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\begin{solution}{2}
\begin{enumerate}
\setcounter{enumi}{1}
\item See figure \ref{fig:9.3b}.
\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 right=of q2] {$q_4$};
\node[state] (q5) [right=of q3] {$q_5$};
\node[state] (q6) [above right=of q2] {$q_6$};
\draw (q0) -- node[above] {$B/B,R$} ++ (q1);
\draw (q1) edge[loop above] node[above,align=left] {$a/a,R$\\$b/b,R$} (q1);
\draw (q1) -- node[above,align=left] {$B/B,L$\\$X/X,L$\\$Y/Y,L$} ++ (q2);
\draw (q2) -- node[above] {$a/X,R$} (q3);
\draw (q2) -- node[below left] {$b/Y,R$} (q4);
\draw (q3) edge[out=30,in=60,looseness=8] node[above,align=left] {$a/a,R$\\$b/b,R$\\$X/X,R$\\$Y/Y,R$} (q3);
\draw (q3) -- node[above] {$B/a,L$} ++ (q5);
\draw (q4) edge[loop below] node[below,align=left] {$a/a,R$\\$b/b,R$\\$X/X,R$\\$Y/Y,R$} (q4);
\draw (q4) -- node[below right] {$B/b,L$} ++ (q5);
\draw (q5) edge[loop above] node[above,align=left] {$a/a,L$\\$b/b,L$\\$X/X,L$\\$Y/Y,L$} (q5);
\draw (q5) edge[out=270,in=270,looseness=2] node[below] {$B/B,R$} (q1);
\draw (q2) -- node[above left] {$B/B,R$} ++ (q6);
\draw (q6) edge[loop above] node[above,align=left] {$X/a,R$\\$Y/b,R$\\$a/a,L$\\$b/b,L$} (q6);
\end{tikzpicture}
\caption{The state machine of exercise 9.3b.}
\label{fig:9.3b}
\end{figure*}
We read $a$s and $b$s from the input string until the end in $q_1$. The last character is replaced by an $X$ or $Y$ for an $a$ or $b$, respectively, by $q_2$. Then, we place an $a$ or $b$ on the first blank on the right, depending on the last character of the input string. This happens in $q_3$ and $q_4$. We go back to the very beginning of the string in $q_5$, and repeat the process starting in $q_2$, but now stop reading at the beforelast character (and the one before that, and so on). If there are no more characters in the input string, we go to $q_6$ where we replace all $X$s by $a$s and all $Y$s by $b$s, after which we terminate.
\item The idea would be to first move the last character to the right, then the last $2$ characters of the new string, then the last $3$ characters of that string, et cetera., until we are back at the leftmost position. This could be recognised by putting a marker there, if needed.
%Todo draw a diagram
\end{enumerate}
\end{solution}
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