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Unformatted text preview: 7.10 Minimally k(Arc)Strong Directed Multigraphs 383 precisely one minimal koutcritical set X u and enters precisely one minimal kincritical set Y u . Here minimal means with respect to inclusion. Lemma 7.10.2 If X,Y are crossing kincritical sets in D , then X ∩ Y and X ∪ Y are also kincritical sets and d ( X,Y ) = 0 . Proof: Suppose X,Y are crossing and kincritical. Using (7.2) we get k + k = d ( X ) + d ( Y ) = d ( X ∪ Y ) + d ( X ∩ Y ) + d ( X,Y ) ≥ k + k, implying that X ∩ Y and X ∩ Y are both kincritical and d ( X,Y ) = 0. ut Intuitively, Lemma 7.10.2 implies that minimally karcstrong directed multigraphs have vertices of small indegree and vertices small outdegree. The next result by Mader shows that this is indeed the case. In fact, a much stronger statement holds. Theorem 7.10.3 [535] Every minimally karcstrong directed multigraph has at least two vertices x,y with d + ( x ) = d ( x ) = d + ( y ) = d ( y ) = k . Proof: We give a proof due to Frank [260]. Let R be a family of kincritical sets with the property that every arc in D enters at least one member of R . (7.27) By our remark above such a family exists since D is minimally karcstrong. Our first goal is to make R crossfree (that is, we want to replace R by a new family R * of kincritical sets such that R * still satisfies (7.27) and no two members of R * are crossing). To do this we apply the socalled uncrossing technique which is quite useful in several proofs. If there are crossing members X,Y in R , then by Lemma 7.10.2, X ∩ Y and X ∪ Y are kin critical and d ( X,Y ) = 0. Hence every arc entering X or Y also enters X ∪ Y , or X ∩ Y . Thus we can replace the sets X,Y by X ∩ Y,X ∪ Y in R (we only add sets if they are not already there). Since  X ∩ Y  2 +  X ∪ Y  2 >  X  2 +  Y  2 and the number of sets in R does not increase, we will end up with a family R which is crossfree. Note that we could have obtained such a family directly by choosing the members in R as the unique minimal kincritical sets entered by the arcs of A . However, this choice would make the proof more complicated, since we lose the freedom of just working with a crossfree family satisfying (7.27). We shall use this freedom in Case 2 below. Assume below that R is crossfree . (7.28) Now the trick is to consider an arbitrary fixed vertex s and show that V s contains a vertex with indegree and outdegree k . This will imply the theorem. 384 7. Global Connectivity Let s be fixed and define the families S and U as follows S = { X ∈ R : s 6∈ X } , U = { V X : s ∈ X ∈ R} . (7.29) Let L = L ( R ) = S ∪ U . Claim A: The family L is laminar....
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This document was uploaded on 08/10/2011.
 Spring '11
 Algorithms

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