Kempe changes in $H$-free graphs

TL;DR

This work characterizes when all $H$-free graphs are Kempe connected, proving the dichotomy that this holds iff $H$ is an induced subgraph of $P_4$. It combines constructive counterexamples showing non-Kempe connectivity in certain hereditary classes (notably $2K_2$-free graphs) with a tight sufficiency argument for $P_4$-free graphs, anchored by Kempe frozen colourings. The authors introduce families of $2K_2$-free graphs with Kempe frozen colourings and extend an augmentation operation to preserve $2K_2$-freeness while increasing the colour count, demonstrating that there is no universal colour-extension bound ensuring Kempe class membership. The results delineate the boundary between Kempe connectivity and reconfiguration-like properties in hereditary graph classes, offering precise guidance for recolouring analysis in constrained graph families.

Abstract

Given a $k$-colouring of a graph $G$ and two of the colours, a $Kempe$ $chain$ is a connected component of the subgraph of $G$ induced by the vertices coloured with one of these two colours. A $Kempe$ $swap$ changes one colouring into another by interchanging the colours of the vertices in a Kempe chain. Two colourings are $Kempe$ $equivalent$ if each can be obtained from the other by a series of Kempe swaps; the set of Kempe equivalent colourings is called a $Kempe$ $class$. For a graph $G$, let $χ(G)$ denote its chromatic number and let $\mathcal{C}_{k}(G)$ denote the set of all $k$-colourings of $G$. We say $G$ is $Kempe$ $connected$ if for all $k\ge χ(G)$, $\mathcal{C}_{k}(G)$ forms a Kempe class. For a graph $H$, graph $G$ is called $H$-$free$ if no induced subgraph of $G$ is isomorphic to $H$. We prove that every $H$-free graph is Kempe connected if and only if $H$ is an induced subgraph of the path on four vertices, $P_4$. The graph 2$K_2$ consists of four vertices and two edges which are not adjacent. We prove that for all $p\ge 0$, there is a $k$-colourable 2$K_2$-free graph $G$ such that $\mathcal{C}_{k+p}(G)$ does not form a Kempe class.

Figures (5)

• Figure 1: Two 3-colourings of a triangle-free graph which are not Kempe equivalent
• Figure 2: Two 3-colourings of a $C_4$-free graph which are not Kempe equivalent
• Figure 3: Two $3$-colourings of the triangular prism which are not Kempe equivalent Heuvel2013.
• Figure 4: A $C_4$-free graph $\overline{D_2}$ with a 4-clique-partition (left) and a Kempe frozen 5-clique-partition (right). The numbers indicate which clique a vertex is in. Equivalently, the numbers indicate a 4-colouring of the complement $D_2$ of the graph shown (left) and a Kempe frozen 5-colouring of $D_2$ (right).
• Figure 6: A $C_4$-free graph $\overline{Y_2}$ with a 4-clique-partition (left) and a Kempe frozen 6-clique-partition (right). Equivalently, a 4-colouring of the complement $Y_2$ (left) and a Kempe frozen 6-colouring (right).

Theorems & Definitions (29)

• Theorem 1: BonamyCographsKempeConn
• Theorem 2
• Theorem 3
• Theorem 4: BCM
• Lemma 1
• proof
• Lemma 2
• proof
• Lemma 3
• proof