\begin{frame}{Exercise}
  \begin{exampleblock}{Understanding the start case}
    \begin{center}
    \begin{tikzpicture}[default,node distance=12mm,->,s/.style={minimum size=5mm}]
      \node (q0) [state,s] {}; \draw ($(q0) + (-8mm,0mm)$) -- (q0); 
      \subnfa{r}{$(q0)+(10mm,0mm)$}
      \node (qf) at ($(rf) + (10mm,0mm)$) [fstate,s] {}; 
  
      \draw (q0) to node [above,pos=.4] {$\lambda$} (rs);
      \draw (rf) to node [above,pos=.4] {$\lambda$} (qf);
  
      \draw (q0) to[out=70,in=120,looseness=.4] node [above,label] {$\lambda$} (qf);
      \draw (qf) to[out=180+60,in=180+130,looseness=.4] node [below,label] {$\lambda$} (q0);
  
      \node [left of=q0,anchor=east,node distance=9mm] {$r^*$:};
    \end{tikzpicture}
    \end{center}
    
    Note that:
    \begin{itemize}
      \item $((a^\ast)\cdot b)^\ast$ shows that the new starting state is needed
      \item $(a\cdot(b^\ast))^\ast$ shows that the new final state is needed
    \end{itemize}
    \bigskip
    
    What goes wrong without introducing the new start/final state?
  \end{exampleblock}
\end{frame}