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\begin{solution}{4}
See figure \ref{fig:9.4} and \ref{fig:9.4-clean}.
\begin{figure*}[p]
\centering
\begin{tikzpicture}[->, node distance=2cm]
\node[state,initial] (q0) {$q_0$};
\node[state] (q1) [right=of q0] {$q_1$};
\node[state] (q2) [right=of q1] {$q_2$};
\node[state] (q3) [right=of q2] {$q_3$};
\node[state] (q4) [right=of q3] {$q_4$};
\node[state] (q5) [right=of q4] {$q_5$};
\draw (q0) -- node[above] {$B/B,R$} ++ (q1);
\draw (q1) -- node[above] {$a/a,R$} ++ (q2);
\draw (q2) -- node[above] {$a/a,R$} ++ (q3);
\draw (q3) -- node[above] {$a/a,R$} ++ (q4);
\draw (q4) -- node[above] {$c/c,L$} ++ (q5);
\draw (q1) edge[loop above] node[above,align=left] {$B/B,R$\\$b/b,R$\\$c/c,R$} (q1);
\draw (q2) edge[bend left] node[below,align=left] {$B/B,R$\\$b/b,R$\\$c/c,R$} (q1);
\draw (q3) edge[bend right] node[above,align=left] {$B/B,R$\\$b/b,R$\\$c/c,R$} (q1);
\draw (q4) edge[bend left] node[below,align=left] {$B/B,R$\\$b/b,R$} (q1);
\draw (q4) edge[loop above] node[above,align=left] {$a/a,R$} (q4);
\draw (q5) edge[loop above] node[above,align=left] {$a/a,L$\\$b/b,L$\\$c/c,L$} (q5);
\end{tikzpicture}
\caption{The state machine of exercise 9.4.}
\label{fig:9.4}
\end{figure*}
\begin{figure}[h]
\begin{minted}[bgcolor=mintedbg,tabsize=0,fontsize=\footnotesize]{clean}
tape_9_4 = [Just c \\ c <- fromString "baaabbaaacabab"]
ex_9_4 :: TuringMachine Char
ex_9_4 = { alphabet = ['a', 'b', 'c'],
inputs = ['a', 'b', 'c'],
transition = f }
where
f :: Int (Maybe Char) -> TuringMachineMove Char
f 0 Nothing = Step 1 Nothing Right
f 1 (Just 'a') = Step 2 (Just 'a') Right
f 1 c = Step 1 c Right
f 2 (Just 'a') = Step 3 (Just 'a') Right
f 2 c = Step 1 c Right
f 3 (Just 'a') = Step 4 (Just 'a') Right
f 3 c = Step 1 c Right
f 4 (Just 'a') = Step 4 (Just 'a') Right
f 4 (Just 'c') = Step 5 (Just 'c') Left
f 4 c = Step 1 c Right
f 5 (Just c) = Step 5 (Just c) Left
f _ _ = Halt
\end{minted}
\caption{The state machine of exercise 9.4, for use with CleanTuringMachines.}
\label{fig:9.4-clean}
\end{figure}
The intuition is that in state $q_i$ we have read $i-1$ consecutive $a$s, for $i\in\{1,2,3,4\}$, or even more in the case of $q_4$. Hence, the only possibility to let the machine terminate should be when we read a $c$ in $q_4$. This is done in $q_5$, which also has the function of bringing back the tape head to position $0$.
\end{solution}
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