math-research/papers/plane_depth/paper.tex

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\begin{document}

\title{Plane Depth}

\author{Eric Bauerfeld}
\address{}
\curraddr{}
\email{}
\thanks{}

\subjclass[2010]{Primary }

\keywords{plane graph, triangulation, plane depth, level edge, deep
embedding, quadrilateral decomposition, $k$-outerplanar graph}

\date{}

\dedicatory{}

\begin{abstract}
Given a plane embedding of a graph with outer cycle $C$, the
\emph{plane depth} of a vertex is its graph distance to $C$.  We
develop this depth function into a layered combinatorial structure on
plane triangulations: the subgraph induced by each depth level is
outerplanar (recovering Baker's notion of a $k$-outerplanar graph);
each triangular face is classified by its depth multiset as
\emph{up}, \emph{down}, or \emph{neutral}; and the \emph{deep
embedding} of a maximal planar graph, obtained by inserting a vertex
into every neutral face (including the outer face), has every face
either up or down.  Pairing adjacent triangles across their unique
level edge yields a \emph{quadrilateral decomposition} of the
spherical deep embedding into three combinatorial types: shallow
diamonds, deep diamonds, and S quads.

This paper isolates the foundational depth-and-decomposition material
that supports several downstream applications --- including the
quadrilateral sequencing of \cite{bauerfeld-pds-seq} and the
nested-tire colouring framework of \cite{bauerfeld-nested-tires}.
\end{abstract}

\maketitle

\section{Definitions}

\begin{definition}
\label{def:plane-depth}
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
\[
    \mathrm{depth}(v) = \min_{u \in V(C)} d(v, u),
\]
where $d(v, u)$ denotes the graph distance between $v$ and $u$ in
$G$.
\end{definition}

\begin{definition}
\label{def:level-edge}
An edge $\{u, v\} \in E(G)$ is a \emph{level edge} if
$\mathrm{depth}(u) = \mathrm{depth}(v)$.
\end{definition}

\begin{definition}
\label{def:triangle-types}
A triangle $\{u, v, w\}$ in $G$ is an \emph{up triangle} if the
multiset of depths of its vertices is $\{d, d+1, d+1\}$ for some
$d \geq 0$, a \emph{down triangle} if the multiset of depths is
$\{d, d, d+1\}$ for some $d \geq 0$, and a \emph{neutral triangle} if
the multiset of depths is $\{d, d, d\}$ for some $d \geq 0$.
\end{definition}

\begin{remark}
We now relate our terminology to existing terminology, namely
$k$-outerplanar graphs \cite{baker1994}.  The following definition and
lemma show that the subgraph induced by any single depth level
relative to any source set on the outer face is outerplanar, i.e.\
$1$-outerplanar in the sense of Baker.
\end{remark}

\begin{definition}
A plane graph is \emph{outerplanar} if every vertex lies on the outer
face.  More generally, a plane graph is \emph{$k$-outerplanar} for
$k \geq 1$ if removing all vertices on the outer face yields a
$(k-1)$-outerplanar graph, where every graph on the empty vertex set
is $0$-outerplanar.
\end{definition}

\section{Outerplanarity of depth levels}

\begin{lemma}
\label{lem:outerplanarity}
Let $G$ be a planar graph with a plane embedding $\Pi$, and let
$S \subseteq V(G)$ be a nonempty set of vertices, every one of which
lies on the boundary of the outer face of $\Pi$.  For each
$d \geq 0$, the subgraph of $G$ induced by
\[
    V_d^S := \{ v \in V(G) : \mathrm{dist}_G(v, S) = d \}
\]
is outerplanar.

The special case $S = V(C)$, where $C$ is the outer cycle, recovers
$V_d^S = V_d$ (depth-$d$ vertices as in
Definition~\ref{def:plane-depth}) and is the form most often used in
applications.
\end{lemma}

\begin{proof}
Let $H = G[V_d^S]$ with the plane embedding inherited from $\Pi$.  It
suffices to show that every vertex of $H$ lies on the outer face of
$H$.

For $d = 0$, $V_0^S = S$, and by hypothesis every vertex of $S$ lies
on the boundary of the outer face of $\Pi$.  Removing the vertices
and edges of $G \setminus H$ from the embedding only enlarges or
merges face regions, so the outer face of $\Pi$ is contained in the
outer face of $H$, and every vertex of $S$ remains on the outer face
of $H$.

For $d \geq 1$, let $U$ be the open subset of the plane obtained by
removing all vertices and edges of $H$.  We show every $v \in V_d^S$
lies on the boundary of the component $U_{\mathrm{out}}$ of $U$
containing the outer face of $\Pi$.

Since every vertex in $V_{<d}^S := \bigcup_{e < d} V_e^S$ has a
shortest path to $S$ passing entirely through $V_{<d}^S$, the
subgraph $G[V_{<d}^S]$ is connected and contains $S$.  Its vertices
and edges lie in $U$ (none belong to $H$), and $S$ borders the outer
face of $\Pi$, so $G[V_{<d}^S]$ and the outer face of $\Pi$ are
connected within $U$, hence both lie in $U_{\mathrm{out}}$.

Now let $v \in V_d^S$.  Since $d \geq 1$, there exists
$u \in V_{d-1}^S$ adjacent to $v$ in $G$.  The edge $\{v, u\}$ is not
in $H$, so it lies in $U$.  Since $u \in V_{d-1}^S \subseteq
U_{\mathrm{out}}$ and $\{v, u\}$ is a connected subset of $U$
containing $u$, the entire edge lies in $U_{\mathrm{out}}$.  The
vertex $v$ is an endpoint of this edge but is not in $U$, so $v$ lies
on the boundary of $U_{\mathrm{out}}$, i.e.\ on the outer face of
$H$.
\end{proof}

\section{Deep embedding}

\begin{definition}
\label{def:deep-embedding}
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 neutral
triangular face $\{u, v, w\}$ of $G$ --- \emph{including the outer
face}, whose vertices are the three vertices of $C$ --- add a new
vertex $x$ placed in that face and adjacent to each of $u$, $v$,
and $w$.  The vertex added inside the outer face is denoted $x^*$
and called the \emph{outer-cap vertex}; the three triangular faces
it induces with the edges of $C$ are the \emph{outer-cap faces}.
We henceforth view $G'$ as embedded on the sphere $S^2$, with no
distinguished outer face.
\end{definition}

\begin{lemma}
\label{lem:up-down-faces}
Let $G'$ be the deep embedding of a maximal planar graph $G$.  Every
face of $G'$ is either an up triangle or a down triangle.
\end{lemma}

\begin{proof}
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 $\mathrm{depth}(p) = d$.

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 a neutral triangle, a down triangle, or an up
triangle.

We now consider each case under the deep embedding.

\emph{Case 1: up triangle or down triangle.}  These triangles are
not modified by the deep embedding, so they remain as faces of $G'$,
satisfying the lemma.

\emph{Case 2: neutral triangle.}  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
is thus a down triangle, satisfying the lemma.  The same argument
applies to the outer face: the outer-cap vertex $x^*$ is adjacent to
all three vertices of $C$ (which lie at depth $0$), so
$\mathrm{depth}(x^*) = 1$, and each of the three outer-cap faces is
a down triangle.

Since every face of $G'$ falls into one of these cases, the result
follows.
\end{proof}

\section{Quadrilateral decomposition}

\begin{lemma}
\label{lem:unique-level-edge}
Every interior face of $G'$ has exactly one level edge.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:up-down-faces}, each interior face is an up
triangle (depths $\{d, d+1, d+1\}$) or a down triangle (depths
$\{d, d, d+1\}$).  In both cases, exactly one of the three vertex
pairs has equal depth.
\end{proof}

\begin{lemma}
\label{lem:shared-level-edge}
Let $e = \{p, q\}$ be any level edge of $G'$.  Then $e$ is the
unique level edge of both faces incident to it.
\end{lemma}

\begin{proof}
On the sphere, both faces $T, T'$ incident to $e$ are triangles.
Since $p$ and $q$ have equal depth, $e$ is a level edge of $T$ and
of $T'$, and by Lemma~\ref{lem:unique-level-edge} each has $e$ as
its unique level edge.
\end{proof}

\begin{definition}
\label{def:quad-decomposition}
The \emph{quadrilateral decomposition} of $G'$ pairs each face of
$G'$ with the face on the other side of its (unique) level edge.
Each pair, together with the four non-level edges of the two
triangles, bounds a \emph{quadrilateral} of the decomposition.
\end{definition}

\begin{remark}
Because $G'$ is taken on the sphere, every edge lies between two
triangular faces, so the pairing above applies uniformly.  In
particular, each edge of $C$ is a level edge shared between one
interior boundary down triangle (depths $\{0, 0, 1\}$, with the
depth-$1$ vertex inside $C$) and one outer-cap down triangle
(depths $\{0, 0, 1\}$, with apex $x^*$).  The three resulting
quadrilaterals, one per edge of $C$, are the \emph{boundary deep
diamonds}; they are the outermost quadrilaterals of the
decomposition.
\end{remark}

\begin{definition}
\label{def:quad-types}
Each quadrilateral is one of three types, classified by the depths
of its two non-level vertices relative to the depth $d$ of the
shared level edge:
\begin{itemize}
    \item a \emph{shallow diamond}, formed by two up triangles, with
          vertex depths $(d-1, d, d-1, d)$ around the boundary;
    \item a \emph{deep diamond}, formed by two down triangles, with
          vertex depths $(d+1, d, d+1, d)$ around the boundary;
    \item an \emph{S quad}, formed by one up and one down triangle,
          with vertex depths $(d-1, d, d+1, d)$ around the boundary.
\end{itemize}
\end{definition}

\begin{thebibliography}{9}

\bibitem{baker1994}
B.~S.~Baker,
\emph{Approximation algorithms for {NP}-complete problems on planar
graphs},
Journal of the ACM, vol.~41, no.~1, pp.~153--180, 1994.

\bibitem{bauerfeld-pds-seq}
E.~Bauerfeld,
\emph{Plane Depth Sequencing},
manuscript (math-research repository), 2026.

\bibitem{bauerfeld-nested-tires}
E.~Bauerfeld,
\emph{Coloring Nested Tire Graphs},
manuscript (math-research repository), 2026.

\end{thebibliography}

\end{document}