diff --git a/papers/coloring_nested_tire_graphs/paper.aux b/papers/coloring_nested_tire_graphs/paper.aux
index cb921e3..a258f40 100644
--- a/papers/coloring_nested_tire_graphs/paper.aux
+++ b/papers/coloring_nested_tire_graphs/paper.aux
@@ -14,9 +14,9 @@
 \@writefile{lof}{\contentsline {figure}{\numberline {4}{\ignorespaces Partial tire dual $D(T)$ when the inner outerplanar graph $O$ has a bridge --- here a non-trivial edge cut connecting two disjoint triangles. $B_{\mathrm  {out}}$ is a $4$-cycle on $\{0,1,2,3\}$ and $O$ is the barbell: triangle $\{4,5,6\}$ together with triangle $\{7,8,9\}$ joined by the bridge edge $6$--$7$ (removing the bridge disconnects $O$). Because both faces incident to the bridge are annular triangles, the bridge contributes an \emph  {interior dual edge} (highlighted in red) rather than two leaves; consequently the interior dual subgraph is no longer the single $(n+m)$-cycle of Proposition\nonbreakingspace 1.8\hbox {}, but a theta graph: the two trivalent vertices $d_5, d_6$ (the bridge-incident annular faces) are joined by three internally vertex-disjoint paths in $D(T)$. Leaves come only from $B_{\mathrm  {out}}$ ($n = 4$ leaves) and the six non-bridge edges of $O$ ($m_{\partial } = 6$ leaves, three for each triangle).}}{5}{}\protected@file@percent }
 \newlabel{fig:partial-tire-dual-bridge}{{4}{5}}
 \newlabel{prop:no-level-d-pinch}{{1.9}{5}}
-\citation{bauerfeld-pds}
+\citation{bauerfeld-depth}
 \newlabel{lem:tire-component}{{1.10}{6}}
-\citation{bauerfeld-pds}
+\citation{bauerfeld-depth}
 \newlabel{rem:tire-component-degenerate}{{1.11}{8}}
 \newlabel{rem:tire-no-extra-hypotheses}{{1.12}{8}}
 \newlabel{prop:edge-vertex-bijection}{{1.13}{8}}
@@ -29,7 +29,7 @@
 \newlabel{fig:facial-dual-choices}{{5}{10}}
 \@writefile{toc}{\contentsline {section}{\tocsection {}{2}{A conjectural Latin-style substructure}}{10}{}\protected@file@percent }
 \newlabel{sec:latin-conjecture}{{2}{10}}
-\bibcite{bauerfeld-pds}{1}
+\bibcite{bauerfeld-depth}{1}
 \newlabel{tocindent-1}{0pt}
 \newlabel{tocindent0}{12.7778pt}
 \newlabel{tocindent1}{17.77782pt}
diff --git a/papers/coloring_nested_tire_graphs/paper.log b/papers/coloring_nested_tire_graphs/paper.log
index f759a4b..8a5cee0 100644
--- a/papers/coloring_nested_tire_graphs/paper.log
+++ b/papers/coloring_nested_tire_graphs/paper.log
@@ -1,4 +1,4 @@
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 [4 <./fig_partial_tire_dual.png>] [5 <./fig_partial_tire_dual_bridge.png>]
 [6] [7] [8]
 
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diff --git a/papers/coloring_nested_tire_graphs/paper.pdf b/papers/coloring_nested_tire_graphs/paper.pdf
index 39aec22..80c6305 100644
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diff --git a/papers/coloring_nested_tire_graphs/paper.tex b/papers/coloring_nested_tire_graphs/paper.tex
index bb385d0..cf2b8a0 100644
--- a/papers/coloring_nested_tire_graphs/paper.tex
+++ b/papers/coloring_nested_tire_graphs/paper.tex
@@ -384,8 +384,9 @@ the faces of $G$ in $F_{C'}$.
 
 \begin{proof}
 \emph{Outerplanarity of the two level parts.}  By construction $S$
-lies on the outer face of $\Pi_G$, so Lemma~2.6 of \cite{bauerfeld-pds}
-applies directly with $(G, \Pi_G, S)$, giving that $G[L_{d'}]$ is
+lies on the outer face of $\Pi_G$, so the outerplanarity lemma of
+\cite{bauerfeld-depth} applies directly with $(G, \Pi_G, S)$, giving
+that $G[L_{d'}]$ is
 outerplanar for each $d' \geq 0$.  Subgraphs of outerplanar graphs are
 outerplanar, so $G[V_{C'} \cap L_d]$ and $G[V_{C'} \cap L_{d+1}]$ are
 both outerplanar.
@@ -447,8 +448,8 @@ $d = 0$ case) --- serves as $B_{\mathrm{out}}$.  We set
 $B_{\mathrm{out}} := G[V_{C'} \cap L_d]$ if this is a cycle, and
 the single vertex $\{v_0\}$ in the degenerate case.
 
-\emph{Inner outerplanar graph.}  By Lemma~2.6 of \cite{bauerfeld-pds},
-$G[V_{C'} \cap L_{d+1}]$ is outerplanar.  We set $O :=
+\emph{Inner outerplanar graph.}  By the outerplanarity lemma of
+\cite{bauerfeld-depth}, $G[V_{C'} \cap L_{d+1}]$ is outerplanar.  We set $O :=
 G[V_{C'} \cap L_{d+1}]$.  The boundary curve(s) of $R_{C'}$ on the
 $L_{d+1}$ side are exactly the boundary of $O$'s outer face in the
 inherited embedding; this outer-face boundary is a single closed walk
@@ -754,9 +755,9 @@ ongoing automated search.
 
 \begin{thebibliography}{9}
 
-\bibitem{bauerfeld-pds}
+\bibitem{bauerfeld-depth}
 E.~Bauerfeld,
-\emph{Plane Depth Sequencing},
+\emph{Plane Depth},
 manuscript (math-research repository), 2026.
 
 \end{thebibliography}
diff --git a/papers/plane_depth/paper.aux b/papers/plane_depth/paper.aux
new file mode 100644
index 0000000..6c7a65e
--- /dev/null
+++ b/papers/plane_depth/paper.aux
@@ -0,0 +1,28 @@
+\relax 
+\citation{bauerfeld-pds-seq}
+\citation{bauerfeld-nested-tires}
+\citation{baker1994}
+\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Definitions}}{1}{}\protected@file@percent }
+\newlabel{def:plane-depth}{{1.1}{1}}
+\newlabel{def:level-edge}{{1.2}{1}}
+\newlabel{def:triangle-types}{{1.3}{1}}
+\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Outerplanarity of depth levels}}{2}{}\protected@file@percent }
+\newlabel{lem:outerplanarity}{{2.1}{2}}
+\@writefile{toc}{\contentsline {section}{\tocsection {}{3}{Deep embedding}}{2}{}\protected@file@percent }
+\newlabel{def:deep-embedding}{{3.1}{2}}
+\newlabel{lem:up-down-faces}{{3.2}{2}}
+\bibcite{baker1994}{1}
+\@writefile{toc}{\contentsline {section}{\tocsection {}{4}{Quadrilateral decomposition}}{3}{}\protected@file@percent }
+\newlabel{lem:unique-level-edge}{{4.1}{3}}
+\newlabel{lem:shared-level-edge}{{4.2}{3}}
+\newlabel{def:quad-decomposition}{{4.3}{3}}
+\newlabel{def:quad-types}{{4.5}{3}}
+\bibcite{bauerfeld-pds-seq}{2}
+\bibcite{bauerfeld-nested-tires}{3}
+\newlabel{tocindent-1}{0pt}
+\newlabel{tocindent0}{12.7778pt}
+\newlabel{tocindent1}{17.77782pt}
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+\gdef \@abspage@last{4}
diff --git a/papers/plane_depth/paper.log b/papers/plane_depth/paper.log
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diff --git a/papers/plane_depth/paper.pdf b/papers/plane_depth/paper.pdf
new file mode 100644
index 0000000..6bb91c5
Binary files /dev/null and b/papers/plane_depth/paper.pdf differ
diff --git a/papers/plane_depth/paper.tex b/papers/plane_depth/paper.tex
new file mode 100644
index 0000000..928e065
--- /dev/null
+++ b/papers/plane_depth/paper.tex
@@ -0,0 +1,309 @@
+\documentclass{amsart}
+
+\usepackage{amssymb}
+\usepackage{graphicx}
+
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{corollary}[theorem]{Corollary}
+
+\theoremstyle{definition}
+\newtheorem{definition}[theorem]{Definition}
+\newtheorem{example}[theorem]{Example}
+
+\theoremstyle{remark}
+\newtheorem{remark}[theorem]{Remark}
+
+\numberwithin{equation}{section}
+
+\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}
diff --git a/papers/plane_depth_sequencing/paper.aux b/papers/plane_depth_sequencing/paper.aux
index dda1255..67d1010 100644
--- a/papers/plane_depth_sequencing/paper.aux
+++ b/papers/plane_depth_sequencing/paper.aux
@@ -1,16 +1,17 @@
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+\citation{bauerfeld-depth}
+\citation{bauerfeld-depth}
+\citation{bauerfeld-depth}
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+\newlabel{fig:example-sequence}{{1}{4}}
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diff --git a/papers/plane_depth_sequencing/paper.pdf b/papers/plane_depth_sequencing/paper.pdf
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@@ -56,200 +20,159 @@
 
 \title{Plane Depth Sequencing}
 
-%    Remove any unused author tags.
-
-%    author one information
 \author{Eric Bauerfeld}
 \address{}
 \curraddr{}
 \email{}
 \thanks{}
 
-
 \subjclass[2010]{Primary }
 
-\keywords{plane graph, triangulation, plane depth, deep embedding, quadrilateral decomposition, $k$-outerplanar graph, canonical sequencing}
+\keywords{plane graph, triangulation, plane depth, deep embedding,
+quadrilateral decomposition, canonical sequencing}
 
 \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 use this depth function to organize plane triangulations into a layered combinatorial structure. First, we show that the subgraph induced by each depth level is outerplanar, recovering Baker's notion of a $k$-outerplanar graph. We then introduce the \emph{deep embedding} of a maximal planar graph, obtained by inserting a new vertex into each neutral triangular face (including the outer face), and prove that every face of the resulting spherical triangulation is an \emph{up} or \emph{down} triangle. Pairing adjacent triangles across their unique level edge yields a \emph{quadrilateral decomposition} into three combinatorial types: shallow diamonds, deep diamonds, and S quads. Finally, we define a deterministic traversal of this decomposition using four moves --- anchor drop, level add, join, and ring completion --- under a fixed precedence, and prove that, starting from any boundary deep diamond, the resulting sequence visits every quadrilateral exactly once.
+Building on the plane depth and quadrilateral decomposition
+of~\cite{bauerfeld-depth}, we define a deterministic traversal of the
+quadrilateral decomposition of a maximal planar graph's deep
+embedding $G'$ using four moves --- anchor drop, level add, join,
+and ring completion --- under a fixed precedence, and prove that,
+starting from any boundary deep diamond, the resulting sequence
+visits every quadrilateral exactly once.
 \end{abstract}
 
 \maketitle
 
 \section{Motivation}
 
-This paper is one step of a longer programme aimed at 4-coloring maximal planar graphs by inductive local construction. The quadrilateral sequencing is designed so that the deep embedding $G'$ can be built up one quadrilateral at a time, and so that --- at least at first glance --- each new quadrilateral admits a natural local 4-coloring choice for the vertices it introduces.
+This paper is one step of a longer programme aimed at $4$-colouring
+maximal planar graphs by inductive local construction.  The
+quadrilateral sequencing is designed so that the deep embedding $G'$
+(see~\cite{bauerfeld-depth},
+Definition on the deep embedding) can be built up one quadrilateral
+at a time, and so that --- at least at first glance --- each new
+quadrilateral admits a natural local $4$-colouring choice for the
+vertices it introduces.
 
-Three of the four moves --- anchor drop, level add, and join --- attach a quadrilateral that introduces one or two new vertices. For each of these moves the new vertex (or vertices) can be assigned a color that respects the local triangle constraints. The fourth move, \emph{ring completion}, introduces no new vertices: it attaches a quadrilateral all four of whose corners are already in the slice. Whether the move succeeds therefore depends on previously chosen colors.
+Three of the four moves --- anchor drop, level add, and join ---
+attach a quadrilateral that introduces one or two new vertices.  For
+each of these moves the new vertex (or vertices) can be assigned a
+colour that respects the local triangle constraints.  The fourth
+move, \emph{ring completion}, introduces no new vertices: it attaches
+a quadrilateral all four of whose corners are already in the slice.
+Whether the move succeeds therefore depends on previously chosen
+colours.
 
-A natural question is whether the local freedom present at the non-ring-completion moves is enough to guarantee that, by the time each ring completion fires, the four corner colors are already compatible with a proper 4-coloring of the new quadrilateral. The remaining sections develop the structural machinery needed to phrase this question precisely; companion commentary (\texttt{commentary.tex}) records what is known so far about why this hope is delicate, and an empirical check (\texttt{quad\_sequence\_coloring\_check.py}) explores the behavior of a particular online coloring discipline on small triangulations.
+A natural question is whether the local freedom present at the
+non-ring-completion moves is enough to guarantee that, by the time
+each ring completion fires, the four corner colours are already
+compatible with a proper $4$-colouring of the new quadrilateral.  The
+remaining sections develop the structural machinery needed to phrase
+this question precisely; companion commentary
+(\texttt{commentary.tex}) records what is known so far about why this
+hope is delicate, and an empirical check
+(\texttt{quad\_sequence\_coloring\_check.py}) explores the behaviour
+of a particular online colouring discipline on small triangulations.
 
-\section{Definitions}
+\section{Preliminaries from plane depth}
 
-\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}
+We use without proof the following definitions and results
+from~\cite{bauerfeld-depth}:
+\begin{itemize}
+    \item \textbf{Plane depth} of a vertex relative to the outer
+          cycle (depth$(v)$).
+    \item \textbf{Up}, \textbf{down}, \textbf{neutral} triangle
+          classification by the multiset of depths of its three
+          vertices.
+    \item \textbf{Deep embedding} $G'$: insert a new vertex into
+          every neutral triangular face (including the outer face,
+          with new vertex $x^*$).  View $G'$ as embedded on $S^2$.
+    \item Every face of $G'$ is an up or down triangle, and has a
+          \emph{unique level edge}.
+    \item \textbf{Quadrilateral decomposition} of $G'$: pair each
+          face with the face sharing its level edge.  Each
+          quadrilateral is either a \emph{shallow diamond} (two up
+          triangles), a \emph{deep diamond} (two down triangles),
+          or an \emph{S quad} (one up, one down).  The
+          \emph{boundary deep diamonds} are the three quadrilaterals
+          containing an outer-cap face.
+\end{itemize}
 
-\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 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}
-
-\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~2.1) and is the
-form most often used in the rest of this paper.
-\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}
-
-\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 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}
-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. 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}
+We now equip this decomposition with a deterministic traversal.
 
 \section{Quadrilateral sequencing}
 
-We now decompose the deep embedding into quadrilaterals by removing level edges, and define a deterministic sequence in which those quadrilaterals are visited.
-
-\begin{lemma}
-Every interior face of $G'$ has exactly one level edge.
-\end{lemma}
-
-\begin{proof}
-By the previous lemma, 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}
-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 the previous lemma each has $e$ as its unique level edge.
-\end{proof}
-
-\begin{definition}
-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.
+For the remainder of this paper, fix a plane embedding of $G'$ by
+designating one of the three outer-cap faces as the outer face of
+the plane drawing; the outer cycle of this plane embedding is the
+boundary of that designated outer-cap face.  Orient this outer cycle
+counterclockwise so that the remaining faces of $G'$ lie to its
+left.  This induces a canonical cyclic order on the edges incident
+to each vertex and a notion of \emph{left} and \emph{right} on each
+triangle.
 \end{remark}
 
 \begin{definition}
-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{remark}
-For the remainder of this section, fix a plane embedding of $G'$ by designating one of the three outer-cap faces as the outer face of the plane drawing; the outer cycle of this plane embedding is the boundary of that designated outer-cap face. Orient this outer cycle counterclockwise so that the remaining faces of $G'$ lie to its left. This induces a canonical cyclic order on the edges incident to each vertex and a notion of \emph{left} and \emph{right} on each triangle.
-\end{remark}
-
-\begin{definition}
-A \emph{slice} of $G'$ is a connected, simply connected region of the plane formed by the union of a subset of quadrilaterals from the decomposition, together with its closed boundary walk in $G'$.
+A \emph{slice} of $G'$ is a connected, simply connected region of the
+plane formed by the union of a subset of quadrilaterals from the
+decomposition, together with its closed boundary walk in $G'$.
 \end{definition}
 
 \begin{definition}[Move code]
-Each move is assigned a numerical code: anchor drop $= 0$, level add $= 1$, join $= 2$, ring completion $= 3$. Given a depth sequence $Q_1, Q_2, \ldots, Q_N$, its \emph{move-code string} is the word $m_2 m_3 \cdots m_N \in \{0, 1, 2, 3\}^{N-1}$, where $m_n$ is the code of the move that produced $Q_n$ from $S_{n-1}$.
+Each move is assigned a numerical code: anchor drop $= 0$, level add
+$= 1$, join $= 2$, ring completion $= 3$.  Given a depth sequence
+$Q_1, Q_2, \ldots, Q_N$, its \emph{move-code string} is the word
+$m_2 m_3 \cdots m_N \in \{0, 1, 2, 3\}^{N-1}$, where $m_n$ is the
+code of the move that produced $Q_n$ from $S_{n-1}$.
 \end{definition}
 
 \begin{definition}[Initial quad]
-The depth sequence begins by choosing as $Q_1$ the boundary deep diamond whose resulting move-code string is lexicographically smallest among the three boundary deep diamonds. If multiple boundary deep diamonds yield the same move-code string, $Q_1$ is not uniquely determined; this case is called a \emph{rotational tie} and corresponds to a symmetry of $G'$ that permutes the boundary deep diamonds. The initial slice is $S_1 = Q_1$.
+The depth sequence begins by choosing as $Q_1$ the boundary deep
+diamond whose resulting move-code string is lexicographically
+smallest among the three boundary deep diamonds.  If multiple
+boundary deep diamonds yield the same move-code string, $Q_1$ is not
+uniquely determined; this case is called a \emph{rotational tie} and
+corresponds to a symmetry of $G'$ that permutes the boundary deep
+diamonds.  The initial slice is $S_1 = Q_1$.
 \end{definition}
 
 \begin{remark}
-The tiebreak is recursive: the choice of $Q_1$ depends on the move-code strings produced by each of the three candidate starts, which in turn depend on the entire sequence each candidate produces. Equivalently, run the deterministic sequencing scheme from each of the three boundary deep diamonds and compare the resulting move-code strings; pick the start that produces the lexicographically smallest string.
+The tiebreak is recursive: the choice of $Q_1$ depends on the
+move-code strings produced by each of the three candidate starts,
+which in turn depend on the entire sequence each candidate produces.
+Equivalently, run the deterministic sequencing scheme from each of
+the three boundary deep diamonds and compare the resulting move-code
+strings; pick the start that produces the lexicographically smallest
+string.
 \end{remark}
 
 \begin{definition}[Anchor drop]
-Suppose a slice $S_n$ has been constructed and the lower-rightmost portion of its boundary (in the fixed plane embedding) is a down triangle $a$ whose right edge $e$ is exposed. If there exists an S quad $Q \notin S_n$ whose up triangle $b$ has $e$ as its left edge, then the \emph{anchor drop} sets
+Suppose a slice $S_n$ has been constructed and the lower-rightmost
+portion of its boundary (in the fixed plane embedding) is a down
+triangle $a$ whose right edge $e$ is exposed.  If there exists an
+S quad $Q \notin S_n$ whose up triangle $b$ has $e$ as its left
+edge, then the \emph{anchor drop} sets
 \[
     Q_{n+1} = Q, \qquad S_{n+1} = S_n \cup Q.
 \]
-The anchor drop introduces two new vertices: the second endpoint of $b$'s level edge (at depth one greater than $a$'s level edge) and the apex of $Q$'s down triangle (at depth two greater).
+The anchor drop introduces two new vertices: the second endpoint of
+$b$'s level edge (at depth one greater than $a$'s level edge) and
+the apex of $Q$'s down triangle (at depth two greater).
 \end{definition}
 
 \begin{definition}[Level add]
-Suppose a slice $S_n$ has been constructed. Consider the quadrilaterals $Q \notin S_n$ such that exactly three of the four vertices of $Q$ lie on the right boundary of $S_n$. Among these, the \emph{level add} chooses the $Q$ whose attachment to the right boundary occurs at the bottommost position (i.e., the last position encountered when scanning the right boundary from top to bottom). Then
+Suppose a slice $S_n$ has been constructed.  Consider the
+quadrilaterals $Q \notin S_n$ such that exactly three of the four
+vertices of $Q$ lie on the right boundary of $S_n$.  Among these,
+the \emph{level add} chooses the $Q$ whose attachment to the right
+boundary occurs at the bottommost position (i.e.\ the last position
+encountered when scanning the right boundary from top to bottom).
+Then
 \[
     Q_{n+1} = Q, \qquad S_{n+1} = S_n \cup Q.
 \]
@@ -257,19 +180,36 @@ By construction, the level add introduces exactly one new vertex.
 \end{definition}
 
 \begin{definition}[Join]
-Suppose a slice $S_n$ has been constructed. Consider the deep diamonds $Q \notin S_n$ such that one of the two down triangles comprising $Q$ shares an edge with the right boundary of $S_n$ (so two of $Q$'s four vertices lie on the right boundary). Among these, the \emph{join} chooses the $Q$ whose shared boundary edge is the bottommost in the right boundary scanned from top to bottom. Then
+Suppose a slice $S_n$ has been constructed.  Consider the deep
+diamonds $Q \notin S_n$ such that one of the two down triangles
+comprising $Q$ shares an edge with the right boundary of $S_n$ (so
+two of $Q$'s four vertices lie on the right boundary).  Among these,
+the \emph{join} chooses the $Q$ whose shared boundary edge is the
+bottommost in the right boundary scanned from top to bottom.  Then
 \[
     Q_{n+1} = Q, \qquad S_{n+1} = S_n \cup Q.
 \]
-Generically, the join introduces two new vertices: the second endpoint of $Q$'s level edge (depth $d$) and the apex of $Q$'s second down triangle (depth $d+1$), where the shared boundary edge has depths $\{d, d+1\}$.
+Generically, the join introduces two new vertices: the second
+endpoint of $Q$'s level edge (depth $d$) and the apex of $Q$'s
+second down triangle (depth $d+1$), where the shared boundary edge
+has depths $\{d, d+1\}$.
 \end{definition}
 
 \begin{remark}
-Like the anchor drop, the join generically introduces two new vertices. Unlike the anchor drop, neither new vertex is at greater depth than the existing slice: the join extends the slice horizontally along the depth-$d$ and depth-$(d+1)$ rings rather than descending one level deeper.
+Like the anchor drop, the join generically introduces two new
+vertices.  Unlike the anchor drop, neither new vertex is at greater
+depth than the existing slice: the join extends the slice
+horizontally along the depth-$d$ and depth-$(d+1)$ rings rather than
+descending one level deeper.
 \end{remark}
 
 \begin{definition}[Ring completion]
-Suppose a slice $S_n$ has been constructed. Consider the quadrilaterals $Q \notin S_n$ such that all four vertices of $Q$ are already vertices of $S_n$. Among these, the \emph{ring completion} chooses the $Q$ whose attachment to the right boundary of $S_n$ is bottommost (i.e., the last attachment encountered when scanning the right boundary from top to bottom). Then
+Suppose a slice $S_n$ has been constructed.  Consider the
+quadrilaterals $Q \notin S_n$ such that all four vertices of $Q$ are
+already vertices of $S_n$.  Among these, the \emph{ring completion}
+chooses the $Q$ whose attachment to the right boundary of $S_n$ is
+bottommost (i.e.\ the last attachment encountered when scanning the
+right boundary from top to bottom).  Then
 \[
     Q_{n+1} = Q, \qquad S_{n+1} = S_n \cup Q.
 \]
@@ -277,72 +217,178 @@ By construction, the ring completion introduces no new vertices.
 \end{definition}
 
 \begin{definition}[Move selection]
-At each step $n \geq 1$, the next quadrilateral $Q_{n+1}$ is chosen by the first applicable move in the following order of precedence:
+At each step $n \geq 1$, the next quadrilateral $Q_{n+1}$ is chosen
+by the first applicable move in the following order of precedence:
 \begin{enumerate}
     \item anchor drop;
     \item level add;
     \item join;
     \item ring completion.
 \end{enumerate}
-That is, each move is consulted only when no higher-precedence move applies.
+That is, each move is consulted only when no higher-precedence move
+applies.
 \end{definition}
 
 \begin{figure}
 \centering
 \includegraphics[width=0.85\textwidth]{example_figure.pdf}
-\caption{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 $01211333$.}
+\caption{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 $01211333$.}
 \label{fig:example-sequence}
 \end{figure}
 
-Let $N$ denote the total number of quadrilaterals in the decomposition of $G'$; equivalently, $N = |F(G')|/2$, where $F(G')$ is the set of triangular faces of $G'$.
+Let $N$ denote the total number of quadrilaterals in the
+decomposition of $G'$; equivalently, $N = |F(G')|/2$, where $F(G')$
+is the set of triangular faces of $G'$.
 
 \begin{theorem}[Termination and coverage]
-The sequence $Q_1, Q_2, \ldots$ generated by repeatedly applying the move-selection rule starting from any choice of initial quadrilateral $Q_1$ terminates after exactly $N$ steps. Moreover, every quadrilateral of the decomposition appears in the sequence exactly once, and $S_N = G'$.
+The sequence $Q_1, Q_2, \ldots$ generated by repeatedly applying the
+move-selection rule starting from any choice of initial
+quadrilateral $Q_1$ terminates after exactly $N$ steps.  Moreover,
+every quadrilateral of the decomposition appears in the sequence
+exactly once, and $S_N = G'$.
 \end{theorem}
 
 \begin{proof}
 We prove the three claims comprising the theorem.
 
-\emph{(1) No quadrilateral is visited twice.} By construction, each move chooses $Q_{n+1} \notin S_n$ and sets $S_{n+1} = S_n \cup Q_{n+1}$. Hence $n \mapsto Q_n$ is injective.
+\emph{(1) No quadrilateral is visited twice.}  By construction, each
+move chooses $Q_{n+1} \notin S_n$ and sets $S_{n+1} = S_n \cup
+Q_{n+1}$.  Hence $n \mapsto Q_n$ is injective.
 
-\emph{(2) Each $S_n$ is a slice.} We proceed by induction on $n$.
+\emph{(2) Each $S_n$ is a slice.}  We proceed by induction on $n$.
 
-\emph{Base case.} $S_1 = Q_1$ is a single quadrilateral, hence a closed topological disk on the sphere with boundary the closed walk along $Q_1$'s four perimeter edges.
+\emph{Base case.}  $S_1 = Q_1$ is a single quadrilateral, hence a
+closed topological disk on the sphere with boundary the closed walk
+along $Q_1$'s four perimeter edges.
 
-\emph{Inductive step.} Suppose $S_n$ is a slice. We show that for each of the four moves, $S_{n+1} = S_n \cup Q_{n+1}$ is again a slice. Topologically, $S_n$ is a closed disk and $Q_{n+1}$ is a closed disk; their union is a closed disk iff their intersection $S_n \cap Q_{n+1}$ is a connected arc on each of their boundaries. We verify this in each case below by identifying the intersection precisely.
+\emph{Inductive step.}  Suppose $S_n$ is a slice.  We show that for
+each of the four moves, $S_{n+1} = S_n \cup Q_{n+1}$ is again a
+slice.  Topologically, $S_n$ is a closed disk and $Q_{n+1}$ is a
+closed disk; their union is a closed disk iff their intersection
+$S_n \cap Q_{n+1}$ is a connected arc on each of their boundaries.
+We verify this in each case below by identifying the intersection
+precisely.
 
-\emph{(2.1) Anchor drop.} The new quadrilateral $Q_{n+1}$ shares exactly one edge $e$ with $S_n$: the right edge of the boundary down triangle $a$, which is also the left edge of the up triangle $b$. The intersection $S_n \cap Q_{n+1} = e$ is a single edge, a connected arc.
+\emph{(2.1) Anchor drop.}  The new quadrilateral $Q_{n+1}$ shares
+exactly one edge $e$ with $S_n$: the right edge of the boundary down
+triangle $a$, which is also the left edge of the up triangle $b$.
+The intersection $S_n \cap Q_{n+1} = e$ is a single edge, a
+connected arc.
 
-\emph{(2.2) Level add.} Three vertices $v_1, v_2, v_3$ of $Q_{n+1}$ lie on the right boundary of $S_n$. By the move's precedence, no anchor drop or higher-priority move applied at step $n+1$. We claim the two perimeter edges of $Q_{n+1}$ connecting these three vertices, namely $\{v_1, v_2\}$ and $\{v_2, v_3\}$, lie on the boundary of $S_n$. Suppose otherwise: then at least one of these edges has both adjacent faces in $S_n^c$, which means the face of $G'$ on the side of that edge opposite $Q_{n+1}$ is also outside $S_n$. By the structure of the move precedence and the inductive assumption, all such configurations would have been resolved by a higher-priority move first; hence at the moment level add fires, the only viable attachment configuration is the 2-edge path $\{v_1, v_2, v_3\}$. The intersection $S_n \cap Q_{n+1}$ is therefore this 2-edge path, a connected arc.
+\emph{(2.2) Level add.}  Three vertices $v_1, v_2, v_3$ of $Q_{n+1}$
+lie on the right boundary of $S_n$.  By the move's precedence, no
+anchor drop or higher-priority move applied at step $n+1$.  We claim
+the two perimeter edges of $Q_{n+1}$ connecting these three
+vertices, namely $\{v_1, v_2\}$ and $\{v_2, v_3\}$, lie on the
+boundary of $S_n$.  Suppose otherwise: then at least one of these
+edges has both adjacent faces in $S_n^c$, which means the face of
+$G'$ on the side of that edge opposite $Q_{n+1}$ is also outside
+$S_n$.  By the structure of the move precedence and the inductive
+assumption, all such configurations would have been resolved by a
+higher-priority move first; hence at the moment level add fires, the
+only viable attachment configuration is the 2-edge path $\{v_1,
+v_2, v_3\}$.  The intersection $S_n \cap Q_{n+1}$ is therefore this
+2-edge path, a connected arc.
 
-\emph{(2.3) Join.} The deep diamond $Q_{n+1}$ has one of its two down triangles sharing an edge $e$ with the right boundary of $S_n$. The intersection $S_n \cap Q_{n+1} = e$, a single edge, is a connected arc. (The second down triangle of $Q_{n+1}$ contributes two new vertices and lies entirely in $S_n^c$.)
+\emph{(2.3) Join.}  The deep diamond $Q_{n+1}$ has one of its two
+down triangles sharing an edge $e$ with the right boundary of $S_n$.
+The intersection $S_n \cap Q_{n+1} = e$, a single edge, is a
+connected arc.  (The second down triangle of $Q_{n+1}$ contributes
+two new vertices and lies entirely in $S_n^c$.)
 
-\emph{(2.4) Ring completion.} All four vertices of $Q_{n+1}$ lie in $V(S_n)$, and by precedence none of the previous three moves applied. We claim that in this case, at least three of $Q_{n+1}$'s four perimeter edges lie on the boundary of $S_n$, and these edges together form a connected arc along the boundary of $Q_{n+1}$. Indeed, any perimeter edge of $Q_{n+1}$ with both endpoints in $S_n$ but neither side in $S_n$ would force one of the higher-priority moves (specifically join or level add, applied with respect to the face on the other side of that edge); since no such move applied, the configuration is constrained so that the intersection $S_n \cap Q_{n+1}$ is a connected arc consisting of three or four of $Q_{n+1}$'s perimeter edges.
+\emph{(2.4) Ring completion.}  All four vertices of $Q_{n+1}$ lie in
+$V(S_n)$, and by precedence none of the previous three moves
+applied.  We claim that in this case, at least three of $Q_{n+1}$'s
+four perimeter edges lie on the boundary of $S_n$, and these edges
+together form a connected arc along the boundary of $Q_{n+1}$.
+Indeed, any perimeter edge of $Q_{n+1}$ with both endpoints in $S_n$
+but neither side in $S_n$ would force one of the higher-priority
+moves (specifically join or level add, applied with respect to the
+face on the other side of that edge); since no such move applied,
+the configuration is constrained so that the intersection $S_n \cap
+Q_{n+1}$ is a connected arc consisting of three or four of
+$Q_{n+1}$'s perimeter edges.
 
-In all four cases the intersection $S_n \cap Q_{n+1}$ is a connected arc, so $S_{n+1}$ is a closed topological disk and hence a slice. The boundary walk of $S_{n+1}$ is obtained from that of $S_n$ by deleting the arc $S_n \cap Q_{n+1}$ and inserting the complementary arc of $Q_{n+1}$'s boundary.
+In all four cases the intersection $S_n \cap Q_{n+1}$ is a connected
+arc, so $S_{n+1}$ is a closed topological disk and hence a slice.
+The boundary walk of $S_{n+1}$ is obtained from that of $S_n$ by
+deleting the arc $S_n \cap Q_{n+1}$ and inserting the complementary
+arc of $Q_{n+1}$'s boundary.
 
-\emph{(3) As long as $S_n \subsetneq G'$, some move applies.} Let $S_n$ be a slice with $S_n \subsetneq G'$. The complement $G' \setminus S_n$ is a non-empty closed disk on the sphere whose boundary coincides with the boundary of $S_n$. Pick an edge $e$ of the right boundary of $S_n$, and let $F$ be the face of $G'$ on the side of $e$ opposite $S_n$; then $F$ lies in $S_n^c$. Let $Q$ be the quadrilateral containing $F$.
+\emph{(3) As long as $S_n \subsetneq G'$, some move applies.}  Let
+$S_n$ be a slice with $S_n \subsetneq G'$.  The complement
+$G' \setminus S_n$ is a non-empty closed disk on the sphere whose
+boundary coincides with the boundary of $S_n$.  Pick an edge $e$ of
+the right boundary of $S_n$, and let $F$ be the face of $G'$ on the
+side of $e$ opposite $S_n$; then $F$ lies in $S_n^c$.  Let $Q$ be
+the quadrilateral containing $F$.
 
-Let $k$ be the number of vertices of $Q$ lying in $V(S_n)$, and let $j$ be the number of perimeter edges of $Q$ lying on the boundary of $S_n$. Since the edge $e$ belongs to $Q$ and lies on the boundary of $S_n$, we have $j \geq 1$ and $k \geq 2$.
+Let $k$ be the number of vertices of $Q$ lying in $V(S_n)$, and let
+$j$ be the number of perimeter edges of $Q$ lying on the boundary
+of $S_n$.  Since the edge $e$ belongs to $Q$ and lies on the
+boundary of $S_n$, we have $j \geq 1$ and $k \geq 2$.
 
-\emph{Case $k = 2$, $j = 1$.} Only the two endpoints of $e$ are in $S_n$.
+\emph{Case $k = 2$, $j = 1$.}  Only the two endpoints of $e$ are in
+$S_n$.
 \begin{itemize}
-    \item If $Q$ is an S quad and $e$ is the left edge of $Q$'s up triangle, then writing $a$ for the down triangle of $S_n$ across $e$, the anchor drop hypothesis holds and the move applies (possibly with a different choice of $a$ at the lower-rightmost position).
-    \item If $Q$ is a deep diamond, then $e$ belongs to one of its down triangles, and the join hypothesis holds.
-    \item If $Q$ is a shallow diamond, then neither anchor drop nor join applies directly to $Q$. In this case, follow the boundary of $S_n$ to find an adjacent face $F'$ also in $S_n^c$ whose containing quadrilateral $Q'$ admits one of the four moves; such an adjacent quadrilateral exists because $S_n^c$ is connected and is bounded by a closed walk in $G'$.
+    \item If $Q$ is an S quad and $e$ is the left edge of $Q$'s up
+          triangle, then writing $a$ for the down triangle of $S_n$
+          across $e$, the anchor drop hypothesis holds and the move
+          applies (possibly with a different choice of $a$ at the
+          lower-rightmost position).
+    \item If $Q$ is a deep diamond, then $e$ belongs to one of its
+          down triangles, and the join hypothesis holds.
+    \item If $Q$ is a shallow diamond, then neither anchor drop nor
+          join applies directly to $Q$.  In this case, follow the
+          boundary of $S_n$ to find an adjacent face $F'$ also in
+          $S_n^c$ whose containing quadrilateral $Q'$ admits one of
+          the four moves; such an adjacent quadrilateral exists
+          because $S_n^c$ is connected and is bounded by a closed
+          walk in $G'$.
 \end{itemize}
 
-\emph{Case $k = 3$, $j = 2$.} The three boundary vertices form a 2-edge path on $Q$'s outline, and the level add hypothesis holds. (No higher-priority move need have applied; if anchor drop also applies, the rule gives precedence to anchor drop.)
+\emph{Case $k = 3$, $j = 2$.}  The three boundary vertices form a
+2-edge path on $Q$'s outline, and the level add hypothesis holds.
+(No higher-priority move need have applied; if anchor drop also
+applies, the rule gives precedence to anchor drop.)
 
-\emph{Case $k = 4$.} All four vertices of $Q$ are in $V(S_n)$, and $j \geq 1$ by assumption. If a higher-priority hypothesis (anchor drop, level add, join) holds, the corresponding move applies. Otherwise the ring completion hypothesis holds and that move applies.
+\emph{Case $k = 4$.}  All four vertices of $Q$ are in $V(S_n)$, and
+$j \geq 1$ by assumption.  If a higher-priority hypothesis (anchor
+drop, level add, join) holds, the corresponding move applies.
+Otherwise the ring completion hypothesis holds and that move
+applies.
 
-In each case, some move adds either $Q$ or an adjacent quadrilateral $Q'$ to the slice, contradicting the assumption that no move applies. Hence as long as $S_n \subsetneq G'$, some move applies and $S_{n+1}$ is well-defined.
+In each case, some move adds either $Q$ or an adjacent quadrilateral
+$Q'$ to the slice, contradicting the assumption that no move
+applies.  Hence as long as $S_n \subsetneq G'$, some move applies
+and $S_{n+1}$ is well-defined.
 
-Combining (1)--(3): the sequence strictly grows $|S_n|$ by exactly one quadrilateral per step, never revisits a quadrilateral, and must continue until $S_n = G'$. The total number of steps is therefore exactly $N$.
+Combining (1)--(3): the sequence strictly grows $|S_n|$ by exactly
+one quadrilateral per step, never revisits a quadrilateral, and
+must continue until $S_n = G'$.  The total number of steps is
+therefore exactly $N$.
 \end{proof}
 
 \begin{remark}
-Two delicate sub-arguments in the proof of (2) and (3) deserve attention: (a) in (2.2) and (2.4), we relied on the move precedence to rule out configurations where the intersection $S_n \cap Q_{n+1}$ is disconnected; and (b) in (3), the shallow-diamond sub-case argues by ``move to an adjacent face'' but does not pin down which face. A more rigorous treatment would prove the equivalence of the following two statements: (i) the four moves cover every attachment configuration arising from a slice; (ii) at each step the move-selection rule produces a unique well-defined next quadrilateral.
+Two delicate sub-arguments in the proof of (2) and (3) deserve
+attention: (a) in (2.2) and (2.4), we relied on the move precedence
+to rule out configurations where the intersection $S_n \cap Q_{n+1}$
+is disconnected; and (b) in (3), the shallow-diamond sub-case argues
+by ``move to an adjacent face'' but does not pin down which face.
+A more rigorous treatment would prove the equivalence of the
+following two statements: (i) the four moves cover every attachment
+configuration arising from a slice; (ii) at each step the
+move-selection rule produces a unique well-defined next
+quadrilateral.
 \end{remark}
 
 % TODO: state and prove a uniqueness result that the sequence
@@ -351,15 +397,11 @@ Two delicate sub-arguments in the proof of (2) and (3) deserve attention: (a) in
 
 \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-depth}
+E.~Bauerfeld,
+\emph{Plane Depth},
+manuscript (math-research repository), 2026.
 
 \end{thebibliography}
 
 \end{document}
-
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