h5a - MATH 4171 Graph Theory SPRING 2006 Homework Set V –...

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Unformatted text preview: MATH 4171 Graph Theory SPRING 2006 Homework Set V – Solutions 1. Theorem 4.8 states that a necessary condition for a graph G = ( V, E ) to be Hamiltonian is that G- S has at most | S | components, for all S ⊆ V . Is this condition sufficient? In other words, must G = ( V, E ) be Hamiltonian if G- S has at most | S | components, for all S ⊆ V ? You need to justify your conclusion. Conclusion. No, the condition is not sufficient. Justification. Let G be the graph on the right. This graph does not have a Hamiltonian cycle since such a cycle would have to include both edges incident with u , both edges incident with v , and both edges incident with w , which is impossible. However, this graph satisfies the required condition: the graph is 2-connected, so deleting a single vertex always results in a connected graph; deleting { a, x } , { a, y } , a x z y u v w or { a, z } results in two components while deleting any other two vertices results in a connected graph; deleting any three vertices leaves a graph with four vertices and at least one edge, which must have at...
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This note was uploaded on 02/17/2009 for the course MATH 4171 taught by Professor Lax,r during the Spring '08 term at LSU.

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