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+++ b/papers/face_monochromatic_pairs/paper.tex
@@ -957,6 +957,169 @@ $\varepsilon \equiv 0 \pmod 3$. But $\varepsilon \in \{+1, -1\}$,
 so $\varepsilon \not\equiv 0 \pmod 3$ --- contradiction.
 \end{proof}
 
+\subsection*{A partial structural proof of
+Conjecture~\ref{conj:deciding-face}}
+
+We single out a specific candidate face --- the ``flank'' face that
+the spike, side-$0$ edge, and one boundary arc of the new pentagonal
+hole together bound --- and prove the deciding-face property for it
+when the adjacent $G'$-face is pentagonal or hexagonal. This is the
+case for every reduced dual where at least one of $v$'s neighbours
+$B_i$ or $B_{i+1}$ in the parent triangulation $G$ has degree $5$ or
+$6$ --- e.g., for the icosahedron and the dodecahedron's other near
+neighbours.
+
+\begin{definition}[Flank face]
+\label{def:flank-face}
+Fix a reduced dual $\widehat{G}'_{v,i}$ and let $F$ be the post-deletion
+pentagonal hole. Order the five outer endpoints
+$A_0, A_1, A_2, A_3, A_4$ in clockwise order along
+$\partial F$ as in Definition~\ref{def:reduced-dual}. The
+\emph{lower flank face} $F_{i, i+1}^{\flat}$ of
+$\widehat{G}'_{v,i}$ is the face of $\widehat{G}'_{v,i}$ whose
+boundary is the closed walk
+\[
+v_n
+\;\xrightarrow{\text{side-}0}\;
+A_i
+\;\xrightarrow{\partial F \text{-arc}}\;
+A_{i+1}
+\;\xrightarrow{\text{spike}}\;
+v_n,
+\]
+i.e.\ the side-$0$ edge, followed by the maximal sub-arc of
+$\partial F$ from $A_i$ to $A_{i+1}$ not crossing $v_n$ or the
+merged edge, followed by the spike edge in reverse.
+The \emph{upper flank face} $F_{i+1, i+2}^{\flat}$ is defined
+analogously with side-$1$ replacing side-$0$ and $A_{i+1} \to A_{i+2}$
+replacing $A_i \to A_{i+1}$.
+\end{definition}
+
+\begin{lemma}[Flank-length formula]
+\label{lem:flank-length}
+$|F_{i, i+1}^{\flat}| \;=\; n_i - 1$, where $n_i$ is the length of
+the $G'$-face $F_i$ adjacent to $F_v$ across the edge
+$(B_i, B_{i+1})$ in the pre-reduction dual $G'$. (Symmetrically,
+$|F_{i+1, i+2}^{\flat}| = n_{i+1} - 1$.)
+\end{lemma}
+
+\begin{proof}
+After step (1) of Definition~\ref{def:reduced-dual} the dual face
+$F_i$, originally a closed walk of length $n_i$ around the boundary
+edges $(B_i, B_{i+1})$, $(B_{i+1}, A_{i+1})$, $\partial F_i$
+interior, and $(A_i, B_i)$, becomes (after deletion of
+$B_i, B_{i+1}$ and their three incident edges) an open arc from
+$A_{i+1}$ to $A_i$ of length $n_i - 3$ in the post-deletion
+graph. The lower flank face $F_{i, i+1}^{\flat}$ adds back the
+side-$0$ edge $(v_n, A_i)$ (length $1$) and the spike edge
+$(v_n, A_{i+1})$ (length $1$). Its total boundary length is therefore
+$(n_i - 3) + 1 + 1 = n_i - 1$.
+\end{proof}
+
+\begin{lemma}[Flank covering, base case $n_i = 5$]
+\label{lem:flank-covering-base}
+If $n_i = 5$ then $\partial F_{i, i+1}^{\flat} \subseteq V(K_b) \cup V(K_c)$.
+\end{lemma}
+
+\begin{proof}
+When $n_i = 5$ the boundary of $F_{i, i+1}^{\flat}$ visits exactly four
+vertices: $v_n$, $A_i$, a single intermediate $P$, and $A_{i+1}$. We
+have $v_n, A_i, A_{i+1} \in V(K_b) \cup V(K_c)$ by
+Lemma~\ref{lem:kempe-spike}, so it remains to show $P \in V(K_b) \cup V(K_c)$.
+
+The vertex $A_{i+1}$ has degree $3$, with one of its three incident
+edges being the spike (colour $c$) and the other two having colours
+$\{c_0, c_1\}$ (one each, by properness). Lemma~\ref{lem:kempe-spike}
+gives $A_{i+1} \in V(K_b) \cap V(K_c)$, so $K_b$ uses the spike
+together with $A_{i+1}$'s colour-$c_0$ edge, and $K_c$ uses the spike
+together with $A_{i+1}$'s colour-$c_1$ edge. The two non-spike edges
+at $A_{i+1}$ are exactly $A_{i+1} P$ (the boundary edge into the lower
+flank) and $A_{i+1} P'$ (the boundary edge into the upper flank,
+$P' \in F_{i+1, i+2}^{\flat}$). Whichever of these is $A_{i+1} P$,
+its colour is in $\{c_0, c_1\}$, so the corresponding Kempe cycle
+($K_b$ if colour $c_0$, $K_c$ if colour $c_1$) walks $A_{i+1} \to P$,
+placing $P$ in that cycle's vertex set.
+\end{proof}
+
+\begin{lemma}[Flank covering, $n_i = 6$]
+\label{lem:flank-covering-hex}
+If $n_i = 6$ then $\partial F_{i, i+1}^{\flat} \subseteq V(K_b) \cup V(K_c)$.
+\end{lemma}
+
+\begin{proof}
+The boundary now visits five vertices: $v_n$, $A_i$, $P_1, P_2$,
+$A_{i+1}$, with $P_1$ adjacent to $A_i$ and $P_2$ adjacent to
+$A_{i+1}$. The named vertices and $P_2$ are in $V(K_b) \cup V(K_c)$
+by Lemma~\ref{lem:flank-covering-base}'s argument applied at
+$A_{i+1}$. It remains to place $P_1$.
+
+\emph{Case (a):} $\varphi(A_i P_1) = c$. Then $K_b$'s walk uses
+$A_i$'s colour-$c$ edge (along with side-$0$), so $K_b$ walks
+$A_i \to P_1$ and $P_1 \in V(K_b)$.
+
+\emph{Case (b):} $\varphi(A_i P_1) = c_1$. Then $A_i$'s colour-$c$
+edge is the \emph{other} non-side-$0$ edge of $A_i$, namely $A_i Q$
+where $Q$ is the intermediate on $\partial F_{i-1, i}^{\flat}$
+adjacent to $A_i$. Now consider $P_2$. By the analysis above we have
+$\varphi(A_{i+1} P_2) \in \{c_0, c_1\}$, and properness at $P_2$
+gives $\varphi(P_2)$'s three incident edges colours $\{c_0, c, c_1\}$
+in some order, with $\varphi(A_{i+1} P_2)$ accounting for one of
+$\{c_0, c_1\}$. The remaining two colours, $c$ and $(c_1$ or $c_0)$,
+are split between $\varphi(P_1 P_2)$ and $\varphi(P_2 O_2)$ where
+$O_2$ is $P_2$'s third (outer) neighbour. Properness at $P_1$ (which
+has $\varphi(A_i P_1) = c_1$ already by Case (b) hypothesis) forbids
+$\varphi(P_1 P_2) = c_1$. Hence
+\[
+\varphi(P_1 P_2) \;\in\; \{c, c_0\}.
+\]
+Now $P_2 \in V(K_b) \cup V(K_c)$ as shown, and at $P_2$ the cycle that
+contains $P_2$ uses two specific edges of $P_2$. If $P_2 \in V(K_b)$
+the cycle uses $P_2$'s colour-$c$ and colour-$c_0$ edges; if
+$P_2 \in V(K_c)$ it uses colour-$c$ and colour-$c_1$. In either case
+the cycle at $P_2$ uses $P_2$'s colour-$c$ edge. By the
+preceding paragraph $\varphi(P_1 P_2) \in \{c, c_0\}$, so the
+$\{c, \varphi(P_2 P_1)\}$-Kempe cycle through $P_2$ passes from $P_2$
+to $P_1$. If this cycle is $K_b$, $P_1 \in V(K_b)$; if it is $K_c$
+(which requires $\varphi(P_1 P_2) = c$, since $K_c$ uses
+$\{c, c_1\}$), then $P_1 \in V(K_c)$.
+
+In either case $P_1 \in V(K_b) \cup V(K_c)$.
+\end{proof}
+
+\begin{theorem}[Partial proof of
+Conjecture~\ref{conj:deciding-face}]
+\label{thm:deciding-face-partial}
+If at least one of $n_i, n_{i+1}, n_{i+2}, n_{i+3}, n_{i+4} \in \{5, 6\}$
+(equivalently, $G$ has at least one neighbour of $v$ of degree $\le 6$
+in the parent triangulation), then $\widehat{G}'_{v,i}$ has a
+deciding face --- the corresponding flank face
+$F_{j, j+1}^{\flat}$.
+\end{theorem}
+
+\begin{proof}
+Without loss of generality $n_i \in \{5, 6\}$. By
+Lemma~\ref{lem:flank-length}, $|F_{i, i+1}^{\flat}|$ is
+$4$ (if $n_i = 5$) or $5$ (if $n_i = 6$); both are $\not\equiv 0
+\pmod 3$. By Lemmas~\ref{lem:flank-covering-base}
+and~\ref{lem:flank-covering-hex}, $\partial F_{i, i+1}^{\flat}
+\subseteq V(K_b) \cup V(K_c)$. Hence $F_{i, i+1}^{\flat}$ is a
+deciding face.
+\end{proof}
+
+\begin{remark}[The remaining structural case]
+\label{rem:deciding-face-remaining-case}
+Theorem~\ref{thm:deciding-face-partial} covers every $\widehat{G}'_{v,i}$
+in which $v$ has at least one neighbour of degree $\le 6$ in the
+parent triangulation. The remaining case is $\widehat{G}'_{v,i}$
+where all five neighbours of $v$ have degree $\ge 7$ in $G$ (so
+$n_i \ge 7$ for every $i$); we have not yet found a uniform
+structural argument for this case. Empirically (Remark~\ref{rem:deciding-face-empirical}
+below), the deciding face does always exist, with the merged-side
+face $F_{i+3, i+4}^{\flat}$ (of length $n_{i+3} - 2$) often
+playing the deciding role when the flank lengths $n_i - 1$ are all
+$\equiv 0 \pmod 3$.
+\end{remark}
+
 \begin{remark}[Empirical verification of
 Conjecture~\ref{conj:deciding-face}]
 \label{rem:deciding-face-empirical}