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\begin{document}

\title{Nested Tire Decompositions of Plane Triangulations}

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

\subjclass[2010]{Primary }

\keywords{plane graph, triangulation, plane depth, level edge, dual graph, tire graph}

\date{}

\dedicatory{}

\begin{abstract}
We establish the foundational structure of nested level-induced tire
decompositions of a plane triangulation $G$.  A \emph{level source} of
$G$ induces a BFS layering of $G$ and endows the inner planar dual
$G'$ with a \emph{dual depth} grading.  The basic object of study is
the \emph{tire graph} $T$ --- a plane graph whose outer and inner
boundaries bound a closed planar region, the \emph{tire tread} $R$,
triangulated by the \emph{annular edges} $E_{\mathrm{ann}}$.  Our main
structural results are the
\emph{tire-component lemma}, the \emph{tire-tread partition theorem},
and the rooted \emph{tire-tree decomposition}, which together organize
the bounded faces of $G$ into nested tire treads.
\end{abstract}

\maketitle

\section{Introduction}

A classical theorem of Tait recasts the Four Colour Theorem in dual,
edge-colouring terms: a plane triangulation $G$ is properly $4$-vertex-colourable
if and only if its dual cubic graph $G'$ is properly $3$-edge-colourable. Thus a
minimal counterexample to the Four Colour Theorem -- a smallest triangulation
admitting no proper $4$-colouring -- corresponds to a smallest cubic plane graph
admitting no proper $3$-edge-colouring.

The structural study of such a minimal counterexample is the
overarching motivation for the present line of work.  This first
paper establishes the foundational vocabulary --- level sources,
dual depth, tire graphs, and partial tire duals --- on which
subsequent papers in the series build.  In particular, the
companion paper \cite{bauerfeld-nested-tire-duals} uses these
definitions to develop nested-cycle structure theorems and
chain-pigeonhole conjectures for tire annular subgraphs of $G'$.

\paragraph{Related work.}
The structural object underlying this programme --- the set of
proper $4$-colourings of a boundary cycle that extend to a colouring
of a bounded planar region --- is classical.  Birkhoff's reducibility
analysis of the diamond configuration~\cite{birkhoff-reducibility} is
the earliest instance of computing such extension sets to attack the
Four Colour Theorem; the chromatic polynomial framework of Birkhoff
and Lewis~\cite{birkhoff-lewis-chromatic} systematised the counting.
Tutte studied how the chromatic polynomial of a rooted planar
triangulation decomposes along its outer
boundary~\cite{tutte-chromatic-sums-1973} and developed an algebraic
theory of graph colourings organised around separating
subgraphs~\cite{tutte-algebraic-colorings, tutte-four-colour-conjecture}.
The most recent and structurally closest parallel is Dvo\v{r}\'ak
and Lidick\'y's analysis of \emph{coloring count
cones}~\cite{dvorak-lidicky-cones}, which characterises the possible
boundary-extension functions on a fixed outer cycle of a
near-triangulation.  The Heesch--Appel--Haken
approach~\cite{heesch-untersuchungen, robertson-sanders-seymour-thomas}
also uses boundary-extension reasoning, but case-by-case on a finite
unavoidable set of local configurations rather than as part of a
global structural induction.

The tire-tree decomposition introduced here differs from each of
these in shape rather than ingredients.  Birkhoff, Tutte, and
Dvo\v{r}\'ak--Lidick\'y all study \emph{one} boundary; Heesch and
the cleaned-up Appel--Haken proof~\cite{robertson-sanders-seymour-thomas}
study a finite collection of local boundaries.  The present framework
organises the entire triangulation into a hierarchy of annular
regions glued along level cycles, and asks whether boundary-extension
constraints compose compatibly up the hierarchy.  To the authors'
knowledge, no prior work on the Four Colour Theorem has been
organised around a global nested-cycle decomposition of this kind.

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 a set $S \subseteq V$ that is either
\begin{itemize}
\item a single vertex $\{v\}$ (a \emph{vertex source}), or
\item a set inducing a simple cycle in $G$ --- i.e.\ $G[S]$ is a
      simple cycle of length $\geq 3$ (a \emph{cycle source}).
\end{itemize}
\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.  We write $L_d := \{v \in V : \ell_G(v) = d\}$ for the
\emph{level-$d$ vertex set}, and abbreviate $L_{<d} := \bigcup_{d' < d}
L_{d'}$ and $L_{\geq d} := \bigcup_{d' \geq d} L_{d'}$ (similarly
$L_{>d}$, $L_{\leq d}$).
\end{definition}

\begin{definition}[Dual]
\label{def:dual}
The \emph{dual} of $G$, written $G'$, is the inner (weak) planar dual of $G$ with
respect to the embedding $\Pi_G$: it has one vertex $d_f$ for each bounded face
$f$ of $G$, and an edge joining $d_f$ and $d_{f'}$ for each edge of $G$ shared by
two bounded faces $f$ and $f'$. The unbounded outer face contributes no vertex,
and edges of $G$ on the outer boundary contribute no dual edge. Since $G$ is a
triangulation, each vertex $d_f \in V(G')$ corresponds to a triangular face $f$
of $G$, and we write $V(f) \subseteq V$ for its three incident vertices.
\end{definition}

\begin{definition}[Dual depth]
\label{def:dual-depth}
Given a level source $S \subseteq V$, the \emph{dual depth} of a dual vertex
$d_f \in V(G')$ is
\[
    \delta_G(d_f) = \min_{v \in V(f)} \ell_G(v)
        = \min_{v \in V(f)} \mathrm{dist}_G(v, S),
\]
the smallest level among the three vertices of $G$ bounding the face $f$.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.7\textwidth]{fig_dual_depth.png}
\caption{Dual depth in a stacked-ring triangulation $G$ with level source
$S = \{0\}$. Each $G$ vertex is labelled by its level $\ell$. Each bounded face
carries a dual vertex (square, joined by dashed dual edges) coloured by its dual
depth $\delta(d_f) = \min_{v \in V(f)} \ell(v)$: the central fan has depth $0$,
the inner annulus depth $1$, and the outer annulus depth $2$. The outer face
(the level-$3$ triangle) is excluded from the inner dual and carries no dual
vertex.}
\label{fig:dual-depth}
\end{figure}

\begin{definition}[Depth-$d$ dual subgraph and its components]
\label{def:dual-component}
For $d \geq 0$, the \emph{depth-$d$ dual subgraph} is
\[
    G'_d \;:=\; G'\bigl[\,\{d_f \in V(G') : \delta_G(d_f) = d\}\,\bigr],
\]
the inner-dual subgraph induced on the dual vertices of dual depth $d$.
For a connected component $C'$ of $G'_d$ we write
\[
    F_{C'} := \{f : d_f \in V(C')\}, \qquad
    V_{C'} := \bigcup_{f \in F_{C'}} V(f),
\]
for its set of faces and the vertices of $G$ bounding them, and
$R_{C'} := \bigcup_{f \in F_{C'}} f \subseteq |\Pi_G|$ for the closed
planar region these faces cover.
\end{definition}

\begin{definition}[Tire graph]
\label{def:tire-graph}
A \emph{tire graph} consists of a plane graph $T$ together with an
\emph{outer boundary} $B_{\mathrm{out}} \subseteq T$ and a
\emph{connected inner outerplanar graph} $O \subseteq T$ with
$V(B_{\mathrm{out}}) \cap V(O) = \emptyset$, where
\begin{itemize}
  \item $B_{\mathrm{out}}$ is either a simple cycle of length $\geq 3$
        or a single vertex (a \emph{degenerate outer boundary});
  \item $O$ is a connected outerplanar graph; its \emph{inner boundary}
        $B_{\mathrm{in}}$ is the closed walk in $O$ that traces the
        boundary of $O$'s outer face in the inherited embedding,
        which is a simple cycle when $O$ is $2$-connected and a
        non-simple closed walk in general (visiting bridges twice and
        cut-vertices multiple times); if $|V(O)| = 1$, we say $T$ has
        a \emph{degenerate inner boundary}.
\end{itemize}
At most one of $B_{\mathrm{out}}, B_{\mathrm{in}}$ may be degenerate.
The vertex and edge sets of $T$ are
\[
    V(T) = V(B_{\mathrm{out}}) \cup V(O),
    \qquad
    E(T) = E(B_{\mathrm{out}}) \cup E(O) \cup E_{\mathrm{ann}},
\]
where $E_{\mathrm{ann}}$ --- the \emph{annular edges} --- has the
property that, in the plane embedding of $T$, the closed planar
region $R$ bounded externally by $B_{\mathrm{out}}$ and internally
by $B_{\mathrm{in}}$ is partitioned into triangular faces of $T$
whose union is $R$.  We call $R$ the \emph{tire tread} of $T$ and write
$F_{\mathrm{ann}}$ for this set of triangular faces (the \emph{annular
faces}).

When $B_{\mathrm{out}}$ is a simple cycle and $O$ is $2$-connected,
the tread is a closed annulus.  More generally, $R$ is a closed
planar region that may fail to be a $2$-manifold at cut-vertices of
$O$ (where two ``lobes'' of the depth-$d$ region meet at a single
vertex); the inner boundary $B_{\mathrm{in}}$ is then a non-simple
closed walk that visits the cut-vertex multiple times.  The relaxed
definition accommodates outerplanar inner graphs with bridges or
cut-vertices, while $O$ itself remains connected.  When either
boundary is degenerate, the tread is a closed disk with that vertex
as apex.

We summarize the data of a tire graph as the triple
$T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$, from which $B_{\mathrm{in}}$,
the annular faces $F_{\mathrm{ann}}$, and the tread $R$ are determined;
we freely identify a tire graph with its underlying plane graph $T$.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.78\textwidth]{fig_tire_example.png}
\caption{A tire graph with non-degenerate boundaries: outer boundary
$B_{\mathrm{out}}$ a $6$-cycle on vertices $0,\dots,5$ (blue), inner
boundary $B_{\mathrm{in}}$ a $4$-cycle on vertices $6,\dots,9$ (red),
inner outerplanar graph $O = B_{\mathrm{in}} \cup \{7\text{--}9\}$
(with one chord, orange), and $E_{\mathrm{ann}}$ (grey) tiling the
annulus between $B_{\mathrm{out}}$ and $B_{\mathrm{in}}$ by ten
triangular faces.}
\label{fig:tire-example}
\end{figure}

\begin{definition}[Medial tire graph]
\label{def:medial-tire-graph}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph.
Let $T^{\circ}$ be the plane graph obtained from $T$ by deleting every
edge of $O$ that is not on the inner boundary walk $B_{\mathrm{in}}$.
Equivalently, $T^{\circ}$ keeps $B_{\mathrm{out}}$, $B_{\mathrm{in}}$,
and the annular edges, but omits the chords of the inner outerplanar
graph $O$.

The \emph{medial tire graph} of $T$, denoted $M_{\mathrm{tire}}(T)$,
is obtained from the medial graph $M(T^{\circ})$ by deleting every
medial edge whose two endpoint-vertices correspond either to two edges
of $B_{\mathrm{out}}$ or to two edges of $B_{\mathrm{in}}$.  Thus
$V(M_{\mathrm{tire}}(T))$ is naturally indexed by the non-chord edges
of $T$, and two such vertices are adjacent exactly when the
corresponding edges are consecutive on the boundary of a face of
$T^{\circ}$, except for consecutive pairs lying wholly along one of
the two boundary walks.  The medial vertices corresponding to edges of
$E_{\mathrm{ann}}$ are called the \emph{annular medial vertices}.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.78\textwidth]{fig_medial_tire_example.png}
\caption{The medial tire graph for the tire in
Figure~\ref{fig:tire-example}.  The chord of $O$ is drawn faintly and
omitted before taking the medial graph; medial edges between consecutive
outer-boundary edges or consecutive inner-boundary edges are also
omitted.  Each medial vertex is placed at the midpoint of its
corresponding retained tire edge.}
\label{fig:medial-tire-example}
\end{figure}

\begin{remark}
\label{rem:tire-counts}
Let $\mu = |V(B_{\mathrm{out}})|$ and $\nu = |V(B_{\mathrm{in}})|$.  By
Euler's formula on the tire tread $R$, the tire graph has $\mu + \nu$
triangular faces inside $R$ and $|E_{\mathrm{ann}}| = \mu + \nu$
annular edges when neither boundary is degenerate; when exactly one
boundary is degenerate (so $\min(\mu, \nu) = 1$), there are $\mu + \nu - 1$
triangular faces and $|E_{\mathrm{ann}}| = \mu + \nu - 1$.
\end{remark}

\begin{proposition}[Source-side simple-cycle property]
\label{prop:no-level-d-pinch}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and
single-vertex source $v_0$.  Let $d \geq 1$, $v \in L_d$, and let
$C'$ be a connected component of $G'_d$ such that $v$ is incident to
some face in $F_{C'}$.  Then the depth-$d$ faces in $F_{C'}$ incident
to $v$ form a single contiguous arc in $v$'s rotation in $\Pi_G$.

Equivalently: for any such component, the source-side boundary of
$R_{C'}$ is a simple cycle in $L_d$ (no cut-vertices at level $d$).
\end{proposition}

\begin{proof}
Suppose for contradiction that the depth-$d$ faces in $F_{C'}$ at $v$
lie in two or more disjoint arcs of $v$'s rotation.  Adjacent vertices
in $G$ differ in level by at most $1$, so a face at $v$ has depth
exactly $d$ iff both other vertices have level $\geq d$, and depth
$\leq d-1$ iff at least one has level $d-1$.  Hence the gaps between
the depth-$d$ arcs at $v$ are populated by level-$(d-1)$ neighbours of
$v$, occurring in at least two disjoint arcs of $v$'s rotation.  Pick
$p$ in one such gap and $q$ in another.

The BFS ball $G[L_{<d}]$ is connected, so there exists a simple path
$P$ in $G[L_{<d}]$ from $p$ to $q$.  Define the closed walk
\[
    W \;:=\; v \to p \to P \to q \to v.
\]
Every vertex of $P$ lies in $L_{<d}$, while $\ell(v) = d$, so $v$ is
distinct from every vertex of $P$; $P$ is simple, so its internal
vertices are distinct; and $p \neq q$ since they lie in different gaps.
Hence $W$ is a simple cycle in $G$.

By the Jordan curve theorem, the planar embedding of $W$ divides $\Pi_G$
into two regions.  In $v$'s rotation, the edges $v-p$ and $v-q$ lie at
two specific positions, and they split the rotation into two arcs;
each arc lies in one of the two regions determined by $W$.  By choice
of $p, q$, the two arcs of depth-$d$ faces at $v$ in $F_{C'}$ lie in
different regions of $W$ (i.e., one arc on each side).

Since $C'$ is connected in $G'$ and contains depth-$d$ faces in both
arcs, there is a dual path $f_1, f_2, \ldots, f_k$ in $G'_d$ with
$f_1, f_k \in F_{C'}$ incident to $v$ in different arcs, and with the
intermediate faces $f_2, \ldots, f_{k-1}$ not incident to $v$ (a
shortest such dual path).  Consecutive faces $f_i, f_{i+1}$ share an
edge $e_i$ of $G$; for $i \geq 2$, both endpoints of $e_i$ lie in
$L_{\geq d}$ (since neither $f_i$ nor $f_{i+1}$ is incident to $v$,
all six vertices of these two triangles lie in $L_{\geq d}$).  In
particular, $e_i$ shares no endpoint with $W$ except possibly $v$ ---
and $v$ is excluded from $f_2, \ldots, f_{k-1}$.

A planar edge with neither endpoint on a simple closed planar curve
$W$ has both of its incident faces on the same side of $W$.  Applying
this to each $e_i$ ($i \geq 2$) inductively: starting from $f_2$ on
the same side of $W$ as $f_1$ (their shared edge $e_1 = w-w'$ opposite
to $v$ in $f_1$ has $w, w' \in L_{\geq d}$ and hence is not on $W$),
the path $f_2 \to f_3 \to \cdots \to f_{k-1} \to f_k$ stays on one
side of $W$.

But $f_1$ and $f_k$ lie on different sides of $W$ (by construction),
contradicting the conclusion that the entire path lies on one side.
\end{proof}

\begin{lemma}[Tire-component lemma]
\label{lem:tire-component}
Let $G$ be a maximal planar graph and let $S \subseteq V(G)$ be a level
source.  Fix a plane embedding $\Pi_G$ of $G$ in which $S$ lies on the
outer face (such an embedding exists for any planar graph and any
single-vertex source).  For $d \geq 0$, let $C'$ be a connected
component of the depth-$d$ dual subgraph $G'_d$, with faces $F_{C'}$,
bounding vertices $V_{C'}$, and region $R_{C'}$ as in
Definition~\ref{def:dual-component}; let $T_{C'} := G[V_{C'}]$ inherit its
embedding from $\Pi_G$.

Then $T_{C'}$, with the inherited embedding, is a tire graph in the sense of Definition~\ref{def:tire-graph}.  Its outer boundary
$B_{\mathrm{out}}$ is the side of $R_{C'}$ closer to $S$ in $\Pi_G$,
namely the level-$d$ subgraph $G[V_{C'} \cap L_d]$ (a simple cycle or
single vertex); its connected inner outerplanar graph is $O = G[V_{C'} \cap
L_{d+1}]$, and its inner boundary $B_{\mathrm{in}}$ is the outer-face
boundary closed walk of $O$ in the inherited embedding (a simple cycle
when $O$ is $2$-connected, a non-simple closed walk in general).  The
triangular faces of $T_{C'}$ inside the closed boundary region are exactly
the faces of $G$ in $F_{C'}$.
\end{lemma}

\begin{proof}
\emph{Outerplanarity of the two level parts.}  By construction $S$
lies on the outer face of $\Pi_G$, so the outerplanarity lemma of
\cite{bauerfeld-depth} applies directly with $(G, \Pi_G, S)$, giving
that $G[L_{d'}]$ is
outerplanar for each $d' \geq 0$.  Subgraphs of outerplanar graphs are
outerplanar, so $G[V_{C'} \cap L_d]$ and $G[V_{C'} \cap L_{d+1}]$ are
both outerplanar.

\emph{Layer containment.}  Each $f \in F_{C'}$ has at least one vertex
at level $d$, and adjacent vertices in $G$ differ in level by at most
$1$; combined with $\delta_G(d_f) = d$, this forces
$V(f) \subseteq L_d \cup L_{d+1}$.  Hence $V_{C'} \subseteq L_d \cup
L_{d+1}$, and $T_{C'}$ has vertex partition
$V_{C'} = (V_{C'} \cap L_d) \sqcup (V_{C'} \cap L_{d+1})$.

\emph{Boundary edges are monochromatic in level.}  Each edge $e$ on
$\partial R_{C'}$ separates a face $f \in F_{C'}$ from a face
$f' \notin F_{C'}$.  Because $f$ and $f'$ share the edge $e$, their
dual vertices are adjacent in $G'$; if both had depth $d$ they would
lie in the same component of $G'_d$, contradicting $d_{f} \in C'$ and
$d_{f'} \notin C'$.  Hence $\delta_G(d_{f'}) \neq d$; combined with
the bounded-step property of $\delta$ across $G'$-adjacent faces,
$\delta_G(d_{f'}) \in \{d-1, d+1\}$.
\begin{itemize}
\item If $\delta_G(d_{f'}) = d - 1$, the third vertex $w$ of
      $f' = \{u, v, w\}$ (where $u, v$ are the endpoints of $e$) has
      $\ell(w) = d - 1$.  Each of $u, v$ has $\ell \in \{d, d+1\}$
      (from $V(f) \subseteq L_d \cup L_{d+1}$) and is adjacent to $w$,
      forcing $\ell(u), \ell(v) \in \{d-2, d-1, d\} \cap \{d, d+1\} =
      \{d\}$.
\item If $\delta_G(d_{f'}) = d + 1$, then all three vertices of $f'$
      lie in $L_{\geq d+1}$, so in particular $\ell(u) = \ell(v) =
      d + 1$.
\end{itemize}
Each connected boundary component thus carries a single type at every
edge: any vertex on a boundary component has two boundary edges
incident to it (by R1, see below), both of the same type, so its
level is fixed.  Therefore each boundary component of $\partial R_{C'}$
is monochromatic in level.

\emph{Boundary structure.}  Each connected component of
$\partial R_{C'}$ traces a closed walk in $G$ that, by the
monochromaticity above, lies entirely in $L_d$ or entirely in
$L_{d+1}$.  By Proposition~\ref{prop:no-level-d-pinch}, the depth-$d$
faces of $F_{C'}$ at any $v \in L_d \cap V_{C'}$ form a single
contiguous arc in $v$'s rotation, so the source-side boundary walk
visits each $L_d$-vertex of $V_{C'}$ exactly once: it is a simple
cycle.  At vertices $v \in L_{d+1} \cap V_{C'}$ the depth-$d$ faces
may split into multiple arcs of $v$'s rotation; this corresponds
exactly to $v$ being a cut-vertex of $O$, and the inner-side
boundary walk visits $v$ correspondingly many times --- which is
already accommodated by Definition~\ref{def:tire-graph} (where
$B_{\mathrm{in}}$ is the outer-face boundary closed walk of $O$, not
necessarily a simple cycle).

\emph{Outer boundary.}  Because $S$ lies on the outer face of $\Pi_G$,
the boundary curve(s) of $R_{C'}$ on the $L_d$ side are closer to $S$
in the embedding.  In the inherited embedding of $T_{C'}$, the unique
unbounded face is the merged region containing the rest of $\Pi_G$
outside $R_{C'}$ on the $S$ side, so its boundary --- a simple cycle
on $L_d$ (or a single vertex when $V_{C'} \cap L_d = \{v_0\}$, the
$d = 0$ case) --- serves as $B_{\mathrm{out}}$.  We set
$B_{\mathrm{out}} := G[V_{C'} \cap L_d]$ if this is a cycle, and
the single vertex $\{v_0\}$ in the degenerate case.

\emph{Inner outerplanar graph.}  By the outerplanarity lemma of
\cite{bauerfeld-depth}, $G[V_{C'} \cap L_{d+1}]$ is outerplanar.  We set $O :=
G[V_{C'} \cap L_{d+1}]$.  The boundary curve(s) of $R_{C'}$ on the
$L_{d+1}$ side are exactly the boundary of $O$'s outer face in the
inherited embedding; this outer-face boundary is a single closed walk
that traces around $O$ from the outside, traversing any bridge edge
twice and visiting cut-vertices multiple times.  This walk is the
inner boundary $B_{\mathrm{in}}$.  No further restriction on $O$'s
internal structure is needed: when the inner side has several lobes
meeting through cut-vertices or bridges of $O$, the outer-face
boundary closed walk of the connected graph $O$ captures them by
revisiting those vertices or traversing those bridges twice.

\emph{Tire structure.}  The triangular faces of $T_{C'}$ inside the closed
boundary region are by construction the depth-$d$ faces in $F_{C'}$,
and the edges of $T_{C'}$ are $E(B_{\mathrm{out}}) \cup E(O) \cup
E_{\mathrm{ann}}$ where $E_{\mathrm{ann}}$ are the edges of $G$
between $V_{C'} \cap L_d$ and $V_{C'} \cap L_{d+1}$ that bound a face
of $F_{C'}$.
\end{proof}

\begin{theorem}[Tire treads partition the bounded faces]
\label{thm:tread-partition}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and
let $S \subseteq V(G)$ be a level source lying on the outer face.
For each $d \ge 0$ and each connected component $C'$ of $G'_d$, let
$T^{(d, C')}$ denote the tire graph supplied by
Lemma~\ref{lem:tire-component}, with tire tread
$R_{C'} \subseteq |\Pi_G|$.  Then the collection of treads
\[
   \mathcal{R}(G, S) \;:=\;
   \bigl\{\, R_{C'} \,:\, d \ge 0,\;
   C' \text{ a connected component of } G'_d \,\bigr\}
\]
partitions the bounded part of $|\Pi_G|$:
\begin{enumerate}
  \item[(i)] every bounded face $f$ of $G$ is contained in exactly
        one tread $R_{C'} \in \mathcal{R}(G, S)$;
  \item[(ii)] distinct treads in $\mathcal{R}(G, S)$ have disjoint
        interiors and may share only boundary edges or vertices.
\end{enumerate}
\end{theorem}

\begin{proof}
\emph{Existence and uniqueness.}  Each bounded face $f \in F(G)$
has a uniquely-defined dual depth $\delta_G(d_f) \in \mathbb{Z}_{\ge
0}$, so the dual vertex $d_f$ lies in $G'_d$ for $d =
\delta_G(d_f)$ and in no other $G'_{d'}$.  Within $G'_d$, the
vertex $d_f$ belongs to exactly one connected component $C'$.  By
Lemma~\ref{lem:tire-component}, $F_{C'}$ is precisely the set of
faces $f' \in F(G)$ with $d_{f'} \in V(C')$; in particular $f \in
F_{C'}$, hence $f \subseteq R_{C'}$.

For any other tread $R_{C''} \in \mathcal{R}(G, S)$, the
component $C''$ is either at a different depth $d' \ne d$ (in
which case $F_{C''}$ consists of depth-$d'$ faces and $f \notin
F_{C''}$) or at depth $d$ but a different component $C'' \ne C'$
(in which case the two components are vertex-disjoint in $G'_d$,
so again $f \notin F_{C''}$).  In both cases $f \notin R_{C''}$
(more precisely, $f$ is not one of the triangular faces of $G$ in
$F_{C''}$, so $f$'s interior is not contained in $R_{C''}$).

\emph{Disjoint interiors.}  Each tread $R_{C'}$ is the union of
its triangular faces $F_{C'} \subseteq F(G)$; distinct treads
correspond to disjoint $F_{C'}$ (by the argument above), and the
interiors of distinct $G$-faces are disjoint.  Hence interiors of
distinct treads are disjoint.

\emph{Coverage.}  Conversely, every bounded $f \in F(G)$ has $d_f
\in V(G')$ with some dual depth $d$, and thus lies in $R_{C'}$
where $C'$ is its component of $G'_d$.  So $\bigcup_{R \in
\mathcal{R}(G, S)} R$ contains every bounded face of $G$.
\end{proof}

\begin{remark}
\label{rem:tire-component-degenerate}
Either boundary part of $T_{C'}$ in Lemma~\ref{lem:tire-component} may be
degenerate.  At $d = 0$ with single-vertex source $S = \{v_0\}$ the
unique component of $G'_0$ has $V_{C'} \cap L_0 = \{v_0\}$ as the
degenerate \emph{outer} boundary and $V_{C'} \cap L_1$ a cycle (the
link of $v_0$ in $G$) as the inner boundary.  Symmetrically, at
$d = D_{\max}$, $V_{C'} \cap L_{D_{\max}+1} = \emptyset$ degenerates
to a single deepest vertex serving as the \emph{inner} boundary, with
the level-$D_{\max}$ cycle as the outer boundary.
\end{remark}

\begin{remark}
\label{rem:tire-no-extra-hypotheses}
Two structural features of $R_{C'}$ that might at first appear to
obstruct the tire-graph conclusion are both already accommodated by
Definition~\ref{def:tire-graph}:

\emph{Cut-vertices of $O$.}  A vertex $v \in V_{C'} \cap L_{d+1}$ may
have the faces of $F_{C'}$ incident to it split into two or more
arcs in $v$'s rotation in $\Pi_G$, separated by faces of higher
depth.  This corresponds exactly to $v$ being a cut-vertex of
$O = G[V_{C'} \cap L_{d+1}]$, and the inner boundary closed walk
$B_{\mathrm{in}}$ then visits $v$ multiple times --- once for each
arc.  No additional hypothesis is needed.

\emph{Non-$2$-connected inner topology.}  Even when the inner side of
$R_{C'}$ has several lobes joined at cut-vertices or across bridges,
the connected inner outerplanar graph $O$ captures this structure as
a non-$2$-connected outerplanar graph, and its outer-face boundary
closed walk serves as $B_{\mathrm{in}}$ by traversing bridges twice
and visiting cut-vertices multiple times.

In the special case $d = 0$ with single-vertex source $S = \{v_0\}$,
$R_{C'}$ is the star of $v_0$, a topological closed disk with one
boundary cycle (the link of $v_0$); the corresponding tire graph has
degenerate outer boundary $\{v_0\}$.
\end{remark}

\begin{theorem}[Inner dual of a tire tread is outerplanar]
\label{thm:inner-dual-outerplanar}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph,
and let $\Gamma$ be the graph on vertex set
$\{d_f : f \in F_{\mathrm{ann}}\}$ with an edge $d_f d_{f'}$ for
each interior annular edge of $T$ (= each edge of $T$ whose two
incident faces both lie in $F_{\mathrm{ann}}$).  Equivalently,
$\Gamma$ is the subgraph induced on $F_{\mathrm{ann}}$ of the
\emph{full tire dual} $D(T)$ --- the dual of $T$ taken over all of
its triangular faces, in which each boundary edge of $R$ contributes
a degree-$1$ vertex.  Then $\Gamma$ is outerplanar.

Moreover, $\Gamma$ admits a planar embedding as a (possibly
non-simple) Hamilton walk through every $d_f$, plus zero or more
non-crossing chords.
\end{theorem}

\begin{proof}
We argue by cases on whether the tire tread $R$ is a disk or an
annulus.

\medskip
\emph{Case 1: $R$ is a closed disk} (one of $B_{\mathrm{out}},
B_{\mathrm{in}}$ degenerate, by Definition~\ref{def:tire-graph}).
Let $v_0$ be the degenerate-boundary vertex (the apex) and let
$k = |B_{\mathrm{non-deg}}|$ be the length of the non-degenerate
boundary cycle.  The triangulation of $R$ is a \emph{fan} of $k$
triangles around $v_0$: each triangle has the form $\{v_0, u_i,
u_{i+1}\}$ where $u_1, \dots, u_k$ are the boundary-cycle vertices
in cyclic order.  Each triangle has two spoke edges (= the two
edges incident to $v_0$, shared with the two neighbouring fan
triangles) and one boundary edge (in $B_{\mathrm{non-deg}}$,
contributing a leaf in $D(T)$ but no edge in $\Gamma$).  Hence
every $d_f$ has $\Gamma$-degree exactly $2$, and $\Gamma$ is a
single cycle of length $k$.  Cycles are outerplanar.

See Figure~\ref{fig:inner-dual-disk-case} for the disk case
($k = 6$).

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=1.4]
  \def\R{1.8}
  % apex
  \node[circle, fill=black, inner sep=1.6pt, label={right:$v_0$}] (apex) at (0, 0) {};
  % boundary vertices (hexagon)
  \foreach \i in {0,...,5} {
    \pgfmathsetmacro{\ang}{60*\i + 90}
    \node[circle, fill=black, inner sep=1.3pt] (u\i) at (\ang:\R) {};
  }
  % boundary cycle edges (non-degenerate boundary)
  \foreach \i in {0,...,5} {
    \pgfmathtruncatemacro{\j}{mod(\i+1,6)}
    \draw[red, thick] (u\i) -- (u\j);
  }
  % spoke edges (annular interior)
  \foreach \i in {0,...,5} {
    \draw[gray] (apex) -- (u\i);
  }
  % dual vertices at triangle centroids + dual cycle
  \foreach \i in {0,...,5} {
    \pgfmathsetmacro{\angmid}{60*\i + 90 + 30}
    \pgfmathsetmacro{\rmid}{0.62*\R}
    \node[circle, fill=blue!70!black, inner sep=1.6pt] (d\i) at (\angmid:\rmid) {};
  }
  \foreach \i in {0,...,5} {
    \pgfmathtruncatemacro{\j}{mod(\i+1,6)}
    \draw[blue!70!black, very thick] (d\i) -- (d\j);
  }
  % Labels
  \node[red] at (0, -\R - 0.3) {\small non-degenerate boundary $B_{\mathrm{non-deg}}$};
  \node[blue!70!black] at (\R + 1.0, 0.5) {\small dual cycle $\Gamma \cong C_6$};
  \node[blue!70!black] at (\R + 1.0, 0.2) {\small (outerplanar)};
  \node[gray] at (-\R - 0.4, 0.3) {\small spokes};
\end{tikzpicture}
\caption{Case 1 ($R$ = disk, $k = 6$).  The apex $v_0$ sits at the
centre; the non-degenerate boundary $B_{\mathrm{non-deg}}$ (red)
is the hexagonal outer cycle; spokes (grey) triangulate the disk
into a fan of $6$ triangles around $v_0$.  Each triangle has two
spoke edges (interior, contributing $\Gamma$-edges) and one
boundary edge (contributing a leaf in $D(T)$, no $\Gamma$-edge).
The inner dual $\Gamma$ (blue) is the cycle $C_6$ formed by the
six annular face centroids, a manifestly outerplanar graph.}
\label{fig:inner-dual-disk-case}
\end{figure}

\medskip
\emph{Case 2: $R$ is an annulus} (both $B_{\mathrm{out}}$ and
$B_{\mathrm{in}}$ non-degenerate).  We construct an explicit
outerplanar embedding of $\Gamma$ as a Hamilton walk plus
non-crossing chords.

\medskip
\noindent\emph{Step 1: Cyclic ordering of $F_{\mathrm{ann}}$.}
The boundary of the annular tread is the disjoint union
$\partial R = B_{\mathrm{out}} \sqcup \overline{B_{\mathrm{in}}}$
(viewing $B_{\mathrm{in}}$ as a closed walk traced in the
appropriate orientation).  Each boundary edge of $R$ is incident to
exactly one annular face: walking around $B_{\mathrm{out}}$ in
cyclic order produces a sequence
$f^{\mathrm{out}}_1, f^{\mathrm{out}}_2, \dots, f^{\mathrm{out}}_\mu$
of (not necessarily distinct) annular faces, one per
$B_{\mathrm{out}}$-edge; similarly walking around
$B_{\mathrm{in}}$ produces a sequence
$f^{\mathrm{in}}_1, \dots, f^{\mathrm{in}}_{\nu_\partial}$ where
$\nu_\partial$ is the length of the inner-boundary walk.  Pick any
spoke $e^\star = u w \in E_{\mathrm{ann}}$ with $u \in
V(B_{\mathrm{out}})$ and $w \in V(B_{\mathrm{in}})$; cut $R$ along
$e^\star$.  This converts the annulus into a closed disk
$\tilde R$ whose boundary walks once around $B_{\mathrm{out}}$,
once along $e^\star$, once around $B_{\mathrm{in}}$ in reverse,
and once back along $e^\star$.  Concatenating the two boundary
sequences (in the order dictated by this disk traversal) yields a
single cyclic sequence
\[
   \mathcal{S} = (f^{\mathrm{out}}_1, \dots, f^{\mathrm{out}}_\mu,
       f^{\mathrm{in}}_1, \dots, f^{\mathrm{in}}_{\nu_\partial})
\]
of annular faces with multiplicities.

\medskip
\noindent\emph{Step 2: The Hamilton walk.}  Consecutive entries of
$\mathcal{S}$ correspond either to the same annular face (when two
adjacent boundary edges meet at a vertex incident to a single
annular face) or to two annular faces sharing an interior edge of
$E_{\mathrm{ann}}$.  In the former case the walk stays at one
$\Gamma$-vertex; in the latter it uses one $\Gamma$-edge.  The
resulting closed walk in $\Gamma$ visits every face that appears
in $\mathcal{S}$ at least once.

If every $f \in F_{\mathrm{ann}}$ appears in $\mathcal{S}$ (i.e.\
every annular face has at least one boundary edge of $R$), the walk
is a Hamilton walk in $\Gamma$, and we are done up to Step 3.  Each
annular face with two boundary edges contributes a vertex visited
twice; each with three contributes a vertex visited three times.

If some $f \in F_{\mathrm{ann}}$ does not appear in $\mathcal{S}$
(i.e.\ has no boundary edge of $R$), then all three edges of $f$
are interior annular edges, so $d_f$ has degree $3$ in $\Gamma$.
Such a face is ``trapped'' in the interior of the dual graph and
appears as the endpoint of a chord.  Extend the walk by:
whenever it crosses an interior annular edge $e$ shared with a
boundary-free face $f$, detour through $f$ and back.  After
finitely many such detours (one per boundary-free face), the walk
becomes a Hamilton walk visiting every $d_f$.

\medskip
\noindent\emph{Step 3: Non-crossing chords.}  The $\Gamma$-edges
not used by the Hamilton walk constructed in Step~2 are the
remaining interior annular edges.  Each such edge $e \in
E_{\mathrm{ann}}$ corresponds to a chord between two non-adjacent
positions of $\mathcal{S}$.  In the inherited planar embedding of
$\Gamma$ in $R$, these chords are drawn as straight segments
between annular triangle centroids; \emph{they do not cross} because
the underlying $E_{\mathrm{ann}}$ edges they cross are themselves
non-crossing in the planar embedding of $T$.

\medskip
\noindent\emph{Step 4: Outerplanar embedding.}  We now lay out
$\Gamma$ as follows: place the $|F_{\mathrm{ann}}|$ vertices on a
circle in the cyclic order given by $\mathcal{S}$ (treating
multiply-visited faces as single circle vertices).  Connect
consecutive vertices on the circle by the Hamilton-walk edges,
which forms the closed walk.  Draw the remaining edges as chords
inside the circle.  Because the chords were non-crossing in $T$'s
planar embedding, they remain non-crossing here.  All vertices lie
on the outer face (the unbounded region outside the circle),
making $\Gamma$ outerplanar.  $\square$
\end{proof}

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=1.25]
  % Outer hexagon vertices u_i at angles 90, 30, -30, -90, -150, 150
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $u_0$}]  (u0) at (0,     2.5)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]30:\scriptsize $u_1$}]  (u1) at (2.17,  1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-30:\scriptsize $u_2$}] (u2) at (2.17, -1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-90:\scriptsize $u_3$}] (u3) at (0,    -2.5)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-150:\scriptsize $u_4$}](u4) at (-2.17,-1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]150:\scriptsize $u_5$}] (u5) at (-2.17, 1.25)  {};
  % Barbell vertices
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]180:\scriptsize $a_1$}] (a1) at (-0.9,  0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]180:\scriptsize $a_2$}] (a2) at (-0.9, -0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $a_3$}]  (a3) at (-0.25, 0)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $b_1$}]  (b1) at (0.25,  0)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]0:\scriptsize $b_2$}]   (b2) at (0.9,   0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]0:\scriptsize $b_3$}]   (b3) at (0.9,  -0.7) {};
  % B_out edges (red)
  \draw[red, thick] (u0) -- (u1) -- (u2) -- (u3) -- (u4) -- (u5) -- (u0);
  % O edges: left triangle, right triangle, bridge (light red)
  \draw[red!55!white, thick] (a1) -- (a2) -- (a3) -- (a1);
  \draw[red!55!white, thick] (b1) -- (b2) -- (b3) -- (b1);
  \draw[red!55!white, thick] (a3) -- (b1);
  % Spokes (gray)
  \foreach \p/\q in {u0/a1, u0/b2, u0/a3, u0/b1,
                     u1/b2, u1/b3,
                     u2/b3,
                     u3/a2, u3/b3, u3/a3, u3/b1,
                     u4/a2,
                     u5/a1, u5/a2} {
    \draw[gray, thin] (\p) -- (\q);
  }
  % Dual centroids -- 14 annular triangles
  \coordinate (T1)  at (barycentric cs:u0=1,u1=1,b2=1);  % outer cap u0-u1
  \coordinate (T2)  at (barycentric cs:u1=1,u2=1,b3=1);  % outer cap u1-u2
  \coordinate (T3)  at (barycentric cs:u2=1,u3=1,b3=1);  % outer cap u2-u3
  \coordinate (T4)  at (barycentric cs:u3=1,u4=1,a2=1);  % outer cap u3-u4
  \coordinate (T5)  at (barycentric cs:u4=1,u5=1,a2=1);  % outer cap u4-u5
  \coordinate (T6)  at (barycentric cs:u5=1,u0=1,a1=1);  % outer cap u5-u0
  \coordinate (I1)  at (barycentric cs:u5=1,a1=1,a2=1);  % inner cap a1-a2
  \coordinate (I2)  at (barycentric cs:u3=1,a2=1,a3=1);  % inner cap a2-a3
  \coordinate (I3)  at (barycentric cs:u0=1,a1=1,a3=1);  % inner cap a1-a3
  \coordinate (I4)  at (barycentric cs:u0=1,b1=1,b2=1);  % inner cap b1-b2
  \coordinate (I5)  at (barycentric cs:u1=1,b2=1,b3=1);  % inner cap b2-b3
  \coordinate (I6)  at (barycentric cs:u3=1,b1=1,b3=1);  % inner cap b1-b3
  \coordinate (Bup) at (barycentric cs:u0=1,a3=1,b1=1);  % bridge cap upper
  \coordinate (Bdn) at (barycentric cs:u3=1,a3=1,b1=1);  % bridge cap lower
  \foreach \p in {T1,T2,T3,T4,T5,T6,I1,I2,I3,I4,I5,I6,Bup,Bdn} {
    \node[circle, fill=blue!70!black, inner sep=1.4pt] at (\p) {};
  }
  % Hamilton cycle of length 14 (going clockwise from Bup)
  \draw[blue!70!black, very thick] (Bup) -- (I4);   % share u0-b1
  \draw[blue!70!black, very thick] (I4)  -- (T1);   % share u0-b2
  \draw[blue!70!black, very thick] (T1)  -- (I5);   % share u1-b2
  \draw[blue!70!black, very thick] (I5)  -- (T2);   % share u1-b3
  \draw[blue!70!black, very thick] (T2)  -- (T3);   % share u2-b3
  \draw[blue!70!black, very thick] (T3)  -- (I6);   % share u3-b3
  \draw[blue!70!black, very thick] (I6)  -- (Bdn);  % share u3-b1
  \draw[blue!70!black, very thick] (Bdn) -- (I2);   % share u3-a3
  \draw[blue!70!black, very thick] (I2)  -- (T4);   % share u3-a2
  \draw[blue!70!black, very thick] (T4)  -- (T5);   % share u4-a2
  \draw[blue!70!black, very thick] (T5)  -- (I1);   % share u5-a2
  \draw[blue!70!black, very thick] (I1)  -- (T6);   % share u5-a1
  \draw[blue!70!black, very thick] (T6)  -- (I3);   % share u0-a1
  \draw[blue!70!black, very thick] (I3)  -- (Bup);  % share u0-a3
  % Chord: bridge dual edge
  \draw[blue!70!black, very thick, dashed] (Bup) -- (Bdn);
  % Labels
  \node[red] at (0, 3.05) {\small $B_{\mathrm{out}}$ (hexagon)};
  \node[red!55!white] at (2.85, 0.0) {\small barbell $O$};
  \node[blue!70!black] at (-3.15, 0.5) {\small Hamilton cycle};
  \node[blue!70!black] at (-3.15, 0.18) {\small (length 14)};
  \node[blue!70!black] at (-3.15, -0.4) {\small chord = bridge};
  \node[blue!70!black] at (-3.15, -0.7) {\small dual edge};
  \draw[->, blue!70!black, thin] (-2.5, -0.4) -- (-0.15, 0);
\end{tikzpicture}
\caption{Case 2 ($R$ = annulus) with $O$ a barbell.
$B_{\mathrm{out}}$ is the outer hexagon (red); $O$ has two
triangles $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ joined by
the bridge $a_3\text{--}b_1$ (all light red).  The annulus is
triangulated by $14$ annular triangles: $6$ ``outer-cap''
triangles (one per outer edge), $6$ ``inner-cap'' triangles (one
per non-bridge edge of $O$), and $2$ ``bridge-cap'' triangles
$\{u_0, a_3, b_1\}$ and $\{u_3, a_3, b_1\}$ adjacent to the
bridge.  Each blue dot sits at the centroid of an annular
triangle; blue edges connect dual vertices whose triangles share
an interior annular edge (spoke or bridge).  The two bridge-cap
vertices have $\Gamma$-degree $3$ (their triangles have no
boundary edge) and are joined by the dashed blue \emph{chord}
corresponding to the bridge; the remaining $13$ edges form the
Hamilton cycle that wraps around the annulus.  All $14$ vertices
lie on the outer face of the cycle-with-chord embedding, so
$\Gamma \cong \Theta(1, 7, 7)$ is outerplanar.}
\label{fig:inner-dual-annulus-case}
\end{figure}

\begin{remark}
\label{rem:hamilton-cycle-spoke-only}
In the \emph{spoke-only} case (Definition~\ref{def:tire-graph} with
$O$ $2$-connected and $E_{\mathrm{ann}}$ consisting only of spokes),
every annular face has exactly one boundary edge, every
$d_f$ has $\Gamma$-degree $2$, and the construction of the
Theorem~\ref{thm:inner-dual-outerplanar} proof reduces to the
classical Hamilton cycle $\Gamma \cong C_{\mu+\nu}$ with zero chords.
\end{remark}

\begin{remark}
\label{rem:bridge-case-theta}
When $O$ has a bridge $e_{\mathrm{br}} \in E(O)$ whose two
incident faces are annular triangles, $e_{\mathrm{br}}$ contributes
an interior annular edge in $\Gamma$ rather than two leaves in
$D(T)$ (see Definition~1.7 of \cite{bauerfeld-nested-tire-duals}).
The two bridge-incident annular triangles have $\Gamma$-degree $3$;
the resulting $\Gamma$ has the structure of a Hamilton cycle of
length $\mu + \nu_\partial$ plus a single chord (length $1$).  This
corresponds to the theta graph $\Theta(1, b, c)$ identified
empirically in \cite{bauerfeld-nested-tire-duals}, which has no
$K_{2,3}$ subdivision (since one of the three paths has length $1$
and so contributes no degree-$2$ branch vertex), hence is
outerplanar as predicted.
\end{remark}

\begin{theorem}[Tire treads form a rooted tree under face containment]
\label{thm:tread-tree}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$
and let $S \subseteq V(G)$ be a single-vertex level source
$\{v_0\}$ lying on the outer face of $\Pi_G$.  The collection
$\mathcal{R}(G, S)$ of tire treads
(Theorem~\ref{thm:tread-partition}) carries a canonical rooted
tree structure $\mathcal{T}(G, S)$ defined as follows.

\begin{itemize}
\item \emph{Root.}  The depth-$0$ tire tread $T_0$ --- the unique
      tire produced by Lemma~\ref{lem:tire-component} at $d = 0$,
      with degenerate outer boundary $B_{\mathrm{out}} = \{v_0\}$
      and inner outerplanar graph $O^{(T_0)} = G[L_1]$ --- is the
      root.
\item \emph{Parent.}  For each tire tread $T_c$ at depth $d \ge 1$,
      its outer boundary $B_{\mathrm{out}}^{(T_c)}$ is a cycle in
      $L_d$.  The \emph{parent} of $T_c$ is the unique tire tread
      $T_p$ at depth $d - 1$ whose inner outerplanar graph
      $O^{(T_p)}$ has $B_{\mathrm{out}}^{(T_c)}$ as the boundary cycle
      of one of its bounded faces.  Equivalently, $R_c$ lies
      inside this bounded face of $O^{(T_p)}$ (which is itself the
      region of the plane cut off by $B_{\mathrm{out}}^{(T_c)}$ on
      the side away from $S$).
\item \emph{Children.}  The children of a tire tread $T_p$ are in
      bijection with those bounded faces of $O^{(T_p)}$ whose
      interiors contain at least one vertex of $G$ at level
      $\ge d + 2$ --- equivalently, with the connected components
      of $G'_{d+1}$ whose tires have outer boundary cycle equal to
      a bounded face of $O^{(T_p)}$.
\end{itemize}

Every tire tread except $T_0$ has exactly one parent; a tire
tread may have zero, one, or several children.
\end{theorem}

\begin{proof}
\emph{Root is well-defined.}  At $d = 0$ with single-vertex source
$S = \{v_0\}$, the dual subgraph $G'_0$ is connected (every face
of $G$ incident to $v_0$ has dual depth $0$, and they form a
single fan around $v_0$).  By
Lemma~\ref{lem:tire-component}, the unique component of $G'_0$
gives the depth-$0$ tire $T_0$ described above.

\emph{Existence of parent.}  Fix a tire tread $T_c$ at depth $d
\ge 1$ arising from a connected component $C'_c$ of $G'_d$.  Its
outer boundary $B_{\mathrm{out}}^{(T_c)} = G[V_{C'_c} \cap L_d]$ is
a simple cycle in $L_d$ (Lemma~\ref{lem:tire-component}; the
source-side boundary of a tire is always a simple cycle, by
Proposition~\ref{prop:no-level-d-pinch}).  The faces of $G$
immediately outside $B_{\mathrm{out}}^{(T_c)}$ on the side facing $S$
have depth $d - 1$ (one of their three vertices lies in $L_{d-1}$,
two in $L_d$).  Let $C'_p$ be the connected component of
$G'_{d-1}$ containing the dual vertex of any such face.

\emph{Uniqueness of parent.}  $B_{\mathrm{out}}^{(T_c)}$ is a single
simple cycle in $G$, with a well-defined ``$S$-side'' (the side
of the cycle closer to $v_0$ in $\Pi_G$).  The depth-$(d-1)$
faces lying on this side form a single contiguous arc around
$B_{\mathrm{out}}^{(T_c)}$ in the dual --- they are all $G'$-adjacent
in sequence (each pair of consecutive arc faces shares an edge in
$B_{\mathrm{out}}^{(T_c)}$).  Hence they all lie in the same
connected component $C'_p$ of $G'_{d-1}$, which is therefore
unique.

\emph{$B_{\mathrm{out}}^{(T_c)}$ bounds a face of $O^{(T_p)}$.}  The
parent tire $T_p$ has $V(O^{(T_p)}) = V_{C'_p} \cap L_d \supseteq
V(B_{\mathrm{out}}^{(T_c)})$.  The cycle $B_{\mathrm{out}}^{(T_c)}$ is
a subgraph of $O^{(T_p)}$ that bounds a face of $O^{(T_p)}$ in the
inherited embedding: the cycle traces around a depth-$\ge d+1$
region (containing $R_c$ and any descendants of $T_c$), which is
exactly a bounded face of $O^{(T_p)}$.

\emph{Children description.}  The bounded faces of $O^{(T_p)}$ are
in bijection with the connected components of $G'_d$ whose
faces lie inside those bounded regions (= one component per
bounded face, by an argument analogous to the existence-and-
uniqueness step above, applied one level deeper).

\emph{Tree property.}  Every non-root $T_c$ has a unique parent at
strictly smaller depth.  Iterating the parent map strictly
decreases depth, terminating at $T_0$.  No cycles can form
(depth is monotone).  Hence $\mathcal{T}(G, S)$ is a rooted tree.
\end{proof}

\begin{remark}
\label{rem:tree-multiple-children}
A parent tire $T_p$ has multiple children precisely when its
inner outerplanar graph $O^{(T_p)}$ has multiple bounded faces with
non-trivial interiors (= containing depth-$\ge d+2$ vertices of
$G$).  This happens, for instance, when $O^{(T_p)}$ has chords or
cut-vertices that subdivide its inner region.
By contrast, if $O^{(T_p)}$ is a simple cycle (the spoke-only case
of Remark~\ref{rem:hamilton-cycle-spoke-only}) with a non-empty
interior, $T_p$ has exactly one child.
\end{remark}

\begin{theorem}[Tire-tree decomposition]
\label{thm:tire-tree-decomposition}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and let
$v_0 \in V(G)$.  The tree of tire treads $\mathcal{T}(G, \{v_0\})$ of
Theorem~\ref{thm:tread-tree} \emph{decomposes $G$ into nested tires}:
it is a finite rooted tree, rooted at the depth-$0$ tread containing
$v_0$, whose nodes (tire treads) partition the bounded faces of $G$
(Theorem~\ref{thm:tread-partition}).

This decomposition is moreover \emph{self-similar}.  For any tread $T$
in $\mathcal{T}(G, \{v_0\})$ at depth $d \ge 1$, with outer-boundary
cycle $C_T := B_{\mathrm{out}}^{(T)}$, let $G_T$ be the sub-graph of $G$
induced by $C_T$ together with all vertices of $G$ lying in the closed
planar region $R_T \subset |\Pi_G|$ bounded by $C_T$ on the side of
$C_T$ away from $v_0$.  Then:
\begin{enumerate}
\item[(D1)] $G_T$, with the embedding inherited from $\Pi_G$, is a
            \emph{triangulated disk}: every bounded face is a triangle,
            and the outer face is bounded by $C_T$.
\item[(D2)] Taking $C_T$ as a cycle source of $G_T$ (so $C_T$ has
            level $0$ in $G_T$ and the BFS-from-$C_T$ levels in $G_T$
            equal $\ell_G(\cdot) - d$ on $V(G_T)$), the construction of
            Theorem~\ref{thm:tread-tree} extends to give a rooted tree
            of tire treads $\mathcal{T}(G_T, C_T)$ whose depth-$0$ root
            tread has $B_{\mathrm{out}} = C_T$ and inner outerplanar
            graph $O = O^{(T)}$.
\item[(D3)] $\mathcal{T}(G_T, C_T)$ is canonically iso to the sub-tree
            of $\mathcal{T}(G, \{v_0\})$ rooted at $T$, preserving
            outer-boundary cycles, inner outerplanar graphs, and the
            parent--child face correspondence.
\end{enumerate}

In short: pick any vertex $v_0 \in V(G)$ to root the global tree
$\mathcal{T}(G, \{v_0\})$ describing the whole graph; pick any tread $T$
in this tree; then $T$ is itself the root of a local tree
$\mathcal{T}(G_T, C_T)$ describing the triangulated disk of $G$ inside
$C_T$, with $C_T$ as cycle source.  Maximal planar graphs decompose
into nested trees of tire treads.
\end{theorem}

\begin{proof}
\emph{Decomposition.}  Theorem~\ref{thm:tread-tree} gives the rooted
tree structure of $\mathcal{T}(G, \{v_0\})$, with root the depth-$0$
tread containing $v_0$; Theorem~\ref{thm:tread-partition} gives that
its tire treads partition the bounded faces of $G$.  Finiteness of
the tree is immediate from finiteness of $G$.

\emph{(D1) $G_T$ is a triangulated disk.}  By
Lemma~\ref{lem:tire-component} applied to the component of $G'_d$ that
gives rise to $T$, the outer boundary $C_T = B_{\mathrm{out}}^{(T)}$ is
a simple cycle in $L_d^G$.  By the Jordan curve theorem, $C_T$
separates $|\Pi_G| \setminus C_T$ into two open regions; $R_T$ is the
closure of the one not containing $v_0$.  The bounded faces of $G_T$ in
its inherited embedding are exactly the bounded faces of $G$ contained
in $R_T$, each of which is a triangle since $G$ is a triangulation.
The unbounded face of $G_T$'s embedding is the complement of $R_T$,
whose boundary is $C_T$.

\emph{(D2) Level shift.}  We show $\mathrm{dist}_{G_T}(v, C_T) =
\ell_G(v) - d$ for every $v \in V(G_T)$.  When $v \in C_T$ both sides
equal $0$, so fix $v \in V(G_T) \setminus C_T$.

\smallskip

\emph{Step 1: $\mathrm{dist}_G(v, C_T) = \ell_G(v) - d$.}  A shortest
$G$-path from $v$ to $v_0$ must visit $C_T$, since $v$ and $v_0$ lie in
different open regions of $|\Pi_G| \setminus C_T$; let $w$ be its first
$C_T$-vertex.  The $v$-to-$w$ sub-path has length $\ge \mathrm{dist}_G(v,
C_T)$ and the $w$-to-$v_0$ sub-path has length $\ell_G(w) = d$, so
$\ell_G(v) \ge \mathrm{dist}_G(v, C_T) + d$.  Conversely, concatenating
a shortest $G$-path from $v$ to a nearest $C_T$-vertex $w'$ with a
shortest $G$-path from $w'$ to $v_0$ gives a $v$-to-$v_0$ path of
length $\mathrm{dist}_G(v, C_T) + d$, so $\ell_G(v) \le
\mathrm{dist}_G(v, C_T) + d$.

\smallskip

\emph{Step 2: $\mathrm{dist}_{G_T}(v, C_T) = \mathrm{dist}_G(v, C_T)$.}
The inequality $\ge$ is automatic since $G_T \subseteq G$.  For $\le$,
pick a shortest $G$-path $\pi$ from $v$ to $C_T$; we may assume $\pi$
has no internal vertex in $C_T$ (truncate otherwise).  Any internal
vertex of $\pi$ then lies in the same open region of $|\Pi_G| \setminus
C_T$ as $v$, i.e.\ in $R_T \setminus C_T \subseteq V(G_T)$; every edge
of $\pi$ has both endpoints in $V(G_T)$ and so lies in $E(G_T)$.  Hence
$\pi$ is a path in $G_T$ realising $\mathrm{dist}_G(v, C_T)$.

\smallskip

Combining the two steps yields $\mathrm{dist}_{G_T}(v, C_T) = \ell_G(v)
- d$, as claimed.

\emph{(D3) Tree iso.}  By (D2), $L_k^{G_T} = L_{d+k}^G \cap V(G_T)$ for
every $k \ge 0$.  For a bounded face $f$ of $G_T$, dual depth in $G_T$
equals $\min_{u \in V(f)} \ell_{G_T}(u) = \min_{u \in V(f)} \ell_G(u) -
d = \delta_G(d_f) - d$.  Hence the inner-dual subgraph $(G_T)'_{k}$ at
depth $k$ in $G_T$ is the induced subgraph of $G'_{d+k}$ on the faces
of $G$ lying in $R_T$, and two such faces are dual-adjacent in $G_T'$
iff they are dual-adjacent in $G'$ (the shared edge is in $E(G_T)$).

\smallskip

\emph{Step 3: components of $(G_T)'_{k}$ are precisely the depth-$(d+k)$
descendants of $T$ in $\mathcal{T}(G, \{v_0\})$.}  We show by induction
on $k$ that a component $C'$ of $G'_{d+k}$ has $F_{C'} \subseteq R_T$
iff $C'$ is a depth-$(d+k)$ descendant of $T$.

For $k = 0$: the components of $G'_d$ are the depth-$d$ treads; the
component giving rise to $T$ has its faces in $T$'s tread region
$R \subseteq R_T$, while any other depth-$d$ tread $T''$ has
$C_{T''}$ disjoint from $C_T$ and lying in a different bounded face of
$O^{(T_p'')}$ at depth $d - 1$, hence $R_{T''} \cap R_T = \emptyset$.

For $k \ge 1$: by Theorem~\ref{thm:tread-tree}, each component $C'$ of
$G'_{d+k}$ has a unique parent $C'_p$ at depth $d+k-1$, with
$B_{\mathrm{out}}^{(C')}$ bounding a face of $O^{(C'_p)}$; equivalently
$R_{C'}$ lies inside that bounded face, hence inside $R_{C'_p}$.  By
the induction hypothesis $R_{C'_p} \subseteq R_T$ iff $C'_p$ is a
descendant of $T$ at depth $d+k-1$, and $R_{C'} \subseteq R_{C'_p}$, so
$R_{C'} \subseteq R_T$ iff $C'$ is a descendant of $T$ at depth $d+k$.

\smallskip

\emph{Step 4: tread data and child--face correspondence.}  The
Tire-component lemma (Lemma~\ref{lem:tire-component}) and the
source-side simple-cycle property
(Proposition~\ref{prop:no-level-d-pinch}) extend verbatim to the
cycle-sourced triangulated disk $(G_T, C_T)$: the proofs use only
the triangular structure of bounded faces, the local arrangement of
faces around each vertex's rotation, and the connectivity of the BFS
ball $G_T[L_{<k}^{G_T}]$ (which holds for every $k \ge 1$ since
$L_0^{G_T} = V(C_T)$ is connected as a cycle and each higher level is
BFS-adjacent to the previous).  Applied to each component of
$(G_T)'_{k}$, the lemma produces a tire graph with outer boundary
$B_{\mathrm{out}}$, inner outerplanar graph $O$, and tread region $R$
identical to those produced by the corresponding component of
$G'_{d+k}$ in $\mathcal{T}(G, \{v_0\})$, since these data depend only
on level-$d+k$ and level-$(d+k+1)$ vertices and the bounded faces in
between --- all of which are unchanged when restricting to $G_T$.

The depth-$0$ case ($k = 0$) gives a single component, namely the one
producing $T$, with root tread $B_{\mathrm{out}} = C_T$ and $O = O^{(T)}$.

The parent--child face correspondence of
Theorem~\ref{thm:tread-tree} is preserved: for any tread $T'$ in
$\mathcal{T}(G_T, C_T)$ at depth $k$, its children correspond to
non-trivial bounded faces of $O^{(T')}$, and the bounded faces of
$O^{(T')}$ together with the descendant-side interior of each are
identical in $G_T$ and in $G$.

Combining Steps 3 and 4: the bijection $C' \leftrightarrow C'$
(component of $(G_T)'_{k}$ to corresponding component of $G'_{d+k}$
inside $R_T$) lifts to a rooted-tree iso $\mathcal{T}(G_T, C_T) \to
\text{sub-tree of } \mathcal{T}(G, \{v_0\}) \text{ rooted at } T$,
preserving outer boundaries, inner outerplanar graphs, and the
parent--child face correspondence.
\end{proof}

\begin{figure}[h]
\centering
\includegraphics[width=0.95\textwidth]{fig_tire_tree_decomposition.png}
\caption{Tire-tree decomposition
(Theorem~\ref{thm:tire-tree-decomposition}) on a $13$-vertex maximal
planar example $G$ with five BFS levels.  $(a)$ $G$ with vertex source
$v_0$ and $\ell_G \in \{0,1,2,3,4\}$; four nested seams are
highlighted, $C_{T_R} = \{a,b,c\}$ (orange), $C_{T_L} = \{a,c,d\}$
(red, including the chord $a$-$c$ shared with $C_{T_R}$),
$C_{T_{LL}} = \{f_1, f_2, f_3\}$ (purple), $C_{T_{LLL}} = \{g_1, g_2, g_3\}$
(teal).  Inset: the rooted tree of tire treads $\mathcal{T}(G, \{v_0\})$
branches at $T_0$ into the leaf $T_R$ (containing $e$) and a chain
$T_L \to T_{LL} \to T_{LLL}$ (the highlighted sub-tree).
$(b)$ The disk $G_{T_L}$ inside the seam $C_{T_L}$, drawn standalone
with $C_{T_L}$ as cycle source and vertex labels rotated to match the
new (cycle-source) role of the boundary triangle.
$\ell_{G_{T_L}}(\cdot) = \ell_G(\cdot) - 1$ on $V(G_{T_L})$ (verified
by the generator script), and $\mathcal{T}(G_{T_L}, C_{T_L})$ is the
chain $T_L \to T_{LL} \to T_{LLL}$, iso to the highlighted sub-tree
of $(a)$.}
\label{fig:tire-tree-decomposition}
\end{figure}

\begin{remark}
\label{rem:tree-coloring-factorisation}
Combining Theorem~\ref{thm:tread-partition} (treads partition
the bounded faces of $G$) with
Theorem~\ref{thm:tread-tree} (treads form a rooted tree), any
proper coloring problem on $G$'s bounded faces factors through:
\begin{itemize}
\item local coloring problems on each tread (the inner dual of
      each tread is outerplanar by
      Theorem~\ref{thm:inner-dual-outerplanar}), plus
\item consistency constraints along parent-child interfaces (the
      cycle $B_{\mathrm{out}}^{(T_c)}$ shared between a child and the
      face of its parent's $O^{(T_p)}$).
\end{itemize}
This is the structural setup underlying the chain-pigeonhole
program for tire treads.
\end{remark}

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\end{document}
