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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.
\end{abstract}

\maketitle

\section{Introduction}

% TODO: introduction.

By Tait's theorem, a plane triangulation $G$ is properly $4$-vertex-colourable if
and only if its dual cubic graph $G'$ is properly $3$-edge-colourable. We propose
to obstruct a minimal counterexample to the Four Colour Theorem by decomposing its
dual along a minimum edge cut, $3$-edge-colouring the two resulting pieces, and
recombining the colourings.

\section{Strategy}

% Outline of the intended argument. Each step below is to be made precise and
% proved in its own section; nothing here is yet established.

The argument proceeds in six steps.

\begin{enumerate}
  \item \textbf{Minimal counterexample.} Assume a smallest maximal planar graph
        $G$ admitting no proper $4$-colouring. Standard reductions let us take $G$
        to be $5$-connected, so its dual $G'$ is a cyclically $5$-edge-connected
        cubic plane graph.

  \item \textbf{Dualise.} Pass to the dual cubic plane graph $G'$, where, by
        Tait, $4$-colourability of $G$ is equivalent to $3$-edge-colourability of
        $G'$.

  \item \textbf{Minimum edge cut.} Take a smallest edge cut of $G'$ separating it
        into two components $A$ and $B$. (The size of the smallest cyclic edge cut
        equals the cyclic edge connectivity of $G'$; record which cut is used and
        why it is minimal.)

  \item \textbf{Cap to cubic.} Add the minimum number of edges (and, if needed,
        vertices) to each of $A$ and $B$ to absorb the severed cut edges, yielding
        two cubic plane graphs $A^{+}$ and $B^{+}$, each smaller than $G'$.

  \item \textbf{Colour the pieces.} Since $A^{+}$ and $B^{+}$ are smaller than the
        minimal counterexample's dual, each admits a proper $3$-edge-colouring.

  \item \textbf{Reconnect.} Prove that the two $3$-edge-colourings can be made
        compatible across the cut --- after a colour permutation and Kempe-type
        adjustments --- so that they glue to a proper $3$-edge-colouring of $G'$,
        contradicting the assumption that $G$ is a counterexample.
\end{enumerate}

\begin{remark}
Steps (1)--(5) are assembled from standard machinery (minimality, Tait duality,
edge-cut/girth duality, and the inductive existence of colourings on smaller cubic
graphs). The load-bearing claim is step (6): the compatibility of the two
boundary colourings across the cut. The parity of colours along an edge cut in a
$3$-edge-colouring is constrained, and the goal is to show these constraints can
always be satisfied simultaneously on both sides.
\end{remark}

\section{Step 1: 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 and
connectivity 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}

For the connectivity reduction we recall that, in a triangulation, the relevant
small separators are short cycles.

\begin{definition}[Separating cycle]
A cycle $C$ in a triangulation $G$ is \emph{separating} if it is not a face
boundary; equivalently, both the open interior and the open exterior of $C$ (with
respect to the embedding) contain at least one vertex of $G$.
\end{definition}

\begin{lemma}[No separating triangle]
\label{lem:no-sep-tri}
A minimal counterexample $G$ has no separating triangle; consequently $G$ is
$4$-connected.
\end{lemma}

\begin{proof}
Suppose $T = xyz$ is a separating triangle. It splits $G$ into the triangulation
$G_{\mathrm{in}}$ induced by $T$ and the vertices interior to $T$, and the
triangulation $G_{\mathrm{out}}$ induced by $T$ and the vertices exterior to $T$;
in each, $T$ bounds a face. Both are triangulations on fewer vertices than $G$,
so by minimality each has a proper $4$-colouring. Permuting colours in
$G_{\mathrm{out}}$, we may assume the two colourings agree on $\{x,y,z\}$ (the
three corners receive distinct colours in each, so a colour permutation aligns
them). The two colourings then glue to a proper $4$-colouring of $G$, a
contradiction. A triangulation with no separating triangle and $\delta \ge 5$ is
$4$-connected, since a minimal vertex cut of size $\le 3$ in a triangulation is
the vertex set of a separating cycle of that length, and a $3$-cut would give a
separating triangle.
\end{proof}

\begin{proposition}[$5$-connectivity]
\label{prop:5conn}
A minimal counterexample $G$ may be taken to be $5$-connected; equivalently, $G$
has no separating triangle and no separating $4$-cycle.
\end{proposition}

\begin{proof}[Discussion]
By Lemma~\ref{lem:no-sep-tri} it remains to eliminate separating $4$-cycles. This
is the classical reducibility step of Birkhoff: a shortest separating $4$-cycle
$C = w x y z$ bounds inner and outer triangulations; one $4$-colours both by
minimality and reconciles their colourings on $C$ using Kempe-chain exchanges,
the only obstruction being the two ways the four corners can be coloured with
three or four colours. Carrying out the exchanges shows the colourings can always
be aligned, so $G$ has no separating $4$-cycle. As with
Lemma~\ref{lem:no-sep-tri}, the absence of separating $3$- and $4$-cycles in a
triangulation with $\delta \ge 5$ is equivalent to $5$-connectivity. We take this
reduction as standard and refer to the literature on the Four Colour Theorem for
the detailed Kempe-chain bookkeeping.
\end{proof}

The output of this step --- a $5$-connected triangulation $G$ with $\delta(G) \ge
5$ and no separating triangle or $4$-cycle --- is exactly the hypothesis the dual
construction of Step~2 consumes.

\section{Step 2: The dual cubic graph}

% TODO.

\section{Step 3: The minimum edge cut}

% TODO.

\section{Step 4: Capping to cubic planar graphs}

% TODO.

\section{Step 5: Colouring the pieces}

% TODO.

\section{Step 6: Reconnecting the colourings}

% TODO. (The crux.)

\end{document}
