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split: extract foundational depth material into new plane_depth paper

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Splits the existing plane_depth_sequencing paper into two:

  papers/plane_depth/paper.tex (NEW, 4 pages):
    - Plane depth definition.
    - Level edge, up/down/neutral triangle classification.
    - Outerplanarity lemma (formerly Lemma 2.6 of PDS).
    - Deep embedding G' definition.
    - "Every face of G' is up or down" lemma.
    - Unique level edge per face; shared level edge between adjacent faces.
    - Quadrilateral decomposition definition with three types
      (shallow diamond, deep diamond, S quad).

  papers/plane_depth_sequencing/paper.tex (slimmed from 11 → 6 pages):
    - Cites plane_depth for all foundational definitions.
    - Keeps: slice, move definitions (anchor drop, level add, join,
      ring completion), move selection, termination theorem.

  papers/coloring_nested_tire_graphs/paper.tex:
    - Bibliography updated: cite bauerfeld-depth instead of bauerfeld-pds.
    - Two in-text references updated to cite the new outerplanarity
      lemma in plane_depth.

Rationale: the outerplanarity / deep-embedding / quadrilateral-
decomposition material is foundational and reused by multiple
papers (and by the proposed level-cycle generalization).  The
quadrilateral-sequencing programme is one specific application.
Splitting lets coloring_nested_tire_graphs cite the foundations
cleanly without dragging in the sequencing machinery.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
didericis
2026-05-26 13:49:44 -04:00
parent 1513922dec
commit 8b6c2b621c
12 changed files with 904 additions and 301 deletions
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\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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