Submodular functions and perfect graphs

Abrishami, Tara; Chudnovsky, Maria; Dibek, Cemil; Vušković, Kristina

doi:10.48550/arxiv.2110.00108

Cited by 1 publication

(1 citation statement)

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“…This proves that G is diamond-free. □ We also need the following folklore result that appeared in [4]:…”

Section: Proof Letmentioning

confidence: 99%

Induced subgraphs and tree decompositions V. one neighbor in a hole

Abrishami,

Alecu,

Chudnovsky

et al. 2023

Journal of Graph Theory

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What are the unavoidable induced subgraphs of graphs with large treewidth? It is well‐known that the answer must include a complete graph, a complete bipartite graph, all subdivisions of a wall and line graphs of all subdivisions of a wall (we refer to these graphs as the “basic treewidth obstructions”). So it is natural to ask whether graphs excluding the basic treewidth obstructions as induced subgraphs have bounded treewidth. Sintiari and Trotignon answered this question in the negative. Their counterexamples, the so‐called “layered wheels,” contain wheels, where a wheel consists of a hole (i.e., an induced cycle of length at least four) along with a vertex with at least three neighbors in the hole. This leads one to ask whether graphs excluding wheels and the basic treewidth obstructions as induced subgraphs have bounded treewidth. This also turns out to be false due to Davies' recent example of graphs with large treewidth, no wheels and no basic treewidth obstructions as induced subgraphs. However, in Davies' example there exist holes and vertices (outside of the hole) with two neighbors in them. Here we prove that a hole with a vertex with at least two neighbors in it is inevitable in graphs with large treewidth and no basic obstruction. Our main result is that graphs in which every vertex has at most one neighbor in every hole (that does not contain it) and with the basic treewidth obstructions excluded as induced subgraphs have bounded treewidth.

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“…This proves that G is diamond-free. □ We also need the following folklore result that appeared in [4]:…”

Section: Proof Letmentioning

confidence: 99%