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coloring_nested_tire_graphs: theorem that inner dual of tire tread is outerplanar

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NEW Theorem 1.12: For any tire graph T, the inner dual Γ of its
tire tread (= subgraph of D(T) induced on interior dual vertices)
is outerplanar.

The theorem also gives a constructive characterization: Γ admits a
planar embedding as a (possibly non-simple) Hamilton walk through
every d_f, plus zero or more non-crossing chords.

Proof structure (constructive):

Case 1 (R is a disk, one boundary degenerate): the polygon
triangulation has no interior vertex, so its dual is a tree
(p-2 vertices, p-3 diagonals). Trees are outerplanar.

Case 2 (R is an annulus, both boundaries non-degenerate):

  Step 1 - Cyclic ordering: cut R along any spoke e* to convert
  the annulus into a closed disk. The disk boundary traverses
  B_out + e* + B_in (reverse) + e*, yielding a cyclic sequence
  S of annular faces with multiplicities (one per boundary edge,
  + detours for boundary-free faces).

  Step 2 - Hamilton walk: consecutive entries of S share an
  interior annular edge or coincide; the resulting closed walk
  in Γ visits every d_f (using detours for the rare interior
  annular triangles with zero boundary edges).

  Step 3 - Non-crossing chords: remaining interior annular edges
  become chords. Since the underlying E_ann edges in T are
  non-crossing in the planar embedding, the chords are
  non-crossing in Γ.

  Step 4 - Outerplanar layout: place the |F_ann| vertices on a
  circle in S-order, draw walk edges as the circle, chords inside.
  All vertices on outer face → outerplanar.

Two remarks following:

Remark 1.13: spoke-only case is the classical Hamilton cycle
Γ ≅ C_{n+m} with zero chords.

Remark 1.14: bridge case (O with a bridge whose 2 incident faces
are annular) gives the theta graph Θ(1, b, c) — Hamilton cycle of
length n + m_∂ plus a single length-1 chord. The length-1 chord
contributes no degree-2 branch vertex to a K_{2,3} subdivision,
explaining why this is outerplanar despite being a theta graph.

Foundational paper grows from 7 to 8 pages.

This theorem unlocks the chain pigeonhole argument over tire
treads: each tread's coloring problem is on an outerplanar dual
graph, where the structure is locally tractable.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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didericis
2026-05-27 01:47:55 -04:00
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\newlabel{thm:tread-partition}{{1.9}{6}}
\newlabel{rem:tire-component-degenerate}{{1.10}{6}}
\newlabel{rem:tire-no-extra-hypotheses}{{1.11}{6}}
\newlabel{thm:inner-dual-outerplanar}{{1.12}{7}}
\citation{bauerfeld-nested-tire-duals}
\citation{bauerfeld-nested-tire-duals}
\bibcite{bauerfeld-depth}{1}
\bibcite{bauerfeld-nested-tire-duals}{2}
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@@ -466,6 +466,146 @@ 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 $F_{\mathrm{ann}}$). 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}).
Then $T \cap R$ triangulates a polygon with no interior vertices:
the polygon's boundary cycles round the unique non-degenerate
boundary plus the degenerate apex, and all vertices of $V(T)
\cap R$ lie on this polygon. The dual graph of such a polygon
triangulation is a tree (a classical fact: a triangulation of a
$p$-gon with no interior vertex has $p - 2$ triangles and $p - 3$
diagonals, and the diagonals' adjacency graph is a tree). Trees
are outerplanar.
\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
$f^{\mathrm{out}}_1, f^{\mathrm{out}}_2, \dots, f^{\mathrm{out}}_n$
of (not necessarily distinct) annular faces, one per
$B_{\mathrm{out}}$-edge; similarly walking around
$B_{\mathrm{in}}$ produces a sequence
$f^{\mathrm{in}}_1, \dots, f^{\mathrm{in}}_{m_\partial}$ where
$m_\partial$ 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
\[
\mathcal{S} = (f^{\mathrm{out}}_1, \dots, f^{\mathrm{out}}_n,
f^{\mathrm{in}}_1, \dots, f^{\mathrm{in}}_{m_\partial})
\]
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{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 $\Gamma \cong C_{n+m}$ 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 $n + m_\partial$ 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{thebibliography}{9}