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\begin{document}
\title{A List Analogue of Equitable Coloring\\ \large Kostochka,
  Pelsmajer, West. J. Graph Th. Vol. 44 No. 3, Nov. 2003}
\author{Dan Ports}
\maketitle


\section{Some preliminary definitions}
\label{sec:definitions}

\begin{description}
\item[list assignment] a set $L(v)$ of allowable colors for every
  vertex $v$ in a graph
\item[$k$-uniform] a list assignment $L$ for a graph $G$ satisfying
  $\abs{L(v)} = k$ for all $v \in V(G)$
\item[equitably $L$-colorable] a $L$-coloring of $G$ exists such that
  each color appears at at most $\left\lceil\frac{n}{k}\right\rceil$
  vertices
\item[equitably $K$-choosable] a graph that is equitably $L$-colorable
  for all $k$-uniform list assignments $L$
\end{description}


\section{The theorem}
\label{sec:theorems}

\begin{description}
\item[\large Theorem 1] \large If $G$ is a graph and $k \ge \max\{\Delta(G),
  \frac{n(G)}{2}\}$, then $G$ is equitably $k$-choosable unless $G$
  contains $K_{k+1}$, or $G$ is $K_{k,k}$ with $k$ odd
\end{description}

\section{More definitions, and some claims}
\label{sec:claims}


  \begin{itemize}
  \item $U = \{f(v) | v \in V(G) - \{y\}\}$
  \item $U_i = $ the set of colors used $i$ times by $f$ ($i = 1$ or
  $2$)
\item $N(v) = \{u: uv \in E(G)\}$
\item $N[v] = N(v) \cup \{v\}$
\item \textbf{switching on $x$:} giving color $f(x)$ to $y$ and making
  $x$ uncolorable
\item \textbf{switchable:} $x$ is switchable if switching on $x$ gives
  a 2-bounded $L$-coloring of $G-x$
\item $Z = \{x \in V(G) \; |\; f(x) \in U_1 \cap L(y)\}$
\item $Z' = \{x \in V(G) \; |\; f(x) \in U_2 \cap L(y)\}$
\item \large{\textbf{Claim 1} If a vertex $x$ is switchable,
  then $L(x) \subseteq U$. Also, $Z \subseteq N(y)$.}
\item \large{\textbf{Claim 2} If $u$ and $v$ are vertices such that
    $u$ is switchable and $v \in Z$, then $uv \in E(G)$. In
    particular, $Z \subseteq N[u]$.}
\item \large{\textbf{Claim 3} $Z$ is non-empty. If $z \in Z$, then
  $d(z) = k$, and $x \in N(z) - \{y\}$ iff $x$ is switchable. $d(y) =
  k$, $f(N(y)) \subseteq L(y)$, and $\abs{N(y) \cap Z'} =
  \frac{\abs{Z'}}{2}$.}
\item \large{\textbf{Claim 4} If $x$ is a switchable vertex of $Z'$,
    then $d(x) = k$ and $f(N(x) - \{y\}) \subseteq L(x)$, and $N[x]$
    contains all switchable vertices in $Z' \cap N(y)$}
\end{itemize}

\end{document}