Replace plane_depth_labelling with plane_depth_sequencing paper and script, remove unused lib modules
Co-Authored-By: Claude Sonnet 4.6 <noreply@anthropic.com>
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%% filename: amsart-template.tex
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%% date: 2014/07/24
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%% ====================================================================
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% AMS-LaTeX v.2 template for use with amsart
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%
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% Remove any commented or uncommented macros you do not use.
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\documentclass{amsart}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\theoremstyle{definition}
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{example}[theorem]{Example}
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\newtheorem{xca}[theorem]{Exercise}
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\theoremstyle{remark}
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\newtheorem{remark}[theorem]{Remark}
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\numberwithin{equation}{section}
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\begin{document}
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\title{Plane Depth Sequencing}
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% Remove any unused author tags.
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% author one information
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\author{Eric Bauerfeld}
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\address{}
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\curraddr{}
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\email{}
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\thanks{}
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\subjclass[2010]{Primary }
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\keywords{}
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\date{}
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\dedicatory{}
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\begin{abstract}
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\end{abstract}
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\maketitle
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\section{Definitions}
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\begin{definition}
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Let $G$ be a graph with a plane embedding, and let $C$ be the outer cycle of that embedding. The \emph{plane depth} of a vertex $v \in V(G)$ relative to the embedding and $C$ is
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\[
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\mathrm{depth}(v) = \min_{u \in V(C)} d(v, u),
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\]
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where $d(v, u)$ denotes the graph distance between $v$ and $u$ in $G$.
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\end{definition}
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\begin{definition}
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Let $G$ be a maximal planar graph with a plane embedding and outer cycle $C$. The \emph{deep embedding} of $G$ is the graph $G'$ obtained from $G$ by the following operation: for every 3-cycle $\{u, v, w\} \subseteq V(G)$ such that
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\[
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\mathrm{depth}(u) = \mathrm{depth}(v) = \mathrm{depth}(w),
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\]
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add a new vertex $x$ to $G$ adjacent to each of $u$, $v$, and $w$.
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\end{definition}
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\begin{lemma}
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Let $G'$ be the deep embedding of a maximal planar graph $G$. For each face of $G'$, the plane depths of its three vertices are either $d, d+1, d+1$ or $d, d, d+1$ for some $d \geq 0$.
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\end{lemma}
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\begin{proof}
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We first establish that for any edge $\{p, q\}$ in $G$, the depths of $p$ and $q$ differ by at most $1$. Suppose for contradiction that $\mathrm{depth}(p) = d$ and $\mathrm{depth}(q) = d + n$ for some $n \geq 2$. Since $\mathrm{depth}(p) = d$, there exists a path of length $d$ from $p$ to some vertex of $C$. Prepending the edge $\{q, p\}$ gives a path of length $d + 1$ from $q$ to $C$, so $\mathrm{depth}(q) \leq d + 1 < d + n$, a contradiction. The case $\mathrm{depth}(q) = d - n$ is handled identically: there exists a path of length $d - n$ from $q$ to some vertex of $C$, and prepending the edge $\{p, q\}$ gives a path of length $d - n + 1 \leq d - 1 < d$ from $p$ to $C$, contradicting $\mathrm{depth}(p) = d$.
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Since $G$ is a triangulation, every interior face of $G$ is a triangle $\{u,v,w\}$ with all three pairs adjacent. By the above, each pair of vertices in a triangle differs in depth by at most $1$, so no triangle can contain vertices of depths $d$ and $d+2$ simultaneously. The possible depth patterns for a triangle in $G$ are therefore exactly $d,d,d$, or $d,d,d+1$, or $d,d+1,d+1$.
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We now consider each case under the deep embedding.
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\textit{Case 1: depths $d,d,d+1$ or $d,d+1,d+1$.} These triangles are not modified by the deep embedding, so they remain as faces of $G'$ with the stated depth patterns, satisfying the lemma.
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\textit{Case 2: depths $d,d,d$.} The deep embedding inserts a new vertex $x$ adjacent to $u$, $v$, and $w$, replacing the face $\{u,v,w\}$ with three new faces $\{u,v,x\}$, $\{v,w,x\}$, and $\{u,w,x\}$. It remains to determine the depth of $x$ in $G'$. Since $x$ is adjacent only to $u$, $v$, and $w$, every path in $G'$ from $x$ to $C$ must pass through one of them, so $x$ has strictly greater depth than $u$, $v$, and $w$. Each of the three new faces thus has depth pattern $d,d,d+1$, satisfying the lemma.
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Since every face of $G'$ falls into one of these cases, the result follows.
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\end{proof}
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\end{document}
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