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Diffstat (limited to 'ex9-6.tex')
| -rw-r--r-- | ex9-6.tex | 34 |
1 files changed, 33 insertions, 1 deletions
@@ -1,4 +1,36 @@ \begin{solution}{6} - This is not much different from 5a. %todo + 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} |