coloring_nested_tire_dual_graphs: tire-tread partition theorem
NEW: Theorem 1.5 (Tire treads partition the bounded faces).
For a maximal planar graph G with level source S on the outer face,
the family of tire treads { R_{C'} : d ≥ 0, C' a connected component
of G'_d } supplied by the tire-component lemma partitions the
bounded part of |Π_G|:
(i) every bounded face of G lies in exactly one tread R_{C'};
(ii) distinct treads have disjoint interiors.
Proof: each bounded face has a unique dual depth d, hence its dual
vertex lies in G'_d alone, and within G'_d in a unique component C'.
By the tire-component lemma, that C' carries the unique tread
containing the face.
This is the first step toward a chain pigeonhole argument that
colorings extend across the nested tire treads induced by a level
source.
Paper grows to 10 pages.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
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\citation{bauerfeld-nested-tires}
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\newlabel{rem:tire-component-degenerate}{{1.5}{6}}
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\newlabel{thm:tread-partition}{{1.5}{6}}
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\newlabel{rem:tire-no-extra-hypotheses}{{1.6}{6}}
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\newlabel{prop:edge-vertex-bijection}{{1.7}{6}}
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\newlabel{rem:edge-vertex-corollary}{{1.8}{7}}
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\newlabel{def:tire-annular-subgraph}{{1.9}{7}}
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\newlabel{def:tire-annular-face-connector}{{1.10}{7}}
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\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces The bridge case: $T'_{\mathrm {ann}} = \theta (1, 3, 3)$ has three faces $A, B, C$ in its inherited embedding, with respective vertex sets $V(A) = \{v_0, \dots , v_5\}$, $V(B) = \{v_0, v_1, v_2, v_3\}$, and $V(C) = \{v_0, v_3, v_4, v_5\}$. In the surrounding maximal planar $G$, the chord endpoints $v_0, v_3$ (the two annular faces sharing the bridge edge) have all three $G'$-edges inside $T'_{\mathrm {ann}}$, while each non-chord vertex $v_i$ ($i \in \{1, 2, 4, 5\}$) contributes one $G'$-edge to an external non-annular neighbor $u_i$. Each panel highlights $T'_{f'}$ (blue) inside $G'$: dark circles are $V(f')$, gray circles are $G'$-neighbors of $V(f')$ within $T'_{\mathrm {ann}}$, and red squares are external $G'$-neighbors $u_i$. The choice of face $f'$ controls which external neighbors $u_i$ are pulled into $T'_{f'}$ (face $A$ pulls in all four; face $B$ pulls in $u_1, u_2$ and face $C$ pulls in $u_4, u_5$).}}{8}{}\protected@file@percent }
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\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces The bridge case: $T'_{\mathrm {ann}} = \theta (1, 3, 3)$ has three faces $A, B, C$ in its inherited embedding, with respective vertex sets $V(A) = \{v_0, \dots , v_5\}$, $V(B) = \{v_0, v_1, v_2, v_3\}$, and $V(C) = \{v_0, v_3, v_4, v_5\}$. In the surrounding maximal planar $G$, the chord endpoints $v_0, v_3$ (the two annular faces sharing the bridge edge) have all three $G'$-edges inside $T'_{\mathrm {ann}}$, while each non-chord vertex $v_i$ ($i \in \{1, 2, 4, 5\}$) contributes one $G'$-edge to an external non-annular neighbor $u_i$. Each panel highlights $T'_{f'}$ (blue) inside $G'$: dark circles are $V(f')$, gray circles are $G'$-neighbors of $V(f')$ within $T'_{\mathrm {ann}}$, and red squares are external $G'$-neighbors $u_i$. The choice of face $f'$ controls which external neighbors $u_i$ are pulled into $T'_{f'}$ (face $A$ pulls in all four; face $B$ pulls in $u_1, u_2$ and face $C$ pulls in $u_4, u_5$).}}{9}{}\protected@file@percent }
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@@ -388,6 +388,59 @@ between $V_{C'} \cap L_d$ and $V_{C'} \cap L_{d+1}$ that bound a face
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of $F_{C'}$.
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of $F_{C'}$.
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\end{proof}
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\end{proof}
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\begin{theorem}[Tire treads partition the bounded faces]
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\label{thm:tread-partition}
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Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and
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let $S \subseteq V(G)$ be a level source lying on the outer face.
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For each $d \ge 0$ and each connected component $C'$ of $G'_d$, let
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$T^{(d, C')}$ denote the tire graph supplied by
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Lemma~\ref{lem:tire-component}, with tire tread
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$R_{C'} \subseteq |\Pi_G|$. Then the collection of treads
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\[
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\mathcal{R}(G, S) \;:=\;
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\bigl\{\, R_{C'} \,:\, d \ge 0,\;
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C' \text{ a connected component of } G'_d \,\bigr\}
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\]
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partitions the bounded part of $|\Pi_G|$:
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\begin{enumerate}
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\item[(i)] every bounded face $f$ of $G$ is contained in exactly
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one tread $R_{C'} \in \mathcal{R}(G, S)$;
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\item[(ii)] distinct treads in $\mathcal{R}(G, S)$ have disjoint
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interiors and may share only boundary edges or vertices.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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\emph{Existence and uniqueness.} Each bounded face $f \in F(G)$
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has a uniquely-defined dual depth $\delta_G(d_f) \in \mathbb{Z}_{\ge
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0}$, so the dual vertex $d_f$ lies in $G'_d$ for $d =
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\delta_G(d_f)$ and in no other $G'_{d'}$. Within $G'_d$, the
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vertex $d_f$ belongs to exactly one connected component $C'$. By
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Lemma~\ref{lem:tire-component}, $F_{C'}$ is precisely the set of
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faces $f' \in F(G)$ with $d_{f'} \in V(C')$; in particular $f \in
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F_{C'}$, hence $f \subseteq R_{C'}$.
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For any other tread $R_{C''} \in \mathcal{R}(G, S)$, the
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component $C''$ is either at a different depth $d' \ne d$ (in
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which case $F_{C''}$ consists of depth-$d'$ faces and $f \notin
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F_{C''}$) or at depth $d$ but a different component $C'' \ne C'$
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(in which case the two components are vertex-disjoint in $G'_d$,
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so again $f \notin F_{C''}$). In both cases $f \notin R_{C''}$
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(more precisely, $f$ is not one of the triangular faces of $G$ in
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$F_{C''}$, so $f$'s interior is not contained in $R_{C''}$).
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\emph{Disjoint interiors.} Each tread $R_{C'}$ is the union of
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its triangular faces $F_{C'} \subseteq F(G)$; distinct treads
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correspond to disjoint $F_{C'}$ (by the argument above), and the
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interiors of distinct $G$-faces are disjoint. Hence interiors of
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distinct treads are disjoint.
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\emph{Coverage.} Conversely, every bounded $f \in F(G)$ has $d_f
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\in V(G')$ with some dual depth $d$, and thus lies in $R_{C'}$
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where $C'$ is its component of $G'_d$. So $\bigcup_{R \in
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\mathcal{R}(G, S)} R$ contains every bounded face of $G$.
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\end{proof}
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\begin{remark}
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\begin{remark}
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\label{rem:tire-component-degenerate}
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\label{rem:tire-component-degenerate}
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Either boundary part of $C$ in Lemma~\ref{lem:tire-component} may be
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Either boundary part of $C$ in Lemma~\ref{lem:tire-component} may be
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