chore: SAVEPOINT

paper.tex

@@ -258,6 +258,44 @@ Let $\psi \in \Phi_4(G'_{ij})$, so $|\psi(N_{ij}(v_0))| = 4$. By definition of t
 The first equality expands $N(v_0)$; the second uses $\widetilde{\psi}(a_i) = \widetilde{\psi}(a_j) = \psi(a_{ij})$ and $\widetilde{\psi}(u) = \psi(u)$ for $u \notin \{a_i, a_j\}$; the third uses $N_{ij}(v_0) = \{a_{ij}\} \cup (N(v_0) \setminus \{a_i,a_j\})$. Therefore $|\widetilde{\psi}(N(v_0))| = |\psi(N_{ij}(v_0))| = 4$. Since also $\widetilde{\psi} \in \Phi$ by the previous lemma, we conclude $\widetilde{\psi} \in \Phi_4$.
 \end{proof}
 
+\subsection{Locked Colorings}
+
+\begin{definition}
+Let $a_i, a_j \in N(v_0)$ be non-adjacent in $G'$ (where $G' = G - \{v_0, v_1\}$ as in Section 2.1). A coloring $\phi \in \Phi$ is \emph{locked relative to $\{a_i, a_j\}$} if
+\[
+  |\phi(N_{G'}(a_i))| > 2 \quad \text{or} \quad |\phi(N_{G'}(a_j))| > 2.
+\]
+Denote the set of all such colorings by $\Lambda_{ij} \subseteq \Phi$.
+\end{definition}
+
+Intuitively, a coloring is locked relative to $\{a_i, a_j\}$ when the neighborhood of at least one of the two vertices is colored with enough distinct colors to obstruct a Kempe chain swap that would free a color for $v_0$.
+
+\begin{lemma}
+Let $G'_{ij} \in \mathcal{M}$ and let $\psi \in \Phi(G'_{ij})$. If $|\psi(N_{G'_{ij}}(v_1))| \leq 3$, then $\psi$ is the induced coloring of some locked coloring in $\Phi$.
+\end{lemma}
+
+\begin{proof}
+The extension $\widetilde{\psi} \in \Phi$ (by the lemma in Section 3.1). Suppose for contradiction that $\widetilde{\psi} \notin \Lambda_{ij}$, i.e., $\widetilde{\psi}$ is not locked relative to $\{a_i, a_j\}$. Then
+\[
+  |\widetilde{\psi}(N_{G'}(a_i))| \leq 2 \quad \text{and} \quad |\widetilde{\psi}(N_{G'}(a_j))| \leq 2.
+\]
+Since $\widetilde{\psi}$ agrees with $\psi$ on all vertices of $G'_{ij}$, the neighborhoods of $a_i$ and $a_j$ in $G'$ each use at most 2 colors under $\widetilde{\psi}$. With so few colors in each neighborhood, a Kempe chain swap can be performed in $\widetilde{\psi}$ to give $a_i$ and $a_j$ the same color, freeing a fourth color for $v_0$. Simultaneously, since $|\psi(N_{G'_{ij}}(v_1))| \leq 3$, the vertex $v_1$ can be assigned the remaining color not used by its neighbors. Together these assignments extend $\widetilde{\psi}$ to a proper 4-coloring of all of $G$, contradicting the minimality of $G$ as a counterexample. Therefore $\widetilde{\psi} \in \Lambda_{ij}$, and $\psi$ is the induced coloring of the locked coloring $\widetilde{\psi}$.
+\end{proof}
+
+\begin{lemma}
+  Now prove that all colorings of every merged graph relative to $\{a_i, a_j\}$ must be a locked coloring.
+\end{lemma}
+
+\begin{proof}
+\end{proof}
+
+\begin{lemma}
+If all $\phi \in \Phi$ are locked colorings with respect to all pairs of non adjacent vertices $\{a_i, a_j\} \in N(v_0)$, then all colorings of all merged graphs with respect to ${a_k, a_l} \in N(v_1)$ require 4 colors for $N_(v_0)$.
+\end{lemma}
+
+\begin{proof}
+\end{proof}
+
 \end{document}
 
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