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\title{Cut tires form a tree (under depth nesting)}
\author{}
\date{}

\newtheorem*{obs}{Observation}
\newtheorem*{prop}{Proposition}

\begin{document}
\maketitle

\section*{The claim}

\begin{prop}[Cut tires form a forest]
\label{prop:tree}
For each side $i$ of a $6$-edge cut of $G'$, the cut tires of $G'_i$,
parameterised by pairs $(d, f)$ with $d \ge 1$ and $f$ a face of
$H_d$, form a \emph{forest} under the parent--child relation
\[
   \mathrm{parent}\bigl(T_{d+1}^{(f')}\bigr) := T_d^{(f)}
\]
where $f$ is the unique face of $H_d$ in whose planar interior $f'$
lies in the inherited embedding of $G'_i$.

The forest's roots are the cut tires at depth $1$ (one per face of
$H_1$); their ``virtual parent'' is the cut $C$ itself.
\end{prop}

\begin{proof}[Sketch]
$H_{d+1}$ is a subgraph of $G'_i$ with the inherited planar
embedding.  Each face of $H_{d+1}$ is a maximal connected open
region of $|\Pi| \setminus E(H_{d+1})$ in the plane.

In particular, every face of $H_{d+1}$ lies inside some face of $H_d$
(since $H_d$ has fewer edges and so larger faces).  ``Lies inside''
means: the open face region of $H_{d+1}$ is a subset of an open face
region of $H_d$.  This containment is unique because the faces of
$H_d$ partition $|\Pi| \setminus E(H_d)$.

Hence parent is well-defined and unique.  No face of $H_{d+1}$ is
its own parent (because $d+1 > d$).  The relation defines a forest.

The roots are the depth-$1$ cut tires.  Their ``virtual parent'' is
the depth-$0$ pendant configuration, i.e.\ the cut $C$ itself.
\end{proof}

\section*{Why this matters for the chain half}

Chain pigeonhole asks whether the per-tire $S_3$-orbit structure
composes coherently through the chain.  With a tree structure on
the cut tires, this becomes a \textbf{tree dynamic-programming
problem}, not a general graph compatibility problem:
\begin{itemize}
  \item Process tires from leaves to root.
  \item At each leaf: $\pi(T_{\mathrm{leaf}})$ has known structure
        (e.g.\ $S_3$-orbits) from the per-tire half.
  \item Internal node $T_d^{(f)}$ combines:
        \begin{itemize}
          \item Its own internal $\pi(T)$ structure.
          \item Compatibility with each child $T_{d+1}^{(f')}$
                via the bijection $\{\text{in spokes of } T_d^{(f)}\}
                \leftrightarrow \{\text{face boundary edges of }
                T_{d+1}^{(f')}\}$.
        \end{itemize}
  \item Root: $T_1^{(\cdot)}$ projects its out-spoke colours to
        $\sigma_i \in \mathcal{R}_i$.
\end{itemize}

Tree DP is well-understood: $|\mathcal{R}_i|$ can be computed
exactly in linear time in the tree size (with size-$|\pi|$ tables
at each node).  Whether the resulting $\mathcal{R}_0$ and
$\mathcal{R}_1$ intersect is a finite check at the cut.

The tree structure is also a \textbf{strong topological constraint}
on the chain pigeonhole obstruction: any counterexample to
chain pigeonhole at the cut must come from a tree-DP failure,
which is much narrower than a general-graph obstruction.

\section*{Empirical demonstration on Holton-McKay \#0}

\subsection*{$G'_1$ side ($|S| = 28$, depths $0$ to $7$)}

Two depth-$1$ roots:
\begin{itemize}
  \item Root $(1, 0)$: face length $12$, no children
        (the outer ``shell'' of $H_1$).
  \item Root $(1, 1)$: face length $4$, with substantial subtree:
        \begin{itemize}
          \item $(2, 0)$ \hfill $|f| = 7$
            \begin{itemize}
              \item $(3, 0)$ \hfill $|f| = 2$
                  $\Rightarrow$ $(4, 0)$ \hfill $|f| = 4$
                  $\Rightarrow$ $(5, 0)$ \hfill $|f| = 14$
              \item $(3, 1)$ \hfill $|f| = 2$
                  $\Rightarrow$ $(4, 1)$ \hfill $|f| = 8$
                  $\Rightarrow$ $(5, 1)$ \hfill $|f| = 2$
                  $\Rightarrow$ $(6, 0)$ \hfill $|f| = 12$
                  $\Rightarrow$ $(7, 0)$ \hfill $|f| = 2$
              \item $(3, 2)$ \hfill $|f| = 2$
            \end{itemize}
          \item $(2, 1)$ \hfill $|f| = 7$
        \end{itemize}
\end{itemize}

\subsection*{$G'_0$ side ($|S| = 10$, depths $0$ to $2$)}

Two depth-$1$ roots:
\begin{itemize}
  \item Root $(1, 0)$: face length $9$, with one child $(2, 0)$
        ($|f| = 6$).
  \item Root $(1, 1)$: face length $9$, no children.
\end{itemize}

\section*{Caveats on the empirical parent identification}

The empirical demonstration used a vertex-sharing heuristic to
identify parents: a face $f'$ of $H_{d+1}$ shares vertices with a
face $f$ of $H_d$, and we picked the parent as the one with smallest
face length.  This gives ambiguous candidates in some cases ($8$
ambiguous cases observed in $G'_1$) because vertex sharing does not
fully determine geometric containment.

A rigorous parent test would use \emph{point-in-region} containment:
pick a point in the open face of $H_{d+1}$ (e.g., the centroid of
its boundary walk), determine which face of $H_d$ that point lies
in (via the planar embedding's face structure).  This always gives
a unique answer.

The ambiguity in our empirical run doesn't reflect a violation of
the proposition --- it's an artifact of the heuristic.  Despite the
ambiguity, the resulting tree structure looked sensible in both
$G'_0$ and $G'_1$.

\section*{Consequence: the chain half becomes tractable}

With the tree structure established (or assumed), the chain half of
the loose chain pigeonhole conjecture reduces to:

\paragraph{Reformulated chain half (tree DP form).}
For each leaf cut tire $T_{\mathrm{leaf}}$, $\pi(T_{\mathrm{leaf}})$
is non-empty and $S_3$-closed.  Propagating bottom-up through the
parent--child relation preserves $S_3$-closure and non-emptiness.
At the root depth-$1$ tires, $\mathcal{R}_i$ is the join of the
root tires' out-spoke projections.  If $\mathcal{R}_i$ is
$S_3$-closed and contains a full $S_3$-orbit on each side, then
$\mathcal{R}_0 \cap \mathcal{R}_1 \neq \emptyset$ (containing a
common orbit by $S_3$-equivariance).

The remaining questions:
\begin{enumerate}
  \item Is non-emptiness preserved through parent-child propagation?
  \item Is $S_3$-closure preserved? (Yes, by $S_3$-equivariance of
        the proper edge $3$-colouring constraint.)
  \item Does the join of root projections contain a full $S_3$-orbit?
\end{enumerate}

Each of these is now a finite tree DP claim, much more tractable
than the original ``compose through the chain'' formulation.

\section*{Next step}

\begin{enumerate}
  \item Prove Proposition~\ref{prop:tree} rigorously using the
        point-in-region containment definition of parent.
  \item Implement the tree DP empirically on the Holton-McKay graphs
        and confirm $\mathcal{R}_0 \cap \mathcal{R}_1 \neq \emptyset$
        at the cut.
  \item Attempt an analytical bound: $|\mathcal{R}_i| \ge \mathrm{some
        function of tree size}$, ensuring $\mathcal{R}_0 \cap
        \mathcal{R}_1 \neq \emptyset$ in general.
\end{enumerate}

\end{document}
