coloring_nested_tire_graphs: compare 2-SAT vs König-lift approaches to chain pigeonhole

Adds a side-by-side comparison of the two proof attempts now in
the repo:

  Approach 1 (cyclic 2-SAT, in rainbow_proof.tex):
    Proves π_D = P_m (perms-per-half) for one antipodal-chord SP
    tire when m_1 ≥ m - 1.  Open piece: 2-SAT solvability
    (Conjecture 1.5).

  Approach 2 (König lift, in worst_case_proof_sketch.tex):
    Proves |S_1 ∩ S_2| ≥ 6 for two adjacent SP tires sharing γ
    when both chords are on γ.  Open piece: T_2 induces a
    γ-face partition (Conj t2-induces-partition).

Assessment: Approach 2 is more promising because (a) the hard step
is already proven (König's theorem), (b) it proves exactly what we
need (chain-pigeonhole non-emptiness, not the full π_D
characterisation), and (c) it directly explains the empirical
worst-case |S_1 ∩ S_2| = 6 = single S_3-orbit phenomenon.

Approach 1 still has value if we need finer control over π_D's
shape, but for just establishing non-empty overlap Approach 2
suffices.

Both approaches witness the SAME canonical 6-element worst-case
intersection (the rainbow S_3-orbit at γ=6 = the König-lifted
Latin S_3-orbit).

Recommended next move: attack Conj t2-induces-partition.  Write
down the candidate induced γ-partition explicitly, verify it
computationally, then prove inclusion via transfer-matrix / fibre
lifting.

Note: two_approaches_comparison.tex (3 pages).

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

papers/coloring_nested_tire_graphs/notes/two_approaches_comparison.aux

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papers/coloring_nested_tire_graphs/notes/two_approaches_comparison.tex

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+\documentclass[11pt]{article}
+\usepackage{amsmath,amssymb,amsthm}
+\usepackage{graphicx}
+\usepackage{geometry}
+\usepackage{booktabs}
+\geometry{margin=1in}
+
+\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}