From c3f0c43e5d0bb96896c626fa9c66556ad3a18a31 Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Fri, 4 Sep 2015 12:44:53 +0200
Subject: Ex 9.5/9.6

---
 ex9-5.tex | 93 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++---
 1 file changed, 89 insertions(+), 4 deletions(-)

(limited to 'ex9-5.tex')

diff --git a/ex9-5.tex b/ex9-5.tex
index c3a56b7..183213c 100644
--- a/ex9-5.tex
+++ b/ex9-5.tex
@@ -1,14 +1,99 @@
 \begin{solution}{5}
     \begin{enumerate}
-        \item The intuition would be: 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. 
+        \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*}
             
-            %todo draw machine
+            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 The intuition would be: 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.
+        \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*}
 
-            %todo draw machine
+            \begin{figure}[h]
+                \begin{minted}[bgcolor=mintedbg,tabsize=0,fontsize=\footnotesize]{text}
+					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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