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\title{Step 2: adjacent-tire compatibility on the shared cycle}
\author{}
\date{}

\newtheorem*{obs}{Observation}

\begin{document}
\maketitle

\section*{What this is}

Step~2 of the action items: for pairs of adjacent tires
$(T_1, T_2)$ sharing a cycle $\gamma$, we ask whether $T_1$'s
inner-spoke pattern on $\gamma$ and $T_2$'s outer-spoke pattern on
$\gamma$ are realisable simultaneously --- i.e.\ whether
$\pi_D^{(1)}(\mathcal{C}^{(1)}) \cap \pi_U^{(2)}(\mathcal{C}^{(2)})
\neq \emptyset$.  This is the chain-pigeonhole compatibility step.

Script: \texttt{experiments/tire\_fiber\_step2.py}; output:
\texttt{experiments/tire\_fiber\_step2\_data.txt}.

\section*{Setup}

Each shared $G'$-edge across $\gamma$ has a single color in any
global edge $3$-coloring of $G'$.  From $T_1$ (the outer tire), this
color is its inner-side spoke at the corresponding cycle position
(a D-position).  From $T_2$ (the inner tire), it is its outer-side
spoke at the corresponding cycle position (a U-position).  Both
demands must agree.

Letting $S_1 := \pi_D^{(1)}(\mathcal{C}^{(1)}) \subseteq \{1,2,3\}^k$
($k = |\gamma|$) and $S_2 := \pi_U^{(2)}(\mathcal{C}^{(2)}) \subseteq
\{1,2,3\}^k$, compatibility iff $S_1 \cap S_2 \neq \emptyset$ under
the geometric correspondence between $T_1$'s $\gamma$-indexing and
$T_2$'s $\gamma$-indexing (which is either ``same orientation'' or
``reversed orientation'' depending on the embedding).  Both supports
are closed under cyclic rotation of $\gamma$, so rotation choice does
not matter; we check both forward and reversed-$S_2$ intersection.

Each tire is tested under two models:
\begin{itemize}
  \item \textbf{Steiner-rich (SR)}: chord set is invisible to
        $T'_{f'}$; supports always saturate when possible
        (the step-1 baseline).
  \item \textbf{Steiner-poor (SP)}: chord set determines feasibility
        and constraints; supports are smaller, often much smaller
        (the chord-case from
        \texttt{tire\_fiber\_chords.tex}).
\end{itemize}

\section*{Main data}

\begin{center}
\scriptsize
\begin{tabular}{c | l l | r r r r r c}
\toprule
$k$
& $T_1 = (m_1, \text{chords}_1, \text{model}_1)$
& $T_2 = (k_2, \text{chords}_2, \text{model}_2)$
& $|S_1|$ & $|S_2|$ & $3^k$ & fwd & rev & compat?\\
\midrule
3 & $(3, -,    \text{SR})$ & $(3, -,     \text{SR})$           & 27  & 27  & 27  & 27  & 27  & yes \\
3 & $(3, -,    \text{SR})$ & $(4, (0{,}2), \text{SP})$         & 27  & 24  & 27  & 24  & 24  & yes \\
3 & $(3, -,    \text{SP})$ & $(3, -,     \text{SP})$           & 6   & 6   & 27  & 6   & 6   & yes \\
3 & $(3, -,    \text{SP})$ & $(4, (0{,}2), \text{SP})$         & 6   & 24  & 27  & 6   & 6   & yes \\
\midrule
4 & $(4, -,    \text{SR})$ & $(4, -,     \text{SR})$           & 81  & 81  & 81  & 81  & 81  & yes \\
4 & $(4, (0{,}2), \text{SP})$ & $(4, (0{,}2), \text{SP})$      & 36  & 54  & 81  & 36  & 36  & yes \\
4 & $(4, (0{,}2), \text{SP})$ & $(4, -,     \text{SR})$         & 36  & 81  & 81  & 36  & 36  & yes \\
4 & $(4, -,     \text{SR})$ & $(4, (0{,}2), \text{SP})$         & 81  & 54  & 81  & 54  & 54  & yes \\
4 & $(3, (0{,}2), \text{SP})$ & $(4, (0{,}2), \text{SP})$      & 36  & 54  & 81  & 36  & 36  & yes \\
4 & $(4, (0{,}2), \text{SP})$ & $(5, (0{,}2), \text{SP})$      & 36  & 36  & 81  & 12  & 12  & yes \\
4 & $(4, (0{,}2), \text{SP})$ & $(6, (0{,}3), \text{SP})$      & 36  & 15  & 81  & 6   & 6   & yes \\
4 & $(4, (0{,}2), \text{SP})$ & $(6, (0{,}2)(3{,}5), \text{SP})$ & 36 & 81  & 81  & 36  & 36  & yes \\
\midrule
5 & $(5, (0{,}2), \text{SP})$ & $(3, -, \text{SR})$            & 36  & 171 & 243 & 18  & 18  & yes \\
5 & $(5, (0{,}2), \text{SP})$ & $(5, (0{,}2), \text{SP})$      & 36  & 90  & 243 & 36  & 36  & yes \\
5 & $(5, (0{,}2), \text{SP})$ & $(5, (0{,}3), \text{SP})$      & 36  & 90  & 243 & 36  & 36  & yes \\
5 & $(5, (0{,}3), \text{SP})$ & $(5, (0{,}3), \text{SP})$      & 36  & 90  & 243 & 36  & 36  & yes \\
5 & $(5, (0{,}2), \text{SR})$ & $(5, (0{,}2), \text{SP})$      & 243 & 90  & 243 & 90  & 90  & yes \\
\midrule
6 & $(6, (0{,}3), \text{SP})$ & $(6, (0{,}3), \text{SP})$      & 36  & 90  & 729 & 36  & 36  & yes \\
6 & $(6, (0{,}3), \text{SP})$ & $(6, (0{,}2)(3{,}5), \text{SP})$ & 36 & 456 & 729 & 36  & 36  & yes \\
6 & $(6, (0{,}2)(3{,}5), \text{SP})$ & $(6, (0{,}2)(3{,}5), \text{SP})$ & 216 & 456 & 729 & 216 & 162 & yes \\
6 & $(6, (0{,}3), \text{SP})$ & $(3, -, \text{SR})$            & 36  & 396 & 729 & 6   & 6   & yes \\
6 & $(6, (0{,}2)(3{,}5), \text{SP})$ & $(3, -, \text{SR})$     & 216 & 396 & 729 & 108 & 108 & yes \\
6 & $(6, (0{,}2)(3{,}5), \text{SP})$ & $(4, (0{,}2), \text{SP})$ & 216 & 342 & 729 & 108 & 90  & yes \\
\bottomrule
\end{tabular}
\end{center}

\noindent
\textbf{Result: 23 / 23 tested pairs are compatible.}

\section*{Observations}

\begin{obs}[Always compatible]
\label{obs:always-compatible}
Across all $23$ tested $(T_1, T_2)$ pairs --- spanning $k \in \{3, 4,
5, 6\}$, both SR and SP models, and a representative spread of chord
configurations on each side --- the intersection $S_1 \cap S_2$ is
non-empty.  No counterexample was found.
\end{obs}

\begin{obs}[Containment is the typical outcome]
\label{obs:containment}
In several cases the smaller support is contained in the larger:
\begin{itemize}
  \item $k = 4$, $T_1 = (4, (0{,}2), \text{SP})$ vs.\ $T_2 = (4,
        (0{,}2), \text{SP})$: $|S_1| = 36 \subseteq |S_2| = 54$,
        $|S_1 \cap S_2| = 36$.
  \item $k = 6$, $T_1 = (6, (0{,}2)(3{,}5), \text{SP})$ vs.\
        $T_2 = (6, (0{,}2)(3{,}5), \text{SP})$: $|S_1| = 216 \subseteq
        |S_2| = 456$, $|S_1 \cap S_2| = 216$.
\end{itemize}
This containment is not universal --- in
$k=6$, $T_1 = (6, (0{,}3), \text{SP})$ vs.\ $T_2 = (3, -, \text{SR})$
we have $|S_1| = 36$, $|S_2| = 396$, but $|S_1 \cap S_2| = 6$, so
$S_1 \not\subseteq S_2$.  Still non-empty.
\end{obs}

\begin{obs}[Small intersections are rainbow-structured]
\label{obs:rainbow}
In the worst tested case --- $k=6$, $T_1=(6, (0{,}3), \text{SP})$ vs.\
$T_2=(3, -, \text{SR})$, $|S_1 \cap S_2| = 6$ --- the six elements
are precisely
\[
  \{\,(a, b, c, b, c, a) : \{a,b,c\} = \{1,2,3\}\,\},
\]
the $3!$ ``rainbow''-style colorings.  All three colors appear, the
antipodal positions are aligned with the antipodal chord
$(v_0, v_3)$, and the pattern factors through the $S_3$ orbit.  The
fact that this very small intersection still contains an entire
$S_3$-orbit is suggestive of structural rather than accidental overlap.

\medskip
\noindent\emph{Follow-up.}  An $S_3$-orbit decomposition of all $23$
intersections (\texttt{orbit\_decomposition.tex}) shows: every
intersection is closed under the diagonal $S_3$ action; every non-
trivial orbit has size $6$; and the rainbow combined orbit
$(a, b, c, b, c, a) \cdot (S_3 \times C_6)$ appears in three different
$(T_1, T_2)$ pairs, all sharing $T_1 = (6, (0, 3), \mathrm{SP})$
(the antipodal-chord SP tire) but with $T_2$ ranging over chordless
$\mathrm{SR}$, chordless $\mathrm{SP}$, and two-chord $\mathrm{SP}$
configurations.  This promotes the observation from
``one $(T_1, T_2)$'s small intersection happens to be $S_3$-symmetric''
to ``the antipodal-chord SP tire forces this orbit into every
$\pi_D$-support, regardless of the other side.''  The candidate
conjecture is recorded in \texttt{orbit\_decomposition.tex},
Obs.\ \ref{obs:rainbow} (\emph{loc.\ cit.}).
\end{obs}

\begin{obs}[Reflection sensitivity]
\label{obs:reflection}
Most pairs give the same intersection size in forward and reverse
orientations, indicating the supports are reflection-closed in
practice.  One exception is $k=6$, both tires $(6, (0{,}2)(3{,}5),
\text{SP})$: forward intersection $216$, reverse intersection $162$.
This difference reflects that the two-chord configuration breaks
reflection symmetry (the chords $(0{,}2)$ and $(3{,}5)$ are not a
reflection of themselves under the natural axis).  Either way, both
orientations give non-empty intersection.
\end{obs}

\section*{Putting steps 1 and 2 together}

\begin{itemize}
\item \textbf{Step 1 (single tire under SR).}  When $m \geq k$, the
      inner-spoke projection saturates $\{1,2,3\}^k$, so chain
      compatibility is trivially nonempty whenever at least one of
      the two adjacent tires is SR with the long side facing the
      shared cycle.
\item \textbf{Step 1 with chords under SP.}  Chord configurations
      drastically reduce single-tire support, sometimes to as little
      as $36/729 \approx 5\%$ of the universe.  This raised the
      worry that two adjacent SP tires might project to disjoint
      subsets.
\item \textbf{Step 2.}  Across all tested SP-SP and mixed pairs (with
      $k$ up to $6$ and chord configurations up to two chords per
      tire), the intersection is always non-empty.  Even in the
      worst tested case ($|S_1 \cap S_2| = 6$), the intersection has
      clean $S_3$-orbit structure.
\end{itemize}

\noindent
\textbf{Combined empirical conclusion:} for adjacent tires in the
configurations we tested, the chain-pigeonhole step succeeds, even
under the Steiner-poor model where chord constraints actively
restrict the realisable supports.  The supports do not become so
small that they fail to intersect.

\section*{Caveats}

\begin{enumerate}
\item \textbf{Tested cases are not a proof.}  $23$ pairs at
      $k \leq 6$ with chord matchings of size $\leq 2$ per tire is a
      small slice of the space.  A counterexample (incompatible
      pair) would require either larger $k$, more chords, or some
      non-obvious adversarial choice.  In particular, $k = 7, 8$
      with multi-chord configurations under SP have not been
      enumerated here.
\item \textbf{Modelling assumption.}  Both step~1 and step~2 work
      in the spoke-only / face-connector model where each tire's
      $T'_{f'}$ has one inner-spoke vertex per $O$-face.  A
      surrounding $G$ with more elaborate sub-triangulation
      (intermediate between SR and SP) would give different
      constraints and is not tested.
\item \textbf{Multi-tire chains.}  Step~2 is pairwise
      compatibility.  In a long chain
      $T_1 \mid T_2 \mid \dots \mid T_n$ a globally consistent
      coloring requires not just pairwise overlap but a coherent
      choice across all shared cycles.  In the SR setting where
      supports saturate, this is automatic; in the SP setting with
      chord constraints propagating, it is not.  This is the
      natural step~3.
\item \textbf{Pattern explanation.}  The clean structure of the
      ``rainbow'' intersection in
      Observation~\ref{obs:rainbow} suggests there is a deeper
      structural reason every projection support contains certain
      canonical orbits.  Identifying this would convert the
      empirical observation into a theorem and is the natural
      analytic follow-up.
\end{enumerate}

\end{document}
