Discrete Mathematics with Graph Theory (3rd Edition) 353

Discrete Mathematics with Graph Theory (3rd Edition) 353 -...

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Section 12.5 351 7 8 9 11. (a) [BB] ti (b) 7 11 5 3 1 2 (c) (d) 14 13 ~: 2 5 9 6W'-'<] 3 4 10 4 3 ~ 12. (a) The argument that we used to answer Exercise 3 works here as well. Use the strong form of mathematical induction on k, a vertex label, the result being clearly true for k = 1. Suppose k > 1 and the result is true for all i in the interval 1 ~ i < k; thus, there is a path from vertex 1 to the vertex labeled i which uses only edges required by the algorithm. We must prove the result for k: We must prove that there is a path from 1 to the vertex with label k which uses only edges required by the algorithm. Vertex k was so labeled because the algorithm found this vertex unlabeled but adjacent to a previously labeled vertex i. At the time a vertex is labeled, the algorithm chooses the smallest unused label, so i < k. By the induction hypothesis, there is a path from 1 to i using edges required by the algorithm; also, the algorithm uses the edge ik to label vertex k. Thus, there is a path from 1 to
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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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