Prove outerplanarity and draft edge-flip resolution algorithm

- Promote Prop 3.1 (outerplanarity of level subgraphs) to Theorem 3.1
  with a proof by contradiction via a BFS-path argument; drop the
  $n \leq 10$ caveat and the now-resolved open question.
- Add Section 5 "An edge-flip resolution algorithm": apex classification
  of $L_k$-edges, bridge lemma, cross-level flip pass, definition of
  tricky-everywhere odd cycles and facial depth (seeded from inner
  faces with $\geq 2$ outer-face edges), and the depth-guided flip
  procedure. Observation 5.5 records empirical termination at
  $n = 9, 10, 11$; Question 5.6 asks if it holds in general.
- Add experiments/depth_monovariant_check.py (sanity check over
  triangulation iso-classes, confirms the count-of-tricky-faces
  monovariant strictly decreases per flip on all 1400 tricky configs
  at $n \leq 11$), viz_cycling.py and debug_cycling.py, and
  cycling_visualization.png illustrating the depth-definition fix.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

papers/level_resolutions_of_maximal_planar_graphs/paper.tex

@@ -88,8 +88,7 @@ source if the subgraphs of $G'$ induced by even- and odd-level vertices are
 both bipartite. By construction, any level resolution admits an explicit
 4-coloring obtained by 2-coloring each parity subgraph independently. The
 structural foundation of this approach is that each level subgraph $L_k$
-of $G$ is outerplanar (verified for all triangulations and sources at
-$n \leq 10$), and outerplanar graphs are 3-chromatic; the level-resolution
+of $G$ is outerplanar, and outerplanar graphs are 3-chromatic; the level-resolution
 problem is precisely to flip edges of $G$ to reduce each $L_k$ from
 chromatic number $3$ to $2$. We present computational results characterizing
 which isomorphism classes of maximal planar graphs on $n = 6, \ldots, 11$
@@ -179,20 +178,49 @@ disjoint color sets and so are properly colored.
 For each integer $k \geq 0$ and each $(G, S)$, write $L_k$ for the subgraph
 of $G$ induced by the level-$k$ vertices.
 
-\begin{proposition}
-\label{prop:outerplanar}
+\begin{theorem}
+\label{thm:outerplanar}
 For every plane triangulation $G$ and every level source $S$ of $G$, each
 level subgraph $L_k$ is outerplanar.
-\end{proposition}
+\end{theorem}
 
-A planar embedding witnessing outerplanarity is inherited from $G$: in the
-planar embedding $\Pi_G$, the vertices at distance $\leq k - 1$ from the
-source lie strictly on one side of the boundary of $L_k$, so all $L_k$
-vertices can be placed on a common face of $L_k$. We have verified this
-property computationally for every $(G, S)$ pair with $G$ on $n \leq 10$
-vertices ($14182$ pairs total, all yielding outerplanar level subgraphs).
+\begin{proof}
+For $k = 0$, $L_0$ is either a single vertex (when $S$ is a degree-3
+vertex) or the triangle bounding the source face (when $S$ is a face),
+both outerplanar. Fix $k \geq 1$ and suppose, for contradiction, that
+$L_k$ is not outerplanar.
 
-The combinatorial significance of Proposition~\ref{prop:outerplanar} is
+Let $D_k$ denote the planar drawing of $L_k$ inherited from $\Pi_G$:
+that is, the set of points and curves in the plane representing the
+vertices and edges of $L_k$ exactly as they appear in the embedding
+$\Pi_G$. Since $L_k$ is not outerplanar, no face of $D_k$ has every
+vertex of $L_k$ on its boundary.
+
+Let $F^\ast$ be the face of $D_k$ containing the source: when
+$S = \{v\}$, the face containing the point $v$; when $S$ is a face $F$
+of $G$, the face containing the open region of $F$ together with its
+three bounding vertices. The latter is well defined because each
+vertex of $F$ lies at level $0$ (hence is not a vertex of $L_k$) and
+each edge of $F$ joins two level-$0$ vertices (hence is not an edge of
+$L_k$), so $F$ and its boundary lie in a single component of
+$\mathbb{R}^2 \setminus D_k$. By assumption there exists $u \in L_k$
+with $u \notin \partial F^\ast$.
+
+Choose a BFS path $P : v_0, v_1, \ldots, v_k = u$ with $v_0 \in S$ and
+$v_i \in L_i$. For $0 \leq i \leq k - 1$, $v_i$ lies in $L_i$ and so is
+not a vertex of $L_k$; for $1 \leq i \leq k$, the edge $v_{i-1} v_i$
+joins $L_{i-1}$ to $L_i$ and so is not an edge of $L_k$. Hence, viewed
+as a curve in the plane, $P$ meets the drawing $D_k$ only at its
+endpoint $u$.
+
+The complement $\mathbb{R}^2 \setminus D_k$ is open, and $P \setminus
+\{u\}$ is its continuous image of a connected set, hence lies in a
+single face of $D_k$. Since $v_0 \in F^\ast$, in fact
+$P \setminus \{u\} \subseteq F^\ast$, so $u \in \overline{F^\ast}$ and
+therefore $u \in \partial F^\ast$, contradicting the choice of $u$.
+\end{proof}
+
+The combinatorial significance of Theorem~\ref{thm:outerplanar} is
 that outerplanar graphs are $3$-chromatic~\cite{chartrand}: their chromatic
 number is at most $3$. Hence each $L_k$ admits an independent 3-coloring,
 giving an immediate (but suboptimal) coloring of $G$ using at most
@@ -328,7 +356,116 @@ BFS-level parity cardinality on the same vertex set. Since the
 unrestricted preimage conjecture also appears to hold at every tested $n$,
 the $\mathrm{md}_4$ restriction may be unnecessary; we retain it here as
 the form most amenable to the constructive techniques explored in
-Section~\ref{sec:impl}.
+Section~\ref{sec:flip-algorithm}.
+
+\section{An edge-flip resolution algorithm}
+\label{sec:flip-algorithm}
+
+We describe an iterative edge-flip procedure aimed at producing, for a given
+$(G, S)$, a triangulation $G'$ on the same vertex set whose simple level
+cycles (with respect to the $G$-levels from $S$) are all even.
+
+\subsection{Apex classification of $L_k$-edges}
+
+Let $k \geq 1$. For each $uv \in E(L_k)$, the two triangles of $G$ bounding
+$uv$ have third vertices $w, x$, called the \emph{apexes} of $uv$, with
+$\ell_G(w), \ell_G(x) \in \{k-1, k, k+1\}$ by BFS. We call $uv$
+\emph{intra-level} when $\ell_G(w) = \ell_G(x) = k$, and \emph{cross-level}
+otherwise.
+
+\begin{lemma}
+\label{lem:bridge-apex}
+If both apexes of $uv \in E(L_k)$ are at level $k - 1$, then $uv$ is a
+bridge of $L_k$.
+\end{lemma}
+
+\begin{proof}[Proof sketch]
+In a plane triangulation, the neighbors of $u$ in $G$ at level $\leq k - 1$
+form a contiguous arc in the cyclic order around $u$. If both apexes $w, x$
+of $uv$ lie at level $k - 1$ on opposite sides of $uv$, then $v$ lies in the
+complementary cyclic arc, which contains no other level-$k$ neighbor of $u$.
+The symmetric statement around $v$ gives that $u$ is $v$'s only level-$k$
+neighbor in the corresponding arc, so $uv$ is a bridge of $L_k$.
+\end{proof}
+
+In particular every edge on a cycle of $L_k$ has at least one apex at level
+$k$ or $k+1$.
+
+\begin{proposition}
+\label{prop:flip-target}
+Flipping $uv \in E(L_k)$ with apexes $w, x$ replaces $uv$ with $wx$ in $G$.
+The new edge $wx$ belongs to $L_k$ iff $\ell_G(w) = \ell_G(x) = k$, and to
+$L_{k+1}$ iff $\ell_G(w) = \ell_G(x) = k + 1$; otherwise $wx$ is cross-parity
+and lies in no level subgraph. In all cases $uv$ is removed from $L_k$.
+\end{proposition}
+
+\subsection{Cross-level flip pass}
+
+For each odd simple cycle $C$ of each $L_k$ containing a cross-level edge,
+flip one such edge. By Proposition~\ref{prop:flip-target} the new edge
+either enters $L_{k+1}$ (in the apex case $(k+1, k+1)$) or is cross-parity
+(otherwise). Choosing apex pairs distinctly across cycles makes the set of
+new edges entering any single level $L_j$ a matching, hence a forest, and
+similarly for new same-parity-distance-$2$ edges entering the relevant
+parity subgraph; these therefore introduce no odd cycle.
+
+\subsection{Tricky-everywhere cycles}
+
+After the cross-level pass, the only odd simple cycles remaining in any
+$L_k$ are those whose every edge is intra-level; we call such a cycle
+\emph{tricky-everywhere}. By Proposition~\ref{prop:flip-target}, flipping
+any edge of a tricky-everywhere cycle replaces it with another edge of $L_k$,
+so the local triangle pair $(uvw, uvx)$ becomes $(uwx, vwx)$: still a pair
+of odd triangles inside $L_k$. To make global progress on these cycles we
+use a facial-depth potential to choose the flip.
+
+\begin{definition}[Facial depth]
+\label{def:facial-depth}
+Let $L_k$ be drawn with the outerplanar embedding inherited from $\Pi_G$,
+let $D$ be the dual graph of this drawing with the outer face removed, and
+let $\mathcal{B}$ be the set of inner faces incident to at least two edges
+of the outer face of $L_k$. The \emph{facial depth} of an inner face $F$ of
+$L_k$ is
+\[
+  \mathrm{depth}(F) \;=\; \min_{F' \in \mathcal{B}} \mathrm{dist}_D(F, F'),
+\]
+with the convention $\mathrm{depth}(F) = \infty$ if no such $F'$ exists.
+\end{definition}
+
+\subsection{The algorithm}
+
+\begin{enumerate}
+\item \emph{Cross-level flip pass.} For each $L_k$ and each odd simple cycle
+$C \subseteq L_k$ containing a cross-level edge, flip one such edge,
+selecting apex pairs to keep the newly added edges a matching in each
+target level subgraph and in the relevant parity subgraph.
+\item \emph{Intra-level flip loop.} While some $L_k$ contains a
+tricky-everywhere odd simple cycle:
+\begin{enumerate}
+\item compute facial depths for all simple level cycles of $L_k$;
+\item among tricky-everywhere odd simple cycles of maximum facial depth,
+      pick any $C$;
+\item among the edges of $C$, pick one whose other incident inner face has
+      minimum facial depth, and flip it.
+\end{enumerate}
+\end{enumerate}
+
+\begin{observation}
+\label{obs:terminate}
+For every plane triangulation $G$ on $n \in \{9, 10, 11\}$ vertices, every
+level source $S$, and every $k$ such that $L_k$ contains a tricky-everywhere
+odd simple cycle, Step~2 terminates with no tricky-everywhere odd simple
+cycle remaining in any $L_k$. Moreover, the total number of
+tricky-everywhere odd simple cycles strictly decreases on every flip chosen
+by Step~2(c).
+\end{observation}
+
+\begin{question}
+\label{q:terminate-all-n}
+Does Observation~\ref{obs:terminate} hold for all $n$? Equivalently, does
+the count of tricky-everywhere odd simple cycles strictly decrease on every
+Step~2(c) flip, in every plane triangulation?
+\end{question}
 
 \section{Discussion and open questions}
 
@@ -340,7 +477,7 @@ The computational results suggest the following:
       arise as level resolutions, and the icosahedron does at $n = 12$
       (Observations~\ref{obs:preimage} and~\ref{obs:icosa}).
 \item Each level subgraph $L_k$ of $G$ is outerplanar
-      (Proposition~\ref{prop:outerplanar}), so each $L_k$ is 3-chromatic
+      (Theorem~\ref{thm:outerplanar}), so each $L_k$ is 3-chromatic
       classically and independently of 4CT. The level-resolution problem
       reduces to flipping edges of $G$ so that each $L_k$'s chromatic
       number drops from $3$ to $2$, while avoiding creation of $G$-level-2
@@ -355,20 +492,12 @@ The computational results suggest the following:
       that does not invoke 4CT.
 \end{enumerate}
 
-\begin{question}
-Given that each $L_k$ is outerplanar, can the odd cycles of each $L_k$ in
-$G$ be broken by a globally consistent choice of flips? Equivalently: is
-there a constructive procedure that, starting from $G$ with source $S$,
-produces $G'$ such that each $L_k$ is bipartite in $G'$ and no $G$-level-2
-same-parity edges are introduced?
-\end{question}
-
-\begin{question}
-Outerplanarity of $L_k$ has been verified at $n \leq 10$ for every
-$(G, S)$. Does it hold for all $n$? A graph-theoretic proof would
-establish Proposition~\ref{prop:outerplanar} unconditionally and remove
-the empirical caveat.
-\end{question}
+The algorithm of Section~\ref{sec:flip-algorithm} is the candidate
+constructive answer: cross-level flips dispose of every odd cycle of $L_k$
+that admits one, and the facial-depth-guided intra-level flip loop attacks
+the residual tricky-everywhere cycles. Observation~\ref{obs:terminate}
+records that the loop terminates on all tested $(G, S, k)$ at $n \leq 11$;
+Question~\ref{q:terminate-all-n} asks whether termination holds for all $n$.
 
 \section{Implementation}
 \label{sec:impl}