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\begin{document}

\title{Plane Depth Sequencing}

\author{Eric Bauerfeld}
\address{}
\curraddr{}
\email{}
\thanks{}

\subjclass[2010]{Primary }

\keywords{plane graph, triangulation, plane depth, deep embedding,
quadrilateral decomposition, canonical sequencing}

\date{}

\dedicatory{}

\begin{abstract}
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 $4$-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
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 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{Preliminaries from plane depth}

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}

We now equip this decomposition with a deterministic traversal.

\section{Quadrilateral sequencing}

\begin{remark}
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}
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 $S_{n-1}$.
\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 $S_1 = Q_1$.
\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.
\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
\[
    Q_{n+1} = Q, \qquad S_{n+1} = S_n \cup Q.
\]
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).
\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
\[
    Q_{n+1} = Q, \qquad S_{n+1} = S_n \cup Q.
\]
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
\[
    Q_{n+1} = Q, \qquad S_{n+1} = S_n \cup Q.
\]
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 $\{d, d+1\}$.
\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.
\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
\[
    Q_{n+1} = Q, \qquad S_{n+1} = S_n \cup Q.
\]
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:
\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.
\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 $01211333$.}
\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 $G'$.

\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 $S_N = G'$.
\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{(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{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{(2.2) 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{(2.4) 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.

\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 $k \geq 2$.

\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 $G'$.
\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 $k = 4$.}  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.

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.
\end{remark}

% TODO: state and prove a uniqueness result that the sequence
% $Q_1, \ldots, Q_N$ is completely determined by the choice of $Q_1$ and the
% plane embedding.

\begin{thebibliography}{9}

\bibitem{bauerfeld-depth}
E.~Bauerfeld,
\emph{Plane Depth},
manuscript (math-research repository), 2026.

\end{thebibliography}

\end{document}
