\begin{solution}{6}
    See figure \ref{fig:9.6}.

    \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] (q5) [below=of q4] {$q_5$};
            \node[state] (qf) [right=of q4] {$q_f$};
            \draw (q0) -- node[above] {$B/B,R$} ++ (q1);
            \draw (q1) -- node[above] {$a/X,R$} ++ (q2);
            \draw (q2) edge[loop above] node[above,align=center] {$a/a,R$\\$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$} (q4);
            \draw (q4) -- node[left] {$B/B,R$} ++ (q5);

            \draw (q1) -- node[above right] {$b/b,R$} ++ (qf);
            \draw (q2) -- node[above right,align=left] {$B/B,R$\\$X/X,R$} ++ (qf);
            \draw (q3) -- node[below right,align=left] {$b/b,R$\\$B/B,R$} ++ (qf);
            \draw (q4) -- node[below,align=center] {$a/a,R$\\$X/X,R$\\$Y/Y,R$} ++ (qf);
            \draw (qf) edge[out=300,in=330,looseness=8] node[below right,align=left] {$B/B,R$\\$a/a,R$\\$b/b,R$\\$X/X,R$\\$Y/Y,R$} (qf);
        \end{tikzpicture}
        \caption{The state machine of exercise 9.6.}
        \label{fig:9.6}
    \end{figure*}

    All is needed is to add a state $q_f$ in which we can loop forever for every undefined transition, except those of the final state(s).
\end{solution}