didericis
bd8526eb11
Remove the Introduction and Strategy sections and everything after the separating-cycle definition (no-separating-triangle lemma, 5-connectivity proposition, and the Step 2-6 stubs). Rename the section heading from "Step 1: The minimal counterexample" to "The minimal counterexample", drop the now-unused separating-cycle definition, and adjust the lead-in to mention only the degree reduction. Remaining: reduction-to-triangulations lemma, minimal-counterexample definition, |V|>=12 remark, and minimum-degree-5 lemma. Co-Authored-By: Claude Opus 4.7 (1M context) <noreply@anthropic.com>
119 lines
3.9 KiB
TeX
119 lines
3.9 KiB
TeX
%% filename: amsart-template.tex
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%% American Mathematical Society
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%% ====================================================================
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\documentclass{amsart}
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\usepackage{amssymb}
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\usepackage{graphicx}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{corollary}[theorem]{Corollary}
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\newtheorem{proposition}[theorem]{Proposition}
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\begin{document}
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\title{Dual Decomposition of Minimal Counterexamples}
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% author one information
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\author{Eric Bauerfeld}
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\address{}
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\curraddr{}
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\email{}
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\thanks{}
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\subjclass[2010]{Primary }
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\keywords{four colour theorem, plane triangulation, dual graph, cubic planar
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graph, edge connectivity, cyclic edge cut, Tait colouring, $3$-edge-colouring}
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\date{}
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\dedicatory{}
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\begin{abstract}
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% TODO: abstract.
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\end{abstract}
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\maketitle
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\section{The minimal counterexample}
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Throughout, a \emph{triangulation} is a simple plane graph, with a fixed
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embedding, in which every face --- including the outer face --- is bounded by a
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triangle. We first reduce to triangulations, then record the degree properties a
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smallest counterexample must have.
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\begin{lemma}[Reduction to triangulations]
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\label{lem:triangulate}
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If every triangulation is properly $4$-vertex-colourable, then so is every plane
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graph.
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\end{lemma}
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\begin{proof}
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Let $H$ be a plane graph. Add edges to $H$, maintaining planarity, until no
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further edge can be added; the result is a triangulation $H^{+}$ on the same
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vertex set with $E(H) \subseteq E(H^{+})$. A proper $4$-colouring of $H^{+}$
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restricts to a proper $4$-colouring of $H$, since every edge of $H$ is an edge of
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$H^{+}$.
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\end{proof}
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By Lemma~\ref{lem:triangulate}, if the Four Colour Theorem fails then it fails for
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some triangulation. We may therefore make the following assumption.
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\begin{definition}[Minimal counterexample]
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\label{def:minimal}
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Let $G$ be a triangulation on the fewest vertices that admits no proper
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$4$-vertex-colouring. We call $G$ a \emph{minimal counterexample}. By minimality,
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every triangulation on fewer than $|V(G)|$ vertices is properly
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$4$-colourable.
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\end{definition}
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\begin{remark}
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Since every triangulation on at most four vertices is properly $4$-colourable
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(the largest being $K_4$), a minimal counterexample has $|V(G)| \ge 5$; the degree
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bound below sharpens this to $|V(G)| \ge 12$.
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\end{remark}
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\begin{lemma}[Minimum degree]
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\label{lem:mindeg}
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A minimal counterexample $G$ has minimum degree $\delta(G) \ge 5$.
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\end{lemma}
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\begin{proof}
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Suppose some vertex $v$ has $\deg(v) = d \le 4$.
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If $d \le 3$, let $G' = G - v$. Then $G'$ is a plane graph on fewer vertices, so
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by Definition~\ref{def:minimal} and Lemma~\ref{lem:triangulate} it has a proper
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$4$-colouring. The at most three neighbours of $v$ use at most three colours, so
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a fourth colour is free for $v$, extending the colouring to $G$ --- a
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contradiction.
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If $d = 4$, again $4$-colour $G - v$. If the four neighbours of $v$ use at most
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three colours we extend as before, so assume they receive all four colours; let
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$v_1, v_2, v_3, v_4$ be the neighbours in cyclic order around $v$, coloured
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$1,2,3,4$. Consider the subgraph induced by the colour classes $1$ and $3$, and
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let $K$ be its connected component containing $v_1$. If $v_3 \notin K$, swap
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colours $1$ and $3$ on $K$; now no neighbour of $v$ is coloured $1$, freeing it
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for $v$. If $v_3 \in K$, then a $1$--$3$ Kempe chain joins $v_1$ to $v_3$, and
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this chain together with $v$ encloses exactly one of $v_2, v_4$; hence the
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$2$--$4$ component containing $v_2$ cannot also reach $v_4$, and swapping colours
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$2$ and $4$ on it frees colour $2$ for $v$. Either way the colouring extends to
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$G$, a contradiction.
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Hence $\delta(G) \ge 5$.
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\end{proof}
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\end{document}
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