CO330A3solns3

CO330A3solns3 - C&O 330 - SOLUTIONS #3 PROFESSOR D.M....

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Unformatted text preview: C&O 330 - SOLUTIONS #3 PROFESSOR D.M. JACKSON (1) (15 points) Let K n denote the complete graph on n vertices. (a) (7 points) Show that the number of ways of covering K n with paths of vertex-length 1 or more is bracketleftbigg x n n ! bracketrightbigg exp parenleftbigg x (2- x ) 2 (1- x ) parenrightbigg . Solution: Let S be the set of all such coverings of K n for n = , 1 , 2 , . . .. , and let L be the set of all non-null undirected labelled paths. Then S ↔ U circleasterisk L . Let D be the set of all non-null directed labelled paths. Then, with the exception of the path of (vertex) length 1, any path in L may be obtained from precisely 2 paths in D by removing the orientation of the paths. Thus there is a natural action of S 2 on D to give L , where S 2 (the symmetric group on 2 sumbols) acts neutrally on a path of length 1. Then L ↔ D / S 2 ↔ P / S 2 where P is the set of all permutations on { 1 , . . ., n } for all n ≥ . Let ω ( c ) be the number of labels in c ∈ S . Then S ↔ U circleasterisk ( P / S 2 ) is additively ω-preserving. Let S ( x ) = [( S , ω )] e ( x ) . By the circleasterisk-Product Lemma S = exp ([( P / S 2 , ω )] e ( x )) . But [( P , ω )] e ( x ) = ∑ k ≥ 1 x k , so [( P / S 2 , ω )] e ( x ) = x + 1 2 summationdisplay k ≥ 2 x k = x 2 2- x 1- x , recalling that S 2 acts neutrally on paths of length 1. Thus the desired number is bracketleftbigg x n n ! bracketrightbigg S ( x ) = bracketleftbigg x n n ! bracketrightbigg exp parenleftbigg x 2 2- x 1- x parenrightbigg . (b) (8 points) Show that the number of ways of covering K n with cycles of length 3 or more is bracketleftbigg x n n !...
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This note was uploaded on 11/18/2010 for the course CO 330 taught by Professor R.metzger during the Spring '05 term at Waterloo.

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CO330A3solns3 - C&O 330 - SOLUTIONS #3 PROFESSOR D.M....

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