Discrete Mathematics with Graph Theory (3rd Edition) 351

Discrete Mathematics with Graph Theory (3rd Edition) 351 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 12.5 (c) The final backtracking is 17, 13, 1. (d) The final backtracking is 15, 14, 12, 11, 10, 3, 2, 1. 14 349 15 16 15 1 7 12 13 3 9 8 3. We 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 f in the interval 1 ~ f < k; thus, there is a path from 1 to f 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 k which uses only edges required by the algorithm. Vertex k acquires its label because the algorithm finds this vertex unlabeled but adjacent to a previously labeled vertex f. Since at the time a vertex is labeled, the algorithm always picks the smallest unused label, we must have f < k. So, by the induction hypothesis, there is a path from 1 to f using edges required by the algorithm; also, the algorithm uses the edge fk to label k. Thus, there is a path from 1 to k along edges which the algorithm requires. 4. (a) [BB] Consider the status of the algorithm at the time that
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online