plane_depth_sequencing: generalise Lemma 2.6 to arbitrary outer-face source; tighten the citation in coloring_nested_tire_graphs

In `plane_depth_sequencing/paper.tex`: - Lemma 2.6 now allows any nonempty source S ⊆ V(G) whose vertices all lie on the boundary of the outer face of the chosen embedding, rather than only the outer-cycle case S = V(C). - The proof is the same argument with S in place of C: at d=0 each S-vertex remains on the outer face after restriction; for d ≥ 1 the BFS ball V_{<d}^S is connected and reaches the outer face via S. - The original outer-cycle statement is preserved as a remark inside the lemma. - Adds \label{lem:outerplanarity}. In `coloring_nested_tire_graphs/paper.tex`: - The proof of Lemma 1.7 drops the "extends verbatim" caveat and simply cites the generalised Lemma 2.6, noting that since the level source S is a single vertex (per the local Level-source definition) we may freely choose an embedding placing S on the outer face; outerplanarity is a graph property so the conclusion transfers. Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
2026-05-25 15:24:08 -04:00
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@@ -195,12 +195,14 @@ $F_{C'}$.
 \end{lemma}

 \begin{proof}[Proof sketch]
-By Lemma~2.6 of \cite{bauerfeld-pds} (whose argument, given for an
-outer-cycle source, extends verbatim to an arbitrary level source $S$
-by treating $S$ as the depth-$0$ set), the subgraph $G[L_{d'}]$ is
-outerplanar for each $d' \geq 0$.  Since subgraphs of outerplanar
-graphs are outerplanar, both $G[V_{C'} \cap L_d]$ and
-$G[V_{C'} \cap L_{d+1}]$ are outerplanar.
+Since $S$ is a single vertex, we may choose a plane embedding of $G$
+that places $S$ on the outer face.  By Lemma~2.6 of
+\cite{bauerfeld-pds}, applied with this embedding and source set $S$,
+the subgraph $G[L_{d'}]$ is outerplanar for each $d' \geq 0$;
+outerplanarity is a graph property, so this conclusion is independent
+of the embedding choice.  Since subgraphs of outerplanar graphs are
+outerplanar, both $G[V_{C'} \cap L_d]$ and $G[V_{C'} \cap L_{d+1}]$ are
+outerplanar.

 Layer containment (a consequence of the bounded-step property of BFS
 on a triangulation) gives $V_{C'} \subseteq L_d \cup L_{d+1}$, so $C$
@@ -0,0 +1,16 @@
+\relax 
+\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Motivation}}{1}{}\protected@file@percent }
+\citation{baker1994}
+\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Definitions}}{2}{}\protected@file@percent }
+\newlabel{lem:outerplanarity}{{2.6}{2}}
+\@writefile{toc}{\contentsline {section}{\tocsection {}{3}{Quadrilateral sequencing}}{3}{}\protected@file@percent }
+\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces The deep embedding $G'$ of a small maximal planar graph (drawn with one outer-cap face as the outer face), with each quadrilateral $Q_n$ of the canonical sequence labelled by its index and the move code (AD = anchor drop, LA = level add, J = join, RC = ring completion) of the move that produced it. Solid edges are non-level; dashed edges are level. Background colour encodes quadrilateral type: amber for shallow diamonds, teal for deep diamonds, pink for S quads. Outer-cycle vertices are blue, the outer-cap vertex $x^{*}$ is red. The move-code string for this example is $01211333$.}}{6}{}\protected@file@percent }
+\newlabel{fig:example-sequence}{{1}{6}}
+\bibcite{baker1994}{1}
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@@ -107,7 +107,7 @@ A triangle $\{u, v, w\}$ in $G$ is an \emph{up triangle} if the multiset of dept
 \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.
+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}
@@ -115,19 +115,50 @@ A plane graph is \emph{outerplanar} if every vertex lies on the outer face. More
 \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.
+\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~2.1) and is the
+form most often used in the rest of this paper.
 \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 $H$.
+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 $d = 0$, we have $V_0 = C$, so $H$ is outerplanar.
+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 $v \in V_d$ lies on the boundary of the component $U_{\mathrm{out}}$ of $U$ containing the outer face of $G$.
+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 $V_{\leq d-1}$ 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 $U_{\mathrm{out}}$.
+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 $v \in V_d$. 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 $H$.
+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}

 \begin{definition}