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Packing Directed Cycles Efficiently

机译:有效包装定向循环

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摘要

Let G be a simple digraph. The dicycle packing number of G, denoted V_c(G), is the maximum size of a set of arc-disjoint directed cycles in G. Let G be a digraph with a nonnegative arc-weight function w. A function if) from the set C of directed cycles in G to R+ is a fractional dicycle packing of G if Σ_(e∈C ∈C) ψ(C) ≤ w(e) for each e ∈ E(G). The fractional dicycle packing number, denoted V_c~*(G,w), is the maximum value of Σ_(C ∈C) ψ(C) taken over all fractional dicycle packings ψ. In case w ≡ 1 we denote the latter parameter by v_c~*(G). Our main result is that v_c~*(G) - v_c(G) = o(n~2) where n = |V(G)|. Our proof is algorithmic and generates a set of arc-disjoint directed cycles whose size is at least v_c(G) - o(n~2) in randomized polynomial time. Since computing v_c(G) is an NP-Hard problem, and since almost all digraphs have v_c(G) = Θ(n~2) our result is a FPTAS for computing v_c(G) for almost all digraphs. The latter result uses as its main lemma a much more general result. Let F be any fixed family of oriented graphs. For an oriented graph G, let v_F(G) denote the maximum number of arc-disjoint copies of elements of F that can be found in G, and let v_F~*(G) denote the fractional relaxation. Then, V_F~*(G) - v_F(G) = o(n~2). This lemma uses the recently discovered directed regularity lemma as its main tool. It is well known that v_c~*(G,w) can be computed in polynomial time by considering the dual problem. However, it was an open problem whether an optimal fractional dicycle packing ψ yielding v_2~*(G,w) can be generated in polynomial time. We prove that a maximum fractional dicycle packing yielding v_c~*(G, w) with at most O(n~2) dicycles receiving nonzero weight can be found in polynomial time.
机译:令G为简单的有向图。 G的双环填充数(表示为V_c(G))是G中一组弧不相交的有向环的最大尺寸。令G为具有非负弧权函数w的图。如果每个e∈E(G)的Σ_(e∈C∈C)ψ(C)≤w(e),则从G到R +的有向循环的集合C的函数if是G的分数二环填充。分数二轮车装填数表示为V_c〜*(G,w),是所有分数二轮车装填ψ所取的Σ_(C∈C)ψ(C)的最大值。在w≡1的情况下,我们用v_c〜*(G)表示后一个参数。我们的主要结果是v_c〜*(G)-v_c(G)= o(n〜2),其中n = | V(G)|。我们的证明是算法的,并且在随机多项式时间中生成了一组弧不相交的有向循环,其大小至少为v_c(G)-o(n〜2)。由于计算v_c(G)是一个NP-Hard问题,并且由于几乎所有图都有v_c(G)=Θ(n〜2),因此我们的结果是针对几乎所有图的vPT(G)计算FPTAS。后一个结果将更普遍的结果用作其主要引理。令F为任何固定的定向图族。对于有向图G,令v_F(G)表示可以在G中找到的F元素的弧不交叠副本的最大数量,而令v_F〜*(G)表示分数弛豫。然后,V_F〜*(G)-v_F(G)= o(n〜2)。此引理使用最近发现的有向规则引理作为主要工具。众所周知,通过考虑对偶问题,可以在多项式时间内计算v_c〜*(G,w)。但是,是否可以在多项式时间内生成产生v_2〜*(G,w)的最优分数二轮车装填ψ是一个悬而未决的问题。我们证明在多项式时间内可以找到最大分数二轮车堆积产生v_c〜*(G,w),最多O(n〜2)个二轮车接收非零权重。

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