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coloring_nested_tire_graphs: detailed proof of the cut tire forest proposition

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Replaces the earlier sketch with a more detailed two-stage proof:

Stage 1: BFS level-set lemma.
  Lemma (BFS-adj): adjacent edges in line graph differ in depth by
    at most 1.  Proof: BFS-distance property.
  Lemma (level-set): every face of H_d contains only edges of depth
    <d, or only edges of depth >d.  Proof: if a face contains both,
    the line-graph walk connecting them must pass through a depth-d
    edge (by BFS-adj), contradicting the walk being in the face
    (= R^2 \ H_d).

Stage 2: faces of H_{d+1} embed in faces of H_d.
  Key claim: no H_d edge sits strictly inside any face of H_{d+1}.
  Informal topological argument: any H_d edge intruder into f' must
  already lie on f''s boundary closure.

Stage 3: forest structure follows from unique-parent + strictly
  decreasing depth.

HONESTLY ACKNOWLEDGED GAP: the topological argument in Stage 2 is
informal; a rigorous proof would set up the planar rotation system
and trace boundary walks carefully.  Empirically the conclusion
holds across 1486 tested cases (0 failures), giving very strong
support.

Note grows to 5 pages.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
didericis
2026-05-26 22:24:29 -04:00
parent 57eccad49c
commit 410c2a1b35
4 changed files with 197 additions and 31 deletions
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papers/coloring_nested_tire_graphs/notes/cut_tire_tree_structure.aux
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\@writefile{toc}{\contentsline {paragraph}{Reformulated chain half (tree DP form).}{5}{}\protected@file@percent }
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papers/coloring_nested_tire_graphs/notes/cut_tire_tree_structure.tex
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@@ -32,24 +32,133 @@ 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.
\begin{proof}
We prove the proposition in two stages.
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 $|\Pi| \setminus E(H_d)$.
\medskip
\noindent\textbf{Stage 1: the BFS level-set lemma.}
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.
\begin{lem}[BFS depth differs by at most 1 between adjacent edges]
\label{lem:bfs-adj}
Let $e_1, e_2 \in E(G'_i)$ share a vertex (so they are adjacent in
the line graph). Then $|\mathrm{depth}(e_1) - \mathrm{depth}(e_2)|
\le 1$.
\end{lem}
The roots are the depth-$1$ cut tires. Their ``virtual parent'' is
the depth-$0$ pendant configuration, i.e.\ the cut $C$ itself.
\begin{proof}
By definition of BFS depth, $\mathrm{depth}(e) = $ minimum line-graph
distance from $e$ to any pendant. If $e_1, e_2$ are line-graph
adjacent, then a shortest line-graph path from a pendant to $e_2$
can be extended by the one step from $e_2$ to $e_1$, yielding a path
of length $\mathrm{depth}(e_2) + 1$ from a pendant to $e_1$. So
$\mathrm{depth}(e_1) \le \mathrm{depth}(e_2) + 1$, and symmetrically.
\end{proof}
\begin{lem}[Level-set property of $H_d$]
\label{lem:level-set}
For each $d \ge 1$, every face of $H_d$ in the inherited planar
embedding satisfies one of the following:
\begin{itemize}
\item Every edge of $G'_i$ strictly inside the face has depth $< d$
(a ``low-side'' face), or
\item Every edge of $G'_i$ strictly inside the face has depth $> d$
(a ``high-side'' face).
\end{itemize}
\end{lem}
\begin{proof}
Let $f$ be a face of $H_d$. Suppose for contradiction that $f$
contains an edge $e_a$ of depth $a < d$ and an edge $e_b$ of depth
$b > d$ strictly inside. Since $f$ is a connected open region,
there is a continuous path in $f$ from a point on $e_a$ to a point
on $e_b$ avoiding $H_d$'s edges (since $f \subseteq
\mathbb{R}^2 \setminus H_d$).
Slightly perturbed, this path is realised as a sequence of edges in
$G'_i \setminus H_d$ together with possibly some vertices in
$V(G'_i)$ shared between consecutive edges --- i.e.\ a line-graph
walk in $G'_i \setminus H_d$ from $e_a$ to $e_b$ that stays inside
$\overline{f}$.
By Lemma~\ref{lem:bfs-adj}, consecutive edges along this line-graph
walk differ in depth by at most $1$. Going from depth $a < d$ to
depth $b > d$, the walk must pass through some edge of depth exactly
$d$. But that edge is in $H_d$, contradicting that the walk lies in
$G'_i \setminus H_d$.
Hence $f$ contains only edges of depth $< d$, or only edges of depth
$> d$ (or neither, if $f$ contains no edges of $G'_i$ in its
interior).
\end{proof}
\medskip
\noindent\textbf{Stage 2: faces of $H_{d+1}$ embed in faces of $H_d$.}
Pendants (depth $0$ edges) lie in some specific face of $H_d$; that
face is low-side. All other faces of $H_d$ are high-side and
contain depth-$> d$ edges, which includes all of $H_{d+1}$'s edges.
Let $f'$ be a face of $H_{d+1}$. We claim $f'$ is contained in
exactly one face of $H_d$.
\emph{Containment in at least one face:} $f'$ is an open connected
region of $\mathbb{R}^2 \setminus H_{d+1}$. In particular it is
connected. By Lemma~\ref{lem:level-set}, each face of $H_d$ is
either entirely low-side or entirely high-side, and the two types
are separated topologically by $H_d$. Suppose for contradiction
$f'$ intersects two distinct faces $g_1, g_2$ of $H_d$. Then a
path in $f'$ from a point in $g_1$ to a point in $g_2$ crosses some
edge of $H_d$ (since faces of $H_d$ are separated by $H_d$ edges).
But $H_d \subset E(G'_i) \setminus E(H_{d+1})$, so $H_d$ edges are
in $\mathbb{R}^2 \setminus E(H_{d+1})$; they could in principle lie
within $f'$ \emph{except} that $f'$ is a maximal connected open
component of that complement, which already includes the $H_d$
edges. This is where the elementary topological argument is
subtle: we need the additional constraint that no $H_d$ edge
sits strictly inside $f'$.
\emph{No $H_d$ edge sits strictly inside $f'$:} suppose an $H_d$ edge
$e$ is strictly inside $f'$. Then $e$'s endpoints are inside $f'$
(or on $\partial f'$). An endpoint $v$ of $e$ is also incident to
$H_{d+1}$ edges (since $V(H_d) \cap V(H_{d+1})$ contains vertices
where depth-$d$ and depth-$(d+1)$ edges meet; in cubic $G'_i$, $v$
has $3$ edges with various depths). The $H_{d+1}$ edges incident
to $v$ are on $\partial f'$ (the boundary walk of $f'$), so $v \in
\partial f'$. Then $e$'s other endpoint $w$ is also on or inside
$f'$. But moving from $v$ along $e$ into $w$: this curve segment
is inside $f'$ until it reaches $w$. If $w$ is on $\partial f'$,
the entire edge $e$ lies on the boundary closure $\overline{f'}$,
not strictly inside. If $w$ is strictly inside $f'$, then $w$'s
incident edges (including $e$) project into $f'$ in a way that
should appear on $\partial f'$ --- but $e$ is not in $H_{d+1}$,
contradiction.
\medskip
The careful case analysis shows: no $H_d$ edge sits strictly inside
$f'$, hence $f'$ is contained in a single face of $H_d$ (the unique
face whose interior contains $f'$).
\medskip
\noindent\textbf{Conclusion: forest structure.}
The parent relation $(d+1, f') \mapsto (d, f)$ assigns each
$H_{d+1}$ face $f'$ to a unique $H_d$ face $f$ containing it. Since
parent depth is strictly less than child depth, walking up parent
links strictly decreases depth, terminating at a depth-$1$ root (or
at the ``cut'' for the depth-$1$ roots' virtual parent). No cycles
can form. Hence the parent relation defines a forest. \qed
\end{proof}
\paragraph{Caveat on Stage 2.} The argument that ``no $H_d$ edge sits
strictly inside $f'$'' uses an informal topological case analysis on
how an $H_d$ edge inside $f'$ would have to interact with $f'$'s
boundary. A fully rigorous proof would set up the topological
framework more carefully (e.g.\ via the rotation system of the
planar embedding, tracing the boundary walk of $f'$ around an
``intruder'' $H_d$ edge to show it must already lie in
$\partial f'$). Empirically, the conclusion holds across
\textbf{$1486$ tested cases, $0$ failures} (see broader sweep below).
\section*{Why this matters for the chain half}
Chain pigeonhole asks whether the per-tire $S_3$-orbit structure