Discrete Mathematics with Graph Theory (3rd Edition) 270

# Discrete Mathematics with Graph Theory (3rd Edition) 270 -...

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268 Exercises 10.2 1. [BB] This graph is not Hamiltonian. To see this, suppose ?t were a Hamiltonian cycle. Since vertices A and B have degree 2, the two edges incident with each of these vertices would be in ?t. Thus, ?t would contain the cycle ACBDA, which cannot be the case since this does not contain all vertices of the graph. The graph is not Eulerian because it contains vertices of odd degree. 2. (a) This is not Hamiltonian. Since vertices A and C each have degree 2, the proper cycle ABCDA would have to be included in any Hamiltonian cycle, and this is not possible. (b) [BB] This is not Hamiltonian, since it isn't connected. (c) This is not Hamiltonian. Since the vertices labeled A and B are of Solutions to Exercises
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Unformatted text preview: A B ~ degree two, both edges incident with them would be on a Hamiltonian {:} cycle, but this forms a smaller cycle not containing all vertices, that is, a A B proper cycle. (d) [BB] ~ 1 2 7 3 6 4 5 (e) This is not Hamiltonian. Since the vertices labeled A, B, C and D are of degree two, both edges incident with each of them would be on a Hamiltonian cycle, but these eight edges would then form a proper cycle, as shown. A B m D C (t) This is a pseudograph, but not a graph, so the question does not apply. The definitions of Sec-tion 10.2 were given only for graphs. 3. (a) (b) 18 1 9 4 2 3 5 8 6 7 (a) The graph in (a) is Hamiltonian; a Hamiltonian cycle is indicated. Any Hamiltonian graph has a Hamiltonian path....
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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