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coloring_nested_tire_graphs: rainbow-conjecture proof sketch + threshold counterexample

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Attempts to prove the antipodal-chord rainbow conjecture; result is
nuanced:

1. Upper bound (proven cleanly): pi_D ⊆ {σ : (σ_0..σ_{m/2-1}) and
   (σ_{m/2}..σ_{m-1}) are both perms of {1,2,3}}.  This follows
   from the proper-coloring constraint at the two O-face dual
   vertices, each of degree m/2 in T'_{f'}.

2. Lower bound at m_1 ≥ m - 1 (constructive): every σ in the
   above set extends to a proper edge 3-coloring of T'_{f'}.
   Explicit construction at m=6, m_1=6.  In particular rainbow
   ⊂ pi_D.

3. Counterexample at m_1 ≤ m - 2 (refutes original conjecture):
   at m=6, m_1=4, the rainbow σ = (1,2,3,2,3,1) is NOT in pi_D.
   Explicit forcing-propagation contradiction: two length-1
   inter-D-position gaps on T'_ann force conflicting cycle-edge
   colors at a U-position.  Empirically |pi_D| = 18 (half the
   full set) at m=6, m_1 ∈ {3, 4}.

REVISED conjecture: pi_D equals the full "perm-per-face" set
(containing the rainbow orbit) iff m_1 ≥ m - 1.  The threshold
m_1 ≥ m - 1 is sharp.  Verified for m=4 (all m_1 ≥ 3) and m=6
(m_1 ≥ 5).

Consequence: chain-pigeonhole at γ length m reduces to a smaller
overlap condition under m_1 ≥ m - 1.  The case m_1 < m - 1
remains open -- pi_D still nonempty but the rainbow orbit is
missing; structural characterization of the surviving 18-element
support not addressed.

Note: rainbow_proof_sketch.tex (3 pages).

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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\documentclass[11pt]{article}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{graphicx}
\usepackage{geometry}
\usepackage{booktabs}
\geometry{margin=1in}
\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}