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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 $\pi_D$.''

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