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Drop edge-deletion sections from flip-symmetry paper

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Removes the "Further necessary properties of a minimal counterexample"
framing section and the "Edge-deletion subgraphs" section (definition,
4-colorability theorem, Kempe-chain structure theorem). The intended
empirical follow-up on this material did not produce a useful
discriminator, so the development is being shelved.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
didericis
2026-05-14 00:55:23 -04:00
parent 6c7bd9e0db
commit 389fd56f07
6 changed files with 26 additions and 111 deletions
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papers/flip_symmetric_maximal_planar_graphs/paper.aux
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papers/flip_symmetric_maximal_planar_graphs/paper.tex
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@@ -276,80 +276,6 @@ the minimum-degree-$5$ class --- which already contains every
candidate minimum-order $5$-chromatic graph --- flip-symmetric
examples become a vanishing fraction.
\section{Further necessary properties of a minimal counterexample}
The frequency data of Section~\ref{sec:frequency} look unflattering
only when flip-symmetry is weighed against the full class of maximal
planar graphs. The class that actually matters --- minimum-order
$5$-chromatic triangulations that also resist every Kempe-style
reduction --- is far thinner, and flip-symmetry may exclude a
substantially larger fraction of it if the configurations it removes
overlap those responsible for Kempe reducibility. We therefore turn
to identifying further necessary properties of a minimum-order
$5$-chromatic maximal planar graph, of which flip-asymmetry is the
first.
\section{Edge-deletion subgraphs}
\begin{definition}[Edge-deletion subgraph]\label{def:edge-deletion}
Let $G$ be a maximal planar graph and $uv \in E(G)$. The
\emph{edge-deletion subgraph at $uv$} is the spanning subgraph
$G - uv = (V(G),\,E(G) \setminus \{uv\})$. Write
$\mathcal{D}(G) = \{G - uv : uv \in E(G)\}$.
\end{definition}
\begin{theorem}\label{thm:edge-deletion-4colorable}
Let $G_0$ be a maximal planar graph of minimum order with
$\chi(G_0) \geq 5$. Then every $H \in \mathcal{D}(G_0)$ is
$4$-colorable.
\end{theorem}
\begin{proof}
Fix $uv \in E(G_0)$ and let $G_0 / uv$ denote the simple planar graph
obtained by contracting $uv$ and discarding parallel edges. Since
$|V(G_0/uv)| = |V(G_0)| - 1$, the minimality of $G_0$ supplies a
proper $4$-coloring $c$ of $G_0 / uv$. Let $z$ be the contracted
vertex and define $c'\colon V(G_0) \to \{1,2,3,4\}$ by
$c'(u) = c'(v) = c(z)$ and $c'(y) = c(y)$ for $y \notin \{u, v\}$.
Every edge of $G_0 - uv$ is either disjoint from $\{u, v\}$ or
incident to exactly one of them; in either case the corresponding
edge of $G_0 / uv$ has distinct endpoints under $c$, so $c'$ assigns
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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