Unformatted text preview: iii.) Assume T has exactly two vertices of degree 1. Since T has n -1 edges, the sum of the degrees of its vertices is 2(n -1) = 2n - 2. This means that the remaining n - 2 vertices must have degrees totalling 2n - 4 = 2(n - 2). But each of these has degree at least 2, and it follows that each must have degree exactly 2. (iii. ---+ i.) Assume T has n - 2 vertices of degree 2. Corollary 12.1.7 tells us that the other two vertices have degree 1. Since T is connected and has exactly two vertices of odd degree, Theorem 10.1.5 guarantees an Eulerian trail. 5. (a) 9 is connected because there is a path between any two vertices. If it contained a cycle, there would be two paths between any vertices of this cycle, contrary to fact. So 9 is acyclic and connected, hence a tree. (b) A tree with n vertices has n-l edges and I: degv = 21£1. Here then, 33(1)+6(3) +(n-39)7 = 2( n -1). This gives 5n = 220, so n = 44 and the number of vertices of degree 7 is 44 - 39 = 5....
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- Summer '10
- Graph Theory, WI, vertices, Eulerian path, Hamiltonian path