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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}