split: extract foundational depth material into new plane_depth paper
Splits the existing plane_depth_sequencing paper into two:
papers/plane_depth/paper.tex (NEW, 4 pages):
- Plane depth definition.
- Level edge, up/down/neutral triangle classification.
- Outerplanarity lemma (formerly Lemma 2.6 of PDS).
- Deep embedding G' definition.
- "Every face of G' is up or down" lemma.
- Unique level edge per face; shared level edge between adjacent faces.
- Quadrilateral decomposition definition with three types
(shallow diamond, deep diamond, S quad).
papers/plane_depth_sequencing/paper.tex (slimmed from 11 → 6 pages):
- Cites plane_depth for all foundational definitions.
- Keeps: slice, move definitions (anchor drop, level add, join,
ring completion), move selection, termination theorem.
papers/coloring_nested_tire_graphs/paper.tex:
- Bibliography updated: cite bauerfeld-depth instead of bauerfeld-pds.
- Two in-text references updated to cite the new outerplanarity
lemma in plane_depth.
Rationale: the outerplanarity / deep-embedding / quadrilateral-
decomposition material is foundational and reused by multiple
papers (and by the proposed level-cycle generalization). The
quadrilateral-sequencing programme is one specific application.
Splitting lets coloring_nested_tire_graphs cite the foundations
cleanly without dragging in the sequencing machinery.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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\documentclass{amsart}
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\usepackage{amssymb}
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\usepackage{graphicx}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{proposition}[theorem]{Proposition}
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\newtheorem{corollary}[theorem]{Corollary}
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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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\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}
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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{plane graph, triangulation, plane depth, level edge, deep
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embedding, quadrilateral decomposition, $k$-outerplanar graph}
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\date{}
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\dedicatory{}
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\begin{abstract}
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Given a plane embedding of a graph with outer cycle $C$, the
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\emph{plane depth} of a vertex is its graph distance to $C$. We
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develop this depth function into a layered combinatorial structure on
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plane triangulations: the subgraph induced by each depth level is
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outerplanar (recovering Baker's notion of a $k$-outerplanar graph);
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each triangular face is classified by its depth multiset as
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\emph{up}, \emph{down}, or \emph{neutral}; and the \emph{deep
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embedding} of a maximal planar graph, obtained by inserting a vertex
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into every neutral face (including the outer face), has every face
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either up or down. Pairing adjacent triangles across their unique
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level edge yields a \emph{quadrilateral decomposition} of the
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spherical deep embedding into three combinatorial types: shallow
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diamonds, deep diamonds, and S quads.
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This paper isolates the foundational depth-and-decomposition material
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that supports several downstream applications --- including the
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quadrilateral sequencing of \cite{bauerfeld-pds-seq} and the
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nested-tire colouring framework of \cite{bauerfeld-nested-tires}.
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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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\label{def:plane-depth}
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Let $G$ be a graph with a plane embedding, and let $C$ be the outer
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cycle of that embedding. The \emph{plane depth} of a vertex
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$v \in V(G)$ 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
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$G$.
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\end{definition}
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\begin{definition}
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\label{def:level-edge}
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An edge $\{u, v\} \in E(G)$ is a \emph{level edge} if
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$\mathrm{depth}(u) = \mathrm{depth}(v)$.
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\end{definition}
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\begin{definition}
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\label{def:triangle-types}
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A triangle $\{u, v, w\}$ in $G$ is an \emph{up triangle} if the
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multiset of depths of its vertices is $\{d, d+1, d+1\}$ for some
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$d \geq 0$, a \emph{down triangle} if the multiset of depths is
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$\{d, d, d+1\}$ for some $d \geq 0$, and a \emph{neutral triangle} if
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the multiset of depths is $\{d, d, d\}$ for some $d \geq 0$.
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\end{definition}
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\begin{remark}
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We now relate our terminology to existing terminology, namely
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$k$-outerplanar graphs \cite{baker1994}. The following definition and
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lemma show that the subgraph induced by any single depth level
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relative to any source set on the outer face is outerplanar, i.e.\
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$1$-outerplanar in the sense of Baker.
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\end{remark}
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\begin{definition}
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A plane graph is \emph{outerplanar} if every vertex lies on the outer
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face. More generally, a plane graph is \emph{$k$-outerplanar} for
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$k \geq 1$ if removing all vertices on the outer face yields a
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$(k-1)$-outerplanar graph, where every graph on the empty vertex set
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is $0$-outerplanar.
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\end{definition}
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\section{Outerplanarity of depth levels}
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\begin{lemma}
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\label{lem:outerplanarity}
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Let $G$ be a planar graph with a plane embedding $\Pi$, and let
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$S \subseteq V(G)$ be a nonempty set of vertices, every one of which
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lies on the boundary of the outer face of $\Pi$. For each
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$d \geq 0$, the subgraph of $G$ induced by
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\[
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V_d^S := \{ v \in V(G) : \mathrm{dist}_G(v, S) = d \}
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\]
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is outerplanar.
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The special case $S = V(C)$, where $C$ is the outer cycle, recovers
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$V_d^S = V_d$ (depth-$d$ vertices as in
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Definition~\ref{def:plane-depth}) and is the form most often used in
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applications.
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\end{lemma}
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\begin{proof}
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Let $H = G[V_d^S]$ with the plane embedding inherited from $\Pi$. It
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suffices to show that every vertex of $H$ lies on the outer face of
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$H$.
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For $d = 0$, $V_0^S = S$, and by hypothesis every vertex of $S$ lies
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on the boundary of the outer face of $\Pi$. Removing the vertices
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and edges of $G \setminus H$ from the embedding only enlarges or
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merges face regions, so the outer face of $\Pi$ is contained in the
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outer face of $H$, and every vertex of $S$ remains on the outer face
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of $H$.
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For $d \geq 1$, let $U$ be the open subset of the plane obtained by
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removing all vertices and edges of $H$. We show every $v \in V_d^S$
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lies on the boundary of the component $U_{\mathrm{out}}$ of $U$
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containing the outer face of $\Pi$.
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Since every vertex in $V_{<d}^S := \bigcup_{e < d} V_e^S$ has a
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shortest path to $S$ passing entirely through $V_{<d}^S$, the
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subgraph $G[V_{<d}^S]$ is connected and contains $S$. Its vertices
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and edges lie in $U$ (none belong to $H$), and $S$ borders the outer
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face of $\Pi$, so $G[V_{<d}^S]$ and the outer face of $\Pi$ are
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connected within $U$, hence both lie in $U_{\mathrm{out}}$.
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Now let $v \in V_d^S$. Since $d \geq 1$, there exists
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$u \in V_{d-1}^S$ adjacent to $v$ in $G$. The edge $\{v, u\}$ is not
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in $H$, so it lies in $U$. Since $u \in V_{d-1}^S \subseteq
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U_{\mathrm{out}}$ and $\{v, u\}$ is a connected subset of $U$
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containing $u$, the entire edge lies in $U_{\mathrm{out}}$. The
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vertex $v$ is an endpoint of this edge but is not in $U$, so $v$ lies
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on the boundary of $U_{\mathrm{out}}$, i.e.\ on the outer face of
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$H$.
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\end{proof}
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\section{Deep embedding}
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\begin{definition}
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\label{def:deep-embedding}
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Let $G$ be a maximal planar graph with a plane embedding and outer
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cycle $C$. The \emph{deep embedding} of $G$ is the graph $G'$
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obtained from $G$ by the following operation: for every neutral
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triangular face $\{u, v, w\}$ of $G$ --- \emph{including the outer
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face}, whose vertices are the three vertices of $C$ --- add a new
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vertex $x$ placed in that face and adjacent to each of $u$, $v$,
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and $w$. The vertex added inside the outer face is denoted $x^*$
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and called the \emph{outer-cap vertex}; the three triangular faces
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it induces with the edges of $C$ are the \emph{outer-cap faces}.
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We henceforth view $G'$ as embedded on the sphere $S^2$, with no
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distinguished outer face.
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\end{definition}
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\begin{lemma}
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\label{lem:up-down-faces}
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Let $G'$ be the deep embedding of a maximal planar graph $G$. Every
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face of $G'$ is either an up triangle or a down triangle.
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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
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$p$ and $q$ differ by at most $1$. Suppose for contradiction that
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$\mathrm{depth}(p) = d$ and $\mathrm{depth}(q) = d + n$ for some
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$n \geq 2$. Since $\mathrm{depth}(p) = d$, there exists a path of
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length $d$ from $p$ to some vertex of $C$. Prepending the edge
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$\{q, p\}$ gives a path of length $d + 1$ from $q$ to $C$, so
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$\mathrm{depth}(q) \leq d + 1 < d + n$, a contradiction. The case
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$\mathrm{depth}(q) = d - n$ is handled identically: there exists a
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path of length $d - n$ from $q$ to some vertex of $C$, and prepending
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the edge $\{p, q\}$ gives a path of length $d - n + 1 \leq d - 1 < d$
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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
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triangle $\{u, v, w\}$ with all three pairs adjacent. By the above,
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each pair of vertices in a triangle differs in depth by at most $1$,
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so no triangle can contain vertices of depths $d$ and $d + 2$
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simultaneously. The possible depth patterns for a triangle in $G$
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are therefore exactly a neutral triangle, a down triangle, or an up
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triangle.
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We now consider each case under the deep embedding.
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\emph{Case 1: up triangle or down triangle.} These triangles are
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not modified by the deep embedding, so they remain as faces of $G'$,
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satisfying the lemma.
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\emph{Case 2: neutral triangle.} The deep embedding inserts a new
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vertex $x$ adjacent to $u$, $v$, and $w$, replacing the face
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$\{u, v, w\}$ with three new faces $\{u, v, x\}$, $\{v, w, x\}$, and
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$\{u, w, x\}$. It remains to determine the depth of $x$ in $G'$.
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Since $x$ is adjacent only to $u$, $v$, and $w$, every path in $G'$
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from $x$ to $C$ must pass through one of them, so $x$ has strictly
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greater depth than $u$, $v$, and $w$. Each of the three new faces
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is thus a down triangle, satisfying the lemma. The same argument
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applies to the outer face: the outer-cap vertex $x^*$ is adjacent to
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all three vertices of $C$ (which lie at depth $0$), so
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$\mathrm{depth}(x^*) = 1$, and each of the three outer-cap faces is
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a down triangle.
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Since every face of $G'$ falls into one of these cases, the result
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follows.
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\end{proof}
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\section{Quadrilateral decomposition}
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\begin{lemma}
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\label{lem:unique-level-edge}
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Every interior face of $G'$ has exactly one level edge.
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\end{lemma}
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\begin{proof}
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By Lemma~\ref{lem:up-down-faces}, each interior face is an up
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triangle (depths $\{d, d+1, d+1\}$) or a down triangle (depths
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$\{d, d, d+1\}$). In both cases, exactly one of the three vertex
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pairs has equal depth.
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\end{proof}
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\begin{lemma}
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\label{lem:shared-level-edge}
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Let $e = \{p, q\}$ be any level edge of $G'$. Then $e$ is the
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unique level edge of both faces incident to it.
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\end{lemma}
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\begin{proof}
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On the sphere, both faces $T, T'$ incident to $e$ are triangles.
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Since $p$ and $q$ have equal depth, $e$ is a level edge of $T$ and
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of $T'$, and by Lemma~\ref{lem:unique-level-edge} each has $e$ as
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its unique level edge.
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\end{proof}
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\begin{definition}
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\label{def:quad-decomposition}
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The \emph{quadrilateral decomposition} of $G'$ pairs each face of
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$G'$ with the face on the other side of its (unique) level edge.
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Each pair, together with the four non-level edges of the two
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triangles, bounds a \emph{quadrilateral} of the decomposition.
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\end{definition}
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\begin{remark}
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Because $G'$ is taken on the sphere, every edge lies between two
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triangular faces, so the pairing above applies uniformly. In
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particular, each edge of $C$ is a level edge shared between one
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interior boundary down triangle (depths $\{0, 0, 1\}$, with the
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depth-$1$ vertex inside $C$) and one outer-cap down triangle
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(depths $\{0, 0, 1\}$, with apex $x^*$). The three resulting
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quadrilaterals, one per edge of $C$, are the \emph{boundary deep
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diamonds}; they are the outermost quadrilaterals of the
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decomposition.
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\end{remark}
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\begin{definition}
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\label{def:quad-types}
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Each quadrilateral is one of three types, classified by the depths
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of its two non-level vertices relative to the depth $d$ of the
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shared level edge:
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\begin{itemize}
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\item a \emph{shallow diamond}, formed by two up triangles, with
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vertex depths $(d-1, d, d-1, d)$ around the boundary;
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\item a \emph{deep diamond}, formed by two down triangles, with
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vertex depths $(d+1, d, d+1, d)$ around the boundary;
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\item an \emph{S quad}, formed by one up and one down triangle,
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with vertex depths $(d-1, d, d+1, d)$ around the boundary.
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\end{itemize}
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\end{definition}
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\begin{thebibliography}{9}
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\bibitem{baker1994}
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B.~S.~Baker,
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\emph{Approximation algorithms for {NP}-complete problems on planar
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graphs},
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Journal of the ACM, vol.~41, no.~1, pp.~153--180, 1994.
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\bibitem{bauerfeld-pds-seq}
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E.~Bauerfeld,
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\emph{Plane Depth Sequencing},
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manuscript (math-research repository), 2026.
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\bibitem{bauerfeld-nested-tires}
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E.~Bauerfeld,
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\emph{Coloring Nested Tire Graphs},
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manuscript (math-research repository), 2026.
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\end{thebibliography}
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\end{document}
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