coloring_nested_tire_graphs: theorem that inner dual of tire tread is outerplanar

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>

papers/coloring_nested_tire_graphs/paper.aux

@@ -16,6 +16,9 @@
 \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}
 \newlabel{tocindent-1}{0pt}
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+\gdef \@abspage@last{8}

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papers/coloring_nested_tire_graphs/paper.tex

@@ -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}