Kempe equivalence of 4-colourings of some plane triangulations
Jan Florek
TL;DR
The paper investigates Kempe equivalence of 4-colorings on a structured family of plane triangulations $G_n$ with two poles and $2n$ degree-5 vertices. It introduces Kempe invariants $(a,b,c,d)$ for colorings and proves that, outside the constant case, two colorings are Kempe-equivalent exactly when these invariants match; it also characterizes when a coloring is constant or equivalent to a fixed coloring $Q$. The authors determine the number of Kempe classes $K^{\star}(G_n,4)$ and enumerate the total number of 4-colorings, with explicit formulas depending on $n\bmod 3$ (and a mod-$6$ refinement for $K^{\star}$). Extending the analysis to the subgraph $H_n=G_n-b$, they bound the Kempe-change distance between any two 4-colorings, giving modular-case bounds that quantify the reconfiguration complexity. These results deepen understanding of color-reconfiguration in planar triangulations with pole-induced structure and contribute precise invariants and distance bounds for Kempe moves.
Abstract
Let $G_{n}$, where $n \geqslant 5$, be a simple plane triangulation which has $2$ non-adjacent vertices of degree $n$ (called \textit{poles} of $G_n$) and $2n$ vertices of degree~$5$. A set of Kempe equivalent $4$-colourings of $G_{n}$ is called a \textit{Kempe class}. The number of Kempe classes of $G_{n}$ is enumerated. In particular it is shown that there is at least $\lfloor \frac{n}{6} \rfloor$ Kempe classes of $G_{n}$. We say that $4$-colourings $A, B$ of $G_{n}$ are \textit{equal} if there exists a permutation~$P$ of the set of colours such that $A = P \circ B$. Otherwise, $A$, $B$ are \textit{different}. The number of different $4$-colourings of $G_{n}$ is enumerated. Suppose that $H_{n} = G_{n} - b$, where $b$ is a pole of $G_{n}$. We prove that all $4$-colourings of $H_{n}$ are Kempe equivalent up to $\lfloor \frac{13n}{2} \rfloor$ Kempe changes. %$3n$ ($\lfloor \frac{9n}{2} \rfloor$ and $\lfloor \frac{13n}{2} \rfloor$) Kempe changes, for $n \equiv 0\, (mod\, 3)$ ($n \equiv 2\, (mod\, 3)$ and $n \equiv 1\, (mod\, 3)$, respectively).
