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 nl edges and I: degv = 21£1. Here then, 33(1)+6(3) +(n39)7 = 2( n 1). This gives 5n = 220, so n = 44 and the number of vertices of degree 7 is 44  39 = 5....
View
Full Document
 Summer '10
 any
 Graph Theory, WI, vertices, Eulerian path, Hamiltonian path

Click to edit the document details