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.
\end{abstract}

\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}

\begin{definition}[Edges of the reduced dual]
\label{def:edge-names}
The four edges added in steps (3) and (4) of Definition~\ref{def:reduced-dual}
are named as follows. The chord $A_{i+3}A_{i+4}$ is the \emph{merged edge}; the
edge $A_{i+1}v_n$ is the \emph{spike edge}; and the edges $A_iv_n$ and
$A_{i+2}v_n$ are the \emph{side edges}. In the $i = 0$ case of
Figure~\ref{fig:reduced-dual-steps} these are $\{A_3, A_4\}$, $\{A_1, v_n\}$,
and $\{A_0, v_n\}, \{A_2, v_n\}$ respectively.
\end{definition}

We will use the following structural fact about proper $3$-edge-colourings near
a pentagonal face of a cubic plane graph; it is stated for a generic such graph
$H$, not specifically for the reduced dual.

\begin{lemma}[Pentagonal externals]
\label{lem:pentagonal-externals}
Let $H$ be a cubic plane graph and $F$ a pentagonal face of $H$, with
$\partial F$ traversed clockwise as $u_0, u_1, u_2, u_3, u_4$. For each $i$
let $f_i$ be the unique edge of $H$ incident to $u_i$ that does not lie on
$\partial F$. An assignment $\varphi$ of colours from $\{1, 2, 3\}$ to the ten
edges incident to $\{u_0, \dots, u_4\}$ is proper at every $u_i$ if and only if
there is some index $j$ such that
\[
  \varphi(f_j) = \varphi(f_{j+1}) = \varphi(f_{j+2})
  \quad\text{and}\quad
  \{\varphi(f_{j+3}), \varphi(f_{j+4})\}
    = \{1, 2, 3\} \setminus \{\varphi(f_j)\},
\]
indices mod $5$.
\end{lemma}

\begin{proof}
Write $e_i = u_i u_{i+1}$ for the boundary edges of $\partial F$ (indices mod
$5$). A colouring $\varphi$ is proper at every $u_i$ if and only if at each
$u_i$ the three incident edges $e_{i-1}, e_i, f_i$ receive three distinct
colours; whenever this holds, $\varphi(f_i)$ is forced to be the unique colour
in $\{1, 2, 3\} \setminus \{\varphi(e_{i-1}), \varphi(e_i)\}$, and $\varphi$
restricts to a proper $3$-edge-colouring of the cycle $\partial F$.

\textbf{($\Rightarrow$)} The line graph of $\partial F$ is $C_5$, whose
maximum independent set has size $2$, so no colour appears more than twice on
$\partial F$; and since $\partial F$ is an odd cycle, all three colours appear.
The colour multiset on $(\varphi(e_0), \dots, \varphi(e_4))$ is therefore
$(2, 2, 1)$, with the singleton at a unique position. Cyclically shifting
indices we may place this position at $0$; let $c$ be the singleton colour.
The remaining four edges form the path $e_1 e_2 e_3 e_4$, which by propriety
alternates between the other two colours, so for some labelling
$\{a, b, c\} = \{1, 2, 3\}$,
\[
  (\varphi(e_0), \varphi(e_1), \varphi(e_2), \varphi(e_3), \varphi(e_4))
  = (c, a, b, a, b).
\]
Reading off the forced values of $\varphi(f_i)$,
\[
  \varphi(f_0) = a, \quad
  \varphi(f_1) = b, \quad
  \varphi(f_2) = \varphi(f_3) = \varphi(f_4) = c,
\]
which is the lemma's pattern at $j = 2$ (the cyclic shift maps this back to
the corresponding $j$ in the original indexing). This case is the unique
proper $3$-edge-colouring of $\partial F$ up to cyclic shift and permutation of
$\{1, 2, 3\}$ (since $5 \cdot 3! = 30 = P(C_5, 3)$, the chromatic polynomial of
$C_5$ at $3$), so it exhausts every proper $\varphi$.

\textbf{($\Leftarrow$)} The lemma's hypothesis is invariant under cyclic
shifts of indices and under permutations of $\{1, 2, 3\}$, so we may assume
$j = 2$, $\varphi(f_2) = \varphi(f_3) = \varphi(f_4) = c$, $\varphi(f_0) = a$,
and $\varphi(f_1) = b$, with $\{a, b, c\} = \{1, 2, 3\}$. Propriety at $u_i$
and $u_{i+1}$ requires $\varphi(e_i) \notin \{\varphi(f_i), \varphi(f_{i+1})\}$,
which gives
\[
  \varphi(e_0) = c, \quad
  \varphi(e_1) = a, \quad
  \varphi(e_2) \in \{a, b\}, \quad
  \varphi(e_3) \in \{a, b\}, \quad
  \varphi(e_4) = b.
\]
The remaining propriety condition $\varphi(e_{i-1}) \neq \varphi(e_i)$ holds
automatically at $u_0, u_1, u_4$, forces $\varphi(e_2) = b$ at $u_2$, and then
forces $\varphi(e_3) = a$ at $u_3$. The resulting triples
$(\varphi(e_{i-1}), \varphi(e_i), \varphi(f_i))$ at $u_0, u_1, u_2, u_3, u_4$
are
\[
  (b, c, a), \quad (c, a, b), \quad (a, b, c), \quad (b, a, c), \quad (a, b, c),
\]
each a permutation of $\{1, 2, 3\}$, so $\varphi$ is proper at every $u_i$.
\end{proof}

\begin{remark}
The two-element condition $\{\varphi(f_{j+3}), \varphi(f_{j+4})\}
= \{1,2,3\}\setminus\{\varphi(f_j)\}$ cannot be dropped: a 3-colouring
satisfying $\varphi(f_j) = \varphi(f_{j+1}) = \varphi(f_{j+2})$ alone need not
extend, e.g.\ $(1,1,1,1,2)$.
\end{remark}

Since $\widehat{G}'_{v,i}$ is the dual of a triangulation on fewer vertices than
$G$, it is $3$-edge-colourable by the minimality of $G$. The following lemma
constrains every such colouring.

\begin{lemma}
\label{lem:chord-apex}
Let $G$ be a minimal counterexample, and let $\widehat{G}'_{v,i}$ be a reduced
dual of its dual $G'$. Then in every proper $3$-edge-colouring of
$\widehat{G}'_{v,i}$, the merged edge and the spike edge receive the same
colour.
\end{lemma}

\begin{figure}[h]
\centering
\includegraphics[width=0.7\textwidth]{fig_chord_apex_step1.png}\\[0.4em]
\includegraphics[width=0.48\textwidth]{fig_chord_apex_step2.png}\hfill
\includegraphics[width=0.48\textwidth]{fig_chord_apex_step3.png}
\caption{The proof of Lemma~\ref{lem:chord-apex}, illustrated for $i = 0$ on
$G' = $ the dodecahedron. Top: under the assumption $W \neq Y$, propriety at
$v_n$ forces $W \in \{X, Z\}$. Bottom: in either case the lift to $G'$ has
externals satisfying the hypothesis of
Lemma~\ref{lem:pentagonal-externals}, which colours $\partial F_v$ to extend
$\psi$ to a proper $3$-edge-colouring of $G'$.}
\label{fig:chord-apex-proof}
\end{figure}

\begin{proof}
After cyclically relabelling, assume $i = 0$. Suppose for contradiction that
$\varphi$ is a proper $3$-edge-colouring of $\widehat{G}'_{v,0}$ in which the
merged edge $\{A_3, A_4\}$ and the spike edge $\{A_1, v_n\}$ receive different
colours (Figure~\ref{fig:chord-apex-proof}, top), and write
\[
  X = \varphi(\{A_0, v_n\}), \quad
  Y = \varphi(\{A_1, v_n\}), \quad
  Z = \varphi(\{A_2, v_n\}), \quad
  W = \varphi(\{A_3, A_4\}).
\]
Propriety of $\varphi$ at $v_n$ forces $\{X, Y, Z\} = \{1, 2, 3\}$, and the
assumption $W \neq Y$ leaves $W \in \{X, Z\}$.

We lift $\varphi$ to a colouring $\psi$ of $E(G')$ as follows. Let
$B_0, \dots, B_4$ be the boundary vertices of $\partial F_v$ in $G'$, indexed
so that $f_k = B_k A_k$. On every edge of $G'$ that survived the reduction,
set $\psi = \varphi$. At each $A_k$ the two surviving edges retain their
$\varphi$-colours, so the remaining edge at $A_k$ --- in $G'$ this is the
external $f_k$; in $\widehat{G}'_{v,0}$ this is a $v_n$-edge ($k \in
\{0, 1, 2\}$) or the chord ($k \in \{3, 4\}$) --- is forced to take the third
colour at $A_k$. Since the two-surviving-edge colours at $A_k$ agree in $G'$
and $\widehat{G}'_{v,0}$, the third colour does too, giving
\[
  \psi(f_0) = X, \quad \psi(f_1) = Y, \quad \psi(f_2) = Z, \quad
  \psi(f_3) = \psi(f_4) = W
\]
(the last two equalities holding because the chord is a single edge contributing
its colour at each of $A_3$ and $A_4$).

It remains to assign colours to the five boundary edges $B_k B_{k+1}$ of
$\partial F_v$. Apply Lemma~\ref{lem:pentagonal-externals} to $G'$ at the face
$F_v$ with the $B_k$'s as its boundary vertices and the same indexing. The
external vector $(\psi(f_0), \dots, \psi(f_4)) = (X, Y, Z, W, W)$ falls into
one of two cases (Figure~\ref{fig:chord-apex-proof}, bottom):
\begin{itemize}
  \item if $W = Z$, it is $(X, Y, Z, Z, Z)$: three consecutive $Z$'s at
        positions $2, 3, 4$, with $\{X, Y\} = \{1, 2, 3\} \setminus \{Z\}$;
  \item if $W = X$, it is $(X, Y, Z, X, X)$: three consecutive $X$'s at
        positions $3, 4, 0$, with $\{Y, Z\} = \{1, 2, 3\} \setminus \{X\}$.
\end{itemize}
Each case satisfies the hypothesis of Lemma~\ref{lem:pentagonal-externals};
its $(\Leftarrow)$ direction therefore assigns colours to the boundary edges
$B_k B_{k+1}$ that make $\psi$ proper at every $B_k$.

The resulting $\psi$ is a proper $3$-edge-colouring of $G'$: proper at every
$B_k$ by the lemma, at every $A_k$ by the construction, and at every other
vertex because such a vertex has the same neighbourhood in $G'$ as in
$\widehat{G}'_{v,0}$ with the same incident-edge colours. By Tait's theorem,
$G'$ is $3$-edge-colourable iff $G$ is $4$-vertex-colourable, contradicting
that $G$ is a counterexample. The assumption $W \neq Y$ is therefore false.
\end{proof}

\end{document}