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Unformatted text preview: 7.15 Packing Cuts 403 Let C * = { Z 1 ,Z 2 ,...,Z M } and let A i = ( Z i ,V Z i ), i = 1 , 2 ,...,M be the corresponding arc sets. Construct an undirected graph G ( C * ) = ( V,E ) as follows: V = { v 1 ,v 2 ,...,v M } and there is an edge between v i and v j if and only if A i ∩ A j 6 = ∅ . Since D contains at most k arcdisjoint dicuts, it follows that G ( C * ) has at most k independent vertices. Hence it suffices to show that G ( C * ) is a bipartite graph since then we get C = C *  ≤ 2 k . Let v 1 v 2 ...v s v 1 be an arbitrary cycle in G ( C * ). Note that the arc sets of the corresponding dicuts A 1 ,...,A s must be different, since if ( Z i ,V Z i ) = ( Z j ,V Z j ) for some 1 ≤ i < j ≤ s , then every arc in ( Z i ,V Z i ) is covered twice (by ( Z i ,V Z i ) and by ( Z j ,V Z j )) and hence the vertices v i ,v j each have degree one in G ( C * ), contradicting the fact that they are on a cycle. Note also that if two dicuts ( X,V X ) and ( Y,V Y ) have X ∪ Y = V , then they are arcdisjoint and hence are not adjacent in G ( C * ). X Y Y X Y (b) (a) X X Y Figure 7.18 Illustration of the definition of being to the right and left for cuts. In the two situations in part (a) (part (b)) the dicut ( X,V X ) is to the left (right) of the dicut ( Y,V Y ). In the right part of (a) we have X ∪ Y = V . Since A i ∩ A i +1 6 = ∅ for i = 0 , 1 ,...,s 1, where A = A s , it follows from our remarks above that we have either Z i ⊂ Z i +1 or Z i +1 ⊂ Z i . We prove that the two possibilities occur alternatingly and hence s is even. Suppose not, then without loss of generality we have Z ⊂ Z 1 ⊂ Z 2 . Let us say that a dicut A i is to the left of another dicut A j if either Z i ⊂ Z j , or Z i ∪ Z j = V (which is equivalent to V Z i ⊂ Z j ) and that A i is to the right of A j if Z i ∩ Z j = ∅ (which is equivalent to Z i ⊂ V Z j ), or Z j ⊂ Z i (which is equivalent to V Z i ⊂ V Z j ). See Figure 7.18. Since C * contains no crossing members, each A i 6 = A j is either to the right or to the left of A j . Since A 2 is to the right of A 1 and A = A s is to the left of A 1 , it follows that there is some 2 ≤ j ≤ s 1 such that A j is to the right of A 1 and A j +1 is to the left of A 1 . Suppose first that Z j ∩ Z 1 = ∅ , then we cannot have Z j +1 ⊂ Z 1 as A j +1 and A j have a common arc. So we must have Z 1 ∪ Z j +1 = V , but then any arc a common to A j and A j +1 enters Z 1 , contradicting that d ( Z 1 ) = 0. 404 7. Global Connectivity Hence we must have Z 1 ⊂ Z j . The fact that A j ,A j +1 have a common arc a (and hence either Z j ⊂ Z j +1 or Z j +1 ⊂ Z j ) implies that, by the choice of j , we have Z j +1 ⊂ Z 1 ⊂ Z j +1 . But now the arc a belongs to three dicuts A 1 ,A j and A j +1 , a contradiction. This completes the proof of the lemma and, by the remark above, also the proof of the theorem. ut Combining (7.37) and Theorem 7.15.2, we obtain: Corollary 7.15.4 Let D be a nonstrong directed multigraph whose under...
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This document was uploaded on 08/10/2011.
 Spring '11
 Algorithms

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