CS / Math 240 : Intro to Discrete Math 4 / 14 / 2011 Solutions to Homework 8
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چکیده
We first show that a graph is not bipartite if it contains any cycles of odd length. Starting at an arbitrary vertex u in this cycle, we note that there is a path of length 2k + 1 from u to v. We also note that to be bipartite, we must alternate placing vertices on this path into the partitions L and R. However, if we place u in L, then, we see that after following the path along 2k + 1 edges, we must place u in R, a contradiction. A symmetric argument can be shown for placing u in R initially. Now, we must also show that if a graph contains no cycles of odd length, it will be bipartite. We do so through an inductive proof on the number of vertices in the graph.
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