Reframe the constraint floor honestly as a conjecture
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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@@ -102,10 +102,11 @@ realises $5$ sequences against the fan's $4$, since each interior vertex
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contributes two faces but only one constraint. Thus:
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\begin{obs}
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The outer $n$-cycle cannot be constrained below $2^{n-2}$ achievable
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sequences, and no nested structure is needed to reach the floor: a single
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trivial tire is already maximal. Deep nesting only approaches the floor
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from above.
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The interior-free triangulation already attains $2^{n-2}$, no search disk
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beats it, and deep nesting only approaches this value from above ---
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suggesting it is a floor, with a single trivial tire already maximally
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constraining. Whether $2^{n-2}$ is a genuine lower bound for \emph{all}
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disks is the Conjecture below; it is \emph{not} a proven theorem.
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\end{obs}
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The achievability is transparent: in a fan from $v_0$,
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@@ -34,13 +34,14 @@
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\newlabel{rem:why-clusters}{{3.7}{5}}
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\newlabel{conj:heawood-chain-pigeonhole}{{3.8}{5}}
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\newlabel{conj:heawood-route-fct}{{3.9}{5}}
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\bibcite{Heawood1898}{1}
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\@writefile{toc}{\contentsline {section}{\tocsection {}{4}{The constraint floor}}{6}{}\protected@file@percent }
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\newlabel{sec:constraint-floor}{{4}{6}}
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\newlabel{def:achievable-boundary-set}{{4.1}{6}}
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\newlabel{prop:constraint-floor}{{4.2}{6}}
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\newlabel{rem:freedom-positive}{{4.3}{6}}
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\newlabel{rem:floor-consequences}{{4.4}{6}}
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\newlabel{prop:attainment}{{4.2}{6}}
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\newlabel{lem:unstack}{{4.3}{6}}
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\newlabel{conj:constraint-floor}{{4.4}{6}}
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\newlabel{rem:floor-status}{{4.5}{6}}
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\bibcite{Heawood1898}{1}
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\bibcite{bauerfeld-depth}{2}
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@@ -448,9 +448,9 @@ A nested substructure constrains its outer interface through the set of
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Heawood boundary sequences it can realise. By the self-similarity of the
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tire decomposition (\cite{bauerfeld-nested-tires}), the region $G_T$
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enclosed by a tire's outer cycle, away from the source, is itself a
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triangulated disk; we record how tightly any such disk can constrain its
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boundary. The bound below depends only on the disk triangulation, not on
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a tire-tree labelling.
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triangulated disk; we ask how tightly any such disk can constrain its
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boundary. The achievable set below depends only on the disk
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triangulation, not on a tire-tree labelling.
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\begin{definition}[Achievable boundary set of a disk]
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\label{def:achievable-boundary-set}
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@@ -468,62 +468,75 @@ is
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\]
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\end{definition}
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\begin{proposition}[Constraint floor]
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\label{prop:constraint-floor}
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For every triangulated disk $D$ with boundary an $n$-cycle,
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\[
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|\Phi(D)| \;\ge\; 2^{\,n-2},
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\]
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and the bound is attained --- already by the triangulation of the
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$n$-gon with no interior vertices. Consequently no nested structure
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constrains the outer cycle below $2^{\,n-2}$ achievable Heawood
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sequences; the trivial tire is already maximally constraining.
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\begin{proposition}[Interior-free disks attain $2^{n-2}$]
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\label{prop:attainment}
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If $D$ has no interior vertices then $|\Phi(D)| = 2^{\,n-2}$.
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\end{proposition}
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\begin{proof}[Proof of attainment]
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Triangulate the $n$-gon as a fan from $v_0$, with faces
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$\{v_0, v_i, v_{i+1}\}$ for $1 \le i \le n-2$ and labels
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$\lambda_i := \lambda(\{v_0, v_i, v_{i+1}\})$; there are no interior
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vertices, so every labelling is interior-valid. The induced boundary
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values are
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\[
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\lambda^{*}(v_1) = \lambda_1, \quad
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\lambda^{*}(v_i) = \lambda_{i-1} + \lambda_i \ \ (1 < i < n-1), \quad
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\lambda^{*}(v_{n-1}) = \lambda_{n-2}, \quad
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\lambda^{*}(v_0) = \textstyle\sum_{j} \lambda_j .
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\]
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From $\lambda^{*}(v_1)$ and the relations
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$\lambda_i = \lambda^{*}(v_i) - \lambda_{i-1}$ the tuple
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$(\lambda_1, \dots, \lambda_{n-2}) \in \{+1,-1\}^{n-2}$ is recovered from
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the boundary sequence, so the map $\lambda \mapsto \lambda^{*}|_C$ is
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injective and $|\Phi(D)| = 2^{\,n-2}$.
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\begin{proof}
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A triangulation of the $n$-gon has an \emph{ear}: a face
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$(v_{i-1}, v_i, v_{i+1})$ whose middle vertex $v_i$ has face-degree $1$,
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so $\lambda^{*}(v_i)$ equals that face's label and is read directly off
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the boundary sequence. Deleting the ear leaves a triangulation of the
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$(n-1)$-gon inducing the restricted boundary sequence; inducting, the
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$n-2$ face labels are recovered injectively, so $\lambda \mapsto
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\lambda^{*}|_C$ is a bijection onto a set of size $2^{\,n-2}$.
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\end{proof}
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\begin{remark}[Depth is freedom-positive]
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\label{rem:freedom-positive}
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The lower bound is plausible from a counting balance. A triangulated
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disk with $k$ interior vertices has $2k + n - 2$ faces (Euler) and
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imposes exactly $k$ interior Heawood constraints, one per interior
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vertex. So each interior vertex contributes \emph{two} faces --- two new
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$\{+1,-1\}$ degrees of freedom --- against only \emph{one} constraint,
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and the free dimension $(2k + n - 2) - k = k + n - 2$ \emph{grows} with
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depth. Going deeper is freedom-positive on balance: the boundary
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projection $\Phi(D)$ can only retain or enlarge its options, never drop
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below the interior-free value $2^{\,n-2}$. (Empirically $|\Phi(D)|$ does
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grow with $k$; e.g.\ on the $4$-cycle the central-apex wheel realises $5$
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sequences against the fan's $4$.) The constraints relate only
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interior-incident faces and cannot collapse the $n-2$ degrees of freedom
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carried by the boundary-incident faces --- which is the content the lower
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bound must make precise.
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\begin{lemma}[Un-stacking]
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\label{lem:unstack}
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If $v$ is a degree-$3$ interior vertex of $D$, deleting it and restoring
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its link triangle as a single face yields a disk $D'$ with one fewer
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interior vertex and $\Phi(D') = \Phi(D)$.
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\end{lemma}
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\begin{proof}
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The constraint at $v$ forces its three faces to a common value $s$,
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contributing $2s \equiv -s$ to each of the three link vertices; the
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restored triangle, labelled $-s$, reproduces that contribution at each,
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and $s \mapsto -s$ is a bijection on $\{+1,-1\}$. The resulting map is a
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bijection between interior-valid labellings of $D$ and of $D'$ preserving
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every boundary value, hence $\Phi(D) = \Phi(D')$.
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\end{proof}
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\begin{conjecture}[Constraint floor]
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\label{conj:constraint-floor}
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For every triangulated disk $D$ with boundary an $n$-cycle,
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$|\Phi(D)| \ge 2^{\,n-2}$. Equivalently, no nested structure constrains
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the outer cycle below $2^{\,n-2}$ achievable Heawood sequences.
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\end{conjecture}
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\begin{remark}[Status of Conjecture~\ref{conj:constraint-floor}]
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\label{rem:floor-status}
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Iterating Lemma~\ref{lem:unstack} reduces any disk, $\Phi$-faithfully and
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at fixed $n$, to one with no degree-$3$ interior vertex: either
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interior-free, where Proposition~\ref{prop:attainment} gives exactly
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$2^{\,n-2}$, or \emph{irreducible} (every interior vertex of degree
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$\ge 4$). Thus Conjecture~\ref{conj:constraint-floor} holds for the
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entire stacked (Apollonian) class and reduces to the irreducible case.
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Empirically it holds without exception over more than $10^4$ disks, and
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\emph{strictly}: every irreducible disk satisfies $|\Phi(D)| \ge
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\tfrac54 \cdot 2^{\,n-2}$, with equality at a single minimal-degree
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interior vertex; the wheel $W_n$ gives $|\Phi(W_n)| = \lfloor 2^n/3
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\rfloor$ and is \emph{not} the minimiser. A counting balance makes the
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floor plausible --- a disk with $k$ interior vertices has $2k+n-2$ faces
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(Euler) but only $k$ interior constraints, so the linear free dimension
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$k+n-2$ grows with depth --- but this is only heuristic: $|\Phi(D)|$ is
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\emph{not} monotone in $k$ (inserting a degree-$4$ vertex can shrink it),
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it merely never drops below $2^{\,n-2}$. Two natural elementary proofs,
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a $\Phi$-non-increasing vertex reduction and a direct $(n{-}2)$-face
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transversal, both provably fail; a proof appears to need a global
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argument on the Boolean / mod-$3$ structure of $\Phi$.
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\end{remark}
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\begin{remark}
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\label{rem:floor-consequences}
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Two consequences. First, $\Phi(D)$ is a $\mathbb{Z}/3$ zonotope --- a
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projected cube, sign-closed but not a $\mathrm{GF}(3)$ subspace --- and at
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the floor it has size $2^{\,n-2}$ with affine hull of dimension $n-2$.
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Second, since the floor is exponential in the interface length $n$, a
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maximally-constraining child still offers $2^{\,n-2}$ outer options, so
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the interior-free value it has size $2^{\,n-2}$ with affine hull of
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dimension $n-2$. Second, granting Conjecture~\ref{conj:constraint-floor},
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the floor is exponential in the interface length $n$, so a
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maximally-constraining child still offers $2^{\,n-2}$ outer options, and
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the gluing of Conjecture~\ref{conj:heawood-chain-pigeonhole} has the least
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slack at \emph{short} interfaces (e.g.\ $n = 4$ leaves $4$ options) and is
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easy at long ones; the difficulty of the programme is concentrated at
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