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Reframe the constraint floor honestly as a conjecture

Browse Source 
Section 4 no longer states the floor as a proven Proposition. Now: prove
interior-free disks attain 2^(n-2) (ear-peeling) and the un-stacking
lemma, state |Phi(D)| >= 2^(n-2) as a Conjecture, and give an honest
status remark -- holds for the Apollonian class, reduces to the
irreducible case, empirically strict (5/4), but |Phi| is NOT monotone
(the earlier freedom-positive monotonicity claim was wrong) and both
natural elementary proofs provably fail. Soften the note's observation to
match.

Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>
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didericis
2026-06-17 21:32:57 -04:00
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@@ -102,10 +102,11 @@ realises $5$ sequences against the fan's $4$, since each interior vertex
contributes two faces but only one constraint. Thus: contributes two faces but only one constraint. Thus:
\begin{obs} \begin{obs}
The outer $n$-cycle cannot be constrained below $2^{n-2}$ achievable The interior-free triangulation already attains $2^{n-2}$, no search disk
sequences, and no nested structure is needed to reach the floor: a single beats it, and deep nesting only approaches this value from above ---
trivial tire is already maximal. Deep nesting only approaches the floor suggesting it is a floor, with a single trivial tire already maximally
from above. constraining. Whether $2^{n-2}$ is a genuine lower bound for \emph{all}
disks is the Conjecture below; it is \emph{not} a proven theorem.
\end{obs} \end{obs}
The achievability is transparent: in a fan from $v_0$, The achievability is transparent: in a fan from $v_0$,
papers/heawood_restrictions_on_nested_tire_graph_duals/paper.aux
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\bibcite{bauerfeld-depth}{2} \bibcite{bauerfeld-depth}{2}
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\bibcite{bauerfeld-medial-tires}{4} \bibcite{bauerfeld-medial-tires}{4}
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@@ -448,9 +448,9 @@ A nested substructure constrains its outer interface through the set of
Heawood boundary sequences it can realise. By the self-similarity of the Heawood boundary sequences it can realise. By the self-similarity of the
tire decomposition (\cite{bauerfeld-nested-tires}), the region $G_T$ tire decomposition (\cite{bauerfeld-nested-tires}), the region $G_T$
enclosed by a tire's outer cycle, away from the source, is itself a 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 triangulated disk; we ask how tightly any such disk can constrain its
boundary. The bound below depends only on the disk triangulation, not on boundary. The achievable set below depends only on the disk
a tire-tree labelling. triangulation, not on a tire-tree labelling.
\begin{definition}[Achievable boundary set of a disk] \begin{definition}[Achievable boundary set of a disk]
\label{def:achievable-boundary-set} \label{def:achievable-boundary-set}
@@ -468,62 +468,75 @@ is
\] \]
\end{definition} \end{definition}
\begin{proposition}[Constraint floor] \begin{proposition}[Interior-free disks attain $2^{n-2}$]
\label{prop:constraint-floor} \label{prop:attainment}
For every triangulated disk $D$ with boundary an $n$-cycle, If $D$ has no interior vertices then $|\Phi(D)| = 2^{\,n-2}$.
\[
|\Phi(D)| \;\ge\; 2^{\,n-2},
\]
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} \end{proposition}
\begin{proof}[Proof of attainment] \begin{proof}
Triangulate the $n$-gon as a fan from $v_0$, with faces A triangulation of the $n$-gon has an \emph{ear}: a face
$\{v_0, v_i, v_{i+1}\}$ for $1 \le i \le n-2$ and labels $(v_{i-1}, v_i, v_{i+1})$ whose middle vertex $v_i$ has face-degree $1$,
$\lambda_i := \lambda(\{v_0, v_i, v_{i+1}\})$; there are no interior so $\lambda^{*}(v_i)$ equals that face's label and is read directly off
vertices, so every labelling is interior-valid. The induced boundary the boundary sequence. Deleting the ear leaves a triangulation of the
values are $(n-1)$-gon inducing the restricted boundary sequence; inducting, the
\[ $n-2$ face labels are recovered injectively, so $\lambda \mapsto
\lambda^{*}(v_1) = \lambda_1, \quad \lambda^{*}|_C$ is a bijection onto a set of size $2^{\,n-2}$.
\lambda^{*}(v_i) = \lambda_{i-1} + \lambda_i \ \ (1 < i < n-1), \quad
\lambda^{*}(v_{n-1}) = \lambda_{n-2}, \quad
\lambda^{*}(v_0) = \textstyle\sum_{j} \lambda_j .
\]
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} \end{proof}
\begin{remark}[Depth is freedom-positive] \begin{lemma}[Un-stacking]
\label{rem:freedom-positive} \label{lem:unstack}
The lower bound is plausible from a counting balance. A triangulated If $v$ is a degree-$3$ interior vertex of $D$, deleting it and restoring
disk with $k$ interior vertices has $2k + n - 2$ faces (Euler) and its link triangle as a single face yields a disk $D'$ with one fewer
imposes exactly $k$ interior Heawood constraints, one per interior interior vertex and $\Phi(D') = \Phi(D)$.
vertex. So each interior vertex contributes \emph{two} faces --- two new \end{lemma}
$\{+1,-1\}$ degrees of freedom --- against only \emph{one} constraint,
and the free dimension $(2k + n - 2) - k = k + n - 2$ \emph{grows} with \begin{proof}
depth. Going deeper is freedom-positive on balance: the boundary The constraint at $v$ forces its three faces to a common value $s$,
projection $\Phi(D)$ can only retain or enlarge its options, never drop contributing $2s \equiv -s$ to each of the three link vertices; the
below the interior-free value $2^{\,n-2}$. (Empirically $|\Phi(D)|$ does restored triangle, labelled $-s$, reproduces that contribution at each,
grow with $k$; e.g.\ on the $4$-cycle the central-apex wheel realises $5$ and $s \mapsto -s$ is a bijection on $\{+1,-1\}$. The resulting map is a
sequences against the fan's $4$.) The constraints relate only bijection between interior-valid labellings of $D$ and of $D'$ preserving
interior-incident faces and cannot collapse the $n-2$ degrees of freedom every boundary value, hence $\Phi(D) = \Phi(D')$.
carried by the boundary-incident faces --- which is the content the lower \end{proof}
bound must make precise.
\begin{conjecture}[Constraint floor]
\label{conj:constraint-floor}
For every triangulated disk $D$ with boundary an $n$-cycle,
$|\Phi(D)| \ge 2^{\,n-2}$. Equivalently, no nested structure constrains
the outer cycle below $2^{\,n-2}$ achievable Heawood sequences.
\end{conjecture}
\begin{remark}[Status of Conjecture~\ref{conj:constraint-floor}]
\label{rem:floor-status}
Iterating Lemma~\ref{lem:unstack} reduces any disk, $\Phi$-faithfully and
at fixed $n$, to one with no degree-$3$ interior vertex: either
interior-free, where Proposition~\ref{prop:attainment} gives exactly
$2^{\,n-2}$, or \emph{irreducible} (every interior vertex of degree
$\ge 4$). Thus Conjecture~\ref{conj:constraint-floor} holds for the
entire stacked (Apollonian) class and reduces to the irreducible case.
Empirically it holds without exception over more than $10^4$ disks, and
\emph{strictly}: every irreducible disk satisfies $|\Phi(D)| \ge
\tfrac54 \cdot 2^{\,n-2}$, with equality at a single minimal-degree
interior vertex; the wheel $W_n$ gives $|\Phi(W_n)| = \lfloor 2^n/3
\rfloor$ and is \emph{not} the minimiser. A counting balance makes the
floor plausible --- a disk with $k$ interior vertices has $2k+n-2$ faces
(Euler) but only $k$ interior constraints, so the linear free dimension
$k+n-2$ grows with depth --- but this is only heuristic: $|\Phi(D)|$ is
\emph{not} monotone in $k$ (inserting a degree-$4$ vertex can shrink it),
it merely never drops below $2^{\,n-2}$. Two natural elementary proofs,
a $\Phi$-non-increasing vertex reduction and a direct $(n{-}2)$-face
transversal, both provably fail; a proof appears to need a global
argument on the Boolean / mod-$3$ structure of $\Phi$.
\end{remark} \end{remark}
\begin{remark} \begin{remark}
\label{rem:floor-consequences} \label{rem:floor-consequences}
Two consequences. First, $\Phi(D)$ is a $\mathbb{Z}/3$ zonotope --- a 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 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 $n-2$. the interior-free value it has size $2^{\,n-2}$ with affine hull of
Second, since the floor is exponential in the interface length $n$, a dimension $n-2$. Second, granting Conjecture~\ref{conj:constraint-floor},
maximally-constraining child still offers $2^{\,n-2}$ outer options, so the floor is exponential in the interface length $n$, so a
maximally-constraining child still offers $2^{\,n-2}$ outer options, and
the gluing of Conjecture~\ref{conj:heawood-chain-pigeonhole} has the least 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 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 easy at long ones; the difficulty of the programme is concentrated at