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\title{A proof attempt at the worst-case overlap}
\author{}
\date{}

\newtheorem*{thm}{Theorem}
\newtheorem*{prop}{Proposition}
\newtheorem*{lem}{Lemma}
\newtheorem*{conj}{Conjecture}

\begin{document}
\maketitle

\section*{Setup}

Let $\gamma$ be a cycle of length $k$ (divisible by $3$ for the
``all-3'' case), and let $T_1, T_2$ be two adjacent SP tires sharing
$\gamma$ with the threshold condition
$|B_{\mathrm{out}}^{(1)}| \geq k$ and $|B_{\mathrm{in}}^{(2)}| \geq k$.
Assume both tires' inner outerplanar graphs are configured so that
every $O$-face has exactly $3$ $B_{\mathrm{in}}$ edges (strict Latin
case).

The empirical worst-case overlap is exactly $|S_3| = 6$ elements,
forming a single $S_3$-orbit (see
\texttt{tire\_fiber\_step2\_large.tex}, Obs. on the $S_3$-orbit
structure).  This note sketches a proof.

\section*{What I can prove cleanly: both sides give $\gamma$-face partitions}

\begin{prop}[König $\Rightarrow$ overlap $\geq 6$]
\label{prop:konig-overlap}
Suppose both $T_1$ and $T_2$ induce \emph{direct} all-$3$ face
partitions $\mathcal{F}_1, \mathcal{F}_2$ of $E(\gamma)$.  Then
\[
   |\mathcal{L}(\gamma, \mathcal{F}_1) \cap \mathcal{L}(\gamma,
      \mathcal{F}_2)| \;\geq\; 6.
\]
\end{prop}

\begin{proof}[Proof sketch]
Define a bipartite graph $G$:
\begin{itemize}
  \item Left vertices: $\mathcal{F}_1$-faces (there are $k/3$ of
        them).
  \item Right vertices: $\mathcal{F}_2$-faces (also $k/3$).
  \item One edge from $F \in \mathcal{F}_1$ to $F' \in \mathcal{F}_2$
        for each $\gamma$-edge $e \in F \cap F'$.
\end{itemize}
Since each $\gamma$-edge belongs to exactly one face of each
partition, $\gamma$-edges biject with edges of $G$, and $|E(G)| = k$.
Each face has $3$ $\gamma$-edges, so every vertex of $G$ has degree
exactly $3$.  Thus $G$ is $3$-regular bipartite.

By König's edge-coloring theorem, every bipartite graph admits a
proper $\Delta$-edge-coloring; here $\Delta = 3$.  So $G$ has a
proper $3$-edge-coloring $\chi : E(G) \to \{1,2,3\}$.  Transport
$\chi$ back to $\gamma$-edges via the bijection: define
$\sigma(e) := \chi(e)$.  Then for any face $F \in \mathcal{F}_1$,
the three $\gamma$-edges in $F$ are the three $G$-edges incident
to the $F$-vertex (degree $3$), and $\chi$ being proper forces these
to use all three colors --- so $\sigma|_F$ is a permutation of
$\{1,2,3\}$.  Same for $\mathcal{F}_2$.

Hence $\sigma \in \mathcal{L}(\gamma, \mathcal{F}_1) \cap
\mathcal{L}(\gamma, \mathcal{F}_2)$.  The $S_3$ action on colours
sends this $\sigma$ to $|S_3| = 6$ distinct elements of the
intersection (they're distinct because $\sigma$ uses all three
colours).
\end{proof}

\begin{prop}[Lower bound is achieved]
\label{prop:lower-bound-tight}
The lower bound $|{\cdots}| \geq 6$ in
Prop.~\ref{prop:konig-overlap} is achieved exactly when $G$ has a
\emph{unique} proper $3$-edge-coloring up to $S_3$-relabelling ---
equivalently, when $G$ is a single closed $C_{2k/3}$-cycle (with
two alternating perfect matchings, plus a third perfect matching
forced by the remaining edges).  In particular, the worst case is
$G \cong K_{3,3}$ (when $k = 9$) or $G$ a single cycle, both of
which arise empirically.
\end{prop}

(Sketch: a $3$-regular bipartite graph has $\geq 6$ proper $3$-edge-
colourings, with equality iff the colouring is essentially unique.
The unique-up-to-$S_3$ case is exactly when $G$ is highly symmetric.
This matches the empirical observation that worst-case intersections
are single $S_3$-orbits.)

\section*{What remains: the gap in the actual setting}

In the real chain-pigeonhole setup, $T_1$'s chord is on $B_{\mathrm{in}}^{(1)}
= \gamma$ (so $T_1$ \emph{does} give a direct $\gamma$-face partition
$\mathcal{F}_1$), but $T_2$'s chord is on $B_{\mathrm{in}}^{(2)}$,
\emph{not} on $\gamma$.  So $T_2$ does not directly give a
$\gamma$-face partition.

\subsection*{Conjectural extension to the real case}

\begin{conj}[Induced $\gamma$-partition from $T_2$]
\label{conj:t2-induces-partition}
For any SP tire $T_2$ with $B_{\mathrm{in}}^{(2)}$-side chord
structure $O^{(2)}$ such that every $O^{(2)}$-face has exactly $3$
$B_{\mathrm{in}}^{(2)}$-edges, the outer-spoke support
$\pi_U(\mathcal{C}(T_2)) \subseteq \{1,2,3\}^\gamma$ contains a
Latin subset $\mathcal{L}(\gamma, \widetilde{\mathcal{F}_2})$ for
some induced face partition $\widetilde{\mathcal{F}_2}$ of $\gamma$
into triples, where $\widetilde{\mathcal{F}_2}$ is determined by how
$T_2$'s annular triangulation distributes $B_{\mathrm{in}}^{(2)}$ faces
across $\gamma$-edges.
\end{conj}

If Conjecture~\ref{conj:t2-induces-partition} holds, then
Prop.~\ref{prop:konig-overlap} immediately gives the worst-case
overlap $\geq 6$ for general adjacent tire pairs.

\subsection*{Why this is plausible}

At $\gamma = 6$, $T_1 = (m_1, (0,3), \mathrm{SP})$ vs $T_2 = (m_2,
B_{\mathrm{in}}^{(2)} = C_3, \mathrm{SP})$, the empirical $6$-element
intersection consists of patterns $(a, b, c, b, c, a)$ which:
\begin{itemize}
  \item $T_1$'s $\gamma$-face partition $\{0,1,2\} \mid \{3,4,5\}$:
        $\sigma|_{0,1,2} = (a, b, c)$ is a permutation,
        $\sigma|_{3,4,5} = (b, c, a)$ is a permutation.  $\checkmark$
  \item The pattern uses each colour exactly twice (positions $\{0,5\},
        \{1,3\}, \{2,4\}$).  These pair into $3$ groups of $2$ --- but
        not in a way that's directly a Latin $\gamma$-partition of
        size $3$.
\end{itemize}
The $T_2$-side constraint forces a structural pattern (equal pairs at
specific positions); the empirical fact is that $T_2$'s induced
constraint on $\gamma$ \emph{is} ``something like a Latin partition,''
but I haven't yet found the precise statement.

\subsection*{One concrete attempt at the induced partition}

For $T_2$ with $B_{\mathrm{in}}^{(2)} = C_{k_2}$ and balanced
annular triangulation, the dual cycle of $T_2$ alternates between
D-triangles (one $B_{\mathrm{in}}^{(2)}$-edge each) and U-triangles
(one $\gamma$-edge each).  Each $B_{\mathrm{in}}^{(2)}$-edge has, in
some sense, two ``adjacent'' $\gamma$-edges via the two annular edges
of its D-triangle.

\emph{Candidate induced partition.}  Group the $\gamma$-edges by
``which $B_{\mathrm{in}}^{(2)}$-face's pair of $\gamma$-neighbours
they share an annular edge with.''  When $k_2 = k/3$ (so there are
$k/3$ D-triangles, each accounting for $3$ $\gamma$-edges via its $2$
annular edges), this gives a partition of $E(\gamma)$ into $k/3$
triples --- the right shape.

Verifying that this candidate partition gives the correct Latin
subset of $\pi_U$ is an open computation.

\section*{Where the lower bound of $|S_3| = 6$ comes from independently}

Setting aside the construction, the lower bound of $6$ has a clean
abstract origin:

\begin{lem}[Lower bound from $S_3$-invariance]
Both $S_1$ and $S_2$ are $S_3$-invariant (the $S_3$ action on
colours acts on edge $3$-colourings of $G'$).  Hence $S_1 \cap S_2$
is $S_3$-invariant, so it decomposes into $S_3$-orbits.  Any
non-trivial $S_3$-orbit on $\{1,2,3\}^k$ where $\sigma$ uses all
three colours has size exactly $|S_3| = 6$.  Constant $\sigma$
(single colour) orbits have size $3$, but constants are typically
not in the support under SP chord constraints with $\geq 2$ faces
of size $\geq 2$ (which forces $\sigma$ to use both ``other'' colours
on each face).
\end{lem}

So the absolute floor is $6$ once we rule out emptiness.  The
$\gamma$-partition / König argument is the cleanest way I know to
rule out emptiness.

\section*{Summary of what's proven, conjectured, and open}

\begin{itemize}
  \item \textbf{Proven (Prop.~\ref{prop:konig-overlap}):}
        When both $T_1$ and $T_2$ give direct all-$3$ $\gamma$-face
        partitions, $|S_1 \cap S_2| \geq 6$, witnessed by lifting
        a König $3$-edge-colouring of the $3$-regular bipartite
        ``face-incidence'' graph.

  \item \textbf{Lower-bound origin:}
        Both $S_1$ and $S_2$ are $S_3$-invariant, so any non-empty
        intersection has size at least $6$ (the size of a
        non-constant $S_3$-orbit).

  \item \textbf{Conjectured (Conj.~\ref{conj:t2-induces-partition}):}
        The real chain-pigeonhole setup, where $T_2$'s chord is on
        $B_{\mathrm{in}}^{(2)}$ rather than on $\gamma$, also reduces
        to Prop.~\ref{prop:konig-overlap} via an induced
        $\gamma$-partition $\widetilde{\mathcal{F}_2}$.  Empirical
        data is consistent with this conjecture but I haven't
        constructed $\widetilde{\mathcal{F}_2}$ explicitly.

  \item \textbf{Open:}
        Find the explicit map $T_2 \mapsto \widetilde{\mathcal{F}_2}$,
        prove the support inclusion, and combine with
        Prop.~\ref{prop:konig-overlap} to close the worst-case
        lower bound.
\end{itemize}

\end{document}
