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coloring_nested_tire_graphs: refute antipodal-chord rainbow conjecture as originally stated, add refined version with m_1 ≥ m precondition

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Counterexample search uncovered:
  T1 = (m_1=3, chord=(0,3), SP) at γ=6
has |π_D| = 18, but the conjectured rainbow combined orbit has size 36 —
only 6 of the 36 elements actually lie in π_D, and the literal generator
pattern (1,2,3,2,3,1) is itself missing.

Refined conjecture adds m_1 ≥ m precondition (same threshold from step 1
"saturation iff m ≥ k"). Under this precondition all 44 tested
strict-Latin pairs are still compatible. Pointer to log path included.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
didericis
2026-05-26 03:50:12 -04:00
parent b27a4d20a3
commit c900ec9a70
4 changed files with 92 additions and 43 deletions
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papers/coloring_nested_tire_graphs/notes/orbit_decomposition.aux
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\newlabel{obs:antipodal-rainbow-conjecture}{{}{3}} \newlabel{obs:antipodal-rainbow-conjecture}{{}{3}}
\@writefile{toc}{\contentsline {paragraph}{Direct test.}{3}{}\protected@file@percent } \@writefile{toc}{\contentsline {paragraph}{Direct test.}{3}{}\protected@file@percent }
\@writefile{toc}{\contentsline {paragraph}{Why antipodal?}{3}{}\protected@file@percent } \@writefile{toc}{\contentsline {paragraph}{Why antipodal?}{3}{}\protected@file@percent }
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papers/coloring_nested_tire_graphs/notes/orbit_decomposition.log
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papers/coloring_nested_tire_graphs/notes/orbit_decomposition.tex
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@@ -151,28 +151,71 @@ This is a real upgrade on the step-$2$ data:
\section*{Conjecture suggested by the data} \section*{Conjecture suggested by the data}
\begin{obs}[Antipodal-chord rainbow conjecture] \begin{obs}[Antipodal-chord rainbow conjecture --- \emph{refuted as
originally stated}; refined below]
\label{obs:antipodal-rainbow-conjecture} \label{obs:antipodal-rainbow-conjecture}
\emph{Original (incorrect) statement.}
Let $T = (m, (0, m/2), \mathrm{SP})$ be a Steiner-poor tire whose Let $T = (m, (0, m/2), \mathrm{SP})$ be a Steiner-poor tire whose
inner outerplanar graph $O$ is a cycle of length $m$ together with a inner outerplanar graph $O$ is a cycle of length $m$ together with a
single antipodal chord (so $m$ is even). Conjecture: the projection single antipodal chord (so $m$ is even). Conjecture (originally,
support $\pi_D(\mathcal{C}(T))$ on the $|\gamma| = m$ inner-side without an outer-boundary precondition): the projection support
spokes always contains the combined orbit $\pi_D(\mathcal{C}(T))$ on the $|\gamma| = m$ inner-side spokes
always contains the combined orbit
\[ \[
\mathrm{Orbit}\bigl(\,(a, b, c, b, c, \dots, b, c, a)\,\bigr) \mathrm{Orbit}\bigl(\,(a, b, c, b, c, \dots, b, c, a)\,\bigr)
\] \]
under $S_3 \times C_m$ (color permutation $\times$ cyclic rotation), under $S_3 \times C_m$, with $a$-positions at the two chord endpoints
where the pattern has length $m$ and the $a$-positions are exactly and $b, c$ alternating elsewhere. At $m = 6$ this is the rainbow
the two chord endpoints, with $b$ and $c$ alternating elsewhere. orbit of size $36$ that Obs.~\ref{obs:rainbow-source} witnessed.
At $m = 6$ this is the rainbow orbit of size $36$ that
Obs.~\ref{obs:rainbow-source} witnessed. \medskip
\noindent
\emph{The original statement is false:} the
counterexample-search log
(\texttt{experiments/counterexample\_search.log}) records the pair
\[
T_1 = (m_1 = 3,\ (0, 3),\ \mathrm{SP}),
\qquad
T_2 = (k_2 = 3,\ -,\ \mathrm{SP}),
\qquad
|\gamma| = 6,
\]
where $T_1$ is the antipodal-chord SP tire at $\gamma = 6$ but with
\emph{small outer boundary} $m_1 = 3$. Direct computation gives
$|\pi_D(\mathcal{C}(T_1))| = 18$. The conjectured rainbow combined
orbit has size $36$, so the orbit cannot fit inside $\pi_D$; in fact
only $6$ of the $36$ rainbow-orbit elements lie in $\pi_D$, and the
literal generator pattern $(1, 2, 3, 2, 3, 1)$ is itself among the
$30$ \emph{missing} elements. Furthermore, $\pi_U^{(2)}(\mathcal{C}(T_2))$
has $|\pi_U| = 84$, but $\pi_D(\mathcal{C}(T_1)) \cap
\pi_U(\mathcal{C}(T_2)) = \emptyset$ in both orientations, so this is
also a (strict-Latin) counterexample to step-$2$ compatibility.
\medskip
\noindent
\emph{Refined statement (now consistent with all data).}
The original conjecture omitted an outer-boundary precondition. The
correct conjecture is:
\begin{quote}
\textbf{If additionally $m_1 = |B_{\mathrm{out}}^{(1)}| \geq m$
(i.e.\ the outer boundary is at least as long as the inner
boundary), then $\pi_D(\mathcal{C}(T))$ contains the antipodal
rainbow combined orbit.}
\end{quote}
The threshold $m_1 \geq m$ is the same saturation threshold from
step~$1$: when the dual cycle of $T$ has length $n_1 = m_1 + m
\geq 2m$, the spoke projection saturates the full Latin set on
$\gamma$. Under this precondition all $23$ pairs of the original
step-$2$ tests and all $44$ strict-Latin pairs in the search log
with $m_1 \geq m$ \emph{and} $k_2 \geq m$ confirm the property.
\end{obs} \end{obs}
If true, this is a uniform structural property of the antipodal-chord If the refined statement holds, this is a uniform structural property
SP tire, independent of the outer boundary length. The chain of the antipodal-chord SP tire \emph{with sufficient outer cycle
pigeonhole step at $|\gamma| = m$ on such a tire reduces to length}. The chain pigeonhole step at $|\gamma| = m$ on such a tire
``$\pi_U$ of the other tire intersects this fixed orbit,'' a much reduces to ``$\pi_U$ of the other tire intersects this fixed orbit,''
smaller compatibility claim. a much smaller compatibility claim --- but only once both sides clear
the outer-boundary threshold.
\paragraph{Direct test.} At $m = 4$ ($\theta(1, 2, 2) = K_4 - e$) the \paragraph{Direct test.} At $m = 4$ ($\theta(1, 2, 2) = K_4 - e$) the
antipodal-chord SP tire's $\pi_D$ support has size $36$ at $|\gamma| = antipodal-chord SP tire's $\pi_D$ support has size $36$ at $|\gamma| =