Answers n Solutions to (Some Even Numere) Suggeste Exercises in Chpter 11 o Grimli s Discrete n Comintoril Mthemtics Section 11.1 11.1.4. κ(g) = 2. Let V e = {v : v hs even numer o 1 s} n V o = {v : v hs o numer o 1 s}. Then G e n G o, the sugrphs o G inuce y V e n V o, respectively, re the connecte components o G. To prove this, irst oserve tht two jcent vertices o G must either hve oth even or oth o numer o 1 s. Thereore there is no pth rom vertex in V e to vertex in V o. For economy o nottion we will enote t consecutive 0 s y 0 t n t consecutive 1 s y 1 t. Now we will show tht there is pth rom 0 n to ny vertex v in G e. Suppose v hs 2m 1 s where 2m n. Then we hve pth 0 n 1 2 0 n 2 1 4 0 n 4 1 2m 2 0 n 2m+2 1 2m 0 n 2m = v 0. Now v 0 n v hve the sme numer o 1 s n they ier t 2k, k 2m, positions. I k = 0, then v 0 = v n we re one. I not, then there re i j such tht v 0 hs 1 n v hs 0 t the ith position, n v 0 hs 0 n v hs 1 t the jth position. Let v 1 e the n-tuple otine rom v 0 y chnging the 1 in the ith position to 0, n the 0 in the jth position to 1. Exten the pth ove y v 0 v 1. Now v 1 hs the sme numer o 1 s s v, ut they ier t 2k 2 positions. Repeting this process k 1 more times connects the pth to v. This proves tht G e is connecte. The connecteness o G o is lso shown similrly. 11.1.6. The istnce rom to is 0; to e,, c is 1; to g, k is 2; to j, h, l, m is 3; to i is 4. 11.1.10. K 2. 11.1.14.. i. n ii. n + 1 11.1.16. c. n (nth Ctln numer). Section 11.2 11.2.6. There re 11 nonisomorphic grphs with V = 4, n 6 o these re connecte. 1
11.2.8.. The orere (m + 1)-tuple o istinct vertices long the pth etermines the pth. Since this oth irections count s the sme, the nswer is n(n 1) (n m)/2. 11.2.12.. Since Ē = E n E + Ē = ( ) n 2, E = 1 n 2(.. c. Since E = 1 2( n = n(n 1)/4 is n integer n only one o n n n 1 is even, either n or n 1 is ivisile y 4. 11.2.14.. Either o the exmples in 11.2.12. will o.. I x n y re in the sme component o G, then let z e point in nother component. Then the eges {x,z} n {y,z} re not in G, n x z y is pth in Ḡ. I x n y re not in the sme component o G, then {x,y} is not in G, so x y is pth in Ḡ. 11.2.16.. ( ) 6 3 2 ( 3. ( ) 6 4 2 ( 4 c. 6 k=1 ( 6 k. n k=1 ( n k ) 2 ( k ) 2 ( k Section 11.3 11.3.6. 11.3.10. There re 2 n vertices in Q n n ech hs egree n. So the mipoint o pth cn e chosen in 2 n wys n the enpoints in ( ) ( n 2. So the nswer is 2 n n = 2 n n(n 1). 11.3.12. Since the remining its cn e chosen ritrrily ech equivlence clss contins 2 n 2 vertices. The grph inuce y ny one o these equivlence clsses is isomorphic to Q n 2. Since the kth n lth positions cn e 0 n 0, or 0 n 1, or 1 n 0, or 1 n 1, respectively, there re 4 equivlence clsses. 11.3.16. Let v i, 1 i n, n w j, 1 j n, e the 2n vertices. Let the only eges e etween v i n w j i i + j n + 1. Then eg(v i ) = eg(w i ) = n + 1 i or 1 i n. 2
11.3.20.. h i e i j g j k g c e. h i e i j g j k g c e 11.3.26.. Let v 1 v 2 v m 1 v m v 0 e irecte Euler circuit in G. I = v i, = v j n i < j, then = v i v i+1 v j 1 v j = is irecte tril rom to. I = v i, = v j n i > j, then = v i v i+1 v m v 0 v 1 v j 1 v j = is irecte tril rom to. Since there is irecte tril rom to, there is lso irecte pth rom to. Hence G is strongly connecte. The converse is not true. The ollowing irecte grph is strongly connecte, ut oes not hve irecte Euler circuit. 11.3.30. Let the other couples e X 1 n X 2, Y 1 n Y 2, Z 1 n Z 2. One person ierent rom Crolyn shook hns with everyone except his/her spouse. This cnnot e Richr s there is lso person who shook hns with no one. Let us sy it is X 1. Then the person who shook hns with no one is X 2. Now we remove X 1 n X 2 rom the picture. This reuces everyoy s hnshkes y one. Following the sme resoning we euce tht, sy, Y 1 shook hn with everyoy in this group o six except his/her spouse, n Y 2 shook hns with no one in this group o six. Removing Y 1 n Y 2 lso, n repeting the sme resoning we see tht Z 1 shook hn with oth Crolyn n Richr, n Z 2 with neither o them or the other wy roun. So oth Crolyn n Richr shook hns three times. Section 11.4 11.4.6. The three vertices cn e chosen in ( n 3) wys. The other vertex cn e chosen in n 3 wys. So the nswer is (n 3) ( n 3). 11.4.8.. I V = V 1 V 2, V 1 = m n V 2 = n is the prtition o vertices giving K m,n n m < n, then the longest pth contins m + 1 istinct vertices in V 2 lternting with m vertices in V 1. So it hs length 2m. 11.4.10. I V = V 1 V 2 is the prtition o the vertices or the iprtite grph, then the vertices long cycle re lterntingly in V 1 n V 2. So every cycle hs n even numer o vertices on it. Thereore its length is even. 11.4.12.. K 1,5, K 2,4, K 3,3. 3
. K i,n i, 1 i n/2, re the list o nonisomorphic iprtite grphs with n vertices. The nswer is n/2. 11.4.14.. This grph is not plnr, ecuse it contins the ollowing sugrph homeomorphic to K 3,3. h j g e i c. c e c. This grph is not plnr, ecuse it contins the ollowing sugrph homeomorphic to K 3,3. h e c g. e c 4
e. u v w x y z. This grph is not plnr, ecuse it contins the ollowing sugrph homeomorphic to K 5. c i h 11.4.22.. Suppose tht oth G n its complement Ḡ re plnr. Then E 3 V 6 n Ē 3 V 6. Aing these inequlities n using V = V = n n E + Ē = ( n, we get ( n 6n 12, or in other wors, n 2 13n + 24 0. This inequlity oes not hol or n 11 > (11 + 73)/2.. v 8 v 7 v 1 v 6 v 3 v 4 v 5 v 6 v 8 v 4 v 5 v 2 v 1 v 2 v 3 v 7 11.4.28. I G is ul o G, then V = R n E = E. I G n G re isomorphic, then R = V = V = n. Hence, y Euler s Formul, 2 = V E + R = n E + n, n thereore E = 2n 2. 5
Section 11.5 11.5.2. I grph hs Hmiltonin cycle which is lso n Eulerin circuit, then then it contins every ege n every vertex exctly once. Thereore it is the whole grph. Tht is, the grph is C n. I grph hs Hmiltonin pth which is lso n Eulerin tril, then then gin it contins every ege n every vertex exctly once. Thereore it is the whole grph. So the grph is pth. 11.5.4.. With the vertices lele s in Figure 11.52 o the textook, i vertex is remove, then c e j h i g is Hmiltonin cycle in the remining grph. 11.5.6. n 11.5.10.. I V = V 1 V 2 is the prtition o the vertices or the iprtite grph, then the vertices long ny cycle re lterntingly in V 1 n V 2. Since ll vertices elong to Hmiltonin cycle, V 1 n V 2 must e equl.. Agin y the sme oservtion, i the grph hs Hmiltonin cycle n V 1 V 2, then V 1 = V 2 + 1 or V 2 = V 1 + 1. c. 11.5.12. We will prove tht Q n hs Hmiltonin cycle or n 2. For n = 2, 00 01 11 10 00 is Hmiltonin cycle or Q 2. Now or k 2 ssume tht Q k hs Hmiltonin cycle v 1 v 2 v n 1 v n v 1 where n = 2 k. Then 0v 1 0v 2 0v n 1 0v n 1v n 1v n 1 1v 2 1v 1 0v 1 is Hmiltonin cycle or Q k+1. 11.5.18. We hve vertex with n = 12 vertices n the egree o ech vertex is greter thn equl to n/2 = 6. Thereore this grph hs Hmiltonin cycle. This gives seting t roun tle in which ech person is cquinte with the persons sitting on either sie. 11.5.24.. Any Hmiltonin pth in Q 3 will o. For instnce, 000 001 011 010 110 111 101 100.. 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000. 6
11. Supplementry Exercises 11.Supp.8.. I v 1 v 2 =, then v 1 n v 2 re jcent. I v 1 v 2, then s n 5, {1, 2,...,n} hs two-elements set v which hs empty intersection with v 1 v 2, n v 1 v v 2 is pth connecting these two vertices. c. For ny n 5, G hs the ollowing sugrph homeomorphic to K 3,3. Thereore G is nonplnr y Kurtowski s Theorem. {1, 2} {1, 3} {2, 3} {2, 5} {2, 4} {1, 4} {1, 5} {3, 4} {3, 5} {4, 5}. Let v = {,} e vertex. Since {1, 2,...,n} \ {,} hs n 2 elements, there re ( ) n 2 2 two-elements sets which hve empty intersections with v. In other wors, eg(v) = ( ) n 2 2 or ll v in V. On the other hn V = ( ) ( n 2. Since eg(v)/2 = n 2 ) ( 2 /2 n ) 2 = V or n 8 > (9 + 33)/2, G hs Hmiltonin cycle i n 8. 7