CSCE896-HW3

# CSCE896-HW3 - CSCE 896 Graph Theory Homework 3 Fall 2010 1...

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CSCE 896 Graph Theory Homework 3 Fall 2010 1. Let e be an edge of K n . Use Cayley’s Theorem to prove that K n - e has ( n - 2) n n - 3 spanning trees. Proof: Let V ( K n ) = v 1 ,v 2 ,...... ,v n . We may assume that e = v 1 v 2 , and we count the number N of spanning trees which contain the edge e . Let T be such a tree. Then deg T ( v 1 ) + deg T ( v 2 ) = k + 2 for some integer k 0. If k = 0, then n = 2, and the statements proven. Let k > 1, given such a k , let’s assume there are a k number of trees T that have deg T ( v 1 ) + deg T ( v 2 ) = k + 2. Let T 0 be obtained from T by replacing the pair of vertices { v 1 ,v 2 } by a new vertex u and joining u to all neighbors of v 1 and v 2 in T . Then T 0 is a tree of order n - 1 2 and deg 0 T ( u ) = k 1. It is clear that T 7→ T 0 deﬁnes a 2 k - to - 1 map from the set of trees on { v 1 ,...,v n } with deg T ( v 1 ) + degT ( v 2 ) = k + 2, to the set of trees on { u,v 3 ,...... ,v n } with degT 0 ( u ) = k , since every vertex v i , 3 i n , of T can be connected to either v 1 or v 2 , but not both. We know that there are a k = 2 k ( n - 3 k - 1 )( n - 2) n - k - 2 , and N = n - 2 X k =1 a k = n - 2 X k =1 2 k ( n - 3 k - 1 )( n - 2) n -

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CSCE896-HW3 - CSCE 896 Graph Theory Homework 3 Fall 2010 1...

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