coloring_nested_tire_graphs: add note explaining 2-SAT solvability

Standalone explanation of what "2-SAT solvability" means in the context of the rainbow proof (rainbow_proof.tex, Conjecture 1.5): - Defines 2-SAT in general (boolean variables, 2-variable clauses, P-time decidable). - Maps it onto our rainbow proof: variables = orientation bits o_j at D-positions; clauses = inter-D-position gap constraints; cyclic chain wraps around T'_ann. - "Solvable" ⇔ proper edge 3-coloring with given σ exists, i.e. σ ∈ π_D. - Cyclic 2-SAT can in principle fail; toy example (3-cycle of not-equal clauses = odd-cycle 2-coloring obstruction). - Empirically our system never fails for σ ∈ P_m (6-18 satisfying orientations per σ at m=6), but a structural proof is open. - Why it matters: proving Conjecture 1.5 upgrades the rainbow proof's provisional corollary into a theorem and reduces chain pigeonhole to the perms-per-half overlap. Note: two_sat_solvability.tex (3 pages). Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
2026-05-26 11:12:16 -04:00
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+\documentclass[11pt]{article}
+\usepackage{amsmath,amssymb,amsthm}
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+
+\title{What ``2-SAT solvability'' means in the rainbow proof}
+\author{}
+\date{}
+
+\begin{document}
+\maketitle
+
+\section*{2-SAT in general}
+
+\textbf{2-SAT} is short for \emph{2-satisfiability}.  Standard logical
+problem:
+\begin{itemize}
+  \item You have a set of \emph{boolean variables}
+        $x_1, x_2, \dots, x_n$, each taking value $0$ or $1$
+        (false/true).
+  \item You have a set of \emph{clauses}, each a constraint involving
+        exactly \emph{two} variables.
+  \item \textbf{2-SAT solvability} $=$ ``does there exist an
+        assignment of $0/1$ to all variables such that every clause
+        is satisfied simultaneously?''
+\end{itemize}
+
+2-SAT is polynomial-time decidable (linear time, in fact) using the
+\emph{implication graph} and \emph{strongly-connected components}.
+This is in contrast to 3-SAT (3 variables per clause), which is
+NP-complete.
+
+\section*{In our setup (the rainbow proof)}
+
+\textbf{Variables.}  $o_0, o_1, \dots, o_{m-1} \in \{0, 1\}$, one
+\emph{orientation bit} per $D$-position on the dual annular cycle
+$T'_{\mathrm{ann}}$.  At each $D$-position $p_j$, the two incident
+cycle edges must take the two colors of $\{1, 2, 3\} \setminus
+\{\sigma_j\}$ in some order; $o_j$ decides which color sits on the
+left of $p_j$ versus the right.
+
+\textbf{Clauses.}  One per inter-$D$-position gap on
+$T'_{\mathrm{ann}}$.  Each gap involves two adjacent orientations
+$(o_j, o_{j+1})$ and forbids certain combinations:
+
+\begin{center}
+\begin{tabular}{lll}
+\toprule
+gap type & clause type & combos forbidden \\
+\midrule
+length-$1$, $\sigma_j \ne \sigma_{j+1}$
+  & equality (fixed pair)
+  & 3 of 4 forbidden, leaves one valid combo \\
+length-$1$, $\sigma_j = \sigma_{j+1}$
+  & $o_j = o_{j+1}$
+  & $(0,1)$ and $(1,0)$ forbidden \\
+length-$2$, $\sigma_j \ne \sigma_{j+1}$
+  & one combo forbidden
+  & 1 of 4 forbidden \\
+length-$2$, $\sigma_j = \sigma_{j+1}$
+  & $o_j = o_{j+1}$
+  & $(0,1)$ and $(1,0)$ forbidden \\
+length-$\ge 3$
+  & no clause
+  & gap has full slack \\
+\bottomrule
+\end{tabular}
+\end{center}
+
+Each clause involves exactly two variables, so the system is 2-SAT.
+
+\textbf{Cyclic structure.}  Because the gaps wrap cyclically around
+$T'_{\mathrm{ann}}$, the clauses form a \emph{cyclic} chain ---
+clause on $(o_0, o_1)$, then $(o_1, o_2)$, $\ldots$, finally
+$(o_{m-1}, o_0)$.
+
+\section*{What ``solvable'' means concretely}
+
+For a given $\sigma \in \mathcal{P}_m$:
+\begin{itemize}
+  \item \textbf{Solvable} $\Leftrightarrow$ there is an orientation
+        $(o_0, \dots, o_{m-1})$ satisfying all clauses simultaneously
+        $\Rightarrow$ proper edge $3$-coloring of $T'_{f'}$ inducing
+        $\sigma$ exists $\Rightarrow$ $\sigma \in \pi_D$.
+  \item \textbf{Unsolvable} $\Leftrightarrow$ no such orientation
+        $\Rightarrow$ $\sigma \notin \pi_D$.
+\end{itemize}
+
+So the rainbow proof's \textbf{Conjecture~1.5 (2-SAT solvability)}
+says: for every $\sigma$ with both halves a permutation of
+$\{1, 2, 3\}$, the resulting cyclic 2-SAT system has at least one
+satisfying assignment.
+
+\section*{Why a cyclic 2-SAT chain can fail}
+
+A 2-SAT cycle \emph{can} be unsolvable when the implications chain
+around and force a contradiction.  Toy example with 3 variables on
+a cycle:
+\begin{itemize}
+  \item Clause 1: $o_0 \ne o_1$.
+  \item Clause 2: $o_1 \ne o_2$.
+  \item Clause 3: $o_2 \ne o_0$.
+\end{itemize}
+
+Implications:
+\[
+   o_0 = 0 \;\Rightarrow\; o_1 = 1 \;\Rightarrow\; o_2 = 0
+       \;\Rightarrow\; o_0 = 1.
+\]
+Contradiction.  So this $3$-cycle of ``not-equal'' clauses is
+unsatisfiable (it is the chromatic-number obstruction: a triangle
+has no proper $2$-coloring).
+
+\section*{Why we believe our cyclic chain doesn't fail}
+
+For our $m$-variable cyclic 2-SAT (with clauses determined by
+$\sigma \in \mathcal{P}_m$), the analogous question is: does the
+specific pattern of clauses ever create such a wrap-around
+contradiction?
+
+\textbf{Empirically:} no.  For $m = 6$, $m_1 \in \{5, 6, 7, 8\}$ and
+all $36$ elements of $\mathcal{P}_6$, the cyclic 2-SAT system always
+has between $6$ and $18$ satisfying assignments (computed in
+\texttt{experiments/orbit\_decomposition.py}).
+
+\textbf{Structurally:} the constraints have a special form
+(forbidden combos depend on $\sigma_j, \sigma_{j+1}$ via the third
+color $u = \{1, 2, 3\} \setminus \{\sigma_j, \sigma_{j+1}\}$).
+The hope is that a parity or implication-graph argument shows the
+cyclic chain cannot consistently wrap around to a contradiction
+when $\sigma$ has the perms-per-half shape.  This argument is what
+the proof of Conjecture~1.5 would supply, and what is currently
+missing.
+
+\section*{Why we care}
+
+Proving Conjecture~1.5 would convert
+\texttt{rainbow\_proof.tex}'s ``provisional corollary'' into a
+genuine theorem:
+\[
+   \pi_D(\mathcal{C}(T)) = \mathcal{P}_m
+   \quad \text{whenever } m_1 \ge m - 1,
+\]
+which in turn upgrades the chain-pigeonhole step at $|\gamma| = m$
+to a structural claim of the form ``$\pi_U(\mathcal{C}(T_2))$ meets
+the fixed $36$-element set $\mathcal{P}_m$,'' a strictly smaller and
+more tractable overlap condition than ``$\pi_U$ meets $\pi_D$.''
+
+\end{document}