face_monochromatic_pairs: 28-vertex C28 fullerene is the smallest counterexample to Conj 5.5 (face-length ≥ 5 form)

- Conjecture 5.5 marked FALSE; disproof remark rewritten to use the
  28-vertex C28 fullerene as the primary counterexample.
- Main figure swapped from the ad-hoc 40-vertex graph to the 28-vertex
  fullerene (figures/min-face-5-counterexample.png). The 40-vertex
  graph and K_4 are now mentioned only as smaller counterexamples
  outside the face-length-≥-5 class.
- Added the structural fact that face-length ≥ 5 is already the
  strongest face-length restriction admitting any cubic plane graphs
  on the sphere; face-length ≥ 6 would force 0 ≥ 12 via Euler. So no
  further face-length strengthening can save the conjecture.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

papers/face_monochromatic_pairs/paper.tex

@@ -814,7 +814,7 @@ that proof produces a clauses-(1)--(3) witness without ever needing to
 inspect the other Kempe cycle.
 \end{proof}
 
-\begin{conjecture}[Constant Heawood on two edge-sharing Kempe cycles, large-face cubic plane graphs]
+\begin{conjecture}[Constant Heawood on two edge-sharing Kempe cycles, large-face cubic plane graphs --- \textbf{FALSE}]
 \label{conj:no-two-constant-kempe-cycles}
 Let $H$ be a cubic plane graph in which every face has length at
 least $5$, with a proper $3$-edge-colouring $\varphi$. Fix a colour
@@ -826,52 +826,68 @@ share at least one colour-$a$ edge). If $h_\varphi$ is constant on
 $V(K_0)$, then $h_\varphi$ is \emph{not} constant on $V(K_1)$.
 \end{conjecture}
 
-\begin{remark}[On the face-length hypothesis]
-\label{rem:no-two-constant-kempe-face-length-hypothesis}
-Without the hypothesis that every face has length $\ge 5$, the
-conjecture is \emph{false}, with very small counterexamples:
-\begin{itemize}
-\item \textbf{$K_4$ (the tetrahedron, $n = 4$):} The standard
-proper $3$-edge-colouring has $h_\varphi$ constant (all $-1$ or
-all $+1$, depending on planar embedding orientation) on every vertex,
-and every pair of distinct Kempe cycles shares colour-edges. (All
-four faces of $K_4$ are triangles.)
-\item \textbf{An $n = 8$ cubic plane graph with girth $3$
-(graph6 \texttt{G\}GOW[}):} Found by the brute-force enumeration in
-\texttt{experiments/search\_smaller\_counterexample.py}; both Kempe
-cycles are $8$-cycles visiting every vertex, and $h_\varphi$ is
-constant on each.
-\item \textbf{An ad-hoc $n = 40$ cubic plane graph with girth $3$:}
-The graph in \texttt{constant\_heawood\_counterexample.tikz} (face
-lengths $\{3{:}2,\,4{:}4,\,5{:}10,\,6{:}1,\,7{:}1,\,8{:}2,\,9{:}1,\,
-10{:}1\}$, $|V| = 40$) has an $8$-cycle and a $12$-cycle that share a
-colour-red edge and both have $h_\varphi \equiv -1$, with $24$ of the
-$40$ vertices ($60\%$) lying outside $V(K_0) \cup V(K_1)$ --- so this
-is a structurally non-trivial counterexample, unlike $K_4$ and the
-$n = 8$ example where the two Kempe cycles already cover all
-vertices.
-\end{itemize}
-All three counterexamples have at least one face of length
-$\le 4$. The face-length-$\ge 5$ hypothesis above is the smallest
-restriction that simultaneously kills all currently known
-counterexamples; whether it suffices to make the conjecture true is
-open. The smallest cubic plane graphs satisfying this hypothesis have
-$|V(H)| \ge 20$ (with $|V| = 20$ forced to be the dodecahedron, all
-faces pentagonal); brute-force search up to some moderate $|V|$ is in
-\texttt{experiments/search\_smaller\_counterexample.py}.
+\begin{remark}[Disproof of
+Conjecture~\ref{conj:no-two-constant-kempe-cycles}]
+\label{rem:no-two-constant-kempe-cycles-counterexample}
+Conjecture~\ref{conj:no-two-constant-kempe-cycles} is \emph{false}.
+The smallest counterexample is a cubic plane graph $H$ on $28$
+vertices with $12$ pentagonal and $4$ hexagonal faces (a $C_{28}$
+fullerene). It is the planar dual of the third element (in
+\texttt{Sage}'s order) of
+\texttt{graphs.triangulations(16, minimum\_degree=5)}, with canonical
+\texttt{graph6} string
+\begin{center}
+\small\ttfamily
+[kG[A?\_A?\_?\_?K?D?@\_CO?o?@\_??A??@C??O??AG?C????`???a???W???A\_???F.
+\end{center}
+A proper $3$-edge-colouring of $H$ (colours red/blue/green,
+see Figure~\ref{fig:no-two-constant-kempe-counterexample}) makes both
+\begin{align*}
+K_{\mathrm{red},\mathrm{blue}}
+  &= \text{a $12$-cycle on $H$,} \\
+K_{\mathrm{red},\mathrm{green}}
+  &= \text{a different $12$-cycle on $H$,}
+\end{align*}
+sharing the colour-red edge $(0, 1)$ and satisfying
+$h_\varphi \equiv -1$ on the vertex set of each. Globally
+$h_\varphi$ takes value $+1$ on $4$ vertices and $-1$ on $24$; the
+four $+1$-vertices and a further four lie outside
+$V(K_0) \cup V(K_1)$, which has size $20$.
+The construction, verification, and rendering are in
+\texttt{experiments/verify\_28\_vertex\_counterexample.py}, and the
+exhaustive search that found it is in
+\texttt{experiments/search\_min\_face5\_counterexample.py}.
+
+\smallskip
+The face-length-$\ge 5$ hypothesis is in fact the strongest
+face-length hypothesis admitting any cubic plane graphs at all on the
+sphere: by Euler's formula, if every face had length $\ge 6$ then
+the sum of face lengths would satisfy $3|V(H)| \ge 6(|V(H)|/2 + 2)$,
+i.e.\ $0 \ge 12$, contradiction. So strengthening the conjecture by
+raising the minimum face length further is impossible. Without the
+face-length hypothesis there are far smaller counterexamples,
+including the tetrahedron $K_4$ at $|V| = 4$ (every Kempe cycle is a
+$4$-cycle visiting every vertex, with $h_\varphi$ constant on all of
+them by vertex-transitivity) and an $8$-vertex example (graph6
+\texttt{G\}GOW[}) found by
+\texttt{experiments/search\_smaller\_counterexample.py}; both have
+girth $3$. An ad-hoc $40$-vertex counterexample with the same
+``two intersecting Kempe cycles $\equiv -1$, large region outside''
+flavour is in
+\texttt{constant\_heawood\_counterexample.tikz}.
 \end{remark}
 
 \begin{figure}[h]
 \centering
-\includegraphics[width=0.7\textwidth]{figures/no-two-constant-kempe-counterexample.png}
-\caption{The $n = 40$ counterexample to the unrestricted version of
-Conjecture~\ref{conj:no-two-constant-kempe-cycles} (cf.\
-Remark~\ref{rem:no-two-constant-kempe-face-length-hypothesis}). Both
-the outer red/blue $8$-cycle and the red/green $12$-cycle (outer
-frame plus the upper-left ladder side) have $h_\varphi \equiv -1$ and
-share the colour-red edge $(0, 7)$. The face lengths of this drawing
-are $\{3, 4, 5, 6, 7, 8, 9, 10\}$ with two triangles, so it does not
-satisfy the face-length-$\ge 5$ hypothesis above.}
+\includegraphics[width=0.7\textwidth]{figures/min-face-5-counterexample.png}
+\caption{Smallest counterexample to
+Conjecture~\ref{conj:no-two-constant-kempe-cycles}: a $C_{28}$
+fullerene-style cubic plane graph (12 pentagons + 4 hexagons) with a
+proper $3$-edge-colouring on which $h_\varphi$ is simultaneously
+constant ($\equiv -1$) on the red/blue $12$-cycle and the red/green
+$12$-cycle, which share the colour-red edge $(0, 1)$. Light-shaded
+nodes are on $V(K_0) \cap V(K_1)$; medium-shaded on
+$V(K_0) \cup V(K_1) \setminus V(K_0) \cap V(K_1)$; grey on neither.}
 \label{fig:no-two-constant-kempe-counterexample}
 \end{figure}