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coloring_nested_tire_graphs: Tait correspondence + count formula

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NEW Theorem 1.15 (Tait correspondence for tires):
  #{4-colorings of T} / |S_4| = #{3-edge-colorings of Γ} / |S_3|
That is, the number of 4-vertex-colorings of the tire T up to
color permutation equals the number of 3-edge-colorings of the
inner dual Γ up to color permutation.

Proof: standard Tait. Encode 4 colors as Z_2 × Z_2; define
χ*(e*) = c(u) + c(v) for each interior annular edge. The
triangulation constraint guarantees χ* is a proper 3-edge-coloring
of Γ; the lift c → χ* is 4-to-1 (global Z_2 × Z_2 translation).
Quotienting by |S_4| = 24 and |S_3| = 6 gives the stated equality.

NEW Theorem 1.16 (count formula):
  (i) For spoke-only tires (Γ ≅ C_n):
      #{proper 3-edge-colorings of Γ} = 2^n + 2(-1)^n.
  (ii) For single-chord tires (Γ ≅ Θ(1, b, c), b + c = n):
      #{proper 3-edge-colorings of Γ} = 6(α_b α_c + β_b β_c),
      where α_L = (2^{L-1} + 2(-1)^{L-1})/3,
            β_L = (2^{L-1} - (-1)^{L-1})/3.
  Verification: Θ(1, 2, 2) = K_4 \ e gives 6.

Proofs:
  (i) Standard chromatic polynomial of cycle at k = 3.
  (ii) Transfer matrix on the two non-chord paths with chord
       color fixed and endpoint configurations enumerated.

Remark 1.17: For more chords, the count depends on the chord
arrangement, not just (n, k). Two outerplanar graphs with the
same vertex and chord counts can have different 3-edge-coloring
counts. But linear-time computation via tree decomposition
(treewidth ≤ 2 for outerplanar) is always available.

Added Tait's 1880 paper as bibitem.

Page count: 11 → 12. Theorem 1.18 (tree structure) renumbered from
1.15 to 1.18 to make room.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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\newlabel{fig:inner-dual-annulus-case}{{4}{9}}
\newlabel{rem:hamilton-cycle-spoke-only}{{1.13}{9}}
\newlabel{rem:bridge-case-theta}{{1.14}{9}}
\newlabel{thm:tread-tree}{{1.15}{10}}
\bibcite{bauerfeld-depth}{1}
\bibcite{bauerfeld-nested-tire-duals}{2}
\citation{tait-original}
\citation{bauerfeld-nested-tire-duals}
\newlabel{thm:tait-tire}{{1.15}{10}}
\newlabel{thm:count-formula}{{1.16}{10}}
\newlabel{rem:count-general-outerplanar}{{1.17}{11}}
\newlabel{thm:tread-tree}{{1.18}{11}}
\bibcite{tait-original}{1}
\bibcite{bauerfeld-depth}{2}
\bibcite{bauerfeld-nested-tire-duals}{3}
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outerplanar as predicted.
\end{remark}
\begin{theorem}[Tait correspondence: 4-colorings of a tire vs 3-edge-colorings of its inner dual]
\label{thm:tait-tire}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph
(viewed as an annular triangulation of its tire tread $R$) and let
$\Gamma$ be its inner dual
(Theorem~\ref{thm:inner-dual-outerplanar}). Then
\[
\#\bigl\{\text{proper $4$-vertex-colorings of } T\bigr\} \big/ |S_4|
\;=\;
\#\bigl\{\text{proper $3$-edge-colorings of } \Gamma\bigr\} \big/ |S_3|.
\]
That is, the number of $4$-vertex-colorings of $T$ up to permutation
of the colour set $\{0, 1, 2, 3\}$ equals the number of
$3$-edge-colorings of $\Gamma$ up to permutation of the colour set
$\{1, 2, 3\}$.
\end{theorem}
\begin{proof}
The argument is the classical Tait correspondence
\cite{tait-original} adapted to the annular triangulation $T$.
Encode the four colours of a proper $4$-vertex-coloring $c \colon
V(T) \to \mathbb{Z}_2 \times \mathbb{Z}_2$. For each interior
annular edge $e$ of $T$ (whose two incident faces both lie in
$F_{\mathrm{ann}}$, contributing a $\Gamma$-edge $e^*$), set
\[
\chi^*(e^*) \;:=\; c(u) + c(v) \in \mathbb{Z}_2 \times \mathbb{Z}_2,
\qquad \text{where } u, v \text{ are the endpoints of } e.
\]
Since $c(u) \ne c(v)$, we have $\chi^*(e^*) \ne 00$, so $\chi^*$
takes values in $\{01, 10, 11\}$, which we identify with the
$3$-edge-coloring palette $\{1, 2, 3\}$.
\emph{Properness.} At each $\Gamma$-vertex $d_f$ corresponding to
an annular triangle $f = \{u, v, w\}$, the three incident
$\Gamma$-edges (one per cycle-edge of $f$) carry colours
$c(u) + c(v),\; c(v) + c(w),\; c(u) + c(w)$. These three elements
of $\mathbb{Z}_2 \times \mathbb{Z}_2$ sum to $0$ and are pairwise
distinct (their pairwise differences are
$c(u) - c(w),\; c(v) - c(u),\; c(w) - c(v)$, each nonzero), so
they form a permutation of $\{01, 10, 11\}$ --- a proper edge
colouring at $d_f$.
\emph{Surjectivity onto cosets.} Given a proper $3$-edge-coloring
$\chi^*$ of $\Gamma$, the equation $c(u) + c(v) = \chi^*(e^*)$
admits exactly $|\mathbb{Z}_2 \times \mathbb{Z}_2| = 4$ solutions
$c \colon V(T) \to \mathbb{Z}_2 \times \mathbb{Z}_2$ (a global
translation is the only freedom). Hence the map $c \mapsto
\chi^*$ is $4$-to-$1$.
\emph{Count.} Therefore $\#\{\text{4-colorings of } T\} = 4 \cdot
\#\{\text{3-edge-colorings of } \Gamma\}$. Dividing by $|S_4|
= 24$ on the left and $|S_3| = 6$ on the right (since $S_4$ acts
faithfully on the $4$-colorings and $S_3$ on the $3$-edge-colorings,
and the $4$-to-$1$ map respects the $S_4/S_3 \cong S_3$ quotient
via the natural surjection $S_4 \twoheadrightarrow S_3$) gives the
stated equality.
\end{proof}
\begin{theorem}[Count formula for spoke-only and single-chord tires]
\label{thm:count-formula}
Let $T$ be a tire graph with $|F_{\mathrm{ann}}| = n$ annular
triangles, and let $\Gamma$ be its inner dual.
\begin{enumerate}
\item[(i)] If $\Gamma \cong C_n$ (the spoke-only case
of Remark~\ref{rem:hamilton-cycle-spoke-only}), then
\[
\#\bigl\{\text{proper $3$-edge-colorings of } \Gamma\bigr\}
\;=\; 2^n + 2 \cdot (-1)^n.
\]
\item[(ii)] If $\Gamma \cong \Theta(1, b, c)$ with $b + c = n$
(the single-chord case of
Remark~\ref{rem:bridge-case-theta}), then
\[
\#\bigl\{\text{proper $3$-edge-colorings of } \Gamma\bigr\}
\;=\; 6 \,(\alpha_b \alpha_c + \beta_b \beta_c),
\]
where
$\alpha_L = \bigl(2^{L-1} + 2 (-1)^{L-1}\bigr) / 3$ and
$\beta_L = \bigl(2^{L-1} - (-1)^{L-1}\bigr) / 3$.
\end{enumerate}
\end{theorem}
\begin{proof}
\emph{(i)} Standard chromatic polynomial of the cycle: $P(C_n, k)
= (k - 1)^n + (-1)^n (k - 1)$. At $k = 3$:
$2^n + 2 (-1)^n$ (cf.\ Proposition~1.2 of
\cite{bauerfeld-nested-tire-duals}).
\emph{(ii)} By transfer matrix on the two non-chord paths of
$\Theta(1, b, c)$ with the chord edge colour fixed and the two
endpoint colour assignments enumerated. At each trivalent endpoint,
the two path-incident edges receive the two non-chord colours
($2$ ways). Conditional on these assignments, the path's interior
edges form a $3$-edge-coloring of a path of length $b$ (resp.\ $c$)
with both endpoints' colours fixed; the number of such colorings is
the $(c_a, c_b)$-entry of $T^{L-1}$, where $T = J - I$ is the
$3 \times 3$ adjacency matrix of the colour-difference graph.
$T^{L-1}$ has diagonal entries $\alpha_L$ and off-diagonal entries
$\beta_L$. Summing over the four endpoint configurations
($\{x, y\}, \{x', y'\} = \{2, 3\}$ each in two orderings) and
multiplying by the three chord colour choices gives the stated
formula. Verification: $\Theta(1, 2, 2)$ (= $K_4 \setminus e$,
$\alpha_2 = 0,\; \beta_2 = 1$) yields $6 \cdot (0 + 1) = 6$ proper
$3$-edge-colorings, matching the known count for $K_4 \setminus e$.
\end{proof}
\begin{remark}
\label{rem:count-general-outerplanar}
For an inner dual $\Gamma$ with more than one non-crossing chord,
the count depends on the chord structure, not just on the pair
(number of vertices, number of chords). Two outerplanar graphs
with the same $n$ and $k$ can have different proper $3$-edge-coloring
counts depending on how the chords are arranged (nested,
sequential, sharing vertices, etc.). However, every such count
can be computed in linear time by tree-decomposition methods,
since outerplanar graphs have treewidth at most $2$ and the
edge-chromatic polynomial admits a deletion--contraction recursion
that respects the cycle-plus-chord structure.
\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 $\Pi_G$
@@ -876,6 +996,11 @@ program for tire treads.
\begin{thebibliography}{9}
\bibitem{tait-original}
P.~G.~Tait,
\emph{Remarks on the colouring of maps},
Proc.\ Roy.\ Soc.\ Edinburgh \textbf{10} (1880), 729.
\bibitem{bauerfeld-depth}
E.~Bauerfeld,
\emph{Plane Depth},