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\title{Towards a proof of the antipodal-chord rainbow conjecture\\
       Partial result, counterexample, and revised statement}
\author{}
\date{}

\newtheorem*{thm}{Theorem}
\newtheorem*{prop}{Proposition}
\newtheorem*{conj}{Conjecture}
\newtheorem*{obs}{Observation}

\begin{document}
\maketitle

\section*{Status: partial proof + threshold counterexample}

The original conjecture
(\texttt{orbit\_decomposition.tex}, Obs.~``antipodal-rainbow-conjecture'')
claimed: \emph{for any} antipodal-chord SP tire $T = (m_1, (0, m/2),
\mathrm{SP})$ with $m$ even, the projection $\pi_D(\mathcal{C}(T))$
contains the $S_3$-orbit of $(a, b, c, b, c, \dots, b, c, a)$ on the
$m$ inner-side spokes.

\textbf{What we now know:}
\begin{itemize}
  \item The $\subseteq$ direction (necessity, $\pi_D$ is contained in a
        natural ``permutation-per-face'' set) is immediate.
  \item The $\supseteq$ direction (existence of a coloring realising
        the rainbow) holds for $m_1 \ge m - 1$ but fails for
        $m_1 \le m - 2$.
  \item For $m = 6$ specifically: empirically $\pi_D = $ the full
        $36$-element ``both halves are permutations of $\{1,2,3\}$''
        set whenever $m_1 \ge 5$; for $m_1 \in \{3, 4\}$ only $18$
        configurations survive and the rainbow orbit is \emph{not}
        among them.
\end{itemize}

\section*{Setup}

Throughout, let $T$ be a tire with $|B_{\mathrm{out}}| = m_1$,
$|B_{\mathrm{in}}| = m$ even, and inner outerplanar graph
$O = B_{\mathrm{in}} \cup \{(v_0, v_{m/2})\}$.  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}
