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\title{Fiber distributions for spoke-only tire face connectors:\\
       enumeration data for $n = 3, \dots, 12$}
\author{}
\date{}

\newtheorem*{obs}{Observation}
\newtheorem*{defn}{Definition}

\begin{document}
\maketitle

\section*{What this is}

This note is the output of Step~1 of the action items in
\texttt{birkhoff\_heesch\_reducibility.tex}: a brute-force enumeration
of the fiber distribution $N(T'_{f'}; \sigma)$ for small spoke-only
tire annular face connectors, framed as a $k = 3$ edge-coloring of the
graph
\[
   G_n \;:=\; C_n \;+\; \text{(one pendant edge at each cycle vertex)},
\]
which is the spoke-only $T'_{f'}$ when $n = m + k$ (the boundary
cycle lengths of the underlying tire sum to $n$).  Script:
\texttt{experiments/tire\_fiber\_enumeration.py}; raw data:
\texttt{experiments/fiber\_data.json}.

\section*{Setup recap}

A proper edge $3$-coloring of $G_n$ assigns colors to $n$ cycle edges
and $n$ pendant edges so that the three edges incident to each cycle
vertex $v_i$ get distinct colors.  Since $\deg(v_i) = 3$ and we have
exactly $3$ colors, the pendant color $s_i$ is uniquely determined by
the two cycle-edge colors at $v_i$ as $s_i = \{1,2,3\} \setminus
\{c(e_{i-1}), c(e_i)\}$.  Therefore the map
\[
   \{\text{proper edge $3$-colorings of } G_n\}
   \;\longrightarrow\;
   \Sigma := \{1,2,3\}^n, \qquad (c, s) \mapsto s,
\]
is well-defined, and
\[
   P_e(G_n, 3) \;=\; |\{\text{cycle colorings}\}| \;=\; 2^n + 2(-1)^n
   \;=\; \sum_{\sigma \in \Sigma} N(G_n; \sigma).
\]
The realisable support is $\mathcal{C} := \{\sigma : N(G_n; \sigma) >
0\}$.  ``A-reducibility'' (every $\sigma$ realisable) corresponds to
$\mathcal{C} = \Sigma$.

\section*{Main table}

\begin{center}
\small
\begin{tabular}{r r r r r r r r l}
\toprule
$n$ & $P_e$ & $|\mathcal{C}|$ & $|\Sigma| = 3^n$
    & $|\mathcal{C}|/|\Sigma|$ & A-red.? & max $N$ & min $N$ & fiber size dist.\\
\midrule
$3$  & $6$    & $6$    & $27$     & $0.222$ & no & $1$ & $1$ & $1{\times}6$ \\
$4$  & $18$   & $15$   & $81$     & $0.185$ & no & $2$ & $1$ & $1{\times}12,\;2{\times}3$ \\
$5$  & $30$   & $30$   & $243$    & $0.123$ & no & $1$ & $1$ & $1{\times}30$ \\
$6$  & $66$   & $63$   & $729$    & $0.086$ & no & $2$ & $1$ & $1{\times}60,\;2{\times}3$ \\
$7$  & $126$  & $126$  & $2187$   & $0.058$ & no & $1$ & $1$ & $1{\times}126$ \\
$8$  & $258$  & $255$  & $6561$   & $0.039$ & no & $2$ & $1$ & $1{\times}252,\;2{\times}3$ \\
$9$  & $510$  & $510$  & $19683$  & $0.026$ & no & $1$ & $1$ & $1{\times}510$ \\
$10$ & $1026$ & $1023$ & $59049$  & $0.017$ & no & $2$ & $1$ & $1{\times}1020,\;2{\times}3$ \\
$11$ & $2046$ & $2046$ & $177147$ & $0.012$ & no & $1$ & $1$ & $1{\times}2046$ \\
$12$ & $4098$ & $4095$ & $531441$ & $0.008$ & no & $2$ & $1$ & $1{\times}4092,\;2{\times}3$ \\
\bottomrule
\end{tabular}
\end{center}

\section*{Two empirical findings}

\begin{obs}[Fiber-size dichotomy]
\label{obs:dichotomy}
For every $n \in \{3, \dots, 12\}$ examined, every fiber has size $1$
or $2$.  The only $\sigma$ with $N(\sigma) = 2$ are the three constant
configurations $\sigma = (a, a, \dots, a)$ for $a \in \{1, 2, 3\}$,
and only when $n$ is even.  For odd $n$, no constant $\sigma$ is
realisable.
\end{obs}

This is straightforward from the cycle structure: a constant $\sigma$
forces every cycle edge to lie in $\{1,2,3\} \setminus \{a\}$, so the
cycle alternates between two colors --- possible only for even $n$,
and in two ways.  For non-constant $\sigma$, the cycle coloring is
uniquely determined by $\sigma$ via constraint propagation.

\medskip

\noindent
The much more interesting finding concerns projections of $\mathcal{C}$
onto a $k$-position subset of $\sigma$:

\begin{obs}[Spread-projection threshold]
\label{obs:threshold}
Fix $k \geq 3$ and let $P \subseteq \{0, 1, \dots, n-1\}$ be a set
of $k$ positions chosen as evenly spaced as possible (i.e.\
$P = \{\lfloor i n/k \rfloor : 0 \leq i < k\}$).  Write $\pi_P :
\Sigma \to \{1,2,3\}^k$ for the projection.  Then empirically, for
$n \in \{3, \dots, 12\}$,
\[
   \pi_P(\mathcal{C}) \;=\; \{1,2,3\}^k
   \qquad\Longleftrightarrow\qquad n \;\geq\; 2k.
\]
For $n < 2k$ the deficit $|\{1,2,3\}^k| - |\pi_P(\mathcal{C})|$ is
small (between $1$ and roughly $30$ across the data) and shrinks as
$n \to 2k$ from below.
\end{obs}

\paragraph{Threshold data (spread projection).}
\begin{center}
\small
\begin{tabular}{r | c c c c c c c c c c}
\toprule
& \multicolumn{10}{c}{$n$}\\
$k$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12$ \\
\midrule
$3$ & $6$ & $15$ & $24$ & $\mathbf{27}$ & $27$ & $27$ & $27$ & $27$ & $27$ & $27$ \\
$4$ & --- & $15$ & $30$ & $51$ & $78$ & $\mathbf{81}$ & $81$ & $81$ & $81$ & $81$ \\
$5$ & --- & --- & $30$ & $63$ & $114$ & $171$ & $240$ & $\mathbf{243}$ & $243$ & $243$ \\
$6$ & --- & --- & --- & $63$ & $126$ & $237$ & $384$ & $561$ & $726$ & $\mathbf{729}$ \\
\bottomrule
\end{tabular}
\end{center}
\noindent
(Entries are $|\pi_P(\mathcal{C})|$; bold $=$ first row where the
support reaches the universe $3^k$.  Universe sizes: $3^3 = 27,\;
3^4 = 81,\; 3^5 = 243,\; 3^6 = 729$.)

\paragraph{Contiguous projection (for contrast).}
Projecting onto a \emph{contiguous} block $P = \{0, 1, \dots, k-1\}$
gives a much smaller support that quickly stabilises and never reaches
the universe.  For $k = 3$ and $n \geq 5$, the contiguous projection
plateaus at exactly $21/27$, with the $6$ missing patterns being
\[
   \{(a, b, a) \in \{1,2,3\}^3 : a \neq b\}
   \;=\; \{(1,2,1), (1,3,1), (2,1,2), (2,3,2), (3,1,3), (3,2,3)\}.
\]
These are the ``$aba$ palindromes,'' and the obstruction is a local
one: at three consecutive cycle positions with $\sigma = (a, b, a)$,
the two cycle edges $c(e_0)$ and $c(e_1)$ are both forced to be the
unique color not in $\{a, b\}$, contradicting proper-coloring at the
middle vertex.

\section*{Reducibility interpretation}

Translating through the dictionary of
\texttt{birkhoff\_heesch\_reducibility.tex}:

\begin{itemize}
\item \textbf{No spoke-only tire is A-reducible.} The realisable
      support is a vanishing fraction of $\Sigma = \{1,2,3\}^n$ as $n$
      grows ($\sim\!2^n / 3^n$).  This is unsurprising --- it would
      have been startling if it held.

\item \textbf{Fiber concentration is extreme.} Every fiber has size
      $\leq 2$.  This is the strongest possible concentration of the
      fiber distribution: realisable spoke configurations and
      cycle-edge colorings are essentially in bijection.

\item \textbf{Spread-projection saturation gives a strong positive
      result.} The projection of $\mathcal{C}$ onto any half-cycle (or
      shorter spread block) covers \emph{all} of $\{1,2,3\}^k$ once
      $n \geq 2k$.  In tire language: if the dual annular cycle is at
      least twice as long as the shared boundary cycle, every ring
      coloring on the shared cycle is realisable.

\item \textbf{Tires with $m \geq k$ saturate.} A tire with boundary
      lengths $|B_{\mathrm{out}}| = m$ and $|B_{\mathrm{in}}| = k$ has
      dual cycle of length $n = m + k$ in the spoke-only setting.  The
      $k$ inner-direction spokes occupy $k$ positions among $n$; in a
      balanced triangulation (alternating U/D triangles), these
      positions are spread.  Observation~\ref{obs:threshold} then says
      $\pi(\mathcal{C}) = \{1,2,3\}^k$ whenever $m \geq k$.
\end{itemize}

\paragraph{Implication for the chain-pigeonhole step.}
Two tires sharing a cycle $\gamma$ of length $k$ admit a joint edge
$3$-coloring iff the realisable spoke-projections of the two tires
onto $\gamma$ intersect.  By the above:

\begin{itemize}
  \item If \emph{both} adjacent tires have their non-$\gamma$-boundary
        at least as long as $\gamma$, both projections equal
        $\{1,2,3\}^k$ and the intersection is trivially non-empty.
        Chain pigeonhole succeeds.
  \item If \emph{one} side saturates and the other does not, the
        intersection still equals the smaller (non-saturating) set,
        which is non-empty.  Chain pigeonhole succeeds.
  \item The only potentially troublesome case is when \emph{both}
        sides fail to saturate, i.e.\ both adjacent tires have
        non-$\gamma$ boundaries shorter than $\gamma$ (in particular,
        $\gamma$ is locally the longest cycle).
\end{itemize}

In a level-source BFS decomposition with cycles growing outward from
the source, ``$\gamma$ is locally the longest cycle'' happens at most
at one cycle: the outermost level.  So for a level decomposition with
strictly nested cycles, only the outermost shared boundary is a
candidate failure point.

\section*{Conjecture suggested by the data}

\begin{obs}[Conjectural saturation principle]
\label{obs:conjecture}
For any spoke-only tire $T$ with boundary cycle lengths
$|B_{\mathrm{out}}| = m$ and $|B_{\mathrm{in}}| = k$ satisfying
$m \geq k$, the projection of the realisable spoke support
$\mathcal{C}(T'_{f'})$ onto the $k$ inner-direction spokes equals
$\{1,2,3\}^k$, independently of the choice of triangulation
(provided the inner-spoke positions are not all consecutive).
\end{obs}

\noindent
The data verifies this for $m + k = n \leq 12$ and balanced (spread)
triangulations.  A proof would presumably proceed by induction on $m$
or by an explicit construction of cycle colorings realising any
target boundary configuration.  This is the natural next computational
step (e.g.\ extend the enumeration to $n \leq 18$ and try unbalanced
spoke partitions) and the natural next analytic step (try to prove
the saturation principle from first principles).

\section*{Caveats}

\begin{enumerate}
  \item All data is for the \textbf{spoke-only} case
        (Remark~1.16.1): no chords in the inner outerplanar graph
        $O$.  Tires with non-trivial $O$ have additional structure
        inside $T'_{f'}$ that we have not enumerated.
  \item The ``spread'' position pattern in the enumeration script
        uses
        $P = \{\lfloor in/k \rfloor : 0 \leq i < k\}$, which is one
        canonical evenly-spaced choice.  Other spread patterns may
        give different supports at the same $(n, k)$.
  \item The threshold $n = 2k$ is observed empirically up to $n = 12$.
        It is consistent with the heuristic that ``$k$ spread spoke
        positions on $C_n$ are independent enough to range freely once
        $n \geq 2k$,'' but is not proven here.
  \item Adjacent-tire intersection (the actual chain pigeonhole step)
        is left to a separate computation that takes pairs of fiber
        distributions and intersects projections on the shared cycle.
        That is Step~2 of the action items.
\end{enumerate}

\end{document}
