math-research/papers/even_level_graph_generators/paper.tex

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\begin{document}

\title{Even Level Graph Generators}

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%    author one information
\author{Eric Bauerfeld}
\address{}
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\subjclass[2010]{Primary }

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\date{}

\dedicatory{}

\begin{abstract}
\end{abstract}

\maketitle

\section{Introduction}

\section{Definitions}

Throughout, $G = (V, E)$ is a plane maximal planar graph (a triangulation)
with a fixed planar embedding $\Pi_G$. We write $|V| = n$, so $|E| = 3n - 6$
and $G$ has $2n - 4$ triangular faces.

\begin{definition}[Level source]
A \emph{level source} of $G$ is any vertex $v \in V$; we write
$S = \{v\}$ for the level-0 source.
\end{definition}

\begin{definition}[Levels]
Given a level source $S \subseteq V$, the \emph{level} of $v \in V$ is
$\ell_G(v) = \mathrm{dist}_G(v, S)$, the graph distance from $v$ to the nearest
source vertex.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.55\textwidth]{fig_levels.png}
\caption{BFS levels from the degree-$3$ vertex source $S = \{4\}$.
The source is level $0$, its three neighbours are level $1$, and the
remaining vertices are level $2$. Colour encodes the level.}
\label{fig:levels}
\end{figure}

\begin{definition}[Level cycle]
A \emph{level cycle} of $G$ (with respect to a level source $S$) is a
simple cycle in $G$ all of whose vertices have the same level.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.55\textwidth]{fig_level_cycle.png}
\caption{A level cycle in the triangulation of Figure~\ref{fig:levels}.
The triangle $1\!-\!2\!-\!3$ is a simple cycle whose three vertices all
lie at level $1$, so it is a level cycle at level $1$.}
\label{fig:level-cycle}
\end{figure}

\begin{definition}[Edge switch]
\label{def:edge-switch}
Let $G$ be a triangulation with level source $S$, and let $e = uv$ be an
edge of a level cycle of $G$. The \emph{edge switch} at $e$ is the edge
flip on $e$: writing $uvw$ and $uvx$ for the two triangular faces of $G$
containing $e$, the edge $uv$ is removed and the edge $wx$ is added. As
with any edge flip, the result is a triangulation on the same vertex set
provided $w$ and $x$ are non-adjacent in $G$.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.95\textwidth]{fig_edge_switch.png}
\caption{An edge switch on the level cycle of
Figure~\ref{fig:level-cycle}. The chosen cycle edge $1\!-\!2$ is shared
by the triangular faces $(0,1,2)$ and $(1,2,4)$; the switch deletes
$1\!-\!2$ (red, left) and inserts $0\!-\!4$ (green, right). Vertex
colours indicate the original levels in $G$.}
\label{fig:edge-switch}
\end{figure}

\begin{definition}[Parity subgraph]
Let $G$ be a triangulation with level source $S$, and let $G'$ be a triangulation
on the same vertex set as $G$. The \emph{even parity subgraph} $E_{G,S}(G')$ is
the subgraph of $G'$ induced by $\{v \in V : \ell_G(v) \equiv 0 \pmod 2\}$. The
\emph{odd parity subgraph} is defined analogously for odd $\ell_G$.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=\textwidth]{fig_parity_subgraph.png}
\caption{Parity subgraphs of $G' = T$ with respect to the level structure of
Figure~\ref{fig:levels} (here we take $G = G' = T$). Left: $T$ with vertices
coloured by $\ell_G \bmod 2$ (blue $=$ even, orange $=$ odd). Middle: the
even parity subgraph $E_{G,S}(G')$, induced on $\{0, 4, 5, 6\}$; only
edges with both endpoints even appear. Right: the odd parity subgraph
$O_{G,S}(G')$, induced on $\{1, 2, 3\}$; the highlighted triangle shows
that $O_{G,S}(G')$ is not bipartite for this choice of $G'$.}
\label{fig:parity-subgraph}
\end{figure}

\section{Outerplanarity of level components}
\label{sec:outerplanar-components}

For each integer $k \geq 0$ and each $(G, S)$, write $L_k$ for the
subgraph of $G$ induced by the level-$k$ vertices. A \emph{level
component} of $G$ (with respect to $S$) is a connected component of
some $L_k$.

\begin{theorem}
\label{thm:outerplanar-component}
For every plane triangulation $G$ and every level source $S$ of $G$,
every level component of $G$ is outerplanar.
\end{theorem}

\begin{proof}
Since every subgraph of an outerplanar graph is outerplanar, it suffices
to show that each level subgraph $L_k$ is outerplanar. For $k = 0$,
$L_0 = S$ is a single vertex and is trivially outerplanar.

Fix $k \geq 1$ and let $D_k$ be the drawing of $L_k$ inherited from
$\Pi_G$. Let $F^\ast$ be the face of $D_k$ containing the source.
Suppose for contradiction that some $u \in L_k$ does not lie on
$\partial F^\ast$, so $u$ lies on the boundary of some other face of
$D_k$. Take any path $P$ in $G$ from $v_0 \in S$ to $u$. As a curve in
$\Pi_G$, $P$ starts in $F^\ast$ and ends at a point off $\partial
F^\ast$, so it must transition from $F^\ast$ to a different face of
$D_k$; in a planar embedding this can happen only at a vertex of
$D_k$, that is, at a level-$k$ vertex $w$ on $P$. Either $w \neq u$
(so $P$ has length $\geq \mathrm{dist}_G(S, w) + 1 \geq k + 1$), or
$w = u$ (contradicting $u \notin \partial F^\ast$). Since every
$S$-to-$u$ path has length $\geq k + 1$, $\mathrm{dist}_G(S, u) \geq
k + 1$, contradicting $u \in L_k$.
\end{proof}

\section{Even Level Graphs}
\label{sec:even-level-graphs}

\begin{definition}[Even Level Graph]
\label{def:even-level-graph}
A plane triangulation $G$ with level source $S$ is an \emph{Even Level
Graph} if every level cycle of $G$ has even length.
\end{definition}

\begin{theorem}
\label{thm:even-level-4colorable}
Every Even Level Graph is $4$-colorable.
\end{theorem}

\begin{proof}
Since adjacent vertices in $G$ have levels differing by at most $1$,
any edge between two same-parity endpoints in fact connects two
vertices at the same level. Hence
\[
  E_{G,S}(G) \;=\; \bigsqcup_{i \geq 0} L_{2i},
  \qquad
  O_{G,S}(G) \;=\; \bigsqcup_{i \geq 0} L_{2i+1},
\]
and each $L_k$ is bipartite because its cycles are level cycles of
$G$, which have even length by hypothesis. Choose a $2$-coloring of
$E_{G,S}(G)$ in $\{\text{red}, \text{blue}\}$ and a $2$-coloring of
$O_{G,S}(G)$ in $\{\text{yellow}, \text{green}\}$. Same-parity edges
of $G$ are properly colored by the respective bipartition;
opposite-parity edges connect $\{\text{red}, \text{blue}\}$ to
$\{\text{yellow}, \text{green}\}$. The combined assignment is a
proper $4$-coloring of $G$.
\end{proof}

\begin{definition}[Derived level graph]
\label{def:derived-level-graph}
Let $G$ be an Even Level Graph with level source $S$, and let $E$ and
$O$ denote the edge sets of the even and odd parity subgraphs
$E_{G,S}(G)$ and $O_{G,S}(G)$. A \emph{derived level graph} of $G$ is
a triangulation $G'$ on the same vertex set as $G$ obtained by a
sequence of edge switches (Definition~\ref{def:edge-switch}), each
acting on an edge of $E$ or of $O$. We do not update $E$ or $O$ to
reflect the level structure of intermediate triangulations: throughout
the sequence, an edge is classified as belonging to $E$ (resp.\ $O$) if
and only if both of its endpoints have even (resp.\ odd) level in $G$.

A derived level graph $G'$ is \emph{valid} if both $E_{G,S}(G')$ and
$O_{G,S}(G')$ contain only even cycles.
\end{definition}

\begin{definition}[Intertwining tree]
\label{def:intertwining-tree}
A maximal planar graph $G$ is an \emph{intertwining tree} if its
vertex set can be partitioned into two sets $A$ and $B$ such that
both induced subgraphs $G[A]$ and $G[B]$ are trees.
\end{definition}

\begin{theorem}
\label{thm:intertwining-iff-hamiltonian-dual}
A maximal planar graph $G$ is an intertwining tree if and only if its
dual $G^\ast$ has a Hamiltonian cycle.
\end{theorem}

\begin{proof}
($\Rightarrow$) Let $V(G) = A \sqcup B$ with $G[A]$ and $G[B]$ trees.
Every triangular face $\{x,y,z\}$ of $G$ meets both $A$ and $B$: if all
three vertices were in $A$ the triangle would be a cycle in the tree
$G[A]$, and likewise for $B$. Draw a closed curve through the faces of
$G$ separating the $A$-vertices from the $B$-vertices within each face.
Since every face is split, the curve visits every face exactly once and
crosses an edge of $G$ precisely when that edge joins $A$ to $B$; it is
therefore a Hamiltonian cycle of $G^\ast$.

($\Leftarrow$) Let $H$ be a Hamiltonian cycle of $G^\ast$. Drawn in the
plane, $H$ is a Jordan curve visiting every face of $G$ once; let $A$
and $B$ be the vertices of $G$ interior and exterior to $H$. The
$2n-4$ edges of $H$ cross exactly the edges of $G$ between $A$ and $B$,
leaving $(3n-6)-(2n-4) = n-2$ edges inside $G[A]$ and $G[B]$ together.
The edges inside $A$ lie in the disk bounded by $H$ and span $A$
without enclosing a face (each face is cut by $H$), so $G[A]$ is a tree;
likewise $G[B]$.
\end{proof}

\begin{conjecture}
\label{conj:every-triangulation-derived}
Every maximal planar graph is a valid derived level graph of some Even
Level Graph, an intertwining tree, or both.
\end{conjecture}

By Theorem~\ref{thm:intertwining-iff-hamiltonian-dual}, the
intertwining-tree disjunct fails for $G$ exactly when $G^\ast$ is a
counterexample to Tait's conjecture. The smallest such $G^\ast$ have
$38$ vertices (Holton--McKay~\cite{holton-mckay}, exactly $6$ graphs),
so the smallest triangulations that are not intertwining trees occur at
$n = 21$ and there are exactly $6$ of them. Below $n = 21$ every
maximal planar graph is an intertwining tree, which is why the
disjunction holds trivially in that range.

\subsection*{Empirical status}

For each isomorphism class of maximal planar graphs on $n$ vertices,
we ask whether (i) some isomorphic representative is reachable from
some Even Level Graph via $E/O$-edge switches (``derived''), and/or
(ii) it is an intertwining tree. The conjecture holds for the class
iff at least one of (i), (ii) holds.

\begin{center}
\begin{tabular}{rcccccc}
$n$ & \# iso & derived only & inter.\ only & both & missing & status \\\hline
$6$  & $2$    & $0$ & $0$ & $2$    & $0$ & holds \\
$7$  & $5$    & $0$ & $0$ & $5$    & $0$ & holds \\
$8$  & $14$   & $0$ & $0$ & $14$   & $0$ & holds \\
$9$  & $50$   & $0$ & $1$ & $49$   & $0$ & holds \\
$10$ & $233$  & $0$ & $0$ & $233$  & $0$ & holds \\
$11$ & $1249$ & $0$ & $0$ & $1249$ & $0$ & holds \\
$12$ & $7595$ & $0$ & $1$ & $7594$ & $0$ & holds \\
\end{tabular}
\end{center}

\subsection*{The boundary case $n = 21$}

The first triangulations that are \emph{not} intertwining trees are the
six duals of the Holton--McKay graphs, at $n = 21$. For the disjunction
to survive at $n = 21$, each of these six must be a valid derived level
graph. We find:
\begin{itemize}
\item All six duals are confirmed not intertwining trees (exhaustive
check of all $2^{20}-1$ vertex bipartitions), consistent with
Theorem~\ref{thm:intertwining-iff-hamiltonian-dual}.
\item Two of the six are themselves Even Level Graphs (for a suitable
source vertex), hence trivially valid derived level graphs. So the
disjunction holds for them through the derived-level-graph disjunct --
the first instances where that disjunct does work the intertwining-tree
disjunct cannot.
\item The remaining four are not Even Level Graphs for any source. A
bounded backward $E/O$-orbit search (tens of thousands of states, a
handful of source labellings) found no Even Level Graph in their
orbits, but this is far too shallow relative to the orbit size at
$n = 21$ to be conclusive; their status as derived level graphs is
open.
\end{itemize}
Thus at $n = 21$ the disjunction is confirmed for two of the six
critical iso classes and undetermined for the other four. Settling
those four -- equivalently, deciding $E/O$-orbit reachability from an
Even Level Graph -- is the first genuinely open instance of the
conjecture, and calls for either a better reachability algorithm or a
structural invariant of $E/O$-orbits.

\begin{thebibliography}{9}

\bibitem{holton-mckay}
D.~A. Holton and B.~D. McKay.
\emph{The smallest non-Hamiltonian 3-connected cubic planar graphs have
38 vertices}.
Journal of Combinatorial Theory, Series B, 45(3):305--319, 1988.

\end{thebibliography}

\end{document}