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Scaffold the 2^(n-2) constraint-floor proposition

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Add section 4: define the achievable boundary set Phi(D) of a triangulated
disk and state the constraint-floor proposition |Phi(D)| >= 2^(n-2), with
the attainment direction proved (fan injectivity) and the lower bound left
as a marked gap with strategy. Remark records the zonotope structure and
the short-interface concentration of difficulty.

Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>
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didericis
2026-06-17 02:14:52 -04:00
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commit d2156f06ee
4 changed files with 124 additions and 30 deletions
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papers/heawood_restrictions_on_nested_tire_graph_duals/paper.aux
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\bibcite{bauerfeld-nested-tires}{3} \bibcite{bauerfeld-nested-tires}{3}
\bibcite{bauerfeld-medial-tires}{4} \bibcite{bauerfeld-medial-tires}{4}
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papers/heawood_restrictions_on_nested_tire_graph_duals/paper.tex
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@@ -441,6 +441,84 @@ $G$ is properly $4$-vertex-colourable.
%% pigeonhole bound giving $N(k)$; orientation/reversal bookkeeping on %% pigeonhole bound giving $N(k)$; orientation/reversal bookkeeping on
%% the shared interface. %% the shared interface.
\section{The constraint floor}
\label{sec:constraint-floor}
A nested substructure constrains its outer interface through the set of
Heawood boundary sequences it can realise. By the self-similarity of the
tire decomposition (\cite{bauerfeld-nested-tires}), the region $G_T$
enclosed by a tire's outer cycle, away from the source, is itself a
triangulated disk; we record how tightly any such disk can constrain its
boundary. The bound below depends only on the disk triangulation, not on
a tire-tree labelling.
\begin{definition}[Achievable boundary set of a disk]
\label{def:achievable-boundary-set}
Let $D$ be a triangulated disk whose boundary is a simple $n$-cycle
$C = (v_0, \dots, v_{n-1})$. Call a Heawood face-labelling
$\lambda : F(D) \to \{+1,-1\}$ \emph{interior-valid} if
$\sum_{f \ni w} \lambda(f) \equiv 0 \pmod 3$ at every interior vertex $w$
of $D$ (no condition on $C$). The \emph{achievable boundary set} of $D$
is
\[
\Phi(D) \;:=\; \bigl\{\,
(\lambda^{*}(v_0), \dots, \lambda^{*}(v_{n-1}))
\;:\; \lambda \text{ interior-valid} \,\bigr\}
\;\subseteq\; \{0,1,-1\}^{n} .
\]
\end{definition}
\begin{proposition}[Constraint floor]
\label{prop:constraint-floor}
For every triangulated disk $D$ with boundary an $n$-cycle,
\[
|\Phi(D)| \;\ge\; 2^{\,n-2},
\]
and the bound is attained --- already by the triangulation of the
$n$-gon with no interior vertices. Consequently no nested structure
constrains the outer cycle below $2^{\,n-2}$ achievable Heawood
sequences; the trivial tire is already maximally constraining.
\end{proposition}
\begin{proof}[Proof of attainment]
Triangulate the $n$-gon as a fan from $v_0$, with faces
$\{v_0, v_i, v_{i+1}\}$ for $1 \le i \le n-2$ and labels
$\lambda_i := \lambda(\{v_0, v_i, v_{i+1}\})$; there are no interior
vertices, so every labelling is interior-valid. The induced boundary
values are
\[
\lambda^{*}(v_1) = \lambda_1, \quad
\lambda^{*}(v_i) = \lambda_{i-1} + \lambda_i \ \ (1 < i < n-1), \quad
\lambda^{*}(v_{n-1}) = \lambda_{n-2}, \quad
\lambda^{*}(v_0) = \textstyle\sum_{j} \lambda_j .
\]
From $\lambda^{*}(v_1)$ and the relations
$\lambda_i = \lambda^{*}(v_i) - \lambda_{i-1}$ the tuple
$(\lambda_1, \dots, \lambda_{n-2}) \in \{+1,-1\}^{n-2}$ is recovered from
the boundary sequence, so the map $\lambda \mapsto \lambda^{*}|_C$ is
injective and $|\Phi(D)| = 2^{\,n-2}$.
\end{proof}
%% TODO (lower bound): show |Phi(D)| >= 2^{n-2} for EVERY triangulated
%% disk D. Strategy: the n boundary-incident faces (one per boundary edge)
%% carry n-2 independent binary degrees of freedom after the interior
%% Heawood constraints are imposed; those constraints relate only
%% interior-incident faces and cannot collapse the boundary freedom below
%% 2^{n-2}. (See notes/boundary_restriction_structure.tex.)
\begin{remark}
\label{rem:floor-consequences}
Two consequences. First, $\Phi(D)$ is a $\mathbb{Z}/3$ zonotope --- a
projected cube, sign-closed but not a $\mathrm{GF}(3)$ subspace --- and at
the floor it has size $2^{\,n-2}$ with affine hull of dimension $n-2$.
Second, since the floor is exponential in the interface length $n$, a
maximally-constraining child still offers $2^{\,n-2}$ outer options, so
the gluing of Conjecture~\ref{conj:heawood-chain-pigeonhole} has the least
slack at \emph{short} interfaces (e.g.\ $n = 4$ leaves $4$ options) and is
easy at long ones; the difficulty of the programme is concentrated at
short level cycles.
\end{remark}
\begin{thebibliography}{9} \begin{thebibliography}{9}
\bibitem{Heawood1898} \bibitem{Heawood1898}