This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 9.2 Disjoint Path Problems 483 Theorem 9.2.7 [545] Let D = ( V,A ) be a digraph of order n and let k be an integer such that n ≥ 2 k ≥ 2 . If  A  ≥ n ( n 2) + 2 k then D is klinked. ut The proof of Theorem 9.2.7 in [545] is based on the following lemma. Lemma 9.2.8 [545] If D x is klinked for some vertex x ∈ V which satisfies d + ( x ) ,d ( x ) ≥ 2 k 1 , then D is klinked. Proof: Let x 1 ,x 2 ,...,x k ,y 1 ,y 2 ,...,y k ∈ V ( D ) be an arbitrary collection of terminals. We wish to prove that D contains internally disjoint paths P 1 ,P 2 ,...,P k where P i is an ( x i ,y i )path for i = 1 , 2 ,...,k . By the assump tion that D x is klinked, it suffices to consider the case when x = x i for some i or x = y j for some j . Since x is one of the terminals, it follows that among the 2 k terminals at most 2 k 1 of these are outneighbours (inneighbours) of x . Since a path from x to an outneighbour u of x can be taken to be just the arc xu and hence cannot interfere with the other paths we wish to find, we may assume that, if x = x i for some i , then y i 6∈ N + ( x ) and similarly if x = y j for some j then x j 6∈ N ( x ). Let T denote the set of distinct terminals. Now it is easy to see that for every desired path P i starting at x we may choose a private member u i of N + ( x ) T and replace x i by x i = u i . Similarly for every desired path P j ending at x we may choose a private member v j of N ( x ) T and replace y j by y j = v j . If x r ( y s ) was not introduced by the replacements above we let x r = x r ( y s = y s ). Now the existence of the desired linking follows by taking a klinking in D x for the set of terminals x 1 ,x 2 ,...,x k ,y 1 ,y 2 ,...,y k . ut The requirement on the number of arcs in Theorem 9.2.7 is very strong and hence the result is not very useful. However Manoussakis showed by an example that the number of arcs in Theorem 9.2.7 is best possible [545]. The next result, due to Heydemann and Sotteau, shows that for 2linkings one can also get a sufficient condition in terms of δ ( D ). The proof is easy and is left as Exercise 9.6. See also Theorem 9.2.10 below. Theorem 9.2.9 [426] If a digraph D satisfies δ ( D ) ≥ n/ 2 + 1 , then D is 2linked. ut The condition above is still quite restrictive and one would expect a stronger result to hold. Examples from [426] show that we cannot weaken the degree condition. However, we can strengthen the result in the following way. Theorem 9.2.10 If a digraph D satisfies δ ( D ) ≥ n/ 2 + 1 , then for every choice of distinct vertices x,y,u,v ∈ V , D contains internally disjoint ( x,y ), ( u,v )paths P,Q such that V ( P ) ∪ V ( Q ) = V . 484 9. Disjoint Paths and Trees Proof: Let X = V { x,y,u,v } and construct D from D { x,y,u,v } by adding two new vertices p and q such that N D ( p ) = N D ( v ) ∩ X,N + D ( p ) = N + D ( x ) ∩ X, N D ( q ) = N D ( y ) ∩ X,N + D ( q ) = N + D ( u ) ∩ X....
View
Full Document
 Spring '11
 Algorithms, Graph Theory, Digraph, paths, Semicomplete Digraphs

Click to edit the document details