Unformatted text preview: S x S give a coloring for g. (d) If K,11 were a line-of-sight 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 2-colored since it is a cycle of odd length. If we delete any vertex, however, the resulting graph can be 2-colored. 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 2-colored. Say we delete c, obtaining the graph: a ~ b. Here are g' and one possible floor plan....
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- Summer '10
- Graph Theory, Planar graph, Four color theorem, Draw separate graphs