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Discrete Mathematics with Graph Theory (3rd Edition) 271

Discrete Mathematics with Graph Theory (3rd Edition) 271 -...

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Section 10.2 269 (b) The graph in (b) is not Hamiltonian. Both edges incident with a vertex of degree two must be on any Hamiltonian cycle. Thus, the inner cycle would have to be part of any Hamiltonian cycle. Since this inner cycle excludes some vertices, it would be a proper cycle, a contradiction. The graph does have a Hamiltonian path, however. Start at any outer vertex, go around the outside stopping one short of the vertex where you started, then enter and go round the inner cycle. (c) The graph in (c) is not Hamiltonian. If H were a Hamiltonian cycle, then it would contain vertices A and J, these having degree 2, and hence edges AB, AG, JI and JK. Precisely two of the edges incident with K would be in H, hence, precisely one of DK, EK. Suppose DK were in and EK out. Then edges DE and EF would be part of1t. At this point, two edges incident at D are in H, so DC could not be in H. As the only other edges incident with C, both C B and C I would have to be H. D
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