Abstract
A graph is H-free if it contains no induced subgraph isomorphic to H.We prove new complexity results for the two classical cycle transversal problems Feedback Vertex Set and Odd Cycle Transversal by showing that they can be solved in polynomial time on (sP_1+ P_3)-free graphs for every integer sge 1.We show the same result for the variants Connected Feedback Vertex Set and Connected Odd Cycle Transversal. We also prove that the latter two problems are polynomial-time solvable on cographs; this was already known for Feedback Vertex Set and Odd Cycle Transversal. We complement these results by proving that Odd Cycle Transversal and Connected Odd Cycle Transversal are NP-complete on (P_2+ P_5,P_6)-free graphs.
Highlights
Graph transversal problems play a central role in Theoretical Computer Science
We focus on proving new complexity results for Feedback Vertex Set, Connected Feedback Vertex Set, Odd Cycle Transversal and Connected Odd Cycle Transversal on H-free graphs
We proved polynomial-time solvability of Feedback Vertex Set and Odd Cycle Transversal on H-free graphs when H = sP1 + P3 and polynomial-time solvability of their connected variants on H-free graphs, when H = P4 or H = sP1 + P3 ; see Table 1, where we place these results in the context of known results for these problems on H-free graphs
Summary
Graph transversal problems play a central role in Theoretical Computer Science. To define the notion of a graph transversal, let H be a family of graphs, G = (V, E) be a graph and S ⊆ V be a subset of vertices of G. We say that S is an H-transversal of G if G − S is H-free, that is, if G − S contains no induced subgraph isomorphic to a graph of H. S intersects every induced copy of every graph of H in G. Let Cr and Pr denote the cycle and path on r vertices, respectively. S is a vertex cover, feedback vertex set, or odd cycle transversal if S is an H-transversal for, respectively, H = {P2} (that is, G − S is edgeless), H = {C3, C4, ...} (that is, G − S is a forest), or H = {C3, C5, C7, ...} (that is, G − S is bipartite)
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