Chapter 12 357 (c) Let there be x vertices of degree 3 and y of degree 5. Altogether, then, there are n = 2 + x + y vertices, so L: deg v = 2(n -1) says 2 + 3x + 5y = 2(1 + x + y). This simplifies to x + 3y = 0, so x = y = 0, giving the tree precisely two vertices, a contradiction. 6. (a) Since the sum of the degrees of the vertices is 12, the remaining five vertices must all have degree 1. The vertex of degree 5 must be adjacent to the vertex of degree 2; otherwise, it would be adjacent to all five vertices of degree 1, leaving no vertex which could be adjacent to the other vertex of degree 2. Then the vertex of degree 5 will be adjacent to four of the vertices of degree 1, and the vertex of degree 2 will be adjacent to the other. The only possibility is shown on the left. kBG kGB A DAD F E F E (b) Two labeled nonisomorphic trees are shown above. 7. K8 has 86 spanning trees, by Theorem 12.2.3. 8. With vertices labeled as shown, the matrix defined in Kirchhoff's Theorem is
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.