Unformatted text preview: in (}'. Each odd vertex in (} is the end vertex of precisely one new path constructed by the algorithm. In addition, an odd vertex could be an intermediate vertex on a path between odd vertices, so it is even in (}'. 11. Solution 1. The first odd vertex can be paired with any of the remaining n 1. For each pairing, one of the remaining vertices can be paired with any of n  3 vertices. Then one of the remaining vertices can be paired with any of n  5 and so on. The number of pairs is (n l)(n 3)(n 5) ... 5·3· 1 Note that n is even so the product here really does finish as indicated. Solution 2. Let m = n/2. There are (~) ways to select a pair {Vb wd of odd vertices; for each of these ways, there are (n~2) ways to select another pair {V2' W2}, and so on. Altogether, there are ways to select {VI, WI}, ... , { Vm, wm}. Disregarding the order in which these pairs are selected, there are just 2::~! ways to divide the odd vertices into pairs....
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 Summer '10
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 Graph Theory, Vertex, WI, Shortest path problem, odd vertices

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