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ABC of Graph Theory Answers to Chapter 1: ABC of Graph Theory 1.1 Graphs: Basic Notions 1.1.3. c: The empty graph. 1.1.4. Kn. 1.1.5. C;. 1.1.6. a: No. 1.1.7. No. 1.1.8. a: 9; b: 7; c: There are no such graphs. 1.1.9. Hint: See Exercise 1.1.7 and proceed by contradiction. 1.1.10. No. 1.1.11. a: Hint: Count the number of edges in such graph. b: Yes. Hint: Consider an appropriate bipartite graph. c: Hint: See Exercise 1.1.11a. 1.1.12. a: Hint: Use the fact that if a, b are real numbers and a + b = n then ab:::; n2 j4. b: Kp,p for n = 2p. 1.1.13. Complete bipartite graphs. 1.1.14. Selfcomplementary graphs: 0 1 , P4, C5 • 1.1.16. a: Bipartite r-regular graphs of order n exist only for n = 2m, 0:::; r :::; m. One of them with parts A and B may be defined as follows. Let A = Zm = {O, 1, ... , m - I} be the additive group of residuals modulo m and let B be another copy of this group. If x E B then we set N (x) = {x + yEA: y = 0,1, ... , r - I}. Here x E B denotes the copy of x E A. 183 184 Answers, Hints, Solutions b: There are no such graphs. c: Hint: If 0 is the required graph then IEOI = n - 1. 1.1.17. For even n the graph is unique, for odd n there are no such graphs. 1.1.18.. a: 2c ;; b: C ifm:S; C~, and 0 otherwise. C2n 1.1.20. Hint: Two non isomorphic graphs satisfying conditions a)-c) are shown at Fig. A1.1.1. Figure A1.1.1: To Answer 1.1.20 1.1.21. Hint: If f : VO -+ V H is an isomorphism of graphs 0 and H then for any edge e = uv E EO the set f(e) = {f(u), f(v)} is an edge of H. Prove that the mapping e ...... f( e) specifies an isomorphism of line graphs L( 0) and L(H). 1.1.22. b: 0 1 ~ O2 ~ 0 3 , and there are no other pairs of isomorphic graphs at Fig. 1.1.8. 1.1.24. Hint: The numbers of pairwise nonisomorphic graphs of order n are: 4 for n = 3; 11 for n = 4; 34 for n = 5. 1.1.25. a: The complete graph J{4 is the unique cubic graph of order 4. b: The number of pairwise nonisomorphic cubic graphs of order 6 is 2, see Fig. A1.1.2. Hint: Consider the complementary graphs. Figure A1.1.2: Cubic graphs of order 6 c: The number of pairwise non isomorphic cubic graphs of order 8 is 6, see Fig. A1.1.3. Hint: Let G be the required graph. First prove that G contains either the triangle C3 , i.e., three pairwise adjacent vertices, or the cycle C4 , i.e., four vertices a, b, e, d such that {ab, be, cd, da} C EG. Further, consider the following five possibilities: 1.1: Graphs: Basic Notions 185 Figure A1.1.3: Cubic graphs of order 8 (1) G contains K 4 , i.e., four pairwise adjacent vertices; (2) G does not contain K4, but it contains K4 - e, i.e., four vertices a, b, e, d such that {ab, be, cd, da, ae} ~ EG and bd (j. EG; (3) G does not contain neither K4, nor K4 - e, but it contains two triangles; (4) G contains exactly one triangle; (5) G contains no triangles. Prove that G ~ Gi (Fig. A1.1.3) in the case (i), i = 1, ... ,6. In the case (5) G ~ G5 or G ~ G6 . G5 contains the cycle C5 and G6 does not. 1.1.26. b: The number of pairwise non isomorphic cubic circulant graphs of order 6 is 2. c: Yes. 1.1.27. a: Hint: See Fig. 1.2.4a. b: A regular graph of degree C;. The number of edges is C;~. c: Solution: For n = 1 the statement is trivial. Assume that n ;::: 2. Since the application of a single transposition changes the parity of the permutation, no two permutations of the same parity are adjacent in the n-permutation graph. Hence it is a bipartite graph, one part being all even permutations, the other one being all odd permutations. 1.1.28. Solution: Consider the set of all prime numbers in the ascending or­ der: 2, 3, 5 .... Let Pi be the i-th prime number. Let further G be an 186 Answers, Hints, Solutions (n, m)-graph. Let us label its vertices by numbers PI, .. Pn and let us label its edges by the numbers Pn+1, ... , Pn+m. Afterwards we multiply the label of each vertex by the labels of the edges incident to it. This gives a common divisor graph isomorphic to G. 1.1.30. a: The number of edges of G is C~ - m. The number of edges of a selfcom­ plementary graph of order n is n(n - 1)/4. b: Hint: See Exercise 1.1.30a. 1.1.31. a: P4; b: C5 and the graph shown at Fig. ALIA; c: there are no such graphs. A Figure ALIA: To Exercise 1.1.31 1.1.33. Hint: Use Exercise 1.1.32. 1.1.34. No. Hint: Such graph would contain a single vertex of degree 49. Let u be a vertex of degree 50 and let v be of degree 49. Is there an edge uv in the graph? 1.1.35. Solution: The number of vertices of the graph L(G) is m = IEGI = ~ E~=l di· Two edges of G are adjacent as vertices of L(G) if and only if G has a vertex incident to both of them. Therefore all edges of G incident to the same vertex are pairwise adjacent in L(G). Hence 1.1.37. a: I<3; b: The line graph of the tetrahedron is isomorphic to the octahe­ dron. 1.1.38. a: Yes; b: No. 1.1.40. Solution: Suppose that V = V I<5 = {I, 2, 3,4, 5}. Then VL(I<5) = VL(I<5) = {xy: x,y E V, x =1= y}. The vertices xy and uv are adjacent in L(I<5) ifand only if {x, y}n{u, v} = 0. Hence G = L(I<5) is a 3-regular graph. The induced subgraphs (see Sect. 1.3) G( {12, 34, 15,23, 45}) and G( {13, 24,35,14, 25}) are isomorphic to C5 • Further see Fig. A1.1.5. 1.2: Graphs: Basic Notions 187 12 23 15 Figure Al.1.5: To Exercise 1.1.40 1.2 Walks, Paths, Components 1.2.2. a: ~ I:~=3(k - 1)!C~ if n ~ 3, and 0 for n :S 2; b: ~ I:;~n2(p,q) k!(k - l)!C;C~ if p ~ 2, q ~ 2, and 0 otherwise. Hint: Let A, B be the parts of Kp,q, P :S q, The length of any simple cycle of Kp,q is an even number 2k, 2 :S k :S p. In such cycle the vertices from A and from B alternate. Finally, the number of all simple cycles with the same vertex set V, IVI = 2k, is (k - 1)!k!/2. 1.2.4. a: 3 for n ~ 3, 00 for n = 1,2; b: 4 for p, q ~ 2; 00 for p or q equal to 1; c: 5. 1.2.5. a: Hint: Delete the "excessive" pieces of the walk. b: False; c: False. 1.2.10. Hint: Proof is by contradiction using the Handshake Lemma. 1.2.11. Hint: Starting from any vertex v, move along an incident edge e into the adjacent vertex w. Similarly, move from w further along an edge other than e. Continue this process until some vertex will be visited twice. Select a cycle in the traversed walk. 1.2.12. a: Hint: Suppose that are given paths, UI = VI = U, Uk = Vm = V, U a and Va are the first (counting from u) different vertices of these paths, u{3 and v1' are the first coinciding vertices of these paths after U a and Va. Consider the union of the (Ua-I, u{3)-subpath of P and (Va-I, v1')-subpath of Q. 1.2.13. a: Hint: Use Konig's theorem. b: False. 1.2.14. See Fig. A1.2.1. The graphs G 1 = KI and G6 = P4 are selfcomplementary. 188 Answers, Hints, Solutions • • • II AIINtx1 Gs Figure A1.2.1: To Exercise 1.2.14 1.2.15. Km,n, m, n ;::: 1, m + n ;::: 3. Hint: Demonstrate that after the deletion of any edge e = uv from Km,n either d(u, v) = 3 in the resulting graph or it becomes disconnected. 1.2.16. Hint: Use the previous exercise. 1.2.17. Hint: The number of such graphs is 5. 1.2.18. a: the star K1,n, n ;::: 2; b: Kn; c: K2 ; d: the star K1,n; e: the star K1,n or the triangle C3 ; f: the star K1,n, n ;::: 2; g: the path graph Pn , n ;::: 3; h: the cycle graph Cn. Hint: Using the Handshake Lemma show that G has no vertices of degrees greater than 2. Then notice that WGI = IEGI. 1.2.20. Hint: In any walk that contains the deleted edge one may replace this edge by the remaining edges of the cycle.
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