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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} +\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 and -connectivity properties a smallest counterexample must have. +triangle. We first reduce to triangulations, then record the degree properties a +smallest counterexample must have. \begin{lemma}[Reduction to triangulations] \label{lem:triangulate} @@ -169,78 +115,4 @@ $G$, 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}