This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 vertexlength 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 nonnull undirected labelled paths. Then S ↔ U circleasterisk L . Let D be the set of all nonnull 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 ωpreserving. Let S ( x ) = [( S , ω )] e ( x ) . By the circleasteriskProduct 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 !...
View
Full
Document
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.
 Spring '05
 R.Metzger

Click to edit the document details