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\begin{document}

\title[Even Level Graph Generators]{Even Level Graph Generators:\\
a constructive conjecture stronger than the Four Color Theorem}

%    Remove any unused author tags.

%    author one information
\author{Eric Bauerfeld}
\address{}
\curraddr{}
\email{}
\thanks{}


\subjclass[2010]{Primary }

\keywords{}

\date{}

\dedicatory{}

\begin{abstract}
We investigate whether two constructive families of $4$-colorable
triangulations are, for each $n$, already rich enough to produce every
maximal planar graph on $n$ vertices. The first family is the
\emph{bridge-derived level graphs}: starting from an \emph{Even Level
Graph} -- a triangulation all of whose level cycles, measured by
breadth-first distance from a chosen source vertex, are even -- we apply
\emph{bridge switches}, edge switches that never close a cycle in either
parity subgraph. Every such graph is $4$-colorable, inheriting the parity
$2$-coloring of an Even Level Graph. The second family is the
\emph{intertwining trees}: triangulations whose vertices split into two
sets each inducing a tree, which are $4$-colorable by coloring the two
trees from disjoint pairs of colors. We conjecture that every maximal
planar graph is a bridge-derived level graph, an intertwining tree, or
both. Since both families are $4$-colorable by construction, the
conjecture implies the four color theorem for triangulations, and hence
for all planar graphs; in fact it is \emph{strictly stronger}, demanding
not merely that a $4$-coloring exist but that every triangulation be
assembled by one of these two explicit constructions. A proof would
therefore be a new, constructive proof of the four color theorem -- and
correspondingly the conjecture is at least as hard, and very likely
harder, than that theorem. We show that a
triangulation is an intertwining tree exactly when its dual is
Hamiltonian, so every triangulation on at most $20$ vertices is an
intertwining tree and the first possible failures occur at $n = 21$, at
the six duals of the Holton--McKay graphs. We verify that all six are
bridge-derived level graphs, confirming the conjecture in its first
nontrivial case. Pushing further, we identify by exhaustive generation
the unique $44$-vertex non-Hamiltonian \emph{cyclically $5$-connected}
cubic planar graph -- settling a uniqueness question Holton--McKay left
open -- whose $24$-vertex $5$-connected dual is the first test of the
conjecture outside the $3$-cut family; it too is a bridge-derived level
graph, two bridge switches from an Even Level Graph.
\end{abstract}

\maketitle

\section{Introduction}

The four color theorem states that every planar graph is properly
$4$-colorable. It suffices to prove this for \emph{maximal} planar graphs
(triangulations), since every planar graph is a spanning subgraph of one.
The known proofs proceed by reducing a hypothetical minimal counterexample,
either through computer-checked unavoidable configurations or through
discharging.

We take a constructive view. Instead of coloring an arbitrary
triangulation, we ask which triangulations can be \emph{assembled} by
operations that manifestly preserve $4$-colorability, and whether those
operations reach all of them. We study two such constructions, and our
motivating question is whether the two together are sufficient: does every
maximal planar graph on $n$ vertices arise from one of them?

The first construction builds on the \emph{level} structure of a
triangulation. Fixing a source vertex and taking breadth-first levels, an
\emph{Even Level Graph} (Definition~\ref{def:even-level-graph}) is a
triangulation whose level cycles are all even; equivalently both of its
parity subgraphs are bipartite, and a $2$-coloring of each parity subgraph
-- two colors for the even-level vertices, two for the odd -- is a proper
$4$-coloring (Theorem~\ref{thm:even-level-4colorable}). From an Even Level
Graph we generate further triangulations by edge switches; restricting to
\emph{bridge switches} (Definition~\ref{def:bridge-switch}), which add an
edge to a parity subgraph only when it is a bridge there, guarantees that
no new cycle -- and in particular no odd cycle -- ever appears in a parity
subgraph. The resulting \emph{bridge-derived level graphs}
(Definition~\ref{def:bridge-derived-level-graph}) therefore remain
$4$-colorable by the same parity coloring.

The second construction is purely combinatorial: an \emph{intertwining
tree} (Definition~\ref{def:intertwining-tree}) is a triangulation whose
vertex set partitions into two parts each inducing a tree. Coloring one
tree from $\{1,2\}$ and the other from $\{3,4\}$ is a proper $4$-coloring,
since edges inside a part join differently-colored tree vertices and edges
across the parts join the disjoint color sets.

Our central question is whether these two families exhaust all
triangulations
(Conjecture~\ref{conj:every-triangulation-derived}). As both families
consist of $4$-colorable graphs, an affirmative answer would constitute a
constructive proof of the four color theorem for triangulations, and
hence for all planar graphs.

We emphasize that the conjecture is a \emph{stronger} statement than the
four color theorem, not an equivalent reformulation of it. A proper
$4$-coloring with its colors grouped as $\{1,2\}\mid\{3,4\}$ is exactly a
partition of the vertices into two parts each inducing a bipartite
subgraph, so the four color theorem is precisely the assertion that every
triangulation admits such a partition. The conjecture asserts strictly
more: that the partition can be realized \emph{constructively} -- as the
level parity of an Even Level Graph reached by bridge switches, or as a
split into two induced \emph{trees}. The four color theorem alone supplies
neither construction; bridge-derivability in particular is a reachability
condition well beyond the bare existence of a $4$-coloring, so the
conjecture implies the four color theorem but is not implied by it.
A proof would accordingly be a new, constructive proof of the four color
theorem, and the conjecture is at least as hard to settle -- and, absent
any structural characterization of the bridge-derived family, very likely
harder.

We connect the two constructions through duality: a triangulation is an
intertwining tree if and only if its dual is Hamiltonian
(Theorem~\ref{thm:intertwining-iff-hamiltonian-dual}). Tait's conjecture --
that every $3$-connected cubic planar graph is Hamiltonian -- fails first
at $38$ vertices, where Holton and McKay found exactly six counterexamples;
dually, every triangulation on at most $20$ vertices is an intertwining
tree, and the first triangulations that are not are the six $21$-vertex
duals of the Holton--McKay graphs. These six are therefore the first
nontrivial test of the conjecture, and we verify that all six are
bridge-derived level graphs -- each at most four bridge switches from an
Even Level Graph, and two of them Even Level Graphs already.

\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}

\subsection*{Enumeration for small $n$}

Even Level Graphs are scarce among triangulations. For each $n$ we
enumerated all iso classes of plane triangulations and tested, for every
choice of source vertex, whether all level subgraphs are bipartite
(equivalently, whether every level cycle is even).
Table~\ref{tab:elg-counts} records, for $4 \leq n \leq 11$: the number of
triangulation iso classes; how many of them admit at least one Even Level
Graph source; the number of Even Level Graph iso classes (pairs $(G, S)$
up to isomorphism, i.e.\ valid sources counted up to $\mathrm{Aut}(G)$);
and the number of \emph{flag-rooted} Even Level Graphs,
\[
  \sum_{G} \frac{4E}{|\mathrm{Aut}(G)|}\, s(G),
  \qquad E = 3n - 6,
\]
where $s(G)$ is the number of valid sources of $G$. The flag-rooting is the
automorphism-free count: $\mathrm{Aut}(G)$ acts freely on the $4E$ flags of a
$3$-connected triangulation, so every summand is an integer.

The smallest Even Level Graph is the octahedron at $n = 6$: from any vertex
the four neighbours form a $4$-cycle at level $1$ and the antipode sits alone
at level $2$. Below $n = 6$ every triangulation forces an odd level cycle, so
no Even Level Graph exists.

\begin{table}[ht]
\centering
\begin{tabular}{ccccc}
$n$ & triangulations & with ELG source & ELG iso classes & flag-rooted ELGs \\\hline
$4$  & $1$    & $0$   & $0$   & $0$ \\
$5$  & $1$    & $0$   & $0$   & $0$ \\
$6$  & $2$    & $1$   & $1$   & $6$ \\
$7$  & $5$    & $2$   & $2$   & $45$ \\
$8$  & $14$   & $5$   & $6$   & $186$ \\
$9$  & $50$   & $13$  & $14$  & $651$ \\
$10$ & $233$  & $37$  & $45$  & $2766$ \\
$11$ & $1249$ & $129$ & $169$ & $14346$ \\
\end{tabular}
\caption{Even Level Graph counts for $4 \leq n \leq 11$. The
\emph{triangulations} column is the number of plane-triangulation iso classes
(OEIS A000109). \emph{ELG iso classes} counts pairs $(G, S)$ up to
isomorphism; \emph{flag-rooted ELGs} is the automorphism-free count
$\sum_G \tfrac{4E}{|\mathrm{Aut}(G)|}\,s(G)$.}
\label{tab:elg-counts}
\end{table}

\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}[Bridge switch]
\label{def:bridge-switch}
Let $G'$ be a triangulation reached from an Even Level Graph $G$, with
parity classes inherited from $G$ as in
Definition~\ref{def:derived-level-graph}. An edge switch on an edge
$e \in E \cup O$ of $G'$, replacing $uvw, uvx$ by the edge $wx$, is a
\emph{bridge switch} if either
\begin{itemize}
\item the new edge $wx$ is a cross-parity edge (one endpoint even, the
other odd), so $wx$ enters neither parity subgraph; or
\item $wx$ is a same-parity edge and is a \emph{bridge} in the parity
subgraph it joins -- that is, $w$ and $x$ lie in different connected
components of that parity subgraph, so adding $wx$ creates no new cycle.
\end{itemize}
\end{definition}

\begin{definition}[Bridge-derived level graph]
\label{def:bridge-derived-level-graph}
A \emph{bridge-derived level graph} of an Even Level Graph $G$ is a
triangulation obtained from $G$ by a sequence of bridge switches
(Definition~\ref{def:bridge-switch}).
\end{definition}

Because a bridge switch never closes a cycle in a parity subgraph, it
never introduces an odd cycle there. As an Even Level Graph has
bipartite parity subgraphs (every level cycle is even), every
bridge-derived level graph has bipartite parity subgraphs as well, and
so is automatically a valid derived level graph. Equivalently, the
first Betti number of each parity subgraph is non-increasing along any
sequence of bridge switches.

\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 bridge-derived level graph of some Even
Level Graph, an intertwining tree, or both.
\end{conjecture}

Since a bridge-derived level graph is automatically a valid derived level
graph, this is a stronger statement than the corresponding conjecture
phrased with arbitrary $E/O$ switches; it is also the form that the
evidence below actually supports.

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*{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, and
their full $E/O$-orbits ($\sim\!10^8$ states per source labelling) are
far too large to exhaust. Restricting to \emph{bridge switches}
(Definition~\ref{def:bridge-switch}) shrinks the relevant orbits by
roughly two orders of magnitude and, crucially, keeps every reachable
triangulation valid. A backward bridge-switch search over the valid
parity partitions found an Even Level Graph witness for each of the
four, so all four are \emph{bridge-derived level graphs}
(Definition~\ref{def:bridge-derived-level-graph}) and hence valid
derived level graphs. The witnessing orbits are small -- between a few
hundred and $\sim\!1.7\times 10^5$ states -- even though other parity
partitions of the same triangulations have orbits exceeding $10^6$;
finding one good partition suffices. Each witness is in fact only a
\emph{handful} of bridge switches from its dual: the explicit Even Level
Graph, parity labelling, and bridge-switch sequence are recorded for all
six -- path lengths $3,1,2,4$ for these four and $0$ for the two that are
Even Level Graphs outright -- and each step has been verified to be a
valid bridge switch.
\end{itemize}
Thus at $n = 21$ the disjunction is confirmed for all six critical iso
classes: two are Even Level Graphs outright, and the other four are
bridge-derived level graphs. The bridge-switch restriction is what made
the search tractable -- it both shrinks the orbit and guarantees
validity, so any Even Level Graph located in a backward orbit is an
immediate witness. Table~\ref{tab:n21} records the outcome for each dual.

\begin{table}[ht]
\centering
\begin{tabular}{cccc}
dual & intertwining tree & Even Level Graph source & bridge switches to ELG \\\hline
$0$ & no & --       & $3$ \\
$1$ & no & $10$     & $0$ \\
$2$ & no & $9$      & $0$ \\
$3$ & no & --       & $1$ \\
$4$ & no & --       & $2$ \\
$5$ & no & --       & $4$ \\
\end{tabular}
\caption{The six Holton--McKay duals at $n = 21$, the first triangulations
that are not intertwining trees. Each is a bridge-derived level graph:
duals $1$ and $2$ are Even Level Graphs outright (zero switches), and the
remaining four reach an Even Level Graph in $1$--$4$ bridge switches. All
witnesses are step-verified.}
\label{tab:n21}
\end{table}

\begin{figure}[ht]
\centering
\includegraphics[width=\textwidth]{figures/n21_duals.png}
\caption{The six Holton--McKay duals, drawn as crossing-free planar graphs
and coloured by parity (blue even, orange odd, with respect to the fixed
level-parity labelling). The solid green edges are the bridge edges
introduced by the bridge switches from each dual's witness Even Level
Graph. Each green edge is a bridge of its parity subgraph, so no new cycle
-- and in particular no odd cycle -- is created; duals $1$ and $2$ coincide
with their Even Level Graphs and have no added edge.}
\label{fig:n21-duals}
\end{figure}

\subsection*{The cyclically-$5$-connected case: $n = 24$}

The six $n = 21$ duals all carry non-trivial $3$-cuts in the cubic
picture; dually, each contains a separating triangle, so each is built
from smaller pieces and lies in the most reducible part of the
non-Hamiltonian world. (The famous $46$-vertex Tutte graph is no
improvement here: it too is only cyclically $3$-connected, and its
$25$-vertex dual has separating triangles.) The genuinely new regime is
the \emph{cyclically $5$-connected} one, dual to a $5$-connected
triangulation -- no separating $3$- or $4$-cycle, hence nothing to
decompose along. By Holton--McKay, the smallest non-Hamiltonian
cyclically $5$-connected cubic planar graph has $44$ vertices (Fig.~2.10
of~\cite{holton-mckay}, attributed to Tutte; minimality due to
Faulkner--Younger), and its dual is a $24$-vertex $5$-connected
triangulation.

We obtain this graph by generation rather than transcription. A
$44$-vertex cubic planar graph is the dual of a $24$-vertex
triangulation, and a cubic graph is cyclically $5$-connected if and only
if its dual triangulation is $5$-connected. Enumerating all $5$-connected
triangulations on $24$ vertices (\texttt{plantri -c5}, $6833$ of them)
and testing each dual for Hamiltonicity, we find that \emph{exactly one}
has a non-Hamiltonian dual. This both produces the graph and, granting
the correctness of the generator and the Hamiltonicity test, settles the
uniqueness question Holton--McKay left open: there is a unique
non-Hamiltonian cyclically $5$-connected cubic planar graph on $44$
vertices.

Let $T$ be its dual: a $24$-vertex triangulation with vertex connectivity
$5$ and no separating triangle, and -- since its dual is non-Hamiltonian
-- not an intertwining tree. We find that $T$ is nonetheless a
bridge-derived level graph. Of its $333$ valid parity partitions most are
useless: their backward bridge-orbits exceed $8 \times 10^5$ states with
no Even Level Graph in sight. But one partition has a backward orbit of
only $4678$ states containing an Even Level Graph (source $s = 19$,
maximum level $4$) at depth $2$. The two bridge switches carrying that
Even Level Graph to $T$ are
\[
  \text{remove } \{16,21\},\ \text{add } \{20,22\}
  \quad\text{and}\quad
  \text{remove } \{15,18\},\ \text{add } \{6,19\},
\]
each adding a same-parity edge that is a bridge of the (odd, resp.\ even)
parity subgraph; both steps have been verified to be valid bridge
switches. So the disjunction holds for $T$ through the bridge-derived
disjunct, and the ``one good partition suffices'' phenomenon seen at
$n = 21$ persists into the cyclically $5$-connected regime -- the first
test of the conjecture genuinely outside the $3$-cut family.

\begin{figure}[ht]
\centering
\includegraphics[width=0.7\textwidth]{figures/fig210_dual.png}
\caption{The $24$-vertex dual $T$ of the unique $44$-vertex
non-Hamiltonian cyclically $5$-connected cubic planar graph
(Holton--McKay Fig.~2.10), drawn crossing-free and coloured by the fixed
parity labelling (blue even, orange odd). $T$ is $5$-connected and not an
intertwining tree, yet is a bridge-derived level graph: the two solid
green edges $\{6,19\}$ and $\{20,22\}$ are the bridge edges introduced by
the two bridge switches carrying its witness Even Level Graph (source
$19$) to $T$. Each green edge is a bridge of its parity subgraph -- $\{6,
19\}$ in the even subgraph, $\{20,22\}$ in the odd -- so no new cycle, and
in particular no odd cycle, is created.}
\label{fig:n24-dual}
\end{figure}

\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}