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\title{Step 2 at $k = 9$ and $k = 12$:\\
       extending the adjacent-tire compatibility experiment}
\author{}
\date{}

\newtheorem*{obs}{Observation}

\begin{document}
\maketitle

\section*{What this is}

A continuation of \texttt{tire\_fiber\_step2.tex}: that note tested
adjacent-tire compatibility (the chain-pigeonhole step) at $k \leq 6$,
all $23/23$ pairs compatible.  This note pushes the experiment to
$k \in \{9, 12\}$, enabled by the numpy-optimized fiber enumeration
in \texttt{experiments/tire\_fiber\_chords\_fast.py}.  Together with
the earlier data, every tested pair --- now $23$ at $k \leq 6$ plus
$23$ more at $k \in \{9, 12\}$, $46$ in total --- is compatible.

Script: \texttt{experiments/tire\_fiber\_step2\_large.py}.
Output:  \texttt{experiments/tire\_fiber\_step2\_large\_data.txt}.

\section*{Setup recap}

Two adjacent tires $T_1, T_2$ share a cycle $\gamma$ of length $k$.
We project each tire's $\sigma$-support onto $\gamma$:
\[
   S_1 := \pi_D^{(1)}(\mathcal{C}^{(1)}) \subseteq \{1,2,3\}^k,
   \quad
   S_2 := \pi_U^{(2)}(\mathcal{C}^{(2)}) \subseteq \{1,2,3\}^k,
\]
and ask whether $S_1 \cap S_2 \neq \emptyset$ (forward), or
$S_1 \cap \mathrm{reverse}(S_2) \neq \emptyset$ (reverse).  Either
non-empty intersection means the pair is \emph{compatible}.

\section*{Chord configurations used}

Under the Steiner-poor model, every $O$-face must have at most $3$
$B_{\mathrm{in}}$ edges.  For $k > 3$ this forces specific chord
matchings.  Constrained chord sets used (validated):

\begin{center}
\small
\begin{tabular}{l l l}
\toprule
label & chord set & face sizes \\
\midrule
\texttt{K9\_CHORDS\_3}        & $\{(0,2),(3,5),(6,8)\}$            & $(2, 2, 2, 3)$ \\
\texttt{K9\_CHORDS\_NESTED}   & $\{(0,2),(3,7),(4,6)\}$            & $(2, 2, 2, 3)$ (different arrangement) \\
\texttt{K12\_CHORDS\_3}       & $\{(0,3),(4,7),(8,11)\}$            & $(3, 3, 3, 3)$ (symmetric thirds) \\
\texttt{K12\_CHORDS\_NESTED}  & $\{(0,3),(4,11),(5,7),(8,10)\}$    & $(2, 2, 2, 3, 3)$ \\
\bottomrule
\end{tabular}
\end{center}

\section*{Results at $k = 9$ (13/13 compatible)}

\begin{center}
\scriptsize
\begin{tabular}{l l r r r r r}
\toprule
$T_1$ & $T_2$ & $|S_1|$ & $|S_2|$ & $|S_1 \cap S_2|$ fwd & rev & compat? \\
\midrule
$(9, \texttt{K9-3}, \text{SP})$  & $(9, \texttt{K9-3}, \text{SP})$  & 1296 & 7866 & 1188 & 858 & yes \\
$(9, \texttt{K9-3}, \text{SP})$  & $(9, \texttt{K9-N}, \text{SP})$  & 1296 & 7536 & 1242 & 1110 & yes \\
$(9, \texttt{K9-N}, \text{SP})$  & $(9, \texttt{K9-N}, \text{SP})$  & 1296 & 7536 & 1176 & 1224 & yes \\
$(12, \texttt{K9-3}, \text{SP})$ & $(9, \texttt{K9-3}, \text{SP})$  & 1296 & 7866 & 1188 & 858 & yes \\
$(9, \texttt{K9-3}, \text{SP})$  & $(12, \texttt{K12-3}, \text{SP})$ & 1296 & 1302 & 72   & 90   & yes \\
$(9, \texttt{K9-3}, \text{SP})$  & $(12, \texttt{K12-N}, \text{SP})$ & 1296 & 9456 & 840  & 606  & yes \\
$(9, \texttt{K9-3}, \text{SR})$  & $(9, \texttt{K9-3}, \text{SP})$  & 19683 & 7866 & 7866 & 7866 & yes \\
$(9, \texttt{K9-3}, \text{SP})$  & $(3, -, \text{SR})$              & 1296 & 3681 & 108  & 108  & yes \\
$(9, \texttt{K9-3}, \text{SP})$  & $(4, (0,2), \text{SP})$          & 1296 & 3162 & 276  & 108  & yes \\
$(9, \texttt{K9-3}, \text{SP})$  & $(6, (0,3), \text{SP})$          & 1296 & 942  & 54   & 60   & yes \\
$(9, \texttt{K9-3}, \text{SP})$  & $(6, (0,2)(3,5), \text{SP})$     & 1296 & 6210 & 402  & 324  & yes \\
$(9, \texttt{K9-N}, \text{SP})$  & $(6, (0,3), \text{SP})$          & 1296 & 942  & 54   & 36   & yes \\
$(9, \texttt{K9-N}, \text{SP})$  & $(12, \texttt{K12-N}, \text{SP})$ & 1296 & 9456 & 732  & 768  & yes \\
\bottomrule
\end{tabular}
\end{center}

Universe size at $k = 9$: $3^9 = 19{,}683$.

\section*{Results at $k = 12$ (10/10 compatible)}

\begin{center}
\scriptsize
\begin{tabular}{l l r r r r r}
\toprule
$T_1$ & $T_2$ & $|S_1|$ & $|S_2|$ & $|S_1 \cap S_2|$ fwd & rev & compat? \\
\midrule
$(12, \texttt{K12-3}, \text{SP})$  & $(12, \texttt{K12-3}, \text{SP})$  & 1296 & 12840 & 960 & 564 & yes \\
$(12, \texttt{K12-N}, \text{SP})$  & $(12, \texttt{K12-N}, \text{SP})$  & 7776 & 100938 & 5928 & 3414 & yes \\
$(12, \texttt{K12-3}, \text{SP})$  & $(12, \texttt{K12-N}, \text{SP})$  & 1296 & 100938 & 1128 & 912 & yes \\
$(12, \texttt{K12-3}, \text{SP})$  & $(3, -, \text{SR})$                 & 1296 & 31176 & 192 & 192 & yes \\
$(12, \texttt{K12-3}, \text{SP})$  & $(4, (0,2), \text{SP})$              & 1296 & 27378 & 48 & 60 & yes \\
$(12, \texttt{K12-3}, \text{SP})$  & $(6, (0,3), \text{SP})$              & 1296 & 9882 & \textbf{6} & \textbf{6} & yes \\
$(12, \texttt{K12-3}, \text{SP})$  & $(6, (0,2)(3,5), \text{SP})$         & 1296 & 61224 & 12 & 12 & yes \\
$(12, \texttt{K12-3}, \text{SP})$  & $(9, \texttt{K9-3}, \text{SP})$       & 1296 & 94116 & 18 & 90 & yes \\
$(12, \texttt{K12-N}, \text{SP})$  & $(9, \texttt{K9-3}, \text{SP})$       & 7776 & 94116 & 552 & 852 & yes \\
$(12, \texttt{K12-3}, \text{SR})$  & $(12, \texttt{K12-3}, \text{SP})$  & 531441 & 12840 & 12840 & 12840 & yes \\
\bottomrule
\end{tabular}
\end{center}

Universe size at $k = 12$: $3^{12} = 531{,}441$.

\section*{Observations}

\begin{obs}[Still compatible everywhere]
\label{obs:still-compat}
$46$ of $46$ tested pairs are compatible across $k \in \{3, 4, 5, 6,
9, 12\}$ ($23$ from the earlier note plus $23$ new at $k \in \{9, 12\}$).
No counterexample has been found.
\end{obs}

\begin{obs}[The $S_3$-orbit pattern persists at $k = 12$]
\label{obs:s3-orbit}
The smallest tested intersection at $k = 12$ is again \emph{exactly
$6$ elements}, occurring at $T_1 = (12, \texttt{K12-3}, \text{SP})$ vs.\
$T_2 = (6, (0,3), \text{SP})$.  Direct inspection shows the six
elements are a single $S_3$-orbit of the canonical pattern
\[
   (1, 2, 3, 2, 2, 1, 3, 3, 2, 3, 1, 1).
\]
Decoded against \texttt{K12-3}'s face structure (faces $\{0,1,2\}$,
$\{4,5,6\}$, $\{8,9,10\}$, and the outer face $\{3,7,11\}$ on the
$B_{\mathrm{in}}$ edges):
\begin{itemize}
  \item face $\{0,1,2\}$: $\sigma$-values $(1, 2, 3)$ -- a permutation of $\{1,2,3\}$.
  \item face $\{4,5,6\}$: $\sigma$-values $(2, 1, 3)$ -- a permutation.
  \item face $\{8,9,10\}$: $\sigma$-values $(2, 3, 1)$ -- a permutation.
  \item face $\{3,7,11\}$: $\sigma$-values $(2, 3, 1)$ -- a permutation.
\end{itemize}
Every face receives all three colors exactly once.  This is precisely
the ``Latin-square-flavoured'' structural pattern from the $k = 6$
worst case, now extended to $k = 12$.
\end{obs}

\begin{obs}[Bigger supports come from more chords]
\label{obs:more-chords-bigger}
Nested chord sets give substantially larger supports than symmetric
ones:
\begin{itemize}
  \item At $k = 12$, $\texttt{K12-3}$ (symmetric, faces $3{+}3{+}3{+}3$)
        gives $|S_1| = 1296$.
  \item $\texttt{K12-N}$ (nested, faces $2{+}2{+}2{+}3{+}3$) gives
        $|S_1| = 7776 = 6 \cdot 1296$.
\end{itemize}
The factor of $6$ is suggestive but I have not chased it.
\end{obs}

\begin{obs}[Intersection sizes are multiples of $6$]
\label{obs:multiples-6}
Every observed forward and reverse intersection size in the new data
is a multiple of $6$:
\begin{quote}
$1188, 1242, 1176, 858, 1110, 1224, 72, 90, 840, 606, 7866, 108, 276,
54, 60, 402, 324, 36, 732, 768$,\\
$960, 564, 5928, 3414, 1128, 912, 192, 48, 60, 6, 12, 18, 90, 552, 852,
12840$
\end{quote}
This is consistent with both $S_1$ and $S_2$ being $S_3$-invariant
(closed under color permutations), so their intersection decomposes
into $S_3$-orbits, each of size $6$.
\end{obs}

\section*{Speculative theorem}

\begin{quote}
\textbf{Conjecture.}  For any SP-feasible tire $T$ (i.e.\ every
$O$-face has at most $3$ $B_{\mathrm{in}}$ edges), the projection
$\pi_D(\mathcal{C}(T))$ on the $\gamma$-side contains the
``Latin-flavoured'' subset
\[
   \mathcal{L}(\gamma, O) \;:=\; \{\sigma \in \{1,2,3\}^{|\gamma|}
       : \sigma \text{ restricted to each $O$-face is a permutation
           of $\{1,2,3\}$}\},
\]
which is an $S_3$-invariant set of size at least $6$ (and exactly
$6$ in the maximally constrained case).  Adjacent tires share this
common substructure on $\gamma$, so the chain-pigeonhole intersection
is non-empty.
\end{quote}

The data is consistent with this conjecture and points to a
structural proof: any tire that admits an edge $3$-coloring at all
must admit a globally ``Latin-style'' one, and Latin-style colorings
are dictated entirely by face structure (not by which tire's face it
is).  Adjacent tires that share a cycle $\gamma$ see the same Latin
constraints from each other's side, so their Latin-style supports
agree.

This would be the analog, on the chord side, of step~1's
``saturation iff $m \geq k$'' result on the spoke-only side.

\section*{Performance notes}

The numpy-optimized enumeration in
\texttt{experiments/tire\_fiber\_chords\_fast.py} replaces the
brute-force $3^n$-iteration with direct construction of the
$2^n + 2(-1)^n$ proper edge $3$-colorings of $C_n$.  Speedups
benchmarked against the slow version:
\begin{itemize}
  \item $n = 12$: 0.001s vs 0.21s -- $\sim 146\times$
  \item $n = 15$: 0.013s vs 5.4s -- $\sim 424\times$
  \item $n = 18$: 0.11s vs (extrapolated) $\sim 150$s
  \item $n = 24$: 6.6s vs (extrapolated) $\sim 30$ hours
\end{itemize}
Total wall time for the $k = 9$ block: a few seconds; $k = 12$ block:
$\sim 4$ minutes (dominated by the single SR-vs-SP case at $n = 24$).

\section*{Caveats}

\begin{enumerate}
\item \textbf{Still finite, still not a proof.}  $46$ pairs is still
      a small slice.  Larger $k$ ($\geq 15$) or unusual chord
      configurations could harbour a counterexample.  The conjecture
      above suggests there is no counterexample, but is unproven.
\item \textbf{Multi-tire chains.}  Step~2 is pairwise compatibility.
      A long nested chain of SP tires requires pairwise overlap at
      each shared cycle \emph{plus} mutual consistency across all
      shared cycles simultaneously.  The Latin-style conjecture, if
      true, would imply chain-wide consistency via a common
      Latin coloring of all annular faces at once.
\item \textbf{Model still SP/SR only.}  Intermediate
      sub-triangulations (some Steiner vertices, some not) are not
      enumerated.
\item \textbf{Chord set space is sampled, not enumerated.}  Each $k$
      has many distinct chord matchings; this experiment uses only
      one or two per $k$.  More exhaustive enumeration could
      strengthen the empirical evidence.
\end{enumerate}

\end{document}
