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\title{Closed PDS chains under SR:\\
       outer-triangle absorption and the chain-pigeonhole conclusion}
\author{}
\date{}

\newtheorem*{obs}{Observation}
\newtheorem*{conj}{Conjecture}

\begin{document}
\maketitle

\section*{What this note records}

This is the experimental record of the SR + PDS closed-chain
experiment that followed the realisation (from
\texttt{tire\_fiber\_chunked.py} and \texttt{sr\_chain\_consistency.py})
that the Steiner-poor (SP) model with chord-on-$O$ constraints is
not the right setting for PDS-decomposed maximal planar graphs.  In
the correct SR setting, each $\gamma$-edge has its own inner-spoke
pendant, and the chord-induced face structure of $O$ does not enter
$T'_{f'}$ directly.  Under this regime:

\begin{itemize}
  \item Step 1 (saturation) gives $\pi_D$ on the long-side saturates
        $\{1,2,3\}^k$ when $m \geq k$ (proven empirically up to
        $k = 12$ in \texttt{tire\_fiber\_data.tex}; conjectured in
        general).
  \item Step 2 (pairwise chain compatibility) is automatic from
        step 1 whenever PDS grows outward ($m_1 \geq |\gamma|$ in
        each adjacent tire's outer-side).
  \item Step 3 (chain-wide consistency for open chains): held in
        every tested case in \texttt{sr\_chain\_consistency.py},
        with forward state size growing monotonically.
\end{itemize}

The remaining concern was the \textbf{boundary} of the chain.  A PDS
on a planar triangulated disk has two ends:

\begin{itemize}
  \item Innermost: $L_0$ a single source vertex (degenerate $B_{\text{in}}$
        for $T_1$).  No constraint on $\sigma$ at $L_0$.
  \item Outermost: $L_n$ the boundary of the outer face of $G$.  In
        the standard reduction (Birkhoff: minimal counterexample is
        internally $6$-connected, hence the outer face is a triangle),
        $L_n$ is a $3$-cycle.  Its dual constraint forces $\sigma$ at
        $L_n$ to be a permutation of $\{1,2,3\}$.
\end{itemize}

\section*{The closed-chain experiment}

Script: \texttt{experiments/sr\_closed\_chain.py}.  Output:
\texttt{experiments/sr\_closed\_chain\_data.txt}.

We forward-propagate state through a tire chain $T_1 | T_2 | \dots |
T_n$ with $T_1$'s $B_{\text{in}}$ degenerate (the source vertex) and
$T_n$'s $B_{\text{out}}$ being the outer triangle of length $3$.
Each tire is SR (no chord constraints) and the chain shape (the
sequence of level-cycle sizes) is varied.  At each step, the state
is the set of $\sigma$-patterns realisable on the current shared
cycle.

\begin{figure}[h]
\centering
\includegraphics[width=\textwidth]{fig_outer_triangle_absorption.png}
\caption{(A) A representative closed PDS chain on a triangulated
disk: source $L_0$, level cycles $L_1, L_2, L_3, L_4$ with sizes
$5, 6, 5, 3$.  Tires $T_1, T_2, T_3, T_4$ are the annular regions
between consecutive levels.  (B) The outer triangle's contribution
in $G'$: the outer face of $G$ becomes a single $G'$-vertex of
degree $3$.  Proper edge $3$-colouring requires its three incident
$G'$-edges (the duals of the three outer-triangle edges) to be
distinct.  So $\sigma|_{L_n}$ must be a permutation of $\{1,2,3\}$.
(C) State trajectory for the chain
$(6, 1) | (8, 6) | (10, 8) | (10, 10) | (8, 10) | (5, 8) | (3, 5)$:
state grows where cycles are widest and \emph{collapses to exactly
$6$ at the outer triangle}.}
\label{fig:absorption}
\end{figure}

\section*{Results}

Across all $10$ tested chains (source degrees $5, 6, 7$; chain
lengths $4$ to $7$; various growth/shrink shapes; outer triangle
fixed at size $3$):

\begin{itemize}
  \item Every chain is consistent (state never empties).
  \item In every chain, the final state at $L_n$ has size
        \textbf{exactly $6$}, with all $6$ elements being the
        permutations of $\{1, 2, 3\}$ on the $3$ outer-triangle edges.
  \item The outer-face dual-vertex constraint (degree $3$, distinct
        colours) is satisfied automatically.
\end{itemize}

Sample trajectories:

\begin{center}
\small
\begin{tabular}{l l}
chain & state sizes \\ \hline
$(5,1) | (6,5) | (5,6) | (3,5)$ & $30 \to 132 \to 60 \to 6$ \\
$(5,1) | (7,5) | (5,7) | (3,5)$ & $30 \to 312 \to 60 \to 6$ \\
$(5,1) | (8,5) | (8,8) | (5,8) | (3,5)$ & $30 \to 708 \to 1476 \to 60 \to 6$ \\
$(6,1) | (9,6) | (8,9) | (5,8) | (3,5)$ & $63 \to 1836 \to 1623 \to 60 \to 6$ \\
$(6,1) | (8,6) | (10,8) | (10,10) | (8,10) | (5,8) | (3,5)$ & $63 \to 783 \to 9993 \to 14643 \to 1641 \to 60 \to 6$ \\
$(7,1) | (9,7) | (7,9) | (3,7)$ & $126 \to 2130 \to 546 \to 6$ \\
\end{tabular}
\end{center}

\section*{What ``outer triangle absorption'' means}

The empirical pattern is striking: the final state is always exactly
$6$, and those $6$ are exactly the permutations of $\{1,2,3\}$.  But
this could be one of two distinct things:

\begin{description}
  \item[(H1) Chain-dependent absorption.]
        The state \emph{before} $T_n$ depends on the chain, but every
        such state happens to filter through $T_n$ to the same
        $6$-element set.  In this case the chain-pigeonhole step is
        non-trivial: it succeeds in feeding $T_n$ with the right
        kind of input.
  \item[(H2) $T_n$-only absorption.]
        $T_n$ (the outer tire with $m_n = 3$) has the property that
        its $\sigma_U$-support is intrinsically the $6$ permutations
        of $\{1, 2, 3\}$, \emph{regardless} of what state is fed to
        its $\sigma_D$-side.  In this case the absorption is a local
        property of $T_n$ alone, and the chain-consistency before
        $T_n$ contributes nothing.
\end{description}

These two are testable empirically: under (H1), feeding $T_n$ an
unrelated or random $\sigma_D$ state should not always give $6$
permutations; under (H2), the output is always $6$ permutations
regardless of input.

\begin{conj}[$T_n$-absorption is local]
\label{conj:absorption-local}
For any tire $T_n$ with $m_n = 3$ and $k_n \geq 3$, the $\sigma_U$
projection of $T_n$'s joint support equals exactly the $S_3$-orbit
of $(1, 2, 3)$ on the $3$ outer-spoke positions.  Equivalently:
every proper edge $3$-colouring of $C_{n_n}$, when restricted to the
$3$ U-positions, gives a permutation of $\{1, 2, 3\}$.
\end{conj}

If the conjecture holds, the closed-chain pattern is partly
\emph{trivial}: the outer tire takes care of itself.  The non-trivial
content of the chain consistency lives in (a) chain non-emptiness
(state never collapses) and (b) the final state's intersection with
the $6$-element target being non-empty.

Under (H1), in contrast, the chain-pigeonhole is doing real work,
threading the input to $T_n$ through a specific structure that the
chain enforces.

\section*{Why this matters for 4CT}

A proof of 4CT via this approach would have four ingredients:

\begin{enumerate}
  \item \textbf{PDS exists}: Bauerfeld's PDS gives a tire
        decomposition of every maximal planar $G$ from a level
        source.
  \item \textbf{SR is correct}: for $G$ a triangulation with
        sufficient connectivity (Birkhoff: internally $6$-connected),
        the SR model accurately describes $T'_{f'}$ at each tire.
        (This is what makes the chord-on-$O$ artifacts go away.)
  \item \textbf{Open-chain compatibility}: at every shared $\gamma$,
        $\pi_D$ from the outer tire saturates $\{1,2,3\}^\gamma$, so
        every $\sigma$ from the inner tire admits a continuation.
        (Step-1 saturation + outward PDS.)
  \item \textbf{Closed-chain compatibility}: the forward-propagated
        state at the outer triangle contains a permutation of
        $\{1,2,3\}$.  Empirically yes, in all tested cases.
\end{enumerate}

Items 1 and 2 are structural and depend on the PDS/Birkhoff theory.
Item 3 is step-1 saturation.  Item 4 is what this note's experiment
addresses, with the data above.

If Conjecture~\ref{conj:absorption-local} is true, item 4 is automatic
from local structure of $T_n$.  Combined with item 3 (chain
non-emptiness), the whole pigeonhole story closes for any PDS chain
ending at an outer triangle.

If Conjecture~\ref{conj:absorption-local} is false, item 4 still
holds empirically but for a genuinely chain-dependent reason: the
chain-pigeonhole does real structural work.  Either resolution would
be informative.

\section*{What's next}

\begin{enumerate}
  \item \textbf{Test the absorption conjecture directly.}  Take a
        tire $T_n = (m=3, k)$ for varying $k$ and check whether its
        $\sigma_U$-projection equals exactly the $6$ permutations of
        $\{1,2,3\}$.  This is a single-tire computation, fast.
  \item \textbf{Exhaustive chain enumeration.}  The $10$ tested
        chains are representative but not exhaustive.  Generate all
        valid PDS-shape chains of bounded length (e.g., $\leq 8$
        tires, $\leq 12$ max cycle size) and run forward propagation
        on each.  Any chain where state collapses to $\emptyset$
        before $T_n$, or where the final $L_n$ state misses some
        permutation, would be a real counterexample.
  \item \textbf{Connect to PDS in actual $G$.}  Verify that for an
        internally $6$-connected $G$, every PDS from any source gives
        a chain with the right SR structure (item 2).  This is the
        outstanding gap to make the analysis genuinely a 4CT
        argument.
  \item \textbf{Symbolic proof of absorption.}  If
        Conjecture~\ref{conj:absorption-local} holds, prove it by
        direct cycle-coloring analysis: any proper edge $3$-colouring
        of $C_{m+k}$ (with $m=3$) restricted to $3$ spread positions
        is a permutation.  This should be elementary.
\end{enumerate}

\section*{Caveats}

\begin{itemize}
  \item ``Outer triangle absorption'' is a name I have given to the
        observed phenomenon; it is not yet a theorem.
  \item The $10$ tested chains are hand-picked.  An exhaustive
        enumeration is warranted before claiming strong evidence.
  \item The SR-is-correct claim (item 2 above) is the load-bearing
        modelling assumption.  It is not yet verified in the
        literature in the form I am stating; tying this back to
        actual PDS+G structure is the most important remaining gap.
  \item The reflection-handling in
        \texttt{sr\_closed\_chain.py} picks the orientation that
        maximises forward state at each step.  In a real $G$ with a
        fixed embedding, the orientations are determined, and the
        worst-of-orientations might differ from the best-of.  This
        is unlikely to matter under saturation but is worth checking.
\end{itemize}

\end{document}
