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Unformatted text preview: 8.6 Orientations Achieving High ArcStrong Connectivity 443 8.6 Orientations Achieving High ArcStrong Connectivity Let us recall that an orientation D of a multigraph G = ( V,E ) is obtained by assigning one of the two possible orientations to each edge of G (in particular two parallel edges may receive opposite orientations). By Robbins’ theorem, an undirected multigraph G = ( V,E ) has a strongly connected orientation if and only if G is 2edgeconnected. Below we describe two generalizations of Robbins’ theorem, due to Nash Williams, both of which are much deeper than Robbins’ theorem, especially the one in Theorem 8.6.4. In order to illustrate to usefulness of the splitting technique which was discussed in Chapter 7, we prove Theorem 8.6.3 below using a splitting result for undirected graphs. This theorem, due to Lov´ asz, is analogous to Theorem 7.5.2. The reader is asked to prove this theorem in Exercise 8.37. Analogously to the directed case, we denote by λ ( x,y ) the maximum number of edge disjoint xypaths in G and we say that a graph G = ( V + s,E ) with a special vertex s is kedgeconnected in V if λ ( x,y ) ≥ k holds for all x,y ∈ V . Theorem 8.6.1 (Lov´ asz’s splitting theorem) [522] Let G = ( V + s,E ) be a multigraph with a designated vertex s of even degree and suppose that G is kedgeconnected in V , for some k ≥ 2 . Then for every edge st there exists an edge su such that after splitting off the pair st,su the new graph is still kedgeconnected in V 10 . ut An undirected multigraph G = ( V,E ) is minimally kedgeconnected if G is kedgeconnected ( λ ( G ) = k ), but λ ( G e ) = k 1 for every edge e ∈ E . The following theorem by Mader is analogous to Theorem 7.10.3. The proof is left to the reader as Exercise 8.36. Theorem 8.6.2 [532] Every minimally kedgeconnected multigraph has a vertex of degree k . ut Now we can prove the following famous result of NashWilliams: Theorem 8.6.3 (NashWilliams’ orientation theorem) [583] An undi rected multigraph G = ( V,E ) has a karcstrong orientation D if and only if G is 2 kedgeconnected. Proof: The proof idea used below is due to Lov´ asz [522]. Suppose G has a karcstrong orientation D . Thus for every nonempty proper subset X of V we have d + D ( X ) ,d D ( X ) ≥ k . This implies that in G we have d ( X ) ≥ 2 k and hence, G is 2 kedgeconnected. 10 As for directed graphs (see Section 7.5), splitting off the pair ( su,sv ) means that we replace the edges su,sv by a new edge uv (or a copy of that edge if it already exists). 444 8. Orientations of Graphs To prove the other direction we proceed by induction on the number of edges in G . Let G = ( V,E ) be 2 kedgeconnected. If  E  = 2 k , then G is just two vertices x,y joined by 2 k copies of the edge xy . Clearly this multigraph has a karcstrong orientation. Thus we may proceed to the induction step....
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This document was uploaded on 08/10/2011.
 Spring '11
 Algorithms

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