Unformatted text preview: S x S give a coloring for g. (d) If K,11 were a lineofsight graph, it could be decomposed as in (b). Since K,11 has 55 edges, either £1 :?: 28 or £2 :?: 28. But Theorem 13.1.4 says that £i :s: 3(11)  6 by planarity, which is a contradiction. 10. [BB] It is possible for a short to exist between any pair of nets here. Hence, g ~ K,8, and X(Q) = 8, as required. 11. K,5: [BB] 1 5~2 b a 4 3 1i cannot be 2colored since it is a cycle of odd length. If we delete any vertex, however, the resulting graph can be 2colored. If we delete e, for example, we obtain the graph at the right. Shown are g' and a corresponding floor plan. 5m2 ~ e~: d a Since 1i is K,g, X(1i) = 3. But if we delete any vertex from 1i, the resulting graph can be 2colored. Say we delete c, obtaining the graph: a ~ b. Here are g' and one possible floor plan....
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 Summer '10
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 Graph Theory, Planar graph, Four color theorem, Draw separate graphs

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