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Replace counterexample with minimal n=13 graph found by exhaustive search

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Adds search_counterexample_comprehensive iterating Sage's planar_graphs
generator across all maximal planar graphs of bounded order. Exhaustive
enumeration through order 13 (9150+49566 triangulations) yields exactly
one graph with no plane diamond coloring, at order 13. Updates Theorem
2.6 to assert minimality and uniqueness, and replaces the figure and
edge list with the smaller counterexample.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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didericis
2026-05-09 12:52:30 -04:00
parent 2ae729db1e
commit 88031c9d73
7 changed files with 74 additions and 58 deletions
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@@ -151,35 +151,36 @@ Every maximal planar graph $G$ has a plane diamond coloring.
\end{conjecture} \end{conjecture}
\begin{theorem} \begin{theorem}
Conjecture 2.5 is false. Conjecture 2.5 is false. Moreover, the smallest counterexample has order $13$, and is unique up to isomorphism among triangulations of order at most $13$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
Let $G$ be the maximal planar graph on $16$ vertices with graph6 string\footnote{We use the standard graph6 encoding of McKay; see \cite{mckaygraph6}.} Let $G$ be the maximal planar graph on $13$ vertices with graph6 string\footnote{We use the standard graph6 encoding of McKay; see \cite{mckaygraph6}.}
\[ \[
\texttt{O???IAGKEBEQQYHdplW\{n} \texttt{LQhO?SFIwnTYTi}
\] \]
shown in Figure~\ref{fig:counterexample}. Equivalently, $G$ has edge set shown in Figure~\ref{fig:counterexample}. Equivalently, $G$ has edge set
\begin{align*} \begin{align*}
&\{0,8\}, \{0,10\}, \{0,14\}, \{1,7\}, \{1,10\}, \{1,12\}, \{2,9\}, \{2,11\}, \{2,13\}, \\ &\{0,2\}, \{0,4\}, \{0,11\}, \{1,3\}, \{1,5\}, \{1,12\}, \{2,4\}, \{2,9\}, \{2,11\}, \\
&\{3,9\}, \{3,11\}, \{3,15\}, \{4,8\}, \{4,12\}, \{4,14\}, \{4,15\}, \{5,6\}, \{5,13\}, \\ &\{3,5\}, \{3,10\}, \{3,12\}, \{4,7\}, \{4,9\}, \{4,11\}, \{5,8\}, \{5,10\}, \{5,12\}, \\
&\{5,14\}, \{5,15\}, \{6,11\}, \{6,13\}, \{6,15\}, \{7,10\}, \{7,12\}, \{7,14\}, \{8,10\}, \\ &\{6,7\}, \{6,8\}, \{6,9\}, \{6,10\}, \{6,11\}, \{6,12\}, \{7,8\}, \{7,9\}, \{7,10\}, \\
&\{8,12\}, \{8,14\}, \{9,11\}, \{9,13\}, \{9,15\}, \{10,12\}, \{10,14\}, \{11,13\}, \\ &\{7,11\}, \{8,9\}, \{8,10\}, \{8,12\}, \{9,11\}, \{10,12\}.
&\{11,15\}, \{12,13\}, \{12,14\}, \{12,15\}, \{13,14\}, \{13,15\}, \{14,15\}.
\end{align*} \end{align*}
We have $|V(G)| = 16$ and $|E(G)| = 42 = 3 \cdot 16 - 6$, so $G$ is a triangulation. We have $|V(G)| = 13$ and $|E(G)| = 33 = 3 \cdot 13 - 6$, so $G$ is a triangulation.
By Proposition 2.4, it suffices to verify that for every choice of root $u \in V(G)$ and every pair of distinct colors $c_a, c_b$, no proper $4$-coloring $C$ of $G$ satisfies $C^{-1}(c_a) \subseteq \bigcup_{i \text{ even}} L_i^{(u)}$ and $C^{-1}(c_b) \subseteq \bigcup_{i \text{ odd}} L_i^{(u)}$, where $\{L_i^{(u)}\}$ denotes the distance partition from $u$. By Proposition 2.4, it suffices to verify that for every choice of root $u \in V(G)$ and every pair of distinct colors $c_a, c_b$, no proper $4$-coloring $C$ of $G$ satisfies $C^{-1}(c_a) \subseteq \bigcup_{i \text{ even}} L_i^{(u)}$ and $C^{-1}(c_b) \subseteq \bigcup_{i \text{ odd}} L_i^{(u)}$, where $\{L_i^{(u)}\}$ denotes the distance partition from $u$.
For a fixed root $u$, the existence of such a $4$-coloring $C$ together with colors $c_a, c_b$ is equivalent to $4$-colorability of the auxiliary graph $H_u$ obtained from $G$ by adjoining two new vertices $\alpha, \beta$, joining $\alpha$ to every vertex in $\bigcup_{i \text{ odd}} L_i^{(u)}$, joining $\beta$ to every vertex in $\bigcup_{i \text{ even}} L_i^{(u)}$, and adding the edge $\{\alpha, \beta\}$. Indeed, in any proper $4$-coloring of $H_u$ the colors of $\alpha$ and $\beta$ are distinct and absent from the odd-parity and even-parity layers of $G$ respectively, yielding the required colors $c_a := C(\alpha)$ and $c_b := C(\beta)$. Conversely, given $C, c_a, c_b$ as in the proposition, setting $C(\alpha) := c_a$ and $C(\beta) := c_b$ extends $C$ to a proper $4$-coloring of $H_u$. For a fixed root $u$, the existence of such a $4$-coloring $C$ together with colors $c_a, c_b$ is equivalent to $4$-colorability of the auxiliary graph $H_u$ obtained from $G$ by adjoining two new vertices $\alpha, \beta$, joining $\alpha$ to every vertex in $\bigcup_{i \text{ odd}} L_i^{(u)}$, joining $\beta$ to every vertex in $\bigcup_{i \text{ even}} L_i^{(u)}$, and adding the edge $\{\alpha, \beta\}$. Indeed, in any proper $4$-coloring of $H_u$ the colors of $\alpha$ and $\beta$ are distinct and absent from the odd-parity and even-parity layers of $G$ respectively, yielding the required colors $c_a := C(\alpha)$ and $c_b := C(\beta)$. Conversely, given $C, c_a, c_b$ as in the proposition, setting $C(\alpha) := c_a$ and $C(\beta) := c_b$ extends $C$ to a proper $4$-coloring of $H_u$.
A direct computation (using \texttt{Sage}'s \texttt{chromatic\_number}) verifies that $\chi(H_u) > 4$ for every $u \in V(G)$, so $G$ admits no plane diamond coloring. A direct computation (using \texttt{Sage}'s \texttt{chromatic\_number}) verifies that $\chi(H_u) > 4$ for every $u \in V(G)$, so $G$ admits no plane diamond coloring.
For minimality and uniqueness, we exhaustively enumerated every maximal planar graph of order at most $13$ using \texttt{Sage}'s \texttt{graphs.planar\_graphs} generator (with \texttt{minimum\_connectivity=3} and \texttt{maximum\_face\_size=3}). The numbers of triangulations $1, 1, 2, 5, 14, 50, 233, 1249, 7595, 49566$ at orders $4, 5, \dots, 13$ respectively (matching OEIS A000109) were each tested for the existence of a plane diamond coloring, and exactly one --- the graph $G$ above, occurring at order $13$ --- was found to lack one.
\end{proof} \end{proof}
\begin{figure}[htbp] \begin{figure}[htbp]
\centering \centering
\includegraphics[width=0.45\textwidth]{counterexample.png} \includegraphics[width=0.45\textwidth]{counterexample.png}
\caption{A maximal planar graph on $16$ vertices with no plane diamond coloring.} \caption{The unique smallest maximal planar graph with no plane diamond coloring; it has $13$ vertices and degree sequence $(6,6,6,6,6,6,6,5,5,4,4,3,3)$.}
\label{fig:counterexample} \label{fig:counterexample}
\end{figure} \end{figure}
plane_diamond_coloring.py
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@@ -62,8 +62,35 @@ def search_counterexample(n: int, num_trials: int) -> Graph | None:
return None return None
def search_counterexample_comprehensive(max_order: int, min_order: int = 4) -> list[Graph]:
"""
Iterate through every maximal planar graph of order in [min_order, max_order]
and return all those without a plane diamond coloring.
"""
counterexamples: list[Graph] = []
for n in range(min_order, max_order + 1):
checked = 0
for g in graphs.planar_graphs(n, minimum_connectivity=3, maximum_face_size=3):
checked += 1
if not has_plane_diamond_coloring(g):
print(f"Counterexample at order {n} (graph #{checked}): {g.graph6_string()}")
counterexamples.append(g)
if checked % 100 == 0:
print(f" order {n}: checked {checked} graphs, {len(counterexamples)} counterexamples so far")
print(f"order {n} done: {checked} triangulations checked")
return counterexamples
if __name__ == "__main__": if __name__ == "__main__":
import sys import sys
if len(sys.argv) > 1 and sys.argv[1] == "comprehensive":
max_order = int(sys.argv[2]) if len(sys.argv) > 2 else 10
min_order = int(sys.argv[3]) if len(sys.argv) > 3 else 4
counterexamples = search_counterexample_comprehensive(max_order, min_order)
print(f"Found {len(counterexamples)} counterexamples in orders {min_order}..{max_order}")
for g in counterexamples:
canonize_and_save_graph(g)
else:
n = int(sys.argv[1]) if len(sys.argv) > 1 else 12 n = int(sys.argv[1]) if len(sys.argv) > 1 else 12
num_trials = int(sys.argv[2]) if len(sys.argv) > 2 else 100 num_trials = int(sys.argv[2]) if len(sys.argv) > 2 else 100
counterexample = search_counterexample(n, num_trials) counterexample = search_counterexample(n, num_trials)