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diff --git a/ex9-3.tex b/ex9-3.tex new file mode 100644 index 0000000..8a4b070 --- /dev/null +++ b/ex9-3.tex @@ -0,0 +1,41 @@ +\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} + |