Add Kempe-chain structure of 4-colorings of edge-deletion subgraphs

For G_0 a minimum-order 5-chromatic maximal planar graph and any
4-coloring of G_0 - uv, the endpoints u, v must share a color, and the
color classes pairing that color with each of two other colors must
each induce a u-v path. The Kempe-chain parts follow from a standard
swap-on-component contradiction against the shared-color claim.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
2026-05-14 00:31:40 -04:00
parent f6144b98b5
commit 6c7bd9e0db
5 changed files with 42 additions and 9 deletions
@@ -318,6 +318,38 @@ its endpoints distinct colors. The edge $uv$ itself is absent from
$G_0 - uv$, so $c'$ is a proper $4$-coloring of $G_0 - uv$.
\end{proof}
\begin{theorem}\label{thm:edge-deletion-coloring-structure}
Let $G_0$ be a maximal planar graph of minimum order with
$\chi(G_0) \geq 5$, fix $uv \in E(G_0)$, and let $\varphi$ be any
proper $4$-coloring of $G_0 - uv$. Write $a = \varphi(u)$ and let
$b, c, d$ denote the three remaining colors. Then:
\begin{enumerate}
\item $\varphi(v) = a$;
\item the subgraph of $G_0 - uv$ induced by the vertices of color
$a$ or $b$ contains a path from $u$ to $v$;
\item the subgraph of $G_0 - uv$ induced by the vertices of color
$a$ or $c$ contains a path from $u$ to $v$.
\end{enumerate}
\end{theorem}
\begin{proof}
(1) If $\varphi(v) \neq a$ then $\varphi$ is already a proper
$4$-coloring of $G_0$, since the only edge of $G_0$ absent from
$G_0 - uv$ is $uv$ and its endpoints have distinct colors. This
contradicts $\chi(G_0) \geq 5$, so $\varphi(v) = a$.
(2) Suppose, for contradiction, that $u$ and $v$ lie in distinct
connected components of the subgraph of $G_0 - uv$ induced by the
color classes $a$ and $b$. Let $C$ be the component containing $u$,
and define $\varphi'\colon V(G_0) \to \{a,b,c,d\}$ by swapping colors
$a \leftrightarrow b$ on $C$ and leaving every other vertex
unchanged. Then $\varphi'$ is a proper $4$-coloring of $G_0 - uv$
with $\varphi'(u) = b$ and $\varphi'(v) = a$, contradicting part~(1)
applied to $\varphi'$.
(3) Identical to (2) with $c$ in place of $b$.
\end{proof}
\end{document}
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