Digraphs+Theory,+Algorithms+and+Applications_Part21

Digraphs+Theory,+Algorithms+and+Applications_Part21 - 7.10...

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Unformatted text preview: 7.10 Minimally k-(Arc)-Strong Directed Multigraphs 383 precisely one minimal k-out-critical set X u and enters precisely one minimal k-in-critical set Y u . Here minimal means with respect to inclusion. Lemma 7.10.2 If X,Y are crossing k-in-critical sets in D , then X ∩ Y and X ∪ Y are also k-in-critical sets and d ( X,Y ) = 0 . Proof: Suppose X,Y are crossing and k-in-critical. 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 k-in-critical and d ( X,Y ) = 0. ut Intuitively, Lemma 7.10.2 implies that minimally k-arc-strong directed multigraphs have vertices of small in-degree and vertices small out-degree. 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 k-arc-strong 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 k-in-critical 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 k-arc-strong. Our first goal is to make R cross-free (that is, we want to replace R by a new family R * of k-in-critical sets such that R * still satisfies (7.27) and no two members of R * are crossing). To do this we apply the so-called 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 k-in- 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 cross-free. Note that we could have obtained such a family directly by choosing the members in R as the unique minimal k-in-critical 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 cross-free family satisfying (7.27). We shall use this freedom in Case 2 below. Assume below that R is cross-free . (7.28) Now the trick is to consider an arbitrary fixed vertex s and show that V- s contains a vertex with in-degree and out-degree 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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Digraphs+Theory,+Algorithms+and+Applications_Part21 - 7.10...

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