Colouring Square-Free Graphs without Long Induced Paths

The complexity of {\sc Colouring} is fully understood for $H$-free graphs, but there are still major complexity gaps if two induced subgraphs $H_1$ and $H_2$ are forbidden. Let $H_1$ be the $s$-vertex cycle $C_s$ and $H_2$ be the $t$-vertex path $P_t$. We show that {\sc Colouring} is polynomial-time solvable for $s=4$ and $t\leq 6$, strengthening several known results. Our main approach is to initiate a study into the boundedness of the clique-width of atoms (graphs with no clique cutset) of a hereditary graph class. We first show that the classifications of boundedness of clique-width of $H$-free graphs and $H$-free atoms coincide. We then show that this is not the case if two graphs are forbidden: we prove that $(C_4,P_6)$-free atoms have clique-width at most~18. Our key proof ingredients are a divide-and-conquer approach for bounding the clique-width of a subclass of $C_4$-free graphs and the construction of a new bound on the clique-width for (general) graphs in terms of the clique-width of recursively defined subgraphs induced by homogeneous pairs and triples of sets. As a complementary result we prove that {\sc Colouring} is \NP-complete for $s=4$ and $t\geq 9$, which is the first hardness result on {\sc Colouring} for $(C_4,P_t)$-free graphs. Combining our new results with known results leads to an almost complete dichotomy for \cn restricted to $(C_s,P_t)$-free graphs.