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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}