Discrete Mathematics with Graph Theory (3rd Edition) 272

Discrete Mathematics with Graph Theory (3rd Edition) 272 -...

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270 Solutions to Exercises (g) (h) H~1 Be ! 'J G \ D F E 4. Let Ji be a Hamiltonian cycle. Vertex F has degree 2, so Ji contains both edges incident with F. Since Ji can contain no proper cycles, edge DE is not in Ji and therefore G E is in. The present situation is depicted in the graph on the left below. A B c c c Exactly one of the edges AD, B D is in Ji, so we consider two cases. Case 1: AD is in; BD is out. In this case, since two edges incident with B must be in Ji, both AB and BG are in. So Ji is ABGEF DA as shown in the middle above. Case 2: BD is in; AD is out. In this case, since two edges incident with A must be in Ji,
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Unformatted text preview: both AB and AG are in. So Ji is ABDF EGA as shown on the right above. 5. (a) [BB] Yes, it is Hamiltonian: ABDGEA is a Hamiltonian cycle. (b) Any Hamiltonian graph has a Hamiltonian path; in the graph here, ABDGE is a Hamiltonian path, for instance. (c) [BB] The graph is not Eulerian. Vertices A and E have odd degree. (d) The trail indicated is Eulerian. E B(Jf------"--~C D 6. (a) Yes, it is possible. Here is a Hamiltonian cycle for the knight. 5 64 25 28 41 44 53 46 24 27 4 63 52 47 40 43 CD 6 17 26 29 42 45 54 18 23 62 3 48 51 34 39 7 2 19 16 35 30 55 50 22 13 10 61 58 49 38 33 11 8 15 20 31 36 59 56 14 21 12 9 60 57 32 37...
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