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The complexity of finding arc-disjoint branching flows

Abstract

The concept of arc-disjoint flows in networks was recently introduced in \cite{bangTCSflow}. This is a very general framework within which many well-known and important problems can be formulated. In particular, the existence of arc-disjoint branching flows, that is, flows which send one unit of flow from a given source ss to all other vertices, generalizes the concept of arc-disjoint out-branchings (spanning out-trees) in a digraph. A pair of out-branchings Bs,1+,Bs,2+B_{s,1}^+,B_{s,2}^+ from a root ss in a digraph D=(V,A)D=(V,A) on nn vertices corresponds to arc-disjoint branching flows x1,x2x_1,x_2 (the arcs carrying flow in xix_i are those used in Bs,i+B_{s,i}^+, i=1,2i=1,2) in the network that we obtain from DD by giving all arcs capacity n1n-1.It is then a natural question to ask how much we can lower the capacities on the arcs and still have, say, two arc-disjoint branching flows from the given root ss.We prove that for every fixed integer k2k \geq 2 it is\begin{itemize}\item an NP-complete problem to decide whether a network N=(V,A,u){\cal N}=(V,A,u) where uij=ku_{ij}=k for every arc ijij has two arc-disjoint branching flows rooted at ss.\item a polynomial problem to decide whether a network N=(V,A,u){\cal N}=(V,A,u) on nn vertices and uij=nku_{ij}=n-k for every arc ijij has two arc-disjoint branching flows rooted at ss.\end{itemize}The algorithm for the later result generalizes the polynomial algorithm, due to Lov\'asz, for deciding whether a given input digraph has two arc-disjoint out-branchings rooted at a given vertex.Finally we prove that under the so-called Exponential Time Hypothesis (ETH), for every ϵ>0\epsilon{}>0 and for every k(n)k(n) with (log(n))1+ϵk(n)n2(\log{}(n))^{1+\epsilon}\leq k(n)\leq \frac{n}{2} (and for every large ii we have k(n)=ik(n)=i for some nn) there is no polynomial algorithm for deciding whether a given digraph contains twoarc-disjoint branching flows from the same root so that no arc carries flow larger than nk(n)n-k(n)

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