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Digraphs+Theory,+Algorithms+and+Applications_Part28

# Digraphs+Theory,+Algorithms+and+Applications_Part28 - 9.9...

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Unformatted text preview: 9.9 Arc-Disjoint In- and Out-Branchings 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 NP-complete when we ask for an out-branching and an in-branching that are arc-disjoint 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 2-cycle, 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 2-cycle 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 arc-disjoint branchings F + v ,F- v such that F + v is an out-branching rooted at v and F- v is an in-branching 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 non-existence 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 2-cycle, 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 in-arborescence (out-arborescence ) 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 out-branching induced by the arcs { vw : w ∈ N + ( v ) } ∪ E B ∪ S r j =1 E ( F + j ) and F- v be the in-branching 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 arc-disjoint 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 out-branching or an in-branching 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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