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Discrete Mathematics with Graph Theory (3rd Edition) 375

Discrete Mathematics with Graph Theory (3rd Edition) 375 -...

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Section 13.3 373 8. [BB] False. The graph in Exercise 6 isn't planar since the subgraph with vertices {N 1 , N g , N 4 , N 5 , N 6 , N 7 } and with edge N g N 4 removed is isomorphic to K,g,g. 9. (a) [BB] Imagine the possible shorts between nets as the edges in a graph: They won't cross since they're all parallel. Now compress each net to a point, and we have a plane graph. By the Four Color Theorem, X(Q) :s: 4. (b) [BB] Draw separate graphs g(V, £1), g(V, £2) for horizontal lines of sight [as in (a)] and vertical lines of sight. These graphs have the same vertex set, so combine them. If an edge is repeated (that is, both horizontal and vertical lines exist between nets), then omit one of the original occurrences of it. (c) Color each of g(V, £1) and g(V, £2) in (b) with four colors, from S = {B, G, R, W}. The 16 different ordered pairs in
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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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