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Unformatted text preview: 9.9 ArcDisjoint In and OutBranchings 523 It is easy to reduce (in polynomial time) Problem 9.9.1 for the case when u 6 = v to the case when u = v for arbitrary digraphs (Exercise 9.49). Hence the problem remains NPcomplete when we ask for an outbranching and an inbranching that are arcdisjoint and have the same root. However, Bang Jensen and Huang showed that, if the vertex that is to be the root is adjacent to all other vertices in the digraph and is not in any 2cycle, then the problem becomes polynomially solvable. Theorem 9.9.3 [79] Let D = ( V,A ) be a strongly connected digraph and v a vertex of D such that v is not on any 2cycle and V ( D ) = { v } ∪ N ( v ) ∪ N + ( v ) . Let A = { A 1 ,A 2 ,...A k } ( B = { B 1 ,B 2 ,...,B r } ) denote the set of terminal (initial) components in D h N + ( v ) i ( D h N ( v ) i ). Then D contains a pair of arcdisjoint branchings F + v ,F v such that F + v is an outbranching rooted at v and F v is an inbranching rooted at v if and only if there exist two disjoint arc sets E A ,E B ⊂ A such that all arcs in E A ∪ E B go from N + ( v ) to N ( v ) and every A i ∈ A ( B j ∈ B ) is incident with an arc from E A ( E B ). Furthermore, there exists a polynomial algorithm to find the desired branchings, or demonstrate the nonexistence of such branchings. Proof: We prove the characterization and refer the reader to [79] and Exer cise 9.51 for the algorithmic part. First we note that, if the branchings exist, then the arc sets E A and E B exist. Indeed, if F + v ,F v are such branchings, then there must be an arc from F v ( F + v ) leaving (entering) every terminal (initial) component of D h N + ( v ) i ( D h N ( v ) i ) and since v is not on any 2cycle, all these arcs go from N + ( v ) to N ( v ). Suppose that there exist sets E A and E B as above. Every vertex x ∈ N + ( v ) has a path to one of the terminal components in A and every ver tex in N ( v ) can be reached by a path from one of the initial compo nents in B . Hence, we can choose a family of vertex disjoint arborescences F 1 ,F 2 ,...,F k , F + 1 ,F + 2 ,...,F + r such that F i ( F + j ) is an inarborescence (outarborescence ) rooted at a vertex in A i ( B j ) and S k i =1 V ( F i ) = N + ( v ), S r j =1 V ( F + j ) = N ( v ). Let F + v be the outbranching induced by the arcs { vw : w ∈ N + ( v ) } ∪ E B ∪ S r j =1 E ( F + j ) and F v be the inbranching induced by the arcs { uv : u ∈ N ( v ) }∪ E A ∪ S k i =1 E ( F i ). Then F + v and F v are the desired branchings. ut The following is an easy corollary of Theorem 9.9.3. Corollary 9.9.4 [46] A tournament D = ( V,A ) has arcdisjoint branchings F + v ,F v rooted at a specified vertex v ∈ V if and only if D is strong and for every arc a ∈ A the digraph D a contains either an outbranching or an inbranching with root v . ut There is a small inconsistency in the statement (and the proof) of The orem 9.9.3 in [79] as it was not mentioned thatorem 9....
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This document was uploaded on 08/10/2011.
 Spring '11
 Algorithms

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