328 Solutions to Exercises Solving for kl' we get kl = (3 -2)k3 + (4 -2)k4 + (5 -2)k5 + . .. + 2. This is exactly what we were asked to establish since (i -2)k i is just 2]deg Vj -2], tlte sum over all vertices of degree i. 21. [BB] There are no cycles in tlte subgraph since there are no cycles in CkH2k+2. Also, given any two C vertices, tltere is a patlt between tltem in CkH2k+2 (because CkH2k+2 is connected). Any H vertex on tltis patlt would have degree at least 2. Thus, tltere is none; tlte patlt consists entirely of C vertices and hence lies witltin tlte subgraph. The subgraph is connected, hence a tree. 22. (a) Let x be tlte number of H vertices adjoined. Since Thad k - 1 edges, and one new edge is added for each H, g has (k-1) +x edges. Therefore, E degvi = 2(k-1+x). But Edegvi = 4k+x since each C has degree 4 and each H has degree 1. Therefore, 4k + x = 2k -2 + 2x and x = 2k+2. 23.
This is the end of the preview. Sign up
access the rest of the document.
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.