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Reconfiguring k-path vertex covers (1911.03026v2)

Published 8 Nov 2019 in cs.DS, cs.CC, and cs.DM

Abstract: A vertex subset $I$ of a graph $G$ is called a $k$-path vertex cover if every path on $k$ vertices in $G$ contains at least one vertex from $I$. The \textsc{$k$-Path Vertex Cover Reconfiguration ($k$-PVCR)} problem asks if one can transform one $k$-path vertex cover into another via a sequence of $k$-path vertex covers where each intermediate member is obtained from its predecessor by applying a given reconfiguration rule exactly once. We investigate the computational complexity of \textsc{$k$-PVCR} from the viewpoint of graph classes under the well-known reconfiguration rules: $\mathsf{TS}$, $\mathsf{TJ}$, and $\mathsf{TAR}$. The problem for $k=2$, known as the \textsc{Vertex Cover Reconfiguration (VCR)} problem, has been well-studied in the literature. We show that certain known hardness results for \textsc{VCR} on different graph classes including planar graphs, bounded bandwidth graphs, chordal graphs, and bipartite graphs, can be extended for \textsc{$k$-PVCR}. In particular, we prove a complexity dichotomy for \textsc{$k$-PVCR} on general graphs: on those whose maximum degree is $3$ (and even planar), the problem is $\mathtt{PSPACE}$-complete, while on those whose maximum degree is $2$ (i.e., paths and cycles), the problem can be solved in polynomial time. Additionally, we also design polynomial-time algorithms for \textsc{$k$-PVCR} on trees under each of $\mathsf{TJ}$ and $\mathsf{TAR}$. Moreover, on paths, cycles, and trees, we describe how one can construct a reconfiguration sequence between two given $k$-path vertex covers in a yes-instance. In particular, on paths, our constructed reconfiguration sequence is shortest.

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