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\begin{solution}{5}
\begin{enumerate}
\item See figure \ref{fig:9.5a}.
\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) [below=of q1] {$q_4$};
\node[state,accepting] (q5) [right=of q4] {$q_5$};
\draw (q0) -- node[above] {$B/B,R$} ++ (q1);
\draw (q1) -- node[above] {$a/X,R$} ++ (q2);
\draw (q2) edge[loop above] node[above,align=left] {$a/a,R$\\$Y/Y,R$} (q2);
\draw (q2) -- node[above] {$b/Y,L$} ++ (q3);
\draw (q3) edge[loop right] node[right,align=left] {$a/a,L$\\$Y/Y,L$} (q3);
\draw (q3) edge[bend right, out=300, in=240, looseness=1.3] node[above] {$X/X,R$} (q1);
\draw (q1) -- node[left,align=right] {$B/B,R$\\$Y/Y,R$} ++ (q4);
\draw (q4) edge[loop left] node[left] {$b/b,R$} (q4);
\draw (q4) -- node[above] {$B/B,R$} ++ (q5);
\end{tikzpicture}
\caption{The state machine of exercise 9.5a.}
\label{fig:9.5a}
\end{figure*}
The intuition is: for the first $a$ that we read, mark it with $X$ and mark the next $b$ we see with $Y$. Go back to the beginning and repeat. When we read a $Y$ instead of an $a$, we have finished the $a^i$ part of the string, and we can go to a final state. The machine should check that there is a $b$ for every $a$ and that the $a$s and $b$s are consecutive.
\setcounter{enumi}{2}
\item See figure \ref{fig:9.5c} and \ref{fig:9.5c-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) [below=of q1] {$q_5$};
\node[state,accepting] (q6) [right=of q5] {$q_6$};
\draw (q0) -- node[above] {$B/B,R$} ++ (q1);
\draw (q1) edge[loop above] node[above,align=left] {$b/b,R$\\$X/X,R$} (q1);
\draw (q1) -- node[above] {$a/X,L$} ++ (q2);
\draw (q2) edge[loop above] node[above,align=left] {$b/b,L$\\$X/X,L$} (q2);
\draw (q2) -- node[above] {$B/B,R$} ++ (q3);
\draw (q3) edge[loop above] node[above,align=left] {$a/a,R$\\$X/X,R$} (q3);
\draw (q3) -- node[above] {$b/X,R$} ++ (q4);
\draw (q4) edge[loop right] node[right,align=left] {$a/a,L$\\$b/b,L$\\$X/X,L$} (q4);
\draw (q4) edge[bend right, out=300, in=240, looseness=1.2] node[above] {$B/B,R$} (q1);
\draw (q1) -- node[left] {$B/B,L$} ++ (q5);
\draw (q5) edge[loop left] node[left] {$X/X,L$} (q5);
\draw (q5) -- node[above] {$B/B,R$} ++ (q6);
\end{tikzpicture}
\caption{The state machine of exercise 9.5c.}
\label{fig:9.5c}
\end{figure*}
\begin{figure}[h]
\begin{minted}[bgcolor=mintedbg,tabsize=0,fontsize=\footnotesize]{clean}
tape_9_5_c = [Just c \\ c <- fromString "abbabbabaa"]
ex_9_5_c :: TuringMachine Char
ex_9_5_c = { alphabet = ['a', 'b'],
inputs = ['a', 'b'],
transition = f }
where
f :: Int (Maybe Char) -> TuringMachineMove Char
f 0 Nothing = Step 1 Nothing Right
f 1 (Just 'a') = Step 2 (Just 'X') Left
f 1 (Just c) = Step 1 (Just c) Right
f 1 Nothing = Step 5 Nothing Left
f 2 (Just 'b') = Step 2 (Just 'b') Left
f 2 (Just 'X') = Step 2 (Just 'X') Left
f 2 Nothing = Step 3 Nothing Right
f 3 (Just 'X') = Step 3 (Just 'X') Right
f 3 (Just 'a') = Step 3 (Just 'a') Right
f 3 (Just 'b') = Step 4 (Just 'X') Right
f 4 (Just c) = Step 4 (Just c) Left
f 4 Nothing = Step 1 Nothing Right
f 5 (Just 'X') = Step 5 (Just 'X') Left
f 5 Nothing = Step 6 Nothing Right
f _ _ = Halt
\end{minted}
\caption{The state machine of exercise 9.5c, for use with CleanTuringMachines.}
\label{fig:9.5c-clean}
\end{figure}
The intuition is: find a $b$ for every $a$, mark both with $X$. Repeat until there are no $a$s left. Then check that there are also no $b$s left.
\end{enumerate}
\end{solution}
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