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Discrete Mathematics with Graph Theory (3rd Edition) 362

# Discrete Mathematics with Graph Theory (3rd Edition) 362 -...

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360 Solutions to Exercises 24. (---t) Let 9 be a tree. Then 9 is connected, by definition. Suppose e is an edge of g. We want to show that e is a bridge, that is, that 9 '- {e} is not connected. Suppose to the contrary that 9 '- {e} is connected. Then there is a path P in 9 '- { e } from one end vertex of e to the other. This path, followed bye, is a circuit in g, contradicting the fact that 9 is a tree. ( f-) Suppose 9 is a connected graph and every edge of 9 is a bridge. To show 9 is a tree, we have only to show that 9 contains no cycles. Let C be a cycle and let e be an edge of C. The graph 9 '- {e} is connected because one can pass from one end vertex of e to the other by following the edges of C other than e. So e is not a bridge, a contradiction. 25. No, it is not possible. Let uv be an edge of the graph. Since vertices u and v are part of a Hamiltonian cycle, there are two paths between u and v (either way around the cycle), so deleting
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Unformatted text preview: uv cannot disconnect the graph. 26. (a) No. For example, ~ is not strongly connected. (b) Yes. The underlying graph for a tournament is a complete graph. Since lCn is connected and has no bridges (there are always at least two paths between vertices u and v), it has a strongly connected orientation by Theorem 12.6.3. Exercises 13.1 1. (a) [BB] We draw the graph quickly as a planar graph and then, after some thinking, as a plane with straight edges. k-----(b) [BB] There are seven regions, numbered 1,2, ... ,7, with boundaries aJg, ghe, hbi, icd, bjc, Jedk, and ajk, respectively. (c) [BB] E = 11, V = 6, R = 7, N = 22; so V -E + R = 6 -11 + 7 = 2; N = 22 :::; 22 = 2E and E = 11 :::; 12 = 3V - 6. 2. (a) (b) There are five regions, numbered 1,2,3,4,5, with boundaries abgJ, bchg, cdih, deji and aej f. (c) E = 10, V = 7, R = 5, N = 20; so, V -E + R = 7 -10 + 5 = 2, N = 20 :::; 20 = 2E and E = 10 :::; 15 = 3V - 6....
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