Add outerplanar lemma with Baker citation and relate depth levels to k-outerplanar graphs

Co-Authored-By: Claude Sonnet 4.6 <noreply@anthropic.com>

papers/plane_depth_sequencing/paper.aux

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papers/plane_depth_sequencing/paper.tex

@@ -94,6 +94,30 @@ An edge $\{u, v\} \in E(G)$ is a \emph{level edge} if $\mathrm{depth}(u) = \math
 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
 \[
@@ -120,6 +144,15 @@ We now consider each case under the deep embedding.
 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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