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

Browse Source 
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>
This commit is contained in:
didericis
2026-05-26 11:19:39 -04:00
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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}