Score: 0

Colouring Graphs Without a Subdivided H-Graph: A Full Complexity Classification

Published: December 10, 2025 | arXiv ID: 2512.09859v1

By: Tala Eagling-Vose , Jorik Jooken , Felicia Lucke and more

We consider Colouring on graphs that are $H$-subgraph-free for some fixed graph $H$, i.e., graphs that do not contain $H$ as a subgraph. It is known that even $3$-Colouring is NP-complete for $H$-subgraph-free graphs whenever $H$ has a cycle; or a vertex of degree at least $5$; or a component with two vertices of degree $4$, while Colouring is polynomial-time solvable for $H$-subgraph-free graphs if $H$ is a forest of maximum degree at most $3$, in which each component has at most one vertex of degree $3$. For connected graphs $H$, this means that it remains to consider when $H$ is tree of maximum degree $4$ with exactly one vertex of degree $4$, or a tree of maximum degree $3$ with at least two vertices of degree $3$. We let $H$ be a so-called subdivided "H"-graph, which is either a subdivided $\mathbb{H}_0$: a tree of maximum degree $4$ with exactly one vertex of degree $4$ and no vertices of degree $3$, or a subdivided $\mathbb{H}_1$: a tree of maximum degree $3$ with exactly two vertices of degree $3$. In the literature, only a limited number of polynomial-time and NP-completeness results for these cases are known. We develop new polynomial-time techniques that allow us to determine the complexity of Colouring on $H$-subgraph-free graphs for all the remaining subdivided "H"-graphs, so we fully classify both cases. As a consequence, the complexity of Colouring on $H$-subgraph-free graphs has now been settled for all connected graphs $H$ except when $H$ is a tree of maximum degree $4$ with exactly one vertex of degree $4$ and at least one vertex of degree $3$; or a tree of maximum degree $3$ with at least three vertices of degree $3$. We also employ our new techniques to obtain the same new polynomial-time results for another classic graph problem, namely Stable Cut.

Category
Mathematics:
Combinatorics