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Unformatted text preview: 8.9 Orientations of Mixed Graphs 463 It is not difficult to see that one can formulate the problem of orienting a mixed graph so as to get a karcstrong directed multigraph as a submodular flow problem. We can use the same approach as in Subsection 8.8.4. The only change is that we insist that x ( a ) = 0 for original arcs (Exercise 8.53). Jackson [451] conjectured that Theorem 8.9.1 could be extended to local connectivities and hence providing a generalization of NashWilliams’ strong orientation theorem (Theorem 8.6.4). However, examples by Enni [218] show that this conjecture is false. In the case when the directed part of M = ( V,A,E ) is eulerian such an extension is indeed possible. In [218] Enni shows how to extend Theorem 8.6.5 to the case of mixed graphs when the directed part D = ( V,A ) is eulerian. We remark that there seems to be no easy way of formulating orientation problems concerning local connectivities as submodular flow problems. When we consider orientation problems where the input is a mixed graph M = ( V,A,E ) which we wish to orient so as to satisfy a certain lower bound h ( X ) on the indegree of every subset X of vertices, then we cannot in general apply a theorem like Frank’s orientation theorem (Theorem 8.7.6). The reason for this is that even if the function h ( X ) ‘behaves nicely’, we have to take into account the arcs in A because these will contribute to the indegree of the final oriented graph D . To give an example, consider a mixed graph M = ( V,A,E ) and let h ( X ) = k for all nonempty proper subsets of V and h ( ∅ ) = h ( V ) = 0. That is, we are looking for a karcstrong orientation of M . When we want to apply a theorem like Theorem 8.7.6 we have to consider the revised indegree lower bound h given by h ( X ) = k d D ( X ), where D = ( V,A ) is the directed graph induced by the arcs already oriented in M . The function h is easily seen to be crossing Gsupermodular, where G = ( V,E ) is the undirected part of M (Exercise 8.62). However h is typically negative on certain sets and hence Theorem 8.7.6 cannot be applied. As we mentioned above, for the particular lower bound h ( X ) = k , when ever ∅ 6 = X 6 = V , the problem can be formulated as a submodular flow problem. This is no coincidence as we show below. Let G = ( V,E ) be an undirected graph. Let h : 2 V → Z ∪ {∞} be crossing Gsupermodular with h ( ∅ ) = h ( V ) = 0. Let D = ( V,A ) be an arbitrary but fixed orientation of G . Let x : A ( D ) →{ , 1 } be a vector and define an orientation D = ( V,A ) of G by taking A = { a : a ∈ A,x ( a ) = }∪{ ← a : a ∈ A,x ( a ) = 1 } . Here ← a denotes the opposite orientation of the arc a (compare with Section 8.8.4). Then D will satisfy d D ( U ) ≥ h ( U ) for all U ⊂ V (8.41) if and only if d D ( U ) x ( U ) + x + ( U ) ≥ h ( U ) for all U ⊂ V , or equivalently x ( U ) x + ( U ) ≤ d D ( U ) h ( U ) = b 00 ( U ) for all U ⊂ V. (8.42) 464 8. Orientations of Graphs8....
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This document was uploaded on 08/10/2011.
 Spring '11
 Algorithms

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