Abstract

The three-in-a-tree algorithm of Chudnovsky and Seymour decides in time O(n4)O(n^4) whether three given vertices of a graph belong to an induced tree. Here, we study four-in-a-tree for triangle-free graphs. We give a structural answer to the following question: what does a triangle-free graph look like if no induced tree covers four given vertices? Our main result says that any such graph must have the "same structure", in a sense to be defined precisely, as a square or a cube. We provide an O(nm)O(nm)-time algorithm that given a triangle-free graph GG together with four vertices outputs either an induced tree that contains them or a partition of V(G)V(G) certifying that no such tree exists. We prove that the problem of deciding whether there exists a tree TT covering the four vertices such that at most one vertex of TT has degree at least 3 is NP-complete.

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