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\begin{document}

\title{Plane Depth Sequencing}

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\author{Eric Bauerfeld}
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\begin{abstract}
\end{abstract}

\maketitle

\section{Definitions}

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

\begin{definition}
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 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}

\begin{lemma}
Let $G$ be a graph with a plane embedding and outer cycle $C$. For each $d \geq 0$, the subgraph of $G$ induced by $V_d = \{v \in V(G) : \mathrm{depth}(v) = d\}$ is outerplanar.
\end{lemma}

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

For $d = 0$, we have $V_0 = V(C)$ and $H$ is a subgraph of the cycle $C$, hence a disjoint union of paths, which is outerplanar.

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 $v \in V_d$ lies on the boundary of the component $U_{\mathrm{out}}$ of $U$ containing the outer face of $G$.

Since every vertex in $V_{\leq d-1}$ has a shortest path to $C$ passing entirely through $V_{\leq d-1}$, the subgraph $G[V_{\leq d-1}]$ is connected and contains $C$. Its vertices and edges lie in $U$ (as they are not in $H$), and $C$ borders the outer face of $G$, so $G[V_{\leq d-1}]$ and the outer face of $G$ are connected within $U$, hence both lie in $U_{\mathrm{out}}$.

Now let $v \in V_d$. Since $\mathrm{depth}(v) = d \geq 1$, there exists $u \in V_{d-1}$ adjacent to $v$ in $G$. The edge $\{v, u\}$ is not an edge of $H$, so it lies in $U$. Since $u \in V_{d-1} \subset U_{\mathrm{out}}$ 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}

\begin{definition}
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
\[
    \mathrm{depth}(u) = \mathrm{depth}(v) = \mathrm{depth}(w),
\]
add a new vertex $x$ to $G$ adjacent to each of $u$, $v$, and $w$.
\end{definition}

\begin{lemma}
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.

\textit{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.

\textit{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.

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

\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.

\end{thebibliography}

\end{document}

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