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+ \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}