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Unformatted text preview: there is no 3coloring, start with three colors on a triangle and you will see that eventually a fourth color is required. (f) A 5coloring is shown. Since 9 contains K5 as a sub graph, X(Q) = 5. 3 3 B~Al 2HG C F 2 2 D 3 4 E 1 1 1 * 2 2 2 1 3 Ck::. ..:e4i~_~2 2 1 4 1 2 3 4 ~ 1 3 2 2 1 5~3 1 4 367 6. (a) [BB] The graph 91 on the left has a 3coloring, as shown. Since it contains a triangle, X(91) = 3. The graph 92 on the right has a 3coloring, as shown. By trying to label alternately the vertices of the outer pentagon, we see that two colors will not suffice, so X(92) = 3 too....
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 Summer '10
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 Graph Theory, Angles

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