face_monochromatic_pairs: strengthen Conj 5.5 to face-length ≥ 5, find 28-vertex counterexample

- Paper: Conjecture 5.5 restated with the hypothesis that every face
  of H has length ≥ 5 (= the cubic plane analogue of "no triangles
  or quadrilaterals as faces"). This kills K_4 and the n=8 trivial
  counterexamples (girth 3) and the ad-hoc n=40 counterexample
  (which has 2 triangles and 4 quadrilaterals). A new remark catalogues
  these excluded counterexamples and the smallest cubic plane graphs
  satisfying the hypothesis (dodecahedron at |V| = 20).
- search_smaller_counterexample.py: --min-face=N option to filter
  cubic planar graphs by minimum face length.
- search_min_face5_counterexample.py: enumerates triangulations T
  with min degree ≥ 5 via graphs.triangulations(n, minimum_degree=5),
  takes planar dual (= cubic plane with all faces ≥ 5), and runs the
  Heawood-constancy check.
- Result: smallest counterexample at triangulation order n_T = 16,
  whose dual is a 28-vertex cubic plane graph (graph6
  [kG[A?_A?_?_?K?D?@_CO?o?@_??A??@C??O??AG?C????`???a???W???A_???F).
  Faces: 12 pentagons + 4 hexagons (a C28 fullerene). Both
  K_{red, blue} and K_{red, green} are 12-cycles sharing the
  colour-red edge (0, 1) and both have h_φ ≡ -1. 8 of 28 vertices
  lie outside V(K_0) ∪ V(K_1).
- verify_28_vertex_counterexample.py: reproduces the counterexample,
  verifies all properties, and renders figures/min-face-5-counterexample.png.

Note on the boundary: face-length ≥ 6 is impossible for cubic plane
graphs by Euler (6F = 6(V/2 + 2) > 3V = sum face lengths for V > 4).
So face-length ≥ 5 is the strongest face-length restriction admitting
any cubic plane graphs at all.

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

papers/face_monochromatic_pairs/paper.tex

@@ -814,65 +814,64 @@ 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 --- \textbf{FALSE}]
+\begin{conjecture}[Constant Heawood on two edge-sharing Kempe cycles, large-face cubic plane graphs]
 \label{conj:no-two-constant-kempe-cycles}
-Let $H$ be a cubic plane graph with a proper $3$-edge-colouring
-$\varphi$, fix a colour $a \in \{1, 2, 3\}$, and let $\{b, c\} =
-\{1, 2, 3\} \setminus \{a\}$. Let $K_0$ be an $\{a, b\}$-Kempe cycle
-of $\varphi$ and $K_1$ an $\{a, c\}$-Kempe cycle of $\varphi$ such that
+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
+$a \in \{1, 2, 3\}$ and let $\{b, c\} = \{1, 2, 3\} \setminus \{a\}$.
+Let $K_0$ be an $\{a, b\}$-Kempe cycle of $\varphi$ and $K_1$ an
+$\{a, c\}$-Kempe cycle of $\varphi$ such that
 $E(K_0) \cap E(K_1) \neq \emptyset$ (equivalently, $K_0$ and $K_1$
 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}[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}.
-Figure~\ref{fig:no-two-constant-kempe-counterexample} exhibits a
-concrete counterexample: a cubic plane graph $H$ on $40$ vertices with
-a proper $3$-edge-colouring $\varphi$ (colours red/blue/green) in
-which both
-\begin{align*}
-K_{\mathrm{red},\mathrm{blue}}
-  &= \text{the outer $8$-cycle, and} \\
-K_{\mathrm{red},\mathrm{green}}
-  &= \text{the $12$-cycle (outer frame $+$ upper-left ``ladder'' side)}
-\end{align*}
-share the colour-red edge $(0, 7)$ and satisfy
-$h_\varphi \equiv -1$ on the vertex set of each. Globally $h_\varphi$
-takes value $+1$ on $16$ vertices and $-1$ on $24$ vertices, with all
-of the $+1$-vertices concentrated in the inner ``tilted ladder''
-region, so that both Kempe cycles miss them entirely.
-
-\smallskip
-The underlying graph appears to be a fresh ad-hoc construction: it has
-$60$ edges, vertex- and edge-connectivity $3$, girth $3$ (two
-triangles), is Hamiltonian, is not bipartite, has trivial
-automorphism group, and its $22$ faces have lengths distributed
-as $\{3{:}2,\,4{:}4,\,5{:}10,\,6{:}1,\,7{:}1,\,8{:}2,\,9{:}1,\,10{:}1\}$.
-Its canonical \texttt{graph6} string (via
-\texttt{G.canonical\_label().graph6\_string()}) is
-\begin{center}
-\small\ttfamily
-ggE\_?C?O\_B?B?@C?P???B??\_?@???\_??Sa???W????@a??\_B????G\_???@?????O??@\_???@?\_???K?????\_????K?????o???B?????AG?????B?????G?????AG?????F
-\end{center}
-The construction, the proper $3$-edge-colouring, Kempe-cycle tracing,
-and the Heawood-number computation (via the CW rotation at each
-vertex) are all in \texttt{experiments/counterexample\_conj\_5\_5.py};
-the source drawing is in \texttt{constant\_heawood\_counterexample.tikz}.
+\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}.
 \end{remark}
 
 \begin{figure}[h]
 \centering
-\includegraphics[width=0.85\textwidth]{figures/no-two-constant-kempe-counterexample.png}
-\caption{Counterexample to
-Conjecture~\ref{conj:no-two-constant-kempe-cycles}: a cubic plane
-graph on $40$ vertices with a proper $3$-edge-colouring on which
-$h_\varphi$ is simultaneously constant ($\equiv -1$) on the outer
-red/blue $8$-cycle and on the red/green $12$-cycle (outer frame plus
-the upper-left ladder side), which share the colour-red edge
-$(0, 7)$.}
+\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.}
 \label{fig:no-two-constant-kempe-counterexample}
 \end{figure}