Discrete Mathematics with Graph Theory (3rd Edition) 289

# Discrete Mathematics with Graph Theory (3rd Edition) 289 -...

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Chapter 10 287 7. (a) This graph is Hamiltonian: AFGBCDHEA is a Hamiltonian cycle. (b) This graph is not Hamiltonian. The comer vertices, A, G, E, and C each have degree 2, so any Hamiltonian cycle would have to include the eight edges adjacent with these vertices. Thus any Hamiltonian cycle would have to include the proper cycle ABC D EFG H A, which cannot be. 8. The graph is not Hamiltonian as we explain. Since 1{ were a Hamiltonian circuit. Since vertices B, E, and H have degree 2, 1{ would have to contain edges AB, BC, GH, HI, FE, and ED. Since 1{ cannot contain a proper cycle, not both edges AG and I C could be in 1{. By symmetry, we may assume that IC is not in H. ABC B F E D Then, since 1{ contains precisely two edges incident with C, CD is in 1{. Since DC and DE are now in 1{, DI cannot be in H. But now only one edge incident with I is in 1{, a contradiction. 9. EACBDJGI H F E is a Hamiltonian cycle. 10. (a) Any Hamiltonian cycle must contain precisely two edges incident with
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