math-research/papers/dual_decomposition_minimal_counterexamples/paper.tex

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\begin{document}

\title{Dual Decomposition of Minimal Counterexamples}

%    author one information
\author{Eric Bauerfeld}
\address{}
\curraddr{}
\email{}
\thanks{}

\subjclass[2010]{Primary }

\keywords{four colour theorem, plane triangulation, dual graph, cubic planar
graph, edge connectivity, cyclic edge cut, Tait colouring, $3$-edge-colouring}

\date{}

\dedicatory{}

\begin{abstract}
% TODO: abstract.
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\maketitle

\section{The minimal counterexample}

Throughout, a \emph{triangulation} is a simple plane graph, with a fixed
embedding, in which every face --- including the outer face --- is bounded by a
triangle. We first reduce to triangulations, then record the degree properties a
smallest counterexample must have.

\begin{lemma}[Reduction to triangulations]
\label{lem:triangulate}
If every triangulation is properly $4$-vertex-colourable, then so is every plane
graph.
\end{lemma}

\begin{proof}
Let $H$ be a plane graph. Add edges to $H$, maintaining planarity, until no
further edge can be added; the result is a triangulation $H^{+}$ on the same
vertex set with $E(H) \subseteq E(H^{+})$. A proper $4$-colouring of $H^{+}$
restricts to a proper $4$-colouring of $H$, since every edge of $H$ is an edge of
$H^{+}$.
\end{proof}

By Lemma~\ref{lem:triangulate}, if the Four Colour Theorem fails then it fails for
some triangulation. We may therefore make the following assumption.

\begin{definition}[Minimal counterexample]
\label{def:minimal}
Let $G$ be a triangulation on the fewest vertices that admits no proper
$4$-vertex-colouring. We call $G$ a \emph{minimal counterexample}. By minimality,
every triangulation on fewer than $|V(G)|$ vertices is properly
$4$-colourable.
\end{definition}

\begin{remark}
Since every triangulation on at most four vertices is properly $4$-colourable
(the largest being $K_4$), a minimal counterexample has $|V(G)| \ge 5$; the degree
bound below sharpens this to $|V(G)| \ge 12$.
\end{remark}

\begin{lemma}[Minimum degree]
\label{lem:mindeg}
A minimal counterexample $G$ has minimum degree $\delta(G) \ge 5$.
\end{lemma}

\begin{proof}
Suppose some vertex $v$ has $\deg(v) = d \le 4$.

If $d \le 3$, let $G' = G - v$. Then $G'$ is a plane graph on fewer vertices, so
by Definition~\ref{def:minimal} and Lemma~\ref{lem:triangulate} it has a proper
$4$-colouring. The at most three neighbours of $v$ use at most three colours, so
a fourth colour is free for $v$, extending the colouring to $G$ --- a
contradiction.

If $d = 4$, again $4$-colour $G - v$. If the four neighbours of $v$ use at most
three colours we extend as before, so assume they receive all four colours; let
$v_1, v_2, v_3, v_4$ be the neighbours in cyclic order around $v$, coloured
$1,2,3,4$. Consider the subgraph induced by the colour classes $1$ and $3$, and
let $K$ be its connected component containing $v_1$. If $v_3 \notin K$, swap
colours $1$ and $3$ on $K$; now no neighbour of $v$ is coloured $1$, freeing it
for $v$. If $v_3 \in K$, then a $1$--$3$ Kempe chain joins $v_1$ to $v_3$, and
this chain together with $v$ encloses exactly one of $v_2, v_4$; hence the
$2$--$4$ component containing $v_2$ cannot also reach $v_4$, and swapping colours
$2$ and $4$ on it frees colour $2$ for $v$. Either way the colouring extends to
$G$, a contradiction.

Hence $\delta(G) \ge 5$.
\end{proof}

\section{The reduced dual}

Write $G'$ for the dual of $G$: since $G$ is a triangulation, $G'$ is a cubic
plane graph in which each vertex of $G$ corresponds to a face of $G'$, each face
of $G$ to a vertex of $G'$, and each edge to a dual edge. A vertex of $G$ of
degree $k$ corresponds to a $k$-gonal face of $G'$.

By Lemma~\ref{lem:mindeg}, $\delta(G) \ge 5$, and Euler's formula gives
$\sum_{u \in V(G)}(6 - \deg u) = 12$, so $G$ has a vertex of degree exactly $5$
(indeed at least twelve). Fix such a vertex $v$. Its dual face $F_v$ is a
pentagon, bounded by the five dual vertices corresponding to the five faces of
$G$ incident to $v$.

\begin{definition}[Reduced dual]
\label{def:reduced-dual}
Let $v$ be a degree-$5$ vertex of $G$ with pentagonal dual face $F_v$, and fix an
index $i \in \{0,1,2,3,4\}$. The \emph{reduced dual} $\widehat{G}'_{v,i}$ is the
plane graph obtained from $G'$ as follows.
\begin{enumerate}
  \item Delete the five dual vertices on the boundary of $F_v$, together with all
        edges incident to them. Each deleted vertex is cubic, with two edges on
        $\partial F_v$ and one edge leaving $F_v$; deleting the five boundary
        vertices therefore removes the five external edges as well, dropping their
        five outer endpoints from degree $3$ to degree $2$. These five degree-$2$
        vertices lie on the boundary of a single face $F$ of the resulting graph.
  \item List the five degree-$2$ vertices in clockwise order around $F$ as
        $A = (A_0, A_1, A_2, A_3, A_4)$.
  \item Add a new vertex $v_n$ and join it to $A_i$, $A_{i+1}$, and $A_{i+2}$
        (indices mod $5$) by three new edges.
  \item Add a new edge between $A_{i+3}$ and $A_{i+4}$ (indices mod $5$).
\end{enumerate}
\end{definition}

\begin{remark}
Steps (3) and (4) restore cubicity: $A_i, A_{i+1}, A_{i+2}$ each gain one edge to
$v_n$ and $A_{i+3}, A_{i+4}$ each gain the new edge, so all five return to degree
$3$, and $v_n$ has degree $3$. Since $A_i,\dots,A_{i+2}$ and $A_{i+3}, A_{i+4}$
are each consecutive along $\partial F$, the new vertex and edge can be drawn
inside $F$ without crossings, so $\widehat{G}'_{v,i}$ is again a cubic plane
graph. The construction depends on the choice of $i$ up to the rotational
symmetry of $A$.
\end{remark}

\begin{figure}[h]
\centering
\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step1.png}\hfill
\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step2.png}\\[0.5em]
\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step3.png}\hfill
\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step4.png}
\caption{The four steps of Definition~\ref{def:reduced-dual}, illustrated on
$G' = $ the dodecahedron (dual of the icosahedron) with $F_v$ the inner
pentagon and $i = 0$. Top left: delete the five boundary vertices of $F_v$,
leaving five degree-$2$ vertices on a new face $F$. Top right: order them
clockwise as $A_0,\dots,A_4$. Bottom left: add $v_n$ joined to $A_0, A_1, A_2$.
Bottom right: add the chord $A_3 A_4$, giving the cubic plane graph
$\widehat{G}'_{v,0}$.}
\label{fig:reduced-dual-steps}
\end{figure}

\end{document}