Discrete Mathematics with Graph Theory (3rd Edition) 331

Discrete Mathematics with Graph Theory (3rd Edition) 331 -...

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Section 12.2 329 25. [BB] A tree is a complete bipartite graph if and only if it is K1,n for some n. Proof. Certainly K1,n is a tree. Conversely, if a tree is complete bipartite, then it is Km,n for some m and n. But such a graph has no vertices of degree one unless m or n is 1. The result follows. 26. (a) [BB] Remember that a component in a graph is a maximal connected subgraph. Also, any graph is the disjoint union of its components. If a graph has no cycles, neither does any component, so each component is a tree. (b) By Corollary 12.1.7, each of the c components has at least two vertices of degree 1. So there are at least 2c vertices of degree 1 all together. (c) The graph 0 0 is a forest consisting of two components and two (not four) vertices. (d) Let the number of vertices in the c components be nl, n2, . .. ,nc respectively. We are given that nl + n2 + . .. + nc = n. Since each component is a tree, the number of edges in the ith component
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Unformatted text preview: is ni -1. Thus, the total number of edges is (nl - 1) + (n2 - 1) + . .. + (nc -1) = n - c. 27. (a) [BB] Using Corollary 12.1.7, we have Ldeg(vi) ~ 8, so the tree has at least four edges and hence at least five vertices. If the result is not true, then there are two vertices of degree 3, at most three vertices of degree 1, and the rest have degree at least 2. Then Ldegvi ~ 2(3) + 3(1) + (n - 5)2 = 2n -1, contradicting the fact that L deg Vi = 2 (n - 1). (b) Here is a tree with two vertices of degree 3 and exactly four vertices of degree 1. Exercises 12.2 1. [BB] We show T and two other spanning trees found by adding a and then successively deleting f and 9 are shown to the right. 2. (a) [BB] Here is one possible answer. The second and third trees are isomorphic, but neither is isomorphic to the first. T a 4 5 a...
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