\begin{frame}
    \frametitle{\problemtitle}
    \begin{block}{Problem}
        \begin{itemize}
            \item Place corner-fences into a $100 \times 100$ grid.
            \item Create a connected fenced area of exactly $1 \leq a \leq 5000$ cells.
        \end{itemize}
    \end{block}
    \begin{block}{Solution}
        \begin{itemize}
            \item There are many different constructions of varying implementation difficulty.
            \item Determine minimum $h$ with $(h-1) \cdot 80 \geq a$ (one extra row).
            \item Fence a $h \times 80$ rectangle area.
        \end{itemize}
        \center
        \includegraphics[width=0.3\textwidth]{figure3.pdf}
    \end{block}
\end{frame}

\begin{frame}
    \frametitle{\problemtitle}
    \begin{block}{Case $h = 2$}
        \begin{itemize}
            \item If $h = 2$, then $a \leq 80$, and place a (blue) block as shown below.
            \item Notice that left component is odd, right is even.
            \item Move middle block left or right to increase the relevant component by $2$.
            \item All $a \leq 80$ are achievable this way.
        \end{itemize}
        \center
        \includegraphics[width=0.3\textwidth]{figure1.pdf}
    \end{block}
\end{frame}

\begin{frame}
    \frametitle{\problemtitle}
    \begin{block}{Case $h > 2$}
        \begin{itemize}
            \item By construction, our rectangle has $80 \leq x \leq 160$ cells too much.
            \item Place the right piece (blue) to reduce field by $3$.
            \item Place the left piece (red) to reduce field by $4$.
            \item Use up to $3$ blue pieces, and then as many red pieces as necessary.
            \item One can (easily) show that $80 \leq x \leq 160$ reduction is always possible.
        \end{itemize}
        \center
        \includegraphics[width=0.3\textwidth]{figure2.pdf}
    \end{block}
    % \solvestats
\end{frame}
