Discrete Mathematics with Graph Theory (3rd Edition) 398

Discrete Mathematics with Graph Theory (3rd Edition) 398 -...

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396 Solutions to Review Exercises 8. It is not possible for these people to select different teams. If X = {Bruce, Craig, Camelia, Yuri}, then A(X) = {Russia, Slovakia, Romania}. We have IXI = 4 while IA(X)I = 3: Hall's Marriage Theorem says the desirable cannot be done. 9. Now the desired betting scheme is possible. For example, we could have Albert betting on USA Bruce betting on Canada Craig betting on Slovakia Camelia betting on Romania Oana betting on Finland Yuri betting on Russia. 10. (~) If 9 has a perfect matching, then Proposition 14.4.3 says 9 has an even number of vertices. ( f--) Conversely, suppose 9 has 2n vertices. It is not hard to see that a connected graph in which every
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Unformatted text preview: vertex has degree 2 must be a cycle. But if the vertices of the cycle, written in order, are V!, V2, • .. ,V2n, then edges Vl V2, V3V4, ... , V2n-l Vn give a perfect matching. 11. (a) The sum of the degrees of the vertices ofg is 31VI. This is also 21&1, so 211VI. (b) No, and we present a counterexample at the right. If this graph had a perfect matching, the middle vertex V would have to be joined to an edge which is part of the matching, say vw as shown. Then the five vertices in the leftmost group (or in the bottom group) could not all be saturated, since they could only be joined to each other....
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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.

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