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Published byHarley Alder Modified over 4 years ago

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Coloring Warm-Up

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A graph is 2-colorable iff it has no odd length cycles 1: If G has an odd-length cycle then G is not 2- colorable Proof: Let v 0, …, v 2n+1 be an odd length cycle (n≥1). Suppose G can be colored with two colors red and blue. Without loss of generality [WLOG] color v 0 red. Then v 1 must be colored blue since v 0 and v 1 are adjacent. Then v 2 must be colored red, and in general, all even-indexed vertices must be colored red and all odd-indexed vertices must be colored blue. But v 0 =v 2n+1 must get both colors, contradiction.

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A graph is 2-colorable iff it has no odd length cycles 2: If G has no odd-length cycles then G is 2- colorable. How to define the coloring? Helpful Def. The distance of one vertex from another is the length of the shortest path between them. Proof: Pick any connected component. Pick an arbitrary vertex v. Color all vertices at even distance from v with one color and all vertices at odd distance from v with the other color.

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It suffices to prove that for every n, there are no nodes in G that (a) are distance n apart but (b) also have a path between them of length m, where one of m, n is even and the other is odd. Suppose there were. By WOP there is a least such n, and such a pair of nodes v, w. Then the two paths have no vertices in common except the beginning and end. But the round trip out on one path and returning on the other path is a cycle of length m+n, which is odd, contradiction. QED

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