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Refute inner-boundary conjecture with n=14 counterexample

Browse Source 
The 14-vertex 3-connected triangulation at plantri index 263993 has
no vertex source admitting a witnessing 4-colouring; the level-cycle
conjecture still holds on it via v0=10.  Reorder the section so the
level-cycle conjecture follows the failed inner-boundary refinement.

Co-Authored-By: Claude Sonnet 4.6 <noreply@anthropic.com>
This commit is contained in:
didericis
2026-06-01 21:57:05 -04:00
parent 59410747ea
commit b175f0ab59
8 changed files with 331 additions and 206 deletions
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papers/coloring_nested_tire_graphs/experiments/draw_inner_boundary_counterexample.py
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"""Draw the 14-vertex counterexample to the tire inner-boundary
three-colour conjecture (Conjecture~\\ref{conj:tire-inner-boundary-three-colour}).
This is the graph at index 263993 in the n=14 plantri enumeration: a
3-connected (but not 5-connected) maximal planar graph with degree
sequence [7,7,7,7,7,7,6,6,3,3,3,3,3,3] and exactly 96 proper
4-colourings, none of which witness the inner-boundary restriction
from any vertex source.
"""
from __future__ import annotations
import os
import matplotlib.pyplot as plt
import networkx as nx
from matplotlib.lines import Line2D
OUT_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
EDGES = [
(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8),
(2, 3), (2, 4), (2, 6), (2, 8), (2, 9), (2, 10),
(3, 4), (4, 5), (4, 6), (4, 10),
(5, 6), (6, 7), (6, 9), (6, 10),
(7, 8), (7, 9), (7, 11), (7, 12), (7, 13),
(8, 9), (8, 12), (8, 13), (8, 14),
(9, 11), (9, 12), (9, 14),
(11, 12), (12, 13), (12, 14),
]
def build_graph() -> nx.Graph:
g = nx.Graph()
g.add_edges_from(EDGES)
return g
def main() -> int:
g = build_graph()
is_planar, _ = nx.check_planarity(g)
if not is_planar:
raise RuntimeError("graph should be planar")
pos = nx.planar_layout(g, scale=3.4)
fig, ax = plt.subplots(figsize=(8.2, 7.4))
nx.draw_networkx_edges(g, pos, ax=ax, edge_color="#cbd5e1", width=1.3)
deg3 = [v for v in g.nodes() if g.degree(v) == 3]
deg7 = [v for v in g.nodes() if g.degree(v) == 7]
other = [v for v in g.nodes() if v not in deg3 and v not in deg7]
def draw_nodes(nodes, color):
for v in nodes:
x, y = pos[v]
ax.scatter(
[x], [y],
s=520,
color=color,
edgecolors="black",
linewidths=1.1,
zorder=3,
)
ax.text(
x, y, f"{v}",
ha="center", va="center",
color="white", fontsize=11, fontweight="bold",
zorder=4,
)
draw_nodes(deg7, "#0f172a")
draw_nodes(other, "#475569")
draw_nodes(deg3, "#94a3b8")
legend = [
Line2D([0], [0], marker="o", color="w",
label=r"degree $7$",
markerfacecolor="#0f172a", markeredgecolor="black",
markersize=12),
Line2D([0], [0], marker="o", color="w",
label=r"degree $6$",
markerfacecolor="#475569", markeredgecolor="black",
markersize=12),
Line2D([0], [0], marker="o", color="w",
label=r"degree $3$",
markerfacecolor="#94a3b8", markeredgecolor="black",
markersize=12),
]
ax.legend(handles=legend, loc="lower center", ncol=3, framealpha=0.95)
ax.set_title(
"Counterexample to the inner-boundary three-colour conjecture "
"($n=14$)",
fontsize=13, pad=14,
)
ax.set_aspect("equal")
ax.axis("off")
fig.tight_layout(rect=[0, 0.05, 1, 1])
out = os.path.join(OUT_DIR, "fig_inner_boundary_counterexample.png")
fig.savefig(out, dpi=180, bbox_inches="tight")
plt.close(fig)
print(f"wrote {out}")
return 0
if __name__ == "__main__":
raise SystemExit(main())
papers/coloring_nested_tire_graphs/fig_inner_boundary_counterexample.png
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papers/coloring_nested_tire_graphs/paper.aux
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\bibcite{tait-original}{1}
\bibcite{bauerfeld-depth}{2}
\bibcite{bauerfeld-nested-tire-duals}{3}
papers/coloring_nested_tire_graphs/paper.fdb_latexmk
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@@ -1338,6 +1338,152 @@ $4$-colouring of $G$. Therefore no proper $4$-colouring has the
level-cycle three-colour restriction with respect to $S=\{7\}$.
\end{example}
\subsection*{An inner-boundary refinement}
The level-cycle restriction constrains \emph{every} simple cycle in
every level. For the tire-tree program, the cycles that actually carry
boundary state are fewer: each tire transfers colour information across
its tread between its two boundaries
(Theorem~\ref{thm:tire-chromatic-polynomial-transfer}), so it is the tire
\emph{inner boundaries} $B_{\mathrm{in}}^{(T)}$ --- not all level cycles
--- that one wishes to compress. This motivates a restriction stated
directly in the objects of the decomposition.
\begin{definition}[Tire inner-boundary three-colour restriction]
\label{def:tire-inner-boundary-three-colour}
Let $G$ be a maximal planar graph, let $v_0 \in V(G)$ be a vertex source
on the outer face of $\Pi_G$, and let $c \colon V(G) \to \{1,2,3,4\}$ be
a proper $4$-vertex-colouring of $G$. We say $c$ has the \emph{tire
inner-boundary three-colour restriction} with respect to
$\mathcal{T}(G, \{v_0\})$ if every tire tread $T \in
\mathcal{T}(G, \{v_0\})$ satisfies
\[
|c(V(B_{\mathrm{in}}^{(T)}))| \leq 3,
\]
i.e.\ the inner boundary of every tire omits at least one of the four
colours. (A degenerate inner boundary is a single vertex and the
condition is then vacuous.)
\end{definition}
\begin{conjecture}[Tire inner-boundary three-colour conjecture]
\label{conj:tire-inner-boundary-three-colour}
Every maximal planar graph $G$ admits a vertex source $v_0 \in V(G)$ and
a proper $4$-vertex-colouring $c$ of $G$ such that $c$ has the tire
inner-boundary three-colour restriction with respect to
$\mathcal{T}(G, \{v_0\})$.
\end{conjecture}
\begin{remark}[Relation to the level-cycle conjecture]
\label{rem:inner-boundary-vs-level-cycle}
For a depth-$d$ tire $T$, the inner outerplanar graph satisfies
$O^{(T)} \subseteq G[L_{d+1}]$: a depth-$d$ face has its three vertex
levels in $\{d, d+1\}$ (adjacent vertices differ by at most one level),
so the level-$(d+1)$ vertices of the tire's dual component are exactly
$V(O^{(T)})$. Since $O^{(T)}$ is outerplanar, every one of its vertices
lies on the inner-boundary walk, whence $V(B_{\mathrm{in}}^{(T)}) =
V(O^{(T)}) \subseteq L_{d+1}$ is supported on a single level, and is a
simple level cycle when $O^{(T)}$ is $2$-connected.
Consequently the vertex-source form of
Conjecture~\ref{conj:level-cycle-three-colour} implies
Conjecture~\ref{conj:tire-inner-boundary-three-colour} on every
$2$-connected inner boundary: the witnessing colouring already makes
each such cycle omit a colour. The present conjecture is thus a
\emph{weakening}, constraining only the inner-boundary cycles of one
tire-tree decomposition rather than all level cycles of some level
source. It is no harder than the vertex-source form of
Conjecture~\ref{conj:level-cycle-three-colour}, while targeting exactly
the interface the chromatic-transfer machinery of
Theorem~\ref{thm:tire-chromatic-polynomial-transfer} runs across.
(The non-$2$-connected case --- an
inner boundary whose walk traverses a bridge or cut-vertex of $O^{(T)}$
--- is not covered by the simple-cycle statement of
Conjecture~\ref{conj:level-cycle-three-colour} and must be argued
separately.)
\end{remark}
\subsection*{A counterexample at $n=14$}
Conjecture~\ref{conj:tire-inner-boundary-three-colour} is in fact
false. An exhaustive search over the triangulations enumerated by
\texttt{plantri} at $n=14$ encounters a graph $G^\star$ on $14$ vertices
and $36$ edges --- specifically, the graph at index $263993$ in the
\texttt{plantri} enumeration --- for which no vertex source admits any
witness.
\begin{example}[Counterexample to Conjecture~\ref{conj:tire-inner-boundary-three-colour}]
\label{ex:inner-boundary-counterexample}
Let $G^\star$ be the maximal planar graph with vertex set
$\{1,2,\dots,14\}$ and edge set
\begin{align*}
E(G^\star) = \{
& 12, 13, 14, 15, 16, 17, 18, \\
& 23, 24, 26, 28, 29, 2\,10, \\
& 34, 45, 46, 4\,10, 56, 67, 69, 6\,10, \\
& 78, 79, 7\,11, 7\,12, 7\,13, \\
& 89, 8\,12, 8\,13, 8\,14, \\
& 9\,11, 9\,12, 9\,14, \\
& 11\,12, 12\,13, 12\,14
\}.
\end{align*}
The graph $G^\star$ is a $3$-connected (but not $5$-connected) planar
triangulation with degree sequence
$(7,7,7,7,7,7,6,6,3,3,3,3,3,3)$ and exactly $96$ proper $4$-vertex
colourings. For \emph{every} choice of vertex source
$v_0 \in V(G^\star)$, each of the $96$ proper $4$-colourings of
$G^\star$ has some tire whose inner boundary uses all four colours.
A planar embedding is shown in
Figure~\ref{fig:inner-boundary-counterexample}.
\end{example}
\begin{figure}[ht]
\centering
\includegraphics[width=0.78\textwidth]{fig_inner_boundary_counterexample}
\caption{The $14$-vertex counterexample $G^\star$ to
Conjecture~\ref{conj:tire-inner-boundary-three-colour} in a planar
embedding. The six degree-$3$ vertices split into two triples,
$\{3,5,10\}$ each adjacent to a triangle in the
core $\{1,2,4,6\}$, and $\{11,13,14\}$ each adjacent to a triangle in
the core $\{7,8,9,12\}$; the two cores are joined by the edges
$17,28,69$ together with $12$.}
\label{fig:inner-boundary-counterexample}
\end{figure}
The failure was verified by enumerating, for each of the $14$ vertex
sources, all $96$ proper $4$-colourings of $G^\star$ and computing the
inner boundary $V(B_{\mathrm{in}}^{(T)})$ of every tire $T$ as the
level-$(d+1)$ vertices of the corresponding depth-$d$ dual component
(Remark~\ref{rem:inner-boundary-vs-level-cycle}). Each source has
exactly two non-degenerate inner boundaries (size $\geq 4$), and every
proper $4$-colouring assigns all four colours to at least one of them.
Because Conjecture~\ref{conj:tire-inner-boundary-three-colour} is a
weakening of the vertex-source form of
Conjecture~\ref{conj:level-cycle-three-colour} only \emph{on
$2$-connected inner boundaries}
(Remark~\ref{rem:inner-boundary-vs-level-cycle}), $G^\star$ need not
refute Conjecture~\ref{conj:level-cycle-three-colour}, and in fact does
not: the vertex source $v_0 = 10$ admits a proper $4$-colouring under
which every simple level cycle uses at most three colours. Combining
this with Remark~\ref{rem:inner-boundary-vs-level-cycle}, at least one
tire $T$ under $v_0 = 10$ must have an inner outerplanar graph
$O^{(T)}$ that fails to be $2$-connected --- under the witnessing
colouring, some inner boundary uses all four colours, and if its
$O^{(T)}$ were $2$-connected then by
Remark~\ref{rem:inner-boundary-vs-level-cycle} that inner boundary
would be a simple level cycle, contradicting the level-cycle witness.
The failure of
Conjecture~\ref{conj:tire-inner-boundary-three-colour} is therefore
attributable to the non-$2$-connected case left open by
Remark~\ref{rem:inner-boundary-vs-level-cycle}, not to a deeper failure
of the level-cycle statement on $G^\star$.
\subsection*{The surviving level-cycle conjecture}
The verification on $G^\star$ above is consistent with the broader
empirical picture for the level-cycle restriction, which we record as
the conjecture this section ultimately stands on.
\begin{conjecture}[Level-cycle three-colour conjecture]
\label{conj:level-cycle-three-colour}
Let $G$ be a maximal planar graph. Then there exists a level source
@@ -1410,153 +1556,6 @@ source witnessing the level-cycle three-colour conjecture.}
\label{tab:level-cycle-three-colour-c5-14-16}
\end{table}
\subsection*{An inner-boundary refinement}
The level-cycle restriction constrains \emph{every} simple cycle in
every level. For the tire-tree program, the cycles that actually carry
boundary state are fewer: each tire transfers colour information across
its tread between its two boundaries
(Theorem~\ref{thm:tire-chromatic-polynomial-transfer}), so it is the tire
\emph{inner boundaries} $B_{\mathrm{in}}^{(T)}$ --- not all level cycles
--- that one wishes to compress. This motivates a restriction stated
directly in the objects of the decomposition.
\begin{definition}[Tire inner-boundary three-colour restriction]
\label{def:tire-inner-boundary-three-colour}
Let $G$ be a maximal planar graph, let $v_0 \in V(G)$ be a vertex source
on the outer face of $\Pi_G$, and let $c \colon V(G) \to \{1,2,3,4\}$ be
a proper $4$-vertex-colouring of $G$. We say $c$ has the \emph{tire
inner-boundary three-colour restriction} with respect to
$\mathcal{T}(G, \{v_0\})$ if every tire tread $T \in
\mathcal{T}(G, \{v_0\})$ satisfies
\[
|c(V(B_{\mathrm{in}}^{(T)}))| \leq 3,
\]
i.e.\ the inner boundary of every tire omits at least one of the four
colours. (A degenerate inner boundary is a single vertex and the
condition is then vacuous.)
\end{definition}
\begin{conjecture}[Tire inner-boundary three-colour conjecture]
\label{conj:tire-inner-boundary-three-colour}
Every maximal planar graph $G$ admits a vertex source $v_0 \in V(G)$ and
a proper $4$-vertex-colouring $c$ of $G$ such that $c$ has the tire
inner-boundary three-colour restriction with respect to
$\mathcal{T}(G, \{v_0\})$.
\end{conjecture}
\begin{remark}[Relation to the level-cycle conjecture]
\label{rem:inner-boundary-vs-level-cycle}
For a depth-$d$ tire $T$, the inner outerplanar graph satisfies
$O^{(T)} \subseteq G[L_{d+1}]$: a depth-$d$ face has its three vertex
levels in $\{d, d+1\}$ (adjacent vertices differ by at most one level),
so the level-$(d+1)$ vertices of the tire's dual component are exactly
$V(O^{(T)})$. Since $O^{(T)}$ is outerplanar, every one of its vertices
lies on the inner-boundary walk, whence $V(B_{\mathrm{in}}^{(T)}) =
V(O^{(T)}) \subseteq L_{d+1}$ is supported on a single level, and is a
simple level cycle when $O^{(T)}$ is $2$-connected.
Consequently the vertex-source form of
Conjecture~\ref{conj:level-cycle-three-colour} implies
Conjecture~\ref{conj:tire-inner-boundary-three-colour} on every
$2$-connected inner boundary: the witnessing colouring already makes
each such cycle omit a colour. The present conjecture is thus a
\emph{weakening}, constraining only the inner-boundary cycles of one
tire-tree decomposition rather than all level cycles of some level
source. It is no harder than the vertex-source form of
Conjecture~\ref{conj:level-cycle-three-colour}, while targeting exactly
the interface the chromatic-transfer machinery of
Theorem~\ref{thm:tire-chromatic-polynomial-transfer} runs across.
(The non-$2$-connected case --- an
inner boundary whose walk traverses a bridge or cut-vertex of $O^{(T)}$
--- is not covered by the simple-cycle statement of
Conjecture~\ref{conj:level-cycle-three-colour} and must be argued
separately.)
\end{remark}
\subsection*{Enumeration for the inner-boundary conjecture}
We repeated the exhaustive search of
Conjecture~\ref{conj:level-cycle-three-colour} for the inner-boundary
restriction, testing for each triangulation whether some vertex source
$v_0$ admits a proper $4$-colouring whose tire inner boundaries each omit
a colour. For a depth-$d$ tire the inner-boundary vertex set is computed
directly as the level-$(d+1)$ vertices of the corresponding depth-$d$
dual component, using
Remark~\ref{rem:inner-boundary-vs-level-cycle}. No counterexample
appeared on the full small-$n$ census $4 \leq n \leq 13$
(Table~\ref{tab:inner-boundary-three-colour-counts}) or on the
$5$-connected slice $14 \leq n \leq 24$
(Table~\ref{tab:inner-boundary-three-colour-c5}).
\begin{table}[ht]
\centering
\small
\setlength{\tabcolsep}{4pt}
\begin{tabular}{ccc}
$n$ & triangulations & with witness \\\hline
$4$ & $1$ & $1$ \\
$5$ & $1$ & $1$ \\
$6$ & $2$ & $2$ \\
$7$ & $5$ & $5$ \\
$8$ & $14$ & $14$ \\
$9$ & $50$ & $50$ \\
$10$ & $233$ & $233$ \\
$11$ & $1249$ & $1249$ \\
$12$ & $7595$ & $7595$ \\
$13$ & $49566$ & $49566$ \\
\end{tabular}
\caption{Exhaustive vertex-source search for the tire inner-boundary
three-colour conjecture
(Conjecture~\ref{conj:tire-inner-boundary-three-colour}) on all
triangulation isomorphism classes with $4 \leq n \leq 13$. Every
triangulation in this range admits at least one vertex source
witnessing the conjecture.}
\label{tab:inner-boundary-three-colour-counts}
\end{table}
\begin{table}[ht]
\centering
\small
\setlength{\tabcolsep}{4pt}
\begin{tabular}{ccc}
$n$ & $5$-connected triangulations & with witness \\\hline
$14$ & $1$ & $1$ \\
$15$ & $1$ & $1$ \\
$16$ & $3$ & $3$ \\
$17$ & $4$ & $4$ \\
$18$ & $12$ & $12$ \\
$19$ & $23$ & $23$ \\
$20$ & $71$ & $71$ \\
$21$ & $187$ & $187$ \\
$22$ & $627$ & $627$ \\
$23$ & $1970$ & $1970$ \\
$24$ & $6833$ & $6833$ \\
\end{tabular}
\caption{The $5$-connected triangulations at $14 \leq n \leq 24$
generated by \texttt{plantri -c5 -a}. All $9732$ graphs in this slice
admit a vertex source witnessing the tire inner-boundary three-colour
conjecture.}
\label{tab:inner-boundary-three-colour-c5}
\end{table}
We also re-ran the inner-boundary check on the six dual triangulations
of the Holton--McKay graphs and found a vertex source witnessing the
conjecture for each. In four of the six cases the first source tried
already succeeded; in the remaining two, one or two vertex sources
exhausted all $4320$ proper $4$-colourings before a witnessing source
was found.
Unlike the small-$n$ census, where the first source and colouring tried
typically already witness the restriction, the source choice is
genuinely active in the $5$-connected slice: many vertex sources fail
exhaustively before a witness is found. For instance, in the unique
$n=16$ $5$-connected triangulation two sources exhaust all proper
$4$-colourings with no compatible colouring before a third source
succeeds. This is consistent with the failure of the universal-source
form (Conjecture~\ref{conj:false-universal-level-cycle-three-colour}):
the existential quantifier over the root is doing real work.
\begin{definition}[Seam]
\label{def:seam}
A \emph{seam} of a maximal planar graph $G$ is a simple cycle