Close relatives (of Feedback Vertex Set), revisited

Hugo Jacob, Thomas Bellitto, Oscar Defrain, Marcin Pilipczuk
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引用次数: 4

Abstract

At IPEC 2020, Bergougnoux, Bonnet, Brettell, and Kwon showed that a number of problems related to the classic Feedback Vertex Set (FVS) problem do not admit a $2^{o(k \log k)} \cdot n^{\mathcal{O}(1)}$-time algorithm on graphs of treewidth at most $k$, assuming the Exponential Time Hypothesis. This contrasts with the $3^{k} \cdot k^{\mathcal{O}(1)} \cdot n$-time algorithm for FVS using the Cut&Count technique. During their live talk at IPEC 2020, Bergougnoux et al.~posed a number of open questions, which we answer in this work. - Subset Even Cycle Transversal, Subset Odd Cycle Transversal, Subset Feedback Vertex Set can be solved in time $2^{\mathcal{O}(k \log k)} \cdot n$ in graphs of treewidth at most $k$. This matches a lower bound for Even Cycle Transversal of Bergougnoux et al.~and improves the polynomial factor in some of their upper bounds. - Subset Feedback Vertex Set and Node Multiway Cut can be solved in time $2^{\mathcal{O}(k \log k)} \cdot n$, if the input graph is given as a clique-width expression of size $n$ and width $k$. - Odd Cycle Transversal can be solved in time $4^k \cdot k^{\mathcal{O}(1)} \cdot n$ if the input graph is given as a clique-width expression of size $n$ and width $k$. Furthermore, the existence of a constant $\varepsilon>0$ and an algorithm performing this task in time $(4-\varepsilon)^k \cdot n^{\mathcal{O}(1)}$ would contradict the Strong Exponential Time Hypothesis.
(反馈顶点集)的近亲,重新访问
在IPEC 2020上,Bergougnoux, Bonnet, Brettell和Kwon展示了一些与经典反馈顶点集(FVS)问题相关的问题,在假设指数时间假设的情况下,在树宽的图上最多不允许$2^{o(k \log k)} \cdot n^{\mathcal{O}(1)}$时间算法$k$。这与使用Cut&Count技术的FVS $3^{k} \cdot k^{\mathcal{O}(1)} \cdot n$时间算法形成对比。在IPEC 2020的现场演讲中,Bergougnoux等人提出了一些悬而未决的问题,我们在这项工作中回答了这些问题。-子集偶环截线,子集奇环截线,子集反馈顶点集可以在时间内解决$2^{\mathcal{O}(k \log k)} \cdot n$在树宽最多$k$的图。这与Bergougnoux等人的偶环截线的下界相匹配,并改进了他们的某些上界中的多项式因子。如果输入图以大小为$n$和宽度为$k$的团宽度表达式给出,则子集反馈顶点集和节点多路切割可以及时解决$2^{\mathcal{O}(k \log k)} \cdot n$。如果输入图以大小为$n$,宽度为$k$的团宽表达式给出,奇环截线可以及时求解$4^k \cdot k^{\mathcal{O}(1)} \cdot n$。此外,常数$\varepsilon>0$的存在和在时间$(4-\varepsilon)^k \cdot n^{\mathcal{O}(1)}$中执行此任务的算法将与强指数时间假设相矛盾。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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