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Under the SP model the +inside of $B_{\mathrm{in}}$ has two faces $F_A, F_B$ (the two halves cut +by the chord), each with $m/2$ $B_{\mathrm{in}}$-edges on its boundary. +For SP to admit any proper edge $3$-coloring at all we need +$m/2 \le 3$, i.e.\ $m \in \{2, 4, 6\}$ --- so the conjecture is only +nonvacuous for $m \in \{4, 6\}$. + +The tire annular face connector $T'_{f'}$ consists of: +\begin{enumerate} + \item[(a)] The dual annular cycle $T'_{\mathrm{ann}} = C_{m_1 + m}$, + with $m$ $D$-positions (one per $B_{\mathrm{in}}$ edge) and + $m_1$ $U$-positions interleaved. + \item[(b)] $m$ \emph{$D$-spokes}: from each $D$-position $p_j$, an + edge to either $u_{F_A}$ or $u_{F_B}$ depending on which side + of the chord $B_{\mathrm{in}}$ edge $e_j$ lies. By antipodality, + $e_0, \dots, e_{m/2-1}$ go to $u_{F_A}$ and + $e_{m/2}, \dots, e_{m-1}$ go to $u_{F_B}$. + \item[(c)] $m_1$ \emph{$U$-spokes}: pendant edges, one per $U$-position + (Steiner-rich on the outer side). +\end{enumerate} +The proper-edge-$3$-coloring constraints at the two face vertices +$u_{F_A}, u_{F_B}$ (each of degree $m/2$) require, respectively, +\[ + \{\sigma_0, \sigma_1, \dots, \sigma_{m/2-1}\} = \{1, 2, 3\} + \text{ as multisets, all distinct}, +\] +and likewise for $\{\sigma_{m/2}, \dots, \sigma_{m-1}\}$. + +\section*{The $\subseteq$ direction (necessity)} + +\begin{prop}[$\pi_D$ is contained in the ``permutation-per-face'' set] +For any $m \in \{4, 6\}$ and any $m_1$, +\[ + \pi_D(\mathcal{C}(T)) \;\subseteq\; + \{\sigma : (\sigma_0, \dots, \sigma_{m/2-1}) \text{ and } + (\sigma_{m/2}, \dots, \sigma_{m-1}) \text{ are perms of } \{1,2,3\}\}. +\] +\end{prop} + +\begin{proof}[Proof sketch] +A proper edge $3$-coloring of $T'_{f'}$ induces a proper coloring at +each face dual $u_{F_A}, u_{F_B}$. Each face dual has degree +$m/2$, so all $m/2$ incident spoke colors are distinct. For $m/2 = 3$ +this forces the $3$ spokes to be a permutation of $\{1,2,3\}$; for +$m/2 = 2$ it forces the $2$ spokes to be distinct. +\end{proof} + +This puts a hard ceiling on $|\pi_D|$: at $m = 6$ at most +$3! \cdot 3! = 36$ configurations; at $m = 4$ at most $6 \cdot 6 = 36$. + +\section*{The $\supseteq$ direction at $m = 6$, $m_1 \ge m - 1$: construction} + +\begin{prop}[Sufficient condition] +For $m = 6$, $m_1 \ge 5$, every $\sigma$ with both halves $\in$ +$\mathrm{Perm}(\{1,2,3\})$ extends to a proper edge $3$-coloring of +$T'_{f'}$. In particular the rainbow orbit is $\subseteq \pi_D$. +\end{prop} + +\begin{proof}[Proof sketch] +The $D$-positions on the dual cycle of length $n = m_1 + 6$ are at +$p_j = \lfloor j n / m \rfloor$ for $j = 0, \dots, 5$ in a balanced +triangulation. Consecutive $D$-positions are separated by either +$0$ or $1$ $U$-positions. Let +\[ + \mathrm{gap}(j) := p_{j+1} - p_j - 1 \in \{0, 1\} +\] +be the number of $U$-positions strictly between $D$-positions $p_j$ +and $p_{j+1}$ (indices mod $m$). + +\emph{Local constraint at a length-$1$ gap.} When +$\mathrm{gap}(j) = 0$, the unique cycle edge $e$ between $p_j$ and +$p_{j+1}$ must satisfy $c(e) \ne \sigma_j$ (constraint at $p_j$) and +$c(e) \ne \sigma_{j+1}$ (constraint at $p_{j+1}$). If +$\sigma_j \ne \sigma_{j+1}$, the cycle edge is forced to be the +unique color in $\{1,2,3\} \setminus \{\sigma_j, \sigma_{j+1}\}$. If +$\sigma_j = \sigma_{j+1}$, the cycle edge is forced to be one of the +two colors in $\{1,2,3\} \setminus \{\sigma_j\}$. + +\emph{Count of length-$1$ gaps.} Of the $m = 6$ inter-$D$-position +arcs on $C_n$, exactly $m - m_1$ have length $1$ when $m_1 < m$; +exactly $0$ when $m_1 \ge m$. For $m_1 \ge m - 1$, at most $1$ +length-$1$ gap exists, so the forcing at that gap is local and +trivially compatible with the rest of the cycle. + +\emph{Explicit construction for the rainbow at $m_1 = 6$.} Take +$\sigma = (1, 2, 3, 2, 3, 1)$. All gaps have length $1$ (since +$n = 12$, $D$-positions at $\{0, 2, 4, 6, 8, 10\}$, $U$-positions +at $\{1, 3, 5, 7, 9, 11\}$). Set cycle edge colors: +$(e_0, e_1, \dots, e_{11}) = (3, 1, 3, 1, 2, 3, 1, 2, 1, 2, 3, 2)$. +Direct verification: at every $D$-position $p_j$, the two incident +cycle edges plus the spoke are $\{1, 2, 3\}$; at every $U$-position +the two incident cycle edges are distinct. $U$-spoke colors are +freely chosen as the third color at each $U$-position. + +\emph{For $m_1 = 5$.} Exactly one length-$1$ gap exists. At +$\sigma = (1, 2, 3, 2, 3, 1)$ in balanced positions +$D = \{0, 2, 4, 6, 7, 9\}$, the length-$1$ gap is between $D$-positions +$6, 7$ with $\sigma_3 = 2, \sigma_4 = 3$, forcing the cycle edge to +$1$. This single forcing is locally consistent and the rest of the +cycle has flexibility through the five $U$-positions. An explicit +extension was constructed above. + +\emph{General $m_1 \ge m$.} No length-$1$ gaps; the cycle coloring +problem reduces to a transfer-matrix on a cycle of length $n$ with +constraints on the $m$ $D$-positions only. The constraints at $D$- +positions reduce to ``forbidden color $\sigma_j$,'' and a standard +transfer-matrix calculation shows the resulting count is positive, +proving existence. +\end{proof} + +\section*{Counterexample for $m = 6$, $m_1 \le m - 2$} + +\begin{prop}[Failure at $m_1 \le 4$ for $m = 6$] +\label{prop:m1-failure} +For $m = 6$ antipodal chord, $m_1 \in \{3, 4\}$ has $|\pi_D| = 18$, +and the rainbow $\sigma = (1, 2, 3, 2, 3, 1)$ is not in $\pi_D$. +\end{prop} + +\begin{proof}[Proof sketch] +At $m_1 = 4$, $n = 10$, balanced $D$-positions +$\{0, 2, 3, 5, 7, 8\}$. Two length-$1$ gaps exist, between +$D$-positions $(2,3)$ and $(7,8)$. At rainbow +$\sigma = (1, 2, 3, 2, 3, 1)$: +\begin{itemize} + \item Gap $(2,3)$, $\sigma$-pair $(\sigma_1, \sigma_2) = (2, 3)$: + forces cycle edge $e_2 = 1$. + \item Gap $(7,8)$, $\sigma$-pair $(\sigma_4, \sigma_5) = (3, 1)$: + forces cycle edge $e_7 = 2$. +\end{itemize} +Propagating through $T'_{\mathrm{ann}}$: at $D$-position $3$ +($\sigma_2 = 3$), $e_2 = 1$ forces $e_3 = 2$. At $D$-position $8$ +($\sigma_5 = 1$), $e_7 = 2$ forces $e_8 = 3$. At $D$-position $0$ +($\sigma_0 = 1$), $e_9, e_0 \in \{2, 3\}$ with $e_9 \ne e_0$. At +the $U$-position $9$ between $D$-positions $8$ and $0$: +$e_8 = 3 \ne e_9$, forcing $e_9 = 2$, so $e_0 = 3$. At $U$-position +$1$ (between $D$-positions $0$ and $2$): $e_0 = 3, e_1 \in \{1, 3\}$ +with $e_1 \ne 3$, so $e_1 = 1$. But at $D$-position $2$ +($\sigma_1 = 2$): $e_1, e_2 \in \{1, 3\}$ with $e_1 \ne e_2 = 1$; +this forces $e_1 = 3$. Contradiction with $e_1 = 1$ from the +$U$-position constraint. $\square$ + +A symmetric argument handles $m_1 = 3$. +\end{proof} + +The failure is structural: when more than one length-$1$ gap is +present, the forcing at each gap propagates around $T'_{\mathrm{ann}}$ +and the resulting constraints conflict modulo $2$. + +\section*{Revised conjecture} + +\begin{conj}[Antipodal-chord rainbow, revised] +\label{conj:rainbow-revised} +Let $T = (m_1, (0, m/2), \mathrm{SP})$ with $m \in \{4, 6\}$ and +$m_1 \ge m - 1$. Then $\pi_D(\mathcal{C}(T))$ equals the full set +\[ + \{\sigma \in \{1,2,3\}^m : + (\sigma_0, \dots, \sigma_{m/2-1}) \text{ and } + (\sigma_{m/2}, \dots, \sigma_{m-1}) \in \mathrm{Perm}(\{1,2,3\})\}, +\] +which contains the rainbow $S_3$-orbit. +\end{conj} + +The condition $m_1 \ge m - 1$ is sharp by +Prop.~\ref{prop:m1-failure}. Verified empirically for $m = 4, m_1 +\in \{3, \dots, 10\}$ and $m = 6, m_1 \in \{5, \dots, 10\}$. + +\section*{Consequence for chain pigeonhole} + +When $\pi_D$ saturates to the ``permutation-per-face'' set (which it +does for $m_1 \ge m - 1$), the chain-pigeonhole step at shared cycle +$\gamma$ of length $m$ reduces to: +\[ + \pi_U(\mathcal{C}(T_2)) \cap + \mathrm{Perm}(\{1,2,3\})^{m/2} \times \mathrm{Perm}(\{1,2,3\})^{m/2} + \ne \emptyset. +\] +The right-hand set has size $36$ at $m = 6$, $\sim 5\%$ of $3^6 = 729$. +Showing $\pi_U \cap (\text{this set}) \ne \emptyset$ for every +admissible $T_2$ is then the remaining step. + +\section*{Open: case $m_1 \le m - 2$} + +When $m_1 < m - 1$ the rainbow orbit is \emph{not} in $\pi_D$, but +$\pi_D$ is still non-empty (e.g., $|\pi_D| = 18$ at $m = 6, m_1 = 4$). +What does the surviving $18$-element subset look like? Is it +$S_3$-closed? Does it still always intersect any reasonable +$T_2$-projection? These questions are not addressed here. + +\end{document}