Add the lower-bound proof programme to the note
Section 4: un-stacking lemma (degree-3 removal preserves Phi, proved), ear-peeling base case (k=0 => 2^(n-2)), reduction to the irreducible case, and the irreducible lemma as the sole open conjecture (|Phi| >= 5/4 * 2^(n-2), tight at the degree-4/5 patch; wheel = floor(2^n/3) is not extremal). Records the two dead ends (monotonicity false, universal toggles insufficient) and ties each claim to its experiment. Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>
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\newlabel{lem:unstack}{{}{2}}
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\newlabel{prop:reduction}{{}{3}}
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\newlabel{conj:irreducible}{{}{3}}
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@@ -14,6 +14,7 @@ zonotope structure and the $2^{n-2}$ constraint floor}
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\newtheorem*{prop}{Proposition}
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\newtheorem*{conj}{Conjecture}
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\newtheorem*{lem}{Lemma}
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\newtheorem*{verify}{Empirical check}
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\begin{document}
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\maketitle
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@@ -129,6 +130,87 @@ The minimal set is itself a sign-closed zonotope of size $2^{n-2}$, hull
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dimension $n-2$, not a $\mathrm{GF}(3)$ subspace --- the same fingerprint
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as $\S1$.
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\section*{4. A proof programme for the lower bound}
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The lower bound $|\Phi(D)| \ge 2^{n-2}$ reduces, by an exact
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$\Phi$-preserving reduction, to a single lemma about ``irreducible''
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disks. Two dead ends bound the search first: \emph{monotonicity is false}
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--- inserting a degree-$4$ interior vertex can shrink $|\Phi|$ ($6\to5$,
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$30\to28$; \texttt{monotonicity\_test.py}), so there is no reduce-to-base
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proof by ``adding vertices only grows $\Phi$''; and \emph{universal
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toggles are insufficient} --- a flip preserves feasibility for every
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labelling only if it touches no interior vertex (a \emph{boundary-only}
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face), and an irreducible disk can have none (the wheel has zero). What
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does work:
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\begin{lem}[Un-stacking; degree-$3$ removal preserves $\Phi$]
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\label{lem:unstack}
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Let $v$ be a degree-$3$ interior vertex of $D$, with link triangle
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$abc$ and incident faces $(vab),(vbc),(vca)$. Its constraint
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$\lambda_{vab}+\lambda_{vbc}+\lambda_{vca}\equiv 0 \pmod 3$ over
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$\{+1,-1\}$ forces the three to a common value $s$, so each of $a,b,c$
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receives $2s\equiv -s$ from $v$'s star. Let $D'$ delete $v$ and restore
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$abc$ as one face. Then setting that face to $-s$ reproduces the
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contribution $-s$ at $a,b,c$, and $s\mapsto -s$ is a bijection on
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$\{+1,-1\}$. Hence the map is a bijection between feasible labellings of
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$D$ and of $D'$ preserving every boundary value and interior constraint,
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so
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\[
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\Phi(D) = \Phi(D'), \qquad k(D') = k(D)-1 .
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\]
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\end{lem}
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\begin{verify}\textnormal{(\texttt{monotonicity\_test.py})} Degree-$3$
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insertion gave exact equality in $8884/8884$ trials.\end{verify}
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\begin{lem}[Base case; ear-peeling]
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\label{lem:base}
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If $D$ has no interior vertices ($k=0$) then $|\Phi(D)| = 2^{n-2}$. A
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polygon triangulation has an \emph{ear} $(v_{i-1},v_i,v_{i+1})$ with
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$v_i$ of face-degree $1$, so $\sigma_{v_i}=\lambda_{\mathrm{ear}}$ reads
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the ear label directly; remove it and induct on the $(n-1)$-gon. The
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boundary map is injective, giving $2^{n-2}$.
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\end{lem}
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\begin{prop}[Reduction to the irreducible case]
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\label{prop:reduction}
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Iterating Lemma~\ref{lem:unstack} terminates ($k$ strictly decreases) at
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a residue $D^{\ast}$ with no degree-$3$ interior vertex and the same
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$n$, and $\Phi(D)=\Phi(D^{\ast})$. The residue is either $k=0$, where
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$|\Phi|=2^{n-2}$ by Lemma~\ref{lem:base}, or \emph{irreducible}: $k\ge1$
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with every interior vertex of degree $\ge 4$. Hence
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\[
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|\Phi(D)| \ge 2^{n-2}
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\quad\Longleftarrow\quad
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|\Phi(D^{\ast})| \ge 2^{n-2}\ \text{for every irreducible } D^{\ast}.
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\]
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\end{prop}
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\begin{conj}[Irreducible lemma --- the remaining content]
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\label{conj:irreducible}
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Every irreducible disk satisfies $|\Phi| \ge 2^{n-2}$; in fact
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$|\Phi| \ge \tfrac54\cdot 2^{n-2} = 5\cdot 2^{n-4}$.
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\end{conj}
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\begin{verify}\textnormal{(\texttt{irreducible\_floor.py},
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\texttt{wheel\_extremal.py})} Over $10^4{+}$ irreducible disks
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($n=4,5,6$) there were $0$ floor violations and none sat on the floor.
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The bound $\tfrac54\cdot 2^{n-2}$ is \emph{tight}, attained by a single
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\textbf{minimal-degree} interior vertex (degree $4$ or $5$, which tie):
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the ratio $|\Phi|/2^{n-2}$ rises monotonically with the interior vertex's
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degree, $\tfrac54$ at $d\in\{4,5\}$, $\tfrac{21}{16}$ at $d\in\{6,7\}$,
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$\ldots$, up to $\tfrac43$ at the wheel $d=n$, where
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$|\Phi(W_n)|=\lfloor 2^n/3\rfloor$ exactly. So a proof of
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Conjecture~\ref{conj:irreducible} should be stress-tested against the
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degree-$4$ patch, the tight case --- \emph{not} the wheel.\end{verify}
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\noindent
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\emph{Status.} Lemmas~\ref{lem:unstack}--\ref{lem:base} and
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Proposition~\ref{prop:reduction} are proofs; they settle every disk that
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un-stacks to $k=0$ (the entire Apollonian class). The whole open content
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is Conjecture~\ref{conj:irreducible}, with guaranteed $25\%$ slack and a
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single explicit extremal disk.
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\section*{Consequence for the pigeonhole}
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Even a maximally-constraining child still presents $2^{n-2}$ outer
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