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\title{Two approaches to the chain-pigeonhole step:\\
       cyclic 2-SAT vs.\ K\"onig lift}
\author{}
\date{}

\begin{document}
\maketitle

\section*{Background}

The chain-pigeonhole step asks whether, for any adjacent SP tire pair
$(T_1, T_2)$ sharing cycle $\gamma$ of length $k$, the
projection-supports overlap non-trivially:
\[
   \pi_D(\mathcal{C}(T_1)) \cap \pi_U(\mathcal{C}(T_2)) \ne \emptyset.
\]
Empirically (\texttt{tire\_fiber\_step2.tex} +
\texttt{tire\_fiber\_step2\_large.tex}) the overlap is always
non-empty, and in the worst case has exactly $6$ elements forming a
single $S_3$-orbit.

Two independent proof attempts now exist in the notes, attacking
slightly different statements with very different techniques.  This
note compares them and assesses which is more promising.

\section*{Approach 1: cyclic 2-SAT
(\texttt{rainbow\_proof.tex})}

\subsection*{What it tries to prove}

A single-tire characterisation: for an antipodal-chord SP tire $T$
with $m_1 \ge m - 1$,
\[
   \pi_D(\mathcal{C}(T)) \;=\; \mathcal{P}_m,
\]
where $\mathcal{P}_m \subseteq \{1,2,3\}^m$ is the
``perms-per-face'' set ($(m/2)!^2 \cdot \binom{3}{m/2}^2 = 36$
elements at both $m \in \{4, 6\}$).

\subsection*{Technique}

\begin{enumerate}
  \item \textbf{$\subseteq$ direction:} clean from the
        $O$-face dual proper-colouring constraint.
  \item \textbf{Reduce $\supseteq$ to 2-SAT:} For each
        $\sigma \in \mathcal{P}_m$, define orientation bits
        $o_0, \dots, o_{m-1} \in \{0,1\}$ at the $D$-positions.
        The proper-colouring constraint on $T'_{\mathrm{ann}}$
        becomes a cyclic 2-SAT on these orientations.
  \item \textbf{Open step (Conjecture 1.5):} the cyclic 2-SAT is
        satisfiable for every $\sigma \in \mathcal{P}_m$.
        Empirically true ($6$--$18$ satisfying orientations per
        $\sigma$ at $m = 6$), but a clean structural proof is open.
\end{enumerate}

\subsection*{Status}

The $\subseteq$ direction is fully proven.  The $\supseteq$ direction
reduces to a 2-SAT solvability claim that remains open as
Conjecture~1.5; a naive ``all-zero orientation'' construction fails,
and a correct general argument needs implication-graph or
$S_3$-equivariant case analysis.

\section*{Approach 2: K\"onig lift
(\texttt{worst\_case\_proof\_sketch.tex})}

\subsection*{What it tries to prove}

A two-tire overlap statement: for adjacent SP tires $(T_1, T_2)$
sharing $\gamma$ with both sides above their saturation thresholds,
\[
   |\pi_D(\mathcal{C}(T_1)) \cap \pi_U(\mathcal{C}(T_2))| \;\ge\; 6.
\]
This is the actual chain-pigeonhole question (no need to characterise
$\pi_D$ as a whole).

\subsection*{Technique}

\begin{enumerate}
  \item \textbf{Build a bipartite face-incidence graph $G$:} left
        vertices $=$ $T_1$'s $\gamma$-face partition $\mathcal{F}_1$;
        right vertices $=$ $T_2$'s $\gamma$-face partition
        $\mathcal{F}_2$; edges $=$ $\gamma$-edges (each lies in one
        $\mathcal{F}_1$-face and one $\mathcal{F}_2$-face).  Each
        face has exactly $3$ $\gamma$-edges, so $G$ is $3$-regular
        bipartite.
  \item \textbf{Apply K\"onig's theorem:} every bipartite graph admits
        a proper $\Delta$-edge-colouring.  $G$ gets a proper
        $3$-edge-colouring $\chi : E(G) \to \{1, 2, 3\}$.
  \item \textbf{Lift back to $\gamma$:} define $\sigma(e) := \chi(e)$
        for each $\gamma$-edge.  Then at every $\mathcal{F}_1$-face
        (and $\mathcal{F}_2$-face), the three incident edges have
        three distinct colours, so $\sigma|_F$ is a permutation of
        $\{1,2,3\}$.
  \item Hence $\sigma$ lies in the Latin overlap, and its $S_3$
        orbit (size $6$) is in $\pi_D \cap \pi_U$.
\end{enumerate}

\subsection*{What's still open}

The construction works directly when \emph{both} tires give a chord
on $\gamma$ (so both $\mathcal{F}_1, \mathcal{F}_2$ are direct
$\gamma$-face partitions).  In the actual chain-pigeonhole setup,
$T_2$'s chord is on $B_{\mathrm{in}}^{(2)}$, not on $\gamma$.

\textbf{Open conjecture
(Conj.\ \emph{t2-induces-partition} of the worst-case note):} $T_2$
nonetheless induces a $\gamma$-face partition
$\widetilde{\mathcal{F}_2}$ such that $\pi_U(\mathcal{C}(T_2))
\supseteq \mathcal{L}(\gamma, \widetilde{\mathcal{F}_2})$ (the Latin
subset of $\gamma$-colourings compatible with
$\widetilde{\mathcal{F}_2}$).  A candidate construction exists:
group $\gamma$-edges by which $B_{\mathrm{in}}^{(2)}$-face their
$T_2$-side annular triangles share an edge with.

\section*{Side-by-side comparison}

\begin{center}
\small
\begin{tabular}{l|p{0.4\textwidth}|p{0.4\textwidth}}
\toprule
& \textbf{Approach 1: cyclic 2-SAT} & \textbf{Approach 2: K\"onig lift}\\
\midrule
What it proves
  & Single-tire: $\pi_D = \mathcal{P}_m$ (a full structural
    characterisation, $|\pi_D| = 36$).
  & Two-tire: $|S_1 \cap S_2| \ge 6$ (a lower bound on the
    overlap, witnessed by the rainbow $S_3$-orbit).\\
Hard step
  & Cyclic 2-SAT solvability for every $\sigma \in \mathcal{P}_m$.
  & Showing $T_2$'s side induces a $\gamma$-face partition when
    $T_2$'s chord is on $B_{\mathrm{in}}^{(2)}$ rather than $\gamma$.\\
Tooling
  & Custom orientation-bit machinery; implication-graph
    analysis on a cyclic 2-SAT.
  & Classical: K\"onig's edge-colouring theorem for bipartite
    graphs.\\
Strength of conclusion
  & Stronger than needed: characterises $\pi_D$ exactly.
  & Exactly what's needed for chain pigeonhole.\\
What's proven
  & $\subseteq$ direction (Lemma 1.2);  2-SAT reduction (Prop 1.4);
    sharpness counterexample (Prop 1.7).
  & K\"onig-overlap prop (when both chords on $\gamma$);
    $S_3$-invariance lower-bound argument.\\
What's open
  & 2-SAT solvability (Conj 1.5); empirically verified.
  & ``$T_2$ induces $\gamma$-partition'' (Conj
    \emph{t2-induces-partition}); plausibility check via candidate
    construction.\\
\bottomrule
\end{tabular}
\end{center}

\section*{Which approach is more promising}

\textbf{Approach 2 (K\"onig lift) is more promising}, for three
reasons:

\subsection*{1.\ The hard step is already proven}

K\"onig's theorem is a 100-year-old textbook result.  The K\"onig-
overlap proposition is a clean lift via that theorem and is fully
written down.  The remaining open piece (T_2 induces a
$\gamma$-partition) is a geometric/structural claim about how
$T_2$'s annular triangulation distributes $B_{\mathrm{in}}^{(2)}$ faces
across $\gamma$-edges --- likely amenable to a direct construction.

Approach 1's open piece (2-SAT solvability) is a combinatorial
satisfiability claim with no obvious leveraged tool.  Empirically
true, but the structural reason isn't yet clear.

\subsection*{2.\ Approach 2 proves exactly what we want}

Chain pigeonhole asks: is the overlap non-empty?  Approach 2 directly
addresses this with a lower bound of $6$, which is also empirically
tight.  Approach 1 proves a strictly stronger statement (the full
$\pi_D = \mathcal{P}_m$ characterisation) that is more than chain
pigeonhole needs.  Proving more than necessary is a strictly harder
task and unnecessary for the goal.

\subsection*{3.\ The K\"onig lower bound has the right structure}

Worst-case overlap is empirically a single $S_3$-orbit of size $6$,
which is exactly what the K\"onig lift produces (lift one
3-edge-colouring, then act by $S_3$ on colours).  The proof
mirror the empirical phenomenon.  Approach 1's machinery is more
general and doesn't naturally explain why the worst case is an
$S_3$-orbit.

\subsection*{When Approach 1 still pays off}

Approach 1 \emph{does} give a stronger result: a complete
characterisation of $\pi_D$ for the antipodal-chord SP tire,
including the upper bound $|\pi_D| \le 36$.  This is useful if we
ever need finer control over $\pi_D$'s shape (e.g.\ if we want to
prove that $\pi_D$ does not contain certain $S_3$-orbits, or if we
want to control the chain-pigeonhole overlap above the floor of
$6$).  For just establishing non-empty overlap, Approach 2 suffices.

\section*{Reconciliation: the $6$ is the same $6$}

Both approaches witness the same canonical $6$-element worst-case
intersection:
\begin{itemize}
  \item Approach 1: the rainbow $S_3$-orbit $(a, b, c, b, c, a) \cdot
        S_3$ sits inside $\pi_D \cap \pi_U$ when $\pi_D = \mathcal{P}_m$
        contains it and $\pi_U$ does too.
  \item Approach 2: the K\"onig-lifted Latin colouring's $S_3$-orbit
        sits inside the intersection directly, by construction.
\end{itemize}
For the case $T_1$ antipodal-chord SP, $T_2$ chordless SR, $k = 6$
(the worst tested case), these are literally the same $6$ elements.

\section*{Recommended next move}

Attack the open conjecture in Approach 2 (Conj
\emph{t2-induces-partition}):
\begin{enumerate}
  \item Write down the candidate induced $\gamma$-partition
        $\widetilde{\mathcal{F}_2}$ explicitly (the
        ``group $\gamma$-edges by which
        $B_{\mathrm{in}}^{(2)}$-face's neighbours they share
        annular edges with'' construction).
  \item Verify $\pi_U(\mathcal{C}(T_2)) \supseteq
        \mathcal{L}(\gamma, \widetilde{\mathcal{F}_2})$ computationally
        for $k \in \{3, 4, 5, 6\}$ and several $T_2$ structures.
  \item If empirical fit is exact (as in the worst-case data),
        prove inclusion via a transfer-matrix / fibre-lifting
        argument.
\end{enumerate}
This is more leveraged than continuing to refine the 2-SAT
implication-graph analysis, which would prove a stronger statement
that we do not need.

\end{document}
