\begin{frame}{Turing Machine Configuration}
  \begin{block}{}
    A \emph{configuration} $(q,c)$ of a Turing machine consists of
    \begin{itemize}
      \item a state $q \in Q$, and
      \item a function $c : \mathbb{Z} \to \Gamma$, the \emph{tape content}.
    \end{itemize}
    The non-blank positions $\{ z \in \mathbb{Z} \mid c(z) \ne \Box \}$ are finite.
    \smallskip
    
    The head of the machine stands on $c(0)$.
  \end{block}

  \begin{center}
    \def\cellwidth{10mm}
    \def\cellheight{6mm}
    \def\cellfrom{-3}
    \def\cellto{3}
    \begin{tikzpicture}[default,-,thin]
      \draw [draw=none,fill=orange!15] (\cellfrom*\cellwidth - \cellwidth,\cellheight/2) rectangle (\cellto*\cellwidth + \cellwidth,-\cellheight/2);
      
      \foreach \i in {\cellfrom,...,\cellto} {
        \draw ({\cellwidth*(\i-0.5)},\cellheight/2) -- ({\cellwidth*(\i-0.5)},-\cellheight/2);
        \node (cell\i) at ({\cellwidth*\i},0) {};
      }
      \draw ({\cellwidth*(\cellto+0.5)},\cellheight/2) -- ({\cellwidth*(\cellto+0.5)},-\cellheight/2);

      \draw (\cellfrom*\cellwidth - \cellwidth,\cellheight/2) -- (\cellto*\cellwidth + \cellwidth,\cellheight/2);
      \draw (\cellfrom*\cellwidth - \cellwidth,-\cellheight/2) -- (\cellto*\cellwidth + \cellwidth,-\cellheight/2);

      \node [anchor=east] at (\cellfrom*\cellwidth - \cellwidth,0) {$\dots$};
      \node [anchor=west] at (\cellto*\cellwidth + \cellwidth,0) {$\dots$};

      \node (cu) [rectangle,rounded corners=2mm,draw,inner sep=2mm,fill=orange!15] at ($(cell0) + (0,2*\cellheight)$) {$q$};
      \draw [<->] (cu) to ($(cell0) + (0,\cellheight/2)$);
      
      
      \foreach \i/\t in {-3/c(-3),-2/c(-2),-1/c(-1),0/c(0),1/c(1),2/c(2),3/c(3)} {
        \node [scale=0.9] at (cell\i) {$\t$};
      }
    \end{tikzpicture}\vspace{-1ex}
  \end{center}
  \pause
  
  \begin{goal}{}
    Let $n,m \in \mathbb{N}$ (exist for every configuration) such that 
    \begin{talign}
      \forall i < -n.\; c(i) = \Box && \text{ and } &&
      \forall i > m.\; c(i) = \Box 
    \end{talign}
    Then we \emph{denote the configuration by the finite word}
    \begin{talign}
      c(-n) c(-n+1) \cdots c(-1) \;\alert{q}\; c(0) c(1) \cdots c(m)
    \end{talign}
  \end{goal}
\end{frame}