Discrete Mathematics with Graph Theory (3rd Edition) 364

# Discrete Mathematics with Graph Theory (3rd Edition) 364 -...

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362 (0 This is not planar. It has a subgraph homeomorphic to Ks as shown. Solutions to Exercises 6. Let N be as described in the proof of Corollary 13.1.3. Then 5R ~ N ~ 2E, so 5R ~ 2E. Thus, 5(E - V + 2) ~ 2E, giving 3E ~ 5V - 10. 7. [BB] We know from Theorem 13.1.4 that E ~ 3V - 6. Substituting in E = V + R - 2, we obtain V + R - 2 ~ 3V - 6, or R ~ 2V - 4, as required. 8. (a) [BB] A E = 3 = 3(3) - 6 = 3V - 6. (b) Look at the proof of Theorem 13.1.4. We can only have E = 3V - 6 if 3R = N = 2E, and 3R = N means that every region is a triangle. 9. (a) If g has three vertices, it must have at least two edges since it's connected. It can't have three edges since it is not a triangle. Thus, E = 2 ~ 2 = 2V - 4 as required. If g has four vertices, it has at least three edges. Since it contains no triangles, it is either a tree or a 4-cycle. In either case, E ~ 4 = 2V - 4 as required. Now assume V ~ 5, so E ~ 4. As in the proof of Theorem 13.1.4, count the edges on the boundary of each region and sum over all regions. Denoting the sum N, we have N ~ 2E.
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