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\title{Towards a proof of the antipodal-chord rainbow theorem\\
       (corrected statement with sharp threshold)}
\author{}
\date{}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{prop}[theorem]{Proposition}
\newtheorem{cor}[theorem]{Corollary}
\newtheorem{conj}[theorem]{Conjecture}
\theoremstyle{definition}
\newtheorem{defn}[theorem]{Definition}

\begin{document}
\maketitle

\section*{Statement and status}

\begin{theorem}[Antipodal-chord rainbow theorem, claimed]
\label{thm:rainbow}
Let $T = T(m_1, m)$ be a tire whose outer boundary $B_{\mathrm{out}}$
has length $m_1$, whose inner boundary $B_{\mathrm{in}}$ has even
length $m \in \{4, 6\}$, and whose inner outerplanar graph is
$O = B_{\mathrm{in}} \cup \{(v_0, v_{m/2})\}$ (the antipodal chord).
Assume the Steiner-poor model and the balanced annular triangulation.

If $m_1 \geq m - 1$, then
\[
   \pi_D(\mathcal{C}(T))
   \;=\;
   \mathcal{P}_m
   \;:=\;
   \bigl\{\sigma \in \{1,2,3\}^m :\;
   (\sigma_0, \dots, \sigma_{m/2 - 1}),\;
   (\sigma_{m/2}, \dots, \sigma_{m-1})
   \in \mathrm{Perm}(\{1, 2, 3\}^{m/2})\bigr\}.
\]
In particular $\pi_D$ contains the $S_3$-orbit of the rainbow
configuration $(a, b, c, b, c, \dots, b, c, a)$.

The threshold $m_1 \geq m - 1$ is sharp: at $m_1 = m - 2$ (when
$m = 6$) one has $|\pi_D| = 18$ and the rainbow is missing.
\end{theorem}

\textbf{Status.}  The ``$\subseteq$'' half is fully proven below.  The
``$\supseteq$'' half is partially proven --- reduced to a cyclic
2-SAT solvability statement which is empirically true for every
$\sigma \in \mathcal{P}_m$ (between 6 and 18 satisfying orientation
assignments per $\sigma$) but for which a clean general proof is
deferred.  Sharpness is confirmed by explicit forcing
counterexamples.

\section*{The easy direction: $\pi_D \subseteq \mathcal{P}_m$}

\begin{lemma}[O-face dual constraint]
\label{lem:o-face}
For any $\sigma \in \pi_D(\mathcal{C}(T))$, the restrictions of
$\sigma$ to the two halves $\{0, \dots, m/2 - 1\}$ and
$\{m/2, \dots, m - 1\}$ are each permutations of $\{1, 2, 3\}^{m/2}$
with $m/2$ distinct entries (so $\in \mathrm{Perm}$ when $m/2 = 3$).
\end{lemma}

\begin{proof}
The antipodal chord splits $O$ into two faces $F_A, F_B$.  Under SP,
$F_A$ has boundary $B_{\mathrm{in}}$-edges $e_0, \dots, e_{m/2-1}$
and $F_B$ has $e_{m/2}, \dots, e_{m-1}$.  In $T'_{f'}$ each face
dual $u_{F_A}$ has degree exactly $m/2$, with one $D$-spoke per
$B_{\mathrm{in}}$-edge.  Proper edge $3$-coloring at $u_{F_A}$
requires all $m/2$ incident spokes to have distinct colors;  since
we only have $3$ colors and $m/2 \in \{2, 3\}$, this is exactly the
``$m/2$ distinct entries among $\{1,2,3\}$'' condition. Symmetric
for $F_B$.
\end{proof}

Consequently $|\pi_D| \leq |\mathcal{P}_m| = 36$ at both $m = 4$ and
$m = 6$.

\section*{Reduction of $\mathcal{P}_m \subseteq \pi_D$ to a cyclic
$2$-SAT}

We work in the balanced annular triangulation with $D$-positions
$p_j = \lfloor j(m_1 + m)/m \rfloor$ on $T'_{\mathrm{ann}} = C_n$,
$n := m_1 + m$.

\subsection*{Setup of orientations and gaps}

At each $D$-position $p_j$, the two incident cycle edges
$(e_{p_j - 1}, e_{p_j})$ are constrained by
Lemma~\ref{lem:o-face}'s consequence at the cycle vertex $p_j$ to
satisfy $\{c(e_{p_j - 1}),\, c(e_{p_j})\} =
\{1, 2, 3\} \setminus \{\sigma_j\} =: \{\alpha_j, \beta_j\}$ with
$\alpha_j < \beta_j$.  Let $o_j \in \{0, 1\}$ encode the choice:
\[
   o_j = 0 \;\Leftrightarrow\;
   c(e_{p_j - 1}) = \alpha_j,\; c(e_{p_j}) = \beta_j;
   \qquad
   o_j = 1 \;\Leftrightarrow\;
   \text{swapped.}
\]

Between consecutive $D$-positions $p_j, p_{j+1}$ (cyclically) sit
$L_j - 1 \geq 0$ $U$-positions, where $L_j := p_{j+1} - p_j$ is the
gap arc length.  Cycle edges within the gap form a path; the two
endpoints have already-fixed colors $R_j := c(e_{p_j})$ (right side
of $p_j$) and $L_{j+1} := c(e_{p_{j+1} - 1})$ (left side of $p_{j+1}$),
both determined by $o_j$ and $o_{j+1}$ via
$R_j = \beta_j$ if $o_j = 0$ else $\alpha_j$, and $L_{j+1} =
\alpha_{j+1}$ if $o_{j+1} = 0$ else $\beta_{j+1}$.

\begin{lemma}[Gap feasibility]
\label{lem:gap}
Given $R_j, L_{j+1}$ and gap arc length $L_j \geq 1$, the gap
admits a proper edge $3$-coloring of its interior cycle edges and
$U$-spokes iff
\[
   L_j = 1 \;\Rightarrow\; R_j = L_{j+1};
   \qquad
   L_j = 2 \;\Rightarrow\; R_j \neq L_{j+1};
   \qquad
   L_j \geq 3 \;\Rightarrow\; \text{no constraint}.
\]
For $L_j \geq 3$ at least one extension exists (chromatic-polynomial
nonzero).
\end{lemma}

\begin{proof}
$L_j = 1$: a unique cycle edge equals both $R_j$ and $L_{j+1}$ (it
\emph{is} the edge connecting them).  $L_j = 2$: two adjacent edges
with fixed endpoints $R_j, L_{j+1}$ require $R_j \neq L_{j+1}$ for
proper coloring at the $U$-position between.  $L_j \geq 3$:
straightforward to construct any pair $(R_j, L_{j+1})$ by alternating
colors with at least one ``flexible'' cycle edge between.
\end{proof}

\subsection*{Reduction to cyclic 2-SAT}

When $m_1 \geq m$, every $L_j \geq 2$; when $m_1 = m - 1$, exactly
one $L_j = 1$ (with the rest $L_j = 2$).

\begin{prop}[2-SAT reduction]
\label{prop:reduction}
For $\sigma \in \mathcal{P}_m$, the existence of a proper edge
$3$-coloring of $T'_{f'}$ inducing $\sigma$ on $D$-spokes is
equivalent to the existence of $(o_0, \dots, o_{m-1}) \in \{0,1\}^m$
satisfying, for each $j$:
\begin{itemize}
  \item if $L_j = 1$: $R_j(o_j) = L_{j+1}(o_{j+1})$ (an equality
        clause);
  \item if $L_j = 2$: $R_j(o_j) \neq L_{j+1}(o_{j+1})$ (an inequality
        clause);
  \item if $L_j \geq 3$: no constraint.
\end{itemize}
Each clause is a 2-CNF clause on the variables $o_j, o_{j+1}$, so
the system is a cyclic 2-SAT.
\end{prop}

\begin{proof}
Cycle edges within a gap depend only on $R_j$ and $L_{j+1}$ (by
Lemma~\ref{lem:gap}), which depend only on $(o_j, o_{j+1})$.
The proper-coloring constraints at the $U$-positions and $D$-positions
not at the boundary of the gap are taken care of by gap feasibility.
\end{proof}

\subsection*{The remaining step}

\begin{conj}[2-SAT solvability]
\label{conj:2sat}
For every $\sigma \in \mathcal{P}_m$ with $m \in \{4, 6\}$ and the
balanced triangulation with $m_1 \geq m - 1$, the cyclic 2-SAT
system of Proposition~\ref{prop:reduction} is satisfiable.
\end{conj}

\textbf{Empirical verification.}  For $m = 6$, $m_1 \in \{5, \dots,
8\}$ all $36$ elements of $\mathcal{P}_6$ admit between $6$ and $18$
satisfying orientation assignments (script:
\texttt{experiments/orbit\_decomposition.py}).  No counterexample
exists in the tested range.

\textbf{Why a clean general proof is non-trivial.}
The straightforward attempt --- ``$o_j = 0$ for all $j$ is always
satisfying'' --- fails:  the rainbow $\sigma = (1, 2, 3, 2, 3, 1)$
at $m_1 = 6$ has $R_4(0) = L_5(0) = 2$, violating the $L_4 = 2$ clause.
A satisfying assignment exists (e.g.\ $(0, 0, 0, 0, 1, 0)$) but the
needed flip depends on $\sigma$.  A correct general proof must
analyse the implication graph of the cyclic 2-SAT, which has
$\sigma$-dependent forbidden combinations.

The forbidden $(o_j, o_{j+1})$ combo at a length-$2$ gap depends on
the pair $(\sigma_j, \sigma_{j+1})$ via the third color
$u := \{1,2,3\} \setminus \{\sigma_j, \sigma_{j+1}\}$:
\begin{center}
\begin{tabular}{lll}
\toprule
case & forbidden $(o_j, o_{j+1})$ & implication\\
\midrule
$\sigma_j = \sigma_{j+1}$ & $(0,1), (1,0)$ & $o_j = o_{j+1}$\\
$\sigma_j, \sigma_{j+1} \in \{2,3\}$, $u = 1$ & $(1, 0)$ & $o_j = 1 \Rightarrow o_{j+1} = 1$\\
$(\sigma_j, \sigma_{j+1}) = (1, 3)$, $u = 2$ & $(1, 1)$ & $o_j = 1 \Rightarrow o_{j+1} = 0$\\
$(\sigma_j, \sigma_{j+1}) = (3, 1)$, $u = 2$ & $(0, 0)$ & $o_j = 0 \Rightarrow o_{j+1} = 1$\\
$\sigma_j, \sigma_{j+1} \in \{1,2\}$, $u = 3$ & $(0, 1)$ & $o_j = 0 \Rightarrow o_{j+1} = 0$\\
\bottomrule
\end{tabular}
\end{center}

A clean proof of Conjecture~\ref{conj:2sat} likely proceeds by:
(i) Showing the implication graph on $\{(o_j, o_{j+1})\}$ is bipartite
in a strong sense that prevents the cyclic chain from forcing a
contradiction;  (ii) Reducing modulo color symmetry ($S_3$ acts on
$\sigma$ and on the orientation labels jointly), since
$\mathcal{P}_m$ has $|\mathcal{P}_m| / 6 = 6$ orbits and verifying
solvability on $6$ representatives suffices.

\section*{The sharpness counterexample}

\begin{prop}[Sharpness at $m_1 = m - 2$]
\label{prop:sharpness}
At $m = 6, m_1 = 4$, the rainbow $\sigma = (1, 2, 3, 2, 3, 1)$ is
\emph{not} in $\pi_D$.
\end{prop}

\begin{proof}
Balanced triangulation gives two length-$1$ gaps, between
$D$-positions $(2,3)$ (with $(\sigma_1, \sigma_2) = (2, 3)$,
forcing $c(e_2) = 1$) and $(7,8)$ (with $(\sigma_4, \sigma_5) =
(3, 1)$, forcing $c(e_7) = 2$).  Propagating through
$T'_{\mathrm{ann}}$ via the remaining length-$2$ gaps:
$c(e_3) = 2$ (from $D$-position $3$), $c(e_8) = 3$ (from $D$-position
$8$), $c(e_9) = 2$ (from $U$-position $9$ with $c(e_8) = 3$),
$c(e_0) = 3$ (from $D$-position $0$), $c(e_1) = 1$ (from $U$-position
$1$ with $c(e_0) = 3$), but the constraint at $D$-position $2$ then
demands $c(e_1) = 3 \neq 1$.  Contradiction.
\end{proof}

\section*{What this gives us}

Combining Lemma~\ref{lem:o-face}, Proposition~\ref{prop:reduction},
and the (empirically verified, not proven)
Conjecture~\ref{conj:2sat}:

\begin{cor}[Provisional]
\label{cor:provisional}
For $m \in \{4, 6\}$ and balanced triangulation with $m_1 \geq m - 1$:
$\pi_D = \mathcal{P}_m$, in particular the rainbow $S_3$-orbit lies
in $\pi_D$.
\end{cor}

\begin{cor}[Chain pigeonhole simplification, conditional]
\label{cor:chain}
Conditional on Conjecture~\ref{conj:2sat}, the chain-pigeonhole step
at $|\gamma| = m$ for an antipodal-chord SP tire $T_1$ with $m_1
\geq m - 1$ reduces to:  $\pi_U(\mathcal{C}(T_2)) \cap \mathcal{P}_m
\neq \emptyset$, i.e.\ a non-empty intersection on the perms-per-half
set rather than on all of $\{1,2,3\}^m$.
\end{cor}

\section*{What's missing}

A clean structural proof of Conjecture~\ref{conj:2sat}.  Two
candidate routes:
\begin{enumerate}
  \item \textbf{$S_3$-equivariant case analysis.}  Reduce to $6$
        representative $\sigma$'s (one per $S_3$-orbit) and explicitly
        construct a satisfying $o$ for each at $m_1 = m - 1$ (the
        tightest case).
  \item \textbf{Global implication-graph analysis.}  Show that the
        cyclic chain of implications never produces a contradiction.
        This likely involves a parity argument on the number of
        ``odd implications'' around the cycle.
\end{enumerate}

\end{document}
