dual_decomposition: reduced-dual definition, verification, and step figures

Add Definition 2.1 (reduced dual) and a remark on cubicity/planarity, plus an
experiment verifying it on the icosahedron/dodecahedron and four figures, one
per construction step.

reduced_dual.py builds G' = dodecahedron (dual of the icosahedron), applies the
construction, and confirms the result is a cubic, planar, simple graph whose
dual is a simple triangulation. Finding: the construction is an n -> n-2
reduction (12 -> 10 here), not n-1, since the single apex v_n collapses one more
vertex than a standard pentagon re-triangulation; the result also re-introduces
degree-3 and degree-4 vertices (degree seq [7,5,5,5,5,5,5,4,4,3]).

draw_reduced_dual_steps.py renders fig_reduced_dual_step1..4.png, embedded as a
2x2 grid after the definition.

Co-Authored-By: Claude Opus 4.7 (1M context) <noreply@anthropic.com>

papers/dual_decomposition_minimal_counterexamples/paper.tex

@@ -115,4 +115,63 @@ $G$, a contradiction.
 Hence $\delta(G) \ge 5$.
 \end{proof}
 
+\section{The reduced dual}
+
+Write $G'$ for the dual of $G$: since $G$ is a triangulation, $G'$ is a cubic
+plane graph in which each vertex of $G$ corresponds to a face of $G'$, each face
+of $G$ to a vertex of $G'$, and each edge to a dual edge. A vertex of $G$ of
+degree $k$ corresponds to a $k$-gonal face of $G'$.
+
+By Lemma~\ref{lem:mindeg}, $\delta(G) \ge 5$, and Euler's formula gives
+$\sum_{u \in V(G)}(6 - \deg u) = 12$, so $G$ has a vertex of degree exactly $5$
+(indeed at least twelve). Fix such a vertex $v$. Its dual face $F_v$ is a
+pentagon, bounded by the five dual vertices corresponding to the five faces of
+$G$ incident to $v$.
+
+\begin{definition}[Reduced dual]
+\label{def:reduced-dual}
+Let $v$ be a degree-$5$ vertex of $G$ with pentagonal dual face $F_v$, and fix an
+index $i \in \{0,1,2,3,4\}$. The \emph{reduced dual} $\widehat{G}'_{v,i}$ is the
+plane graph obtained from $G'$ as follows.
+\begin{enumerate}
+  \item Delete the five dual vertices on the boundary of $F_v$, together with all
+        edges incident to them. Each deleted vertex is cubic, with two edges on
+        $\partial F_v$ and one edge leaving $F_v$; deleting the five boundary
+        vertices therefore removes the five external edges as well, dropping their
+        five outer endpoints from degree $3$ to degree $2$. These five degree-$2$
+        vertices lie on the boundary of a single face $F$ of the resulting graph.
+  \item List the five degree-$2$ vertices in clockwise order around $F$ as
+        $A = (A_0, A_1, A_2, A_3, A_4)$.
+  \item Add a new vertex $v_n$ and join it to $A_i$, $A_{i+1}$, and $A_{i+2}$
+        (indices mod $5$) by three new edges.
+  \item Add a new edge between $A_{i+3}$ and $A_{i+4}$ (indices mod $5$).
+\end{enumerate}
+\end{definition}
+
+\begin{remark}
+Steps (3) and (4) restore cubicity: $A_i, A_{i+1}, A_{i+2}$ each gain one edge to
+$v_n$ and $A_{i+3}, A_{i+4}$ each gain the new edge, so all five return to degree
+$3$, and $v_n$ has degree $3$. Since $A_i,\dots,A_{i+2}$ and $A_{i+3}, A_{i+4}$
+are each consecutive along $\partial F$, the new vertex and edge can be drawn
+inside $F$ without crossings, so $\widehat{G}'_{v,i}$ is again a cubic plane
+graph. The construction depends on the choice of $i$ up to the rotational
+symmetry of $A$.
+\end{remark}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step1.png}\hfill
+\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step2.png}\\[0.5em]
+\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step3.png}\hfill
+\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step4.png}
+\caption{The four steps of Definition~\ref{def:reduced-dual}, illustrated on
+$G' = $ the dodecahedron (dual of the icosahedron) with $F_v$ the inner
+pentagon and $i = 0$. Top left: delete the five boundary vertices of $F_v$,
+leaving five degree-$2$ vertices on a new face $F$. Top right: order them
+clockwise as $A_0,\dots,A_4$. Bottom left: add $v_n$ joined to $A_0, A_1, A_2$.
+Bottom right: add the chord $A_3 A_4$, giving the cubic plane graph
+$\widehat{G}'_{v,0}$.}
+\label{fig:reduced-dual-steps}
+\end{figure}
+
 \end{document}