math-research/papers/dual_decomposition_iterated_reduction/paper.tex

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\begin{document}

\title{Iterated Reduction of Dual Minimal Counterexamples}

\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{Setup and background}
\label{sec:background}

This paper is a follow-up to \emph{Dual Decomposition of Minimal
Counterexamples}~\cite{parent}, which introduced the reduced-dual
construction: given a minimal counterexample $G$ to the Four Colour
Theorem, a degree-$5$ vertex $v$ of $G$ (equivalently a pentagonal face
$F_v$ of $G' = \mathrm{dual}(G)$), and an index $i \in \{0,1,2,3,4\}$, the
\emph{reduced dual} $\widehat{G}'_{v,i}$ is the cubic plane graph obtained
from $G'$ by deleting the five boundary vertices of $F_v$, listing the
resulting five degree-$2$ vertices clockwise as $A_0,\dots,A_4$ along the
new face $F$, attaching a new apex vertex $v_n$ to $A_i, A_{i+1}, A_{i+2}$
by three new edges, and adding the chord $A_{i+3} A_{i+4}$. The four edges
added by steps~(3) and~(4) are named the \emph{side-$0$ edge}
($v_n A_i$), the \emph{spike edge} ($v_n A_{i+1}$), the \emph{side-$1$
edge} ($v_n A_{i+2}$), and the \emph{merged edge} ($A_{i+3} A_{i+4}$). The
parent paper also proves two structural lemmas about every proper
$3$-edge-colouring $\varphi$ of $\widehat{G}'_{v,i}$:
\begin{itemize}
  \item the \emph{chord-apex lemma}, asserting
        $\varphi(\mathrm{spike}) = \varphi(\mathrm{merged})$;
  \item the \emph{Kempe-cycle lemma}, asserting that
        the spike and merged edge lie on a common
        $\{\varphi(\mathrm{spike}), \varphi(\mathrm{side-}j)\}$-Kempe
        cycle through the side-$j$ edge for both $j = 0, 1$.
\end{itemize}
We refer the reader to~\cite{parent} for the precise definitions, proofs,
and the pentagonal-externals lemma we will reuse below.

\section{The iterated reduction}
\label{sec:iterated-reduction}

The reduced-dual construction can be iterated: starting from a proper
$3$-edge-colouring $\varphi_1$ of a reduced dual $\widehat{G}'_{v,i}$, we
apply the construction again to that graph at a pentagonal face whose ten
incident edges avoid the four named edges from the first reduction,
extending $\varphi_1$ across the new reduction. The protected edges
accumulate into a set $E$ that grows by four per iteration, and the
process terminates when $E$ has blocked every pentagonal face.

\begin{algorithm}[Iterated reduction with protected edges]
\label{alg:iterated-reduction}
Let $G$ be a triangulation we assume to be a minimal counterexample to the
Four Colour Theorem. The algorithm produces a sequence $H_1, H_2, \dots$ of
cubic plane graphs, proper $3$-edge-colourings $\varphi_t$ of $H_t$, and a
growing set $E$ of protected edges.

\begin{enumerate}
  \item[(0)] Form $G' := \mathrm{dual}(G)$, a cubic plane graph.

  \item[(1)] Choose a degree-$5$ vertex $v$ of $G$ (equivalently a
        pentagonal face $F_v$ of $G'$) and an index
        $i_1 \in \{0, \dots, 4\}$. Apply the reduced-dual construction
        of~\cite{parent} to form $H_1 := \widehat{G'}_{v, i_1}$, and fix
        any proper $3$-edge-colouring $\varphi_1$ of $H_1$ (one exists
        by the minimality of $G$).

  \item[(2)] Initialise
        $E := \{\text{spike}, \text{side-}0, \text{side-}1,
        \text{merged}\}$, the four named edges of the reduction in~(1).

  \item[(3)] (Iterate.) At step $t \geq 2$, given $H_{t-1}$,
        $\varphi_{t-1}$, and $E \subseteq E(H_{t-1})$:
        \begin{enumerate}
          \item[(a)] Find a pentagonal face $F$ of $H_{t-1}$ whose ten
                incident edges --- the five boundary edges of $\partial F$
                and the five external edges at $\partial F$ --- are all
                outside $E$. If no such $F$ exists, terminate.
          \item[(b)] By the pentagonal-externals lemma of~\cite{parent}
                applied to $H_{t-1}$ at $F$ under $\varphi_{t-1}$, the
                external vector has shape $(a, b, c, c, c)$ up to cyclic
                rotation. Choose an index $i_t$ for which
                $\varphi_{t-1}(f_{i_t + 3}) = \varphi_{t-1}(f_{i_t + 4})$
                and $\varphi_{t-1}(f_{i_t})$, $\varphi_{t-1}(f_{i_t + 1})$,
                $\varphi_{t-1}(f_{i_t + 2})$ are three distinct colours.
          \item[(c)] Apply the reduced-dual construction
                of~\cite{parent} to $H_{t-1}$ at $(F, i_t)$ to form $H_t$.
          \item[(d)] Extend $\varphi_{t-1}$ to a proper
                $3$-edge-colouring $\varphi_t$ of $H_t$: every surviving
                edge keeps its $\varphi_{t-1}$-colour, and each new edge
                takes the unique colour completing the palette at its
                endpoint (consistent across both endpoints of the chord
                by the choice of $i_t$).
          \item[(e)] Add the four named edges of the step-$t$ reduction
                to $E$.
        \end{enumerate}

  \item[(4)] Repeat (3) until termination.
\end{enumerate}
\end{algorithm}

\begin{remark}
\label{rem:alg-invariants}
At each iteration, $|V(H_t)| = |V(H_{t-1})| - 4$ and
$|E(H_t)| = |E(H_{t-1})| - 6$, so $H_t$ shrinks at a fixed rate; the
protected set $|E|$ grows by exactly four; and every protected edge
survives all subsequent reductions. Since the graph is finite,
termination is guaranteed. By the pentagonal-externals lemma
of~\cite{parent}, step~(b) never fails: some valid $i_t$ always exists
for any pentagonal face under any proper colouring. Termination is
therefore combinatorial: it occurs precisely when $E$ touches every
pentagonal face of $H_{t-1}$.
\end{remark}

\begin{remark}
\label{rem:alg-chord-apex}
The chord-apex lemma of~\cite{parent} applies only at $t = 1$, when $H_1$
is a reduced dual of $G'$. For $t \geq 2$, $H_t$ is a reduced dual of
$H_{t-1}$ rather than of $G'$, and $H_{t-1}$ is itself
$3$-edge-colourable, so the non-$3$-edge-colourability argument that
drives the chord-apex lemma does not carry over. Whether the constraints
accumulated in $E$ propagate any further structure to $\varphi_t$ for
$t \geq 2$ is left open.
\end{remark}

\begin{figure}[h]
\centering
\includegraphics[width=0.32\textwidth]{fig_alg_step0.png}\hfill
\includegraphics[width=0.32\textwidth]{fig_alg_step1.png}\hfill
\includegraphics[width=0.32\textwidth]{fig_alg_step2.png}
\caption{Algorithm~\ref{alg:iterated-reduction} on
$G'=\mathrm{dual}(G)$, where $G$ is the first min-degree-$5$ plantri
triangulation on $14$ vertices and $\varphi_1$ is a specific proper
$3$-edge-colouring of $H_1$ that satisfies both the chord-apex and
Kempe-cycle conditions of~\cite{parent}, found by
\texttt{experiments/search\_kempe\_property.py}. \emph{Left:} $G'$
($24$ vertices, $36$ edges) with the chosen pentagonal face shaded.
\emph{Centre:} $H_1$ ($20$ vertices, $30$ edges) after step~(1) with
$i_1 = 1$, $3$-edge-coloured by $\varphi_1$; the four edges around
$v_n^{(1)}$ in $E$ are drawn thicker, and the spike and merged edges
share the colour green. \emph{Right:} $H_2$ ($16$ vertices, $24$ edges)
after step~(3) with $i_t = 3$; eight edges are protected, and the
algorithm terminates one step later (no remaining safe pentagonal face
in $H_2$). The generating script is
\texttt{experiments/draw\_iterated\_reduction\_n14.py}; layouts are
Tutte barycentric embeddings with the outer face picked to keep
$v_n^{(1)}, v_n^{(2)}$ in the interior.}
\label{fig:iterated-reduction-trace}
\end{figure}

\section{Structural lemmas on the algorithm's output}
\label{sec:structural-lemmas}

\begin{lemma}[Exactly one matching pair in the algorithm's output]
\label{lem:exactly-one-match}
Let $G$ be a minimal counterexample to the Four Colour Theorem, and let
$(H_{t^*}, \varphi_{t^*})$ be the final graph-and-colouring produced by
some terminating execution of
Algorithm~\ref{alg:iterated-reduction} on $G$, with named pairs
$(\mathrm{spike}_t, \mathrm{merged}_t)$ for $t = 1, \dots, t^*$. Then
there is exactly one $t$ with
$\varphi_{t^*}(\mathrm{spike}_t) = \varphi_{t^*}(\mathrm{merged}_t)$, and
it is $t = 1$.
\end{lemma}

\begin{proof}
The algorithm never re-colours an existing edge: at each iteration
step~(3d) every surviving edge keeps its $\varphi_{t-1}$-colour, and the
four new edges receive fresh colours forced by propriety. Hence for every
$1 \leq k \leq t \leq t^*$,
\[
  \varphi_t(\mathrm{spike}_k) = \varphi_k(\mathrm{spike}_k),
  \qquad
  \varphi_t(\mathrm{merged}_k) = \varphi_k(\mathrm{merged}_k);
\]
the colours of the step-$k$ named edges, once written, are permanent. It
suffices to compare $\varphi_k(\mathrm{spike}_k)$ and
$\varphi_k(\mathrm{merged}_k)$ at the step where each pair is introduced.

\textbf{Case $k = 1$.} Since $G$ is a minimal counterexample, $H_1$ is a
reduced dual of $G'$. The chord-apex lemma of~\cite{parent} applied to
$\varphi_1$ gives
$\varphi_1(\mathrm{spike}_1) = \varphi_1(\mathrm{merged}_1)$.

\textbf{Case $k \geq 2$.} At step $k$ the algorithm picks an index $i_k$
for which $f_{i_k+3} = f_{i_k+4}$ (chord consistency) and
$f_{i_k}, f_{i_k+1}, f_{i_k+2}$ are pairwise distinct (propriety at the
new $v_n$), where $f$ is the external vector of the chosen pentagonal
face of $H_{k-1}$ under $\varphi_{k-1}$. Step~(3d) then assigns
\[
  \varphi_k(\mathrm{spike}_k) = f_{i_k+1},
  \qquad
  \varphi_k(\mathrm{merged}_k) = f_{i_k+3}.
\]
By the pentagonal-externals lemma of~\cite{parent}, $f$ has the $(2,2,1)$
pattern: a block of three consecutive positions $\{p, p+1, p+2\}$
(mod $5$) on which it is constantly some colour $c$, while the remaining
two positions $\{p+3, p+4\}$ hold the two non-$c$ colours, one each. The
condition $f_{i_k+3} = f_{i_k+4}$ forces $(i_k+3, i_k+4)$ to be either
$(p, p+1)$ or $(p+1, p+2)$ --- the two consecutive pairs inside the
block --- and correspondingly $i_k + 1 \in \{p+3, p+4\}$,
\emph{outside} the block. So $f_{i_k+1}$ is not $c$, whereas
$f_{i_k+3} = c$, and hence
$\varphi_k(\mathrm{spike}_k) \neq \varphi_k(\mathrm{merged}_k)$.

Combining the two cases, exactly one $t \in \{1, \dots, t^*\}$ --- namely
$t = 1$ --- has $\varphi_{t^*}(\mathrm{spike}_t) =
\varphi_{t^*}(\mathrm{merged}_t)$.
\end{proof}

\begin{lemma}[All-distinct colouring exists on a 4-colourable graph]
\label{lem:all-distinct-exists}
Let $G$ be a $4$-colourable maximal planar graph of minimum degree
$\geq 5$ (equivalently, a maximal planar graph that is \emph{not} a
minimal counterexample to the Four Colour Theorem). Then there is an
execution of Algorithm~\ref{alg:iterated-reduction} on $G$ whose final
colouring $\varphi_{t^*}$ satisfies
$\varphi_{t^*}(\mathrm{spike}_t) \neq \varphi_{t^*}(\mathrm{merged}_t)$
for every $t \in \{1, \dots, t^*\}$. In particular, there exists a proper
$3$-edge-colouring of $H_{t^*}$ under which every spike-merged pair has
distinct colours.
\end{lemma}

\begin{proof}
The argument mirrors Lemma~\ref{lem:exactly-one-match}, but extends a
colouring \emph{downward} from $G'$ rather than carrying one forward from
$H_1$.

Since $G$ is $4$-colourable, by Tait's theorem
$G' = \mathrm{dual}(G)$ admits a proper $3$-edge-colouring $\xi$. Apply
the pentagonal-externals lemma of~\cite{parent} to $\xi$ at the
pentagonal face $F_v$ selected in step~(1): the external vector
$f = (f_0, \dots, f_4)$ at $F_v$ under $\xi$ has the $(3,1,1)$
cyclic-consecutive shape, with a block of three consecutive positions
$\{p, p+1, p+2\}$ (mod $5$) holding a common colour $c$, and the
remaining two positions $\{p+3, p+4\}$ holding the two non-$c$ colours,
one each. The algorithm's choice of $i_1$ forces $\{i_1+3, i_1+4\}$
inside the $c$-block (so the chord is consistently coloured) and the
three positions $\{i_1, i_1+1, i_1+2\}$ pairwise distinct; in particular
$i_1+1$ lies \emph{outside} the $c$-block.

Choose $\varphi_1$ to be the proper $3$-edge-colouring of $H_1$ that
agrees with $\xi$ on every surviving edge and assigns each new edge at
$A_j$ the unique third colour at $A_j$. Then
$\varphi_1(\mathrm{spike}_1) = f_{i_1+1}$, a value not equal to $c$,
while $\varphi_1(\mathrm{merged}_1) = f_{i_1+3} = c$, so
$\varphi_1(\mathrm{spike}_1) \neq \varphi_1(\mathrm{merged}_1)$.

The same argument repeats at every step $k \geq 2$: the external vector
at the chosen pentagonal face under $\varphi_{k-1}$ has the $(3,1,1)$
cyclic-consecutive shape (pentagonal-externals lemma of~\cite{parent}),
the algorithm's index choice $i_k$ puts $i_k+3, i_k+4$ inside the colour
block and $i_k+1$ outside, and step~(3d) thus assigns
$\varphi_k(\mathrm{spike}_k) \neq \varphi_k(\mathrm{merged}_k)$. The
algorithm preserves these colours through every later step, so
$\varphi_{t^*}(\mathrm{spike}_t) \neq \varphi_{t^*}(\mathrm{merged}_t)$
for every $t \in \{1, \dots, t^*\}$.
\end{proof}

\begin{thebibliography}{9}
\bibitem{parent}
E.~Bauerfeld, \emph{Dual Decomposition of Minimal Counterexamples}.
Companion paper.
\end{thebibliography}

\end{document}