 2 there is vlue 2,it mens tht two prllel edges etween v 1 nd v 2 nd sme ondition ours etween v 2 nd v 1 nd the vlue 1 represent the loops of v 2 nd v 3 ). v v v v v v v v v v 2 v 3 v 1 DIRECTED GRAPH: A direted grph or digrph, onsists of two finite sets: set V(G) of verties nd set D(G) of direted edges, where eh edge is ssoited with n ordered pir of verties lled its end points. If edge e is ssoited with the pir (v, w) of verties, then e is sid to e the direted edge from v to w nd is represented y drwing n rrow from v to w. EXAMPLE OF A DIGRAPH: Virtul University of Pkistn Pge 270

70 v 2 e 1 v e 3 1 e e 2 5 v 4 e 6 e 4 v 3 ADJACENCY MATRIX OF A DIRECTED GRAPH: Let G e grph with ordered verties v 1, v 2,, v n. The djeny mtrix of G is the mtrix A = [ ij ] over the set of non-negtive integers suh tht ij = the numer of rrows from v i to v j for ll i, j = 1, 2,, n. A direted grph with its djeny mtrix is shown e 1 e v 3 1 e e 2 5 v 4 e 6 v 2 e 4 v 3 v 1 v 1 1 v 2 = 0 A v 3 1 v 4 0 Find direted grph tht hs the djeny mtrix v v v is the djeny mtrix The 4 4 djeny mtrix shows tht the grph hs 4 verties sy v 1, v 2, v 3 nd v 4 leled ross the top nd down the left side of the mtrix. A orresponding direted grph is v v v v v v A = v v v 1 v 2 v 4 v 3 It mens tht loop exists from v 1 nd v 3, two rrows go from v 1 to v 4 nd two from v 3 nd v 2 nd one rrow go from v 1 to v 3, v 2 to v 3, v 3 to v 4, v 4 to v 2 nd v 3. THEOREM If G is grph with verties v 1, v 2,,v m nd A is the djeny mtrix of G, then for eh positive integer n, the ijth entry of A n = the numer of wlks of length n from v i to v j for ll integers i,j = 1, 2,, n PROBLEM: Let A = Virtul University of Pkistn Pge 271

71 e the djeny mtrix of grph G with verties v 1, v 2, nd v 3. Find () the numer of wlks of length 2 from v 2 to v 3 () the numer of wlks of length 3 from v 1 to v 3 Drw grph G nd find the wlks y visul inspetion for () () it shows the entry (2,3) from v 2 to v 3 A = AA= = Hene, numer of wlks of length 2(mens multiply mtrix A two times ) from v 2 to v 3 = the entry t (2,3) of A 2 = 2 () A = AA = = it shows the entry (1,3) from v 1 to v 3 Hene, numer of wlks of length 3 from v 1 to v 3 = the entry t (1,3) of A 3 = 15 Wlks from v 2 to v 3 y visul inspetion of grph is e e 2 = v A v e e e 5 4 so in prt ()two Wlks of length 2 from v 2 to v 3 re (i) v 2 e 2 v 1 e 3 v 3 (y using the ove theorem). (ii) v 2 e 2 v 1 e 4 v 3 INCIDENCE MATRIX OF A SIMPLE GRAPH: Let G e grph with verties v 1, v 2,, v n nd edges e 1, e 2,, e n. The inidene mtrix of G is the mtrix M = [m ij ] of size n m defined y v 3 m ij 1 if the vertex vi is inident on the edge e = 0 otherwise j A grph with its inidene mtrix is shown. v 1 e 1 v 4 e 2 e 3 e 4 v 2 e 5 v 3 e e e e e v v M = v v REMARK: In the inidene mtrix 1. Multiple edges re represented y olumns with identil entries (in this mtrix e 4 & e 5 re multiple edges). Virtul University of Pkistn Pge 272

72 2. Loops re represented using olumn with extly one entry equl to 1, orresponding to the vertex tht is inident with this loop nd other zeros (here e 2 is only loop). Virtul University of Pkistn Pge 273

73 LECTURE 42 ISOMORPHISM OF GRAPHS Here we hve grph v 1 v 5 e 1 e 4 e 5 e v 2 2 Whih n lso e defined s v 4 e 3 v 3 v 1 e 3 v 3 e 1 e 4 v 4 Its verties nd edges n e written s: V(G) = {v 1, v 2, v 3, v 4, v 5 }, E(G) = {e 1, e 2, e 3, e 4, e 5 } Edge endpoint funtion is: e 5 v 5 e 2 v 2 Edge Endpoints E 1 {v 1,v 2 } E 2 {v 2,v 3 } E 3 {v 3,v 4 } E 4 {v 4,v 5 } E 5 {v 5,v 1 } G v 5 e 5 v 1 e 1 v 2 e2 Another grph G is G e 5 v 1 v 5 e 1 e 2 v 2 e 4 v 4 e 3 v 3 e 4 v 4 e 3 v 3 Edge endpoint funtion of G is: Edge Endpoints e 1 {v,v } 1 2 e 2 {v,v } 2 3 e 3 {v,v } 3 4 e 4 {v,v } 4 5 e 5 {v,v } 5 1 Edge endpoint funtion of G is: Edge Endpoints e 1 {v,v } 1 3 e 2 {v,v } 2 4 e 3 {v,v } 3 5 e 4 {v,v } 1 4 e 5 {v,v } 2 5 Two grphs (G nd G ) tht re the sme exept for the leling of their verties re not onsidered different. GRAPHS OF EDGE POINT FUNCTIONS Edge point funtion of G is: Edge point funtion of G is: Virtul University of Pkistn Pge 274

74 Edge Endpoints e 1 {v 1,v 2 } e 2 {v 2,v 3 } e 3 {v 3,v 4 } e 4 {v 4,v 5 } e 5 {v 5,v 1 } EdgeEndpoints e 1 {v 1,v 3 } e 2 {v 2,v 4 } e 3 {v 3,v 5 } e 4 {v 1,v 4 } e 5 {v 2,v 5 } Note it tht the grphs G nd G re looking different euse in G the end points of e 1 re v 1,v 2 ut in G re v 1, v 3 et. Buts G is very similr to G,if the verties nd edges of G re releled y the funtion shown elow, then G eomes sme s G: Verties of G Verties of G edges of G edges of G v 1 v 1 v2 e 1 e 1 v 2 e 2 e 2 v 3 v 3 e 3 e4 e 3 v 4 v 4 e 4 v 5 v 5 e 5 e 5 It shows tht if there is one-one orrespondene etween the verties of G nd G, then lso one-one orrespondene etween the edges of G nd G. ISOMORPHIC GRAPHS: Let G nd G e grphs with vertex sets V(G) nd V(G ) nd edge sets E(G) nd E(G ), respetively. G is isomorphi to G if, nd only if, there exist one-to-one orrespondenes g: V(G) V(G ) nd h: E(G) E(G ) tht preserve the edge-endpoint funtions of G nd G in the sense tht for ll v V(G) nd e E(G). v is n endpoint of e g(v) is n endpoint of h(e). EQUIVALENCE RELATION: Grph isomorphism is n equivlene reltion on the set of grphs. 1. Grphs isomorphism is Reflexive (It mens tht the grph should e isomorphi to itself). 2. Grphs isomorphism is Symmetri (It mens tht if G is isomorphi to G then G is lso isomorphi to G). 3. Grphs isomorphism is Trnsitive (It mens tht if G is isomorphi to G nd G is isomorphi to G, then G is isomorphi to G ). ISOMORPHIC INVARIANT: A property P is lled n isomorphi invrint if, nd only if, given ny grphs G nd G, if G hs property P nd G is isomorphi to G, then G hs property P. THEOREM OF ISOMORPHIC INVARIANT: Eh of the following properties is n invrint for grph isomorphism, where n, m nd k re ll non-negtive integers, if the grph: 1. hs n verties. 2. hs m edges. 3. hs vertex of degree k. 4. hs m verties of degree k. 5. hs iruit of length k. 6. hs simple iruit of length k. 7. hs m simple iruits of length k. 8. is onneted. 9. hs n Euler iruit. Virtul University of Pkistn Pge 275

75 10. hs Hmiltonin iruit. DEGREE SEQUENCE: The degree sequene of grph is the list of the degrees of its verties in non-inresing order. Find the degree sequene of the following grph. d Degree of = 2, Degree of = 3, Degree of = 1, Degree of d = 2, Degree of e = 0 By definition, degree of the verties of given grph should e in deresing (noninresing) order. Therefore Degree sequene is: 3, 2, 2, 1, 0 GRAPH ISOMORPHISM FOR SIMPLE GRAPHS: If G nd G re simple grphs (mens the grphs whih hve no loops or prllel edges ) then G is isomorphi to G if, nd only if, there exists one-to-one orrespondene (1-1 nd onto funtion) g from the vertex set V (G) of G to the vertex set V (G ) of G tht preserves the edge-endpoint funtions of G nd G in the sense tht for ll verties u nd v of G, {u, v} is n edge in G {g(u), g(v)} is n edge in G. OR You n sy tht with the property of one-one orrespondene, u nd v re djent in grph G if g (u) nd g (v) re djent in G. Note: It should e noted tht unfortuntely, there is no effiient method for heking tht whether two grphs re isomorphi(methods re there ut tke so muh time in lultions).despite tht there is simple ondition. Two grphs re isomorphi if they hve the sme numer of verties(s there is 1-1 orrespondene etween the verties of oth the grphs) nd the sme numer of edges(lso verties should hve the sme degree. Determine whether the grph G nd G given elow re isomorphi. e G G m n q o e d p As oth the grphs hve the sme numer of verties. But the grph G hs 7 edges nd the grph G hs only 6 edges. Therefore the two grphs re not isomorphi. Note: As the edges of oth the grphs G nd G re not sme then how the one-one orrespondene is possile,tht the reson the grphs G nd G re not isomorphi. Virtul University of Pkistn Pge 276

76 Determine whether the grph G nd G given elow re isomorphi. G G m n o e d q p Both the grphs hve 5 verties nd 7 edges. The vertex q of G hs degree 5. However G does not hve ny vertex of degree 5 (so one-one orrespondene is not possile). Hene, the two grphs re not isomorphi. Determine whether the grph G nd G given elow re isomorphi. G G u v w f d z x e y Clerly the verties of oth the grphs G nd G hve the sme degree (i.e 2 ) nd hving the sme numer of verties nd edges ut isomorphism is not possile. As the grph G is onneted grph ut the grph G is not onneted due to hve two omponents (e nd df). Therefore the two grphs re non isomorphi. Determine whether the grph G nd G given elow re isomorphi. G G v u w f d z x e y Clerly G hs six verties,g lso hs six verties. And the grph G hs two simple iruits of length 3; one is nd the other is defd.but G does not hve ny simple iruit of length 3(s one simple iruit in G is uxwv of length 4).Therefore the two grphs re non-isomorphi. Note: A simple iruit is iruit tht does not hve ny other repeted vertex exept the first nd lst. Virtul University of Pkistn Pge 277

77 Determine whether the grph G nd G given elow re isomorphi. G G s e f w x t h g z y d v u Both the grph G nd G hve 8 verties nd 12 edges nd oth re lso lled regulr grph(s eh vertex hs degree 3).The grph G hs two simple iruits of length 5; fe(i.e strts nd ends t ) nd dhgf(i.e strts nd ends t ). But G does not hve ny simple iruit of length 5 (it hs simple iruit tyxut,vwxuv of length 4 et). Therefore the two grphs re non-isomorphi. Determine whether the grph G nd G given elow re isomorphi. G G v w f u x e d z y We note tht ll the isomorphism invrints seems to e true. We shll prove tht the grphs G nd G re isomorphi. Here G hs four verties of degree 2 nd two verties of degree 3. Similr se in G. Also G nd G hve iruits of length 4.As is djent to nd f in grph G.In grph G u is djent to v nd z. And s nd u hs degree 2 so oth re mpped. And mpped with v, f mpped with z(s oth hve the sme degree lso is djent to f nd u is to z), nd s we moves further we get the 1-1 orrespondene. Define funtion f: V(G) V(G ) s follows. d e f u v w x y z Clerly the ove funtion is one nd onto tht is ijetive mpping. Note tht I write the ove mpping y keeping in mind the invrints of isomorphism s well s the ft tht the mpping should preserve edge end point funtion. Also you should note tht the mpping is not unique. f is lerly ijetive funtion. The ft tht f preserves the edge endpoint funtions of G nd G is shown elow. Virtul University of Pkistn Pge 278

78 Edges of G Edges of G {, } {, } {, d} {d, e} {e, f} {, f} {, f} {u, v} = {g(), g()} {v, y} = {g(), g()} {y, x} = {g(), g(d)} {x, w} = {g(d), g(e)} {w, z} = {g(e), g(f)} {u, z} = {g(), g(f)} {y, z} = {g(), g(f)} ALTERNATIVE We shll prove tht the grphs G nd G re isomorphi. Define funtion f: V(G) V(G ) s follows. G G v w f u x e d y d e f Determine whether the grph G nd G given elow re isomorphi. u v w x y z G d e h f g G s v w z x y t u We shll prove tht the grphs G nd G re isomorphi. Clerly the isomorphism invrints seems to e true etween G nd G. Virtul University of Pkistn Pge 279

79 Define funtion f: V(G) V(G ) s follows. d e f g h s t u v w x y z f is lerly ijetive funtion(s it stisfies onditions the one-one nd onto funtion lerly). The ft tht f preserves the edge endpoint funtions of G nd G is shown elow. Edges of G Edges of G {, } {, } {, d} {,d} {, f} {, g} {, h} {d, e} {e, f} {f, g} {g, h} {h, e} {s, t} = {f(), f()} {t, u} = {f(), f()} {u, v} = {f(), f(d)} {s, v} = {f(), f(d)} {s, z} = {f(), f(f)} {t, y} = {f(), f(g)} {u, x} = {f(), f(h)} {v, w} = {f(d), f(e)} {w, z} = {f(e), f(f)} {z, y} = {f(f), f(g)} {y, x} = {f(g), f(h)} {x, w} = {f(h), f(e)} Find ll non isomorphi simple grphs with three verties. There re four simple grphs with three verties s given elow(whih re non-isomorphi simple grphs). Virtul University of Pkistn Pge 280

80 Find ll non isomorphi simple onneted grphs with three verties. There re two simple onneted grphs with three verties s given elow(whih re non-isomorphi onneted simple grphs). Find ll non isomorphi simple onneted grphs with four verties. There re six simple onneted grphs with four verties s given elow. Virtul University of Pkistn Pge 281

81 LECTURE 43 PLANAR GRAPHS GRAPH COLORING In this leture, we will study tht whether ny grph n e drwn in the plne (mens flt surfe ) without rossing ny edges. It is grph on 4 verties nd written s K 4. Eh vertex is onneted to every other vertex. Note it tht here edges re rossed. Also the ove grph n lso e drwn s In this grph, note it tht eh vertex is onneted to every other vertex, ut no edge is rossed. Note: The grphs shown ove re omplete grphs with four verties (denoted y K 4 ). DEFINITION: A grph is lled plnr if it n e drwn in the plne without ny edge rossed (rossing mens the intersetion of lines). Suh drwing is lled plne drwing of the grph. OR You n sy tht grph is lled plnr in whih the grph rossing numer is 0. EXAMPLES: The grphs given ove re plnr.in the first figure edges re rossed ut it n e redrwn in seond figure where edges re not rossed, so lled plnr. It is lso grph on 4 verties (written s K 4 ) with no edge rossed, hene lled plnr. Note: The grphs given ove re lso omplete grphs (exept seond; re those where eh vertex is onneted to every other vertex) on 4 verties nd is written s K 4. Note: Complete grphs re plnr only for n 4. Virtul University of Pkistn Pge 282