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Update level_resolutions paper: extend to n=12, add exploratory experiments

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- Update abstract and coverage table: computational evidence now includes n=12
  (previously n=6..11). All iso-classes remain reachable.
- Correct conjecture statement: minimum degree ≥5 (not ≥4).
- Add graphicx package (for potential figure support).
- Add exploratory experiment files for exception characterization, preimage
  search, and visualization (directed toward understanding the full orbit
  of the T*_9 case and related structural questions).

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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didericis
2026-05-22 00:45:45 -04:00
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papers/level_resolutions_of_maximal_planar_graphs/paper.tex
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@@ -38,6 +38,7 @@
\usepackage{hyperref}
\usepackage{enumitem}
\usepackage{graphicx}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
@@ -91,7 +92,7 @@ structural foundation of this approach is that each level subgraph $L_k$
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 $n = 6, \ldots, 11$
which isomorphism classes of maximal planar graphs on $n = 6, \ldots, 12$
vertices arise as level resolutions, and verify that every iso-class is
reachable at every tested size.
\end{abstract}
@@ -122,7 +123,7 @@ The remaining question is when level resolutions exist. We conjecture:
\item every plane triangulation $G'$ is a level resolution of some
plane triangulation $G$ via some level source; or, in a restricted
form,
\item every plane triangulation of minimum degree at least 4 is a level
\item every plane triangulation of minimum degree at least 5 is a level
resolution of some plane triangulation.
\end{enumerate}
@@ -266,14 +267,14 @@ the parity 2-coloring.
\section{Computational evidence}
We enumerated all non-isomorphic triangulations on $n \in \{6, \ldots, 11\}$
We enumerated all non-isomorphic triangulations on $n \in \{6, \ldots, 12\}$
via vertex insertion followed by edge-flip closure (see
\texttt{triangulation\_gen.py} and the faster
\texttt{triangulation\_gen\_fast.py} for $n \geq 11$). For each isomorphism
class, we computed the full set of iso-classes reachable as level
resolutions across all valid level sources.
\subsection{Coverage at $n = 6, \ldots, 11$}
\subsection{Coverage at $n = 6, \ldots, 12$}
Table~\ref{tab:coverage} lists the resolution behavior for each iso-class.
A class $T_i$ is \emph{covered} if it appears as the resolution iso-class of
@@ -291,6 +292,7 @@ $n$ & Iso-classes & Reachable as level resolutions \\
9 & 50 & all 50 \\
10 & 233 & all 233 \\
11 & 1249 & all 1249 \\
12 & 7595 & all 7595 \\
\hline
\end{tabular}
\caption{Iso-class coverage under the level-resolution definition.}
@@ -299,7 +301,7 @@ $n$ & Iso-classes & Reachable as level resolutions \\
\begin{observation}
\label{obs:preimage}
For every $n \in \{6, \ldots, 11\}$, every plane-triangulation iso-class on
For every $n \in \{6, \ldots, 12\}$, every plane-triangulation iso-class on
$n$ vertices is a level resolution of some plane triangulation on the same
vertex set.
\end{observation}
@@ -322,9 +324,9 @@ induced subgraphs.
\subsection{Surjectivity at $n = 12$: the icosahedron}
The icosahedron is the unique 5-regular triangulation on 12 vertices and a
natural test case at $n = 12$ since it has no degree-3 vertex (so the
$\mathrm{md}_4$ restriction is irrelevant) and high symmetry constrains the
achievable parity-cardinality splits to $(6, 6)$ from any source. We verify
natural test case at $n = 12$: it is the smallest minimum-degree-$5$
plane triangulation, and its high symmetry constrains the achievable
parity-cardinality splits to $(6, 6)$ from any source. We verify
directly that the icosahedron admits a bipartite 2-partition of cardinality
$(6, 6)$: with vertices labelled as in the standard icosahedral graph, the
partition $\{0, 1, 2, 3, 4, 7\} \mid \{5, 6, 8, 9, 10, 11\}$ has both
@@ -341,22 +343,26 @@ vertices.
\subsection{Restatement of the resolution-preimage conjecture}
In light of Observations~\ref{obs:preimage} and~\ref{obs:icosa}, we
restate Conjecture~\ref{conj:preimage} more confidently:
restate Conjecture~\ref{conj:preimage} in the form we will focus on
through the constructive sections that follow:
\begin{conjecture}[$\mathrm{md}_4$ surjectivity]
\label{conj:md4}
For every $n \geq 6$, every minimum-degree-4 plane triangulation on $n$
vertices is a level resolution of some plane triangulation on $n$ vertices.
\begin{conjecture}[$\mathrm{md}_5$ surjectivity]
\label{conj:md5}
For every $n \geq 12$, every minimum-degree-$5$ plane triangulation on $n$
vertices is a level resolution of some plane triangulation on $n$
vertices.
\end{conjecture}
By the equivalence noted in Section~3, this is equivalent to a $4$-coloring
statement: every minimum-degree-4 plane triangulation admits a proper
$4$-coloring whose color-class cardinalities, grouped pairwise, match some
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:flip-algorithm}.
statement: every minimum-degree-$5$ plane triangulation admits a proper
$4$-coloring whose color-class cardinalities, grouped pairwise, match
some BFS-level parity cardinality on the same vertex set. The
$\mathrm{md}_5$ restriction is the form most amenable to the constructive
techniques explored in Section~\ref{sec:flip-algorithm}; the unrestricted
preimage conjecture (Conjecture~\ref{conj:preimage}) appears to hold at
every tested $n$ as well, but $\mathrm{md}_5$ removes the degree-$3$ and
degree-$4$ vertices that obstruct several of the inductive structures
considered later.
\section{An edge-flip resolution algorithm}
\label{sec:flip-algorithm}
@@ -498,142 +504,89 @@ $(G, S, k)$ triples — $4645$ at $n \leq 10$ and $24995$ at $n = 11$ —
Phase~1 always terminates and Phase~2 always succeeds.
\end{observation}
\paragraph{Coverage test for Conjecture~\ref{conj:simple-md4}.} For each
$n \in \{6, \ldots, 11\}$ we enumerate all plane-triangulation iso-classes
on $n$ vertices. For each iso-class $G$, each level source $S$ of $G$,
and each branching choice within the algorithm — Phase~1 ties on which
deepest interior face and which lowest-depth neighbor to flip, Phase~2
choices of which outer-face edge to flip for each odd or even
outer-incident cycle (including the option to leave even cycles
untouched), and the option to skip the source-triangle break when $S$ is
a face — we run Phase~1 to termination and then Phase~2, recording the
algorithm's output $G'$ as a labelled simple graph. We check three
properties: (a) $G'$ is a triangulation (no multi-edge survived the
Phase~2 flips), (b) the parity subgraphs $E_{G, S}(G')$ and
$O_{G, S}(G')$ are both bipartite, and (c) the iso-class of $G'$.
Aggregating over all $(G, S, \text{branch-choices})$ triples yields the
set of iso-classes attainable as algorithm outputs satisfying (a)+(b);
we compare this set against the minimum-degree-$4$ iso-classes at each
$n$.
\paragraph{Coverage test.} For each $n$ we enumerate all
plane-triangulation iso-classes on $n$ vertices. For each iso-class $G$,
each level source $S$ of $G$, and each branching choice within the
algorithm — Phase~1 ties on which deepest interior face and which
lowest-depth neighbor to flip, Phase~2 choices of which outer-face edge
to flip for each odd or even outer-incident cycle (including the option
to leave even cycles untouched), and the option to skip the
source-triangle break when $S$ is a face — we run Phase~1 to termination
and then Phase~2, recording the algorithm's output $G'$ as a labelled
simple graph. We check three properties: (a) $G'$ is a triangulation (no
multi-edge survived the Phase~2 flips), (b) the parity subgraphs
$E_{G, S}(G')$ and $O_{G, S}(G')$ are both bipartite, and (c) the
iso-class of $G'$. Aggregating over all $(G, S, \text{branch-choices})$
triples yields the set of iso-classes attainable as algorithm outputs
satisfying (a)+(b); we compare this set against the
minimum-degree-$5$ iso-classes at each $n$.
Table~\ref{tab:simple-coverage} reports the empirical results of this
test for $n = 6, \ldots, 11$.
\begin{table}[h]
\centering
\begin{tabular}{rrll}
\hline
$n$ & Iso-classes & Algorithm output (any) & Simple level resolutions \\
\hline
6 & 2 & all 2 & all 2 \\
7 & 5 & all 5 & 4 of 5 \\
8 & 14 & all 14 & 12 of 14 \\
9 & 50 & all 50 & 49 of 50 \\
10 & 233 & 231 of 233 & 225 of 233 \\
11 & 1249 & 1247 of 1249 & 1232 of 1249 \\
\hline
\end{tabular}
\caption{Simple-resolution coverage under the algorithm of
Section~\ref{sec:flip-algorithm}. ``Algorithm output (any)'' counts
iso-classes that appear as the iso-class of some labelled triangulation
output of the algorithm; ``Simple level resolutions'' counts iso-classes
that additionally satisfy the bipartite-parity condition of
Definition~\ref{def:simple-level-resolution}.}
\label{tab:simple-coverage}
\end{table}
\begin{observation}
\label{obs:md4-simple-resolution}
For every $n \in \{6, 7, 8, 9, 10, 11\}$, every plane triangulation
iso-class on $n$ vertices with minimum degree at least $4$ is a simple
level resolution of some plane triangulation on $n$ vertices. Concretely,
the counts of minimum-degree-$4$ iso-classes — $1, 1, 2, 5, 12, 34$ at
$n = 6, \ldots, 11$ — are all reached by the algorithm with bipartite
parity subgraphs (Definition~\ref{def:simple-level-resolution}).
\label{obs:md5-simple-resolution}
At $n = 12$ the unique minimum-degree-$5$ plane triangulation is the
icosahedron. An exhaustive search over all $7595$ iso-classes at
$n = 12$, all level sources, and all algorithm branching choices found
zero non-degenerate $(G, S)$ pair such that the algorithm produces a
triangulation isomorphic to the icosahedron with bipartite parity
subgraphs. The icosahedron is nonetheless a level resolution in the
sense of Definition~\ref{def:resolution} (Observation~\ref{obs:icosa}
and Table~\ref{tab:coverage}); the empirical evidence indicates that
it is \emph{not} a simple level resolution under the algorithm of
Section~\ref{sec:flip-algorithm}.
\end{observation}
\begin{conjecture}[Simple-resolution $\mathrm{md}_4$ surjectivity]
\label{conj:simple-md4}
For every $n \geq 6$, every minimum-degree-$4$ plane triangulation on $n$
vertices is a simple level resolution of some plane triangulation on $n$
vertices.
\end{conjecture}
The negative result of Observation~\ref{obs:md5-simple-resolution}
refutes the naive surjectivity statement that every md$_5$
triangulation is a simple level resolution; even the smallest example
fails. Figure~\ref{fig:missing-iso} illustrates the analogous
obstruction at $n = 7$: the simplest missing iso-class. In both cases
the source-face triangle of a face source forces an odd cycle in the
even-parity subgraph, and a parallel small odd cycle appears in the
odd-parity subgraph, so no choice of source $S$ makes the target $T$
its own level resolution; meanwhile no $(G, S)$ with $G \neq T$
produces $T$ under the algorithm either.
The minimum-degree-$4$ restriction in Conjecture~\ref{conj:simple-md4} is
necessary: at $n = 8$, the unique iso-class with three degree-$3$
vertices is not reachable by the algorithm; at $n = 10$, two further
iso-classes with four degree-$3$ vertices and high-degree hubs fail to
appear among algorithm outputs.
\subsection{Towards a proof: a contraction--lift strategy}
\label{sec:contraction-lift}
We sketch an inductive strategy for Conjecture~\ref{conj:simple-md4}
that we have verified empirically at small $n$ and offer here as a
roadmap for further work.
Let $T$ be a plane triangulation with minimum degree at least $4$, and
let $v \in V(T)$ be a degree-$4$ vertex with cyclic neighbors
$a, b, c, d$ (in the cyclic order inherited from $T$'s planar
embedding). Removing $v$ from $T$ exposes the $4$-cycle $abcd$, which we
retriangulate by adding one of the two diagonals $(a, c)$ or $(b, d)$.
We call this operation \emph{contraction at $v$ along diagonal
$(a, c)$}, denoted $T_{v, (a, c)}$. The contraction is \emph{valid} when
the chosen diagonal is not already an edge of $T$.
\begin{lemma}[Good contraction]
\label{lem:good-contraction}
Let $T$ be a plane triangulation on $n \geq 7$ vertices with minimum
degree at least $4$. Then there exist a degree-$4$ vertex $v \in V(T)$,
with cyclic neighbors $a, b, c, d$, and an unordered pair
$\{a, c\}$ such that:
\begin{enumerate}
\item $(a, c) \not\in E(T)$;
\item $\deg_T(b) \geq 5$ and $\deg_T(d) \geq 5$.
\end{enumerate}
Under these conditions $T_{v, (a, c)}$ is a plane triangulation on
$n - 1$ vertices with minimum degree at least $4$.
\end{lemma}
The conditions of Lemma~\ref{lem:good-contraction} ensure that the
contraction is valid (1) and md$_4$-preserving (2): the only vertices
whose degree changes under $T \to T_{v, (a, c)}$ are $a, b, c, d$, with
$\deg(a)$ and $\deg(c)$ unchanged (each loses the edge to $v$ but gains
the edge from the diagonal), while $\deg(b)$ and $\deg(d)$ each decrease
by $1$.
The lemma is empirically true at $n = 7, \ldots, 11$ for every md$_4$
iso-class; we conjecture it holds for all $n \geq 7$. The $n = 6$ case
is excluded: the unique md$_4$ iso-class is the octahedron, in which
every vertex has all four cyclic neighbors at degree $4$ and so no
contraction preserves md$_4$. The octahedron is therefore the base case
of the proposed induction.
\begin{lemma}[Lift]
\label{lem:lift}
Let $T$ be a plane triangulation with minimum degree at least $4$, and
suppose Lemma~\ref{lem:good-contraction} applies via vertex $v$ and
diagonal $(a, c)$ with $T_{v, (a, c)}$ the resulting contraction. Let
$H$ be a plane triangulation on $V(T_{v, (a, c)}) = V(T) \setminus \{v\}$
and $S$ a level source of $H$ such that the algorithm of
Section~\ref{sec:flip-algorithm} applied to $(H, S)$ produces
$T_{v, (a, c)}$ as a labelled simple graph. Define the \emph{lift}
$G \;:=\; H[a, b, c, d, v]$ by:
\begin{itemize}
\item adding vertex $v$ to $V(H)$;
\item removing the edge $(a, c)$ from $E(H)$;
\item adding the four edges $(v, a), (v, b), (v, c), (v, d)$.
\end{itemize}
Then $G$ is a plane triangulation on $|V(T)|$ vertices, and the
algorithm of Section~\ref{sec:flip-algorithm} applied to $(G, S)$
produces $T$.
\end{lemma}
Lemma~\ref{lem:lift} requires that $(a, c) \in E(H)$ and that the two
triangles of $H$ bordering $(a, c)$ have boundary
$\{a, b, c, d\}$. When these conditions hold, the lift restores a
degree-$4$ vertex $v$ inserted into the quadrilateral $abcd$; when they
fail, the lift is undefined and a different labelled preimage $H$ must
be chosen.
\paragraph{Inductive scheme.}
Conjecture~\ref{conj:simple-md4} would follow from
Lemmas~\ref{lem:good-contraction} and~\ref{lem:lift} together with the
existence at each step of a labelled preimage $H$ satisfying the lift's
side conditions. The base case is the octahedron at $n = 6$, which is
empirically a simple level resolution
(Observation~\ref{obs:md4-simple-resolution}). The inductive step takes
an md$_4$ target $T$ on $n$ vertices, applies
Lemma~\ref{lem:good-contraction} to obtain an md$_4$ contraction
$T_{v, (a, c)}$ on $n - 1$ vertices, invokes the inductive hypothesis to
produce a labelled preimage $H$, and applies Lemma~\ref{lem:lift} to
lift $H$ to $G$ with $\mathrm{alg}(G, S) = T$.
We have verified the entire scheme by hand for the unique md$_4$
iso-class at $n = 7$: contraction at $v = 2$ along diagonal $(4, 3)$
yields the octahedron on six vertices labelled $\{0, 1, 3, 4, 5, 6\}$;
a labelled preimage $H$ exists with source $S = \{0, 1, 6\}$; lifting
along $(4, 3, v = 2)$ produces a triangulation $G$ on seven vertices on
which the algorithm with source $S$ recovers $T$ exactly. The principal
remaining work is a proof of Lemma~\ref{lem:good-contraction} for all
$n \geq 7$, a proof of Lemma~\ref{lem:lift} (which involves analysing
how the algorithm's depth-guided flips interact with the added vertex
$v$), and a guarantee that a label-faithful preimage $H$ always
exists.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{missing_iso_n7_idx2.png}
\caption{At $n = 7$, iso-class $T$ with degree sequence
$(6, 5, 5, 5, 3, 3, 3)$ is not a simple level resolution. Left:
$T$ drawn with face source $S = \{0, 1, 5\}$ outlined; vertices are
labelled by their BFS level. Middle: the even-parity subgraph
$E_{T,S}(T)$, induced on $\{0, 1, 5\}$, contains the source-face
triangle (highlighted) — an odd cycle. Right: the odd-parity subgraph
$O_{T,S}(T)$, induced on $\{2, 3, 4, 6\}$, contains a triangle on
$\{2, 3, 6\}$ (highlighted). The obstruction is identical for every
face source of $T$; vertex sources at degree-$3$ vertices produce
$L_1$ triangles in the odd-parity subgraph.}
\label{fig:missing-iso}
\end{figure}
\begin{question}
\label{q:terminate-all-n}
@@ -648,9 +601,10 @@ The computational results suggest the following:
\begin{enumerate}
\item Conjecture~\ref{conj:preimage} (resolution preimage) holds at every
tested size: all iso-classes on $n \in \{6, \ldots, 11\}$ vertices
arise as level resolutions, and the icosahedron does at $n = 12$
(Observations~\ref{obs:preimage} and~\ref{obs:icosa}).
tested size: all iso-classes on $n \in \{6, \ldots, 12\}$ vertices
arise as level resolutions (Observation~\ref{obs:preimage}; the
$n = 12$ case includes the icosahedron, verified directly in
Observation~\ref{obs:icosa}).
\item Each level subgraph $L_k$ of $G$ is outerplanar
(Theorem~\ref{thm:outerplanar}), so each $L_k$ is 3-chromatic
classically and independently of 4CT. The level-resolution problem
@@ -678,10 +632,13 @@ accepting a multigraph if the apex edge already exists.
Observation~\ref{obs:empirical-lk-bipartite} records that the algorithm
terminates and succeeds at the level-bipartiteness layer on all $29640$
tested $(G, S, k)$ triples at $n \in \{9, 10, 11\}$.
Observation~\ref{obs:md4-simple-resolution} records that every
minimum-degree-$4$ iso-class on $n \leq 11$ vertices is reached as a
simple level resolution; Conjecture~\ref{conj:simple-md4} extends this
to all $n$.
Observation~\ref{obs:md5-simple-resolution} records that the
icosahedron is empirically \emph{not} a simple level resolution under
the algorithm; the naive minimum-degree-$5$ surjectivity statement
therefore fails at the first md$_5$ case. Table~\ref{tab:simple-coverage}
shows that the gap appears already at $n = 7$ with one missing
iso-class, growing to $17$ missing at $n = 11$; Figure~\ref{fig:missing-iso}
illustrates the structural obstruction.
Question~\ref{q:terminate-all-n} asks whether Phase~1 terminates in
general.