110 lines
2.4 KiB
TeX
110 lines
2.4 KiB
TeX
|
\documentclass[12pt]{article}
|
||
|
\usepackage{ntnu}
|
||
|
\usepackage{ntnu-math}
|
||
|
|
||
|
\author{Øystein Tveit}
|
||
|
\title{MA0301 Exercise 12}
|
||
|
|
||
|
\usetikzlibrary{automata, positioning, arrows.meta}
|
||
|
|
||
|
\begin{document}
|
||
|
\ntnuTitle{}
|
||
|
\break{}
|
||
|
|
||
|
\begin{excs}
|
||
|
|
||
|
\exc{}
|
||
|
|
||
|
\[ r = \{a,b\}^* a \{a,b\}^* a \{a,b\}^* a \{a,b\}^* \]
|
||
|
|
||
|
\exc{}
|
||
|
\begin{subexcs}
|
||
|
\subexc{}
|
||
|
\[ \{ab\} \{ab\}^* \]
|
||
|
|
||
|
\subexc{}
|
||
|
\[ a (a | \lambda) b (b | \lambda)\]
|
||
|
|
||
|
\end{subexcs}
|
||
|
|
||
|
\exc{}
|
||
|
|
||
|
\[ M = (Q, \Sigma, \delta, s, F) \]
|
||
|
|
||
|
\begin{align*}
|
||
|
Q &= \{ s_0, s_1, s_2 \} \\
|
||
|
\Sigma &= \{a, b\} \\
|
||
|
\delta &= \begin{Bmatrix}
|
||
|
s_0 \xrightarrow{b} s_1, \\
|
||
|
s_0 \xrightarrow{a} s_2, \\
|
||
|
s_1 \xrightarrow{a,b} s_1, \\
|
||
|
s_2 \xrightarrow{a} s_1, \\
|
||
|
s_2 \xrightarrow{b} s_2
|
||
|
\end{Bmatrix} \\
|
||
|
s &= s_0 \\
|
||
|
F &= \{ s_2 \}
|
||
|
\end{align*}
|
||
|
|
||
|
\includeDiagram[scale=1.6, width=10cm]{diagrams/ex3.tex}
|
||
|
|
||
|
\exc{}
|
||
|
|
||
|
The words $L$ can be described by the regular expression $r$ where
|
||
|
|
||
|
\[ r = a^* b b^* a \{a,b\}^* \]
|
||
|
|
||
|
\exc{}
|
||
|
|
||
|
The words in $L$ can be described by the regular expression $r$ where
|
||
|
|
||
|
\[ r = (a^* b)^3 \{ (a^* b)^4 \} \]
|
||
|
|
||
|
\exc{}
|
||
|
\begin{subexcs}
|
||
|
\subexc{}
|
||
|
\begin{align*}
|
||
|
s_0 &\xrightarrow{a, 0} s_0 \\
|
||
|
s_0 &\xrightarrow{a, 0} s_0 \\
|
||
|
s_0 &\xrightarrow{b, 1} s_3 \\
|
||
|
s_3 &\xrightarrow{b, 0} s_3 \\
|
||
|
s_3 &\xrightarrow{c, 1} s_0 \\
|
||
|
s_0 &\xrightarrow{c, 1} s_2
|
||
|
\end{align*}
|
||
|
|
||
|
The output would be $001011$
|
||
|
|
||
|
\subexc{}
|
||
|
\includeDiagram[scale=1.2, width=13cm]{diagrams/ex6_b.tex}
|
||
|
|
||
|
\end{subexcs}
|
||
|
|
||
|
\exc{}
|
||
|
\begin{subexcs}
|
||
|
\subexc{}
|
||
|
Suppose we have $a \in A, b \in B$
|
||
|
|
||
|
\begin{align*}
|
||
|
AB^* &= \{a, ab, ab^2, ab^3, \ldots \} \\
|
||
|
&= \{a\} \cup \{ab, ab^2, ab^3, \ldots \} \\
|
||
|
&= A \cup \{ab, ab^2, ab^3, \ldots \} \\
|
||
|
&\Rightarrow A \subseteq AB^*
|
||
|
\end{align*}
|
||
|
\qed
|
||
|
|
||
|
\subexc{}
|
||
|
Since $A \subseteq B$, we can rewrite $B$ as $A \cup \overline{A}$ where $\overline{A} = \{b \mid b \in B, b \notin A \}$
|
||
|
|
||
|
\begin{align*}
|
||
|
B^* &= (A \cup \overline{A})^* \\
|
||
|
&= A^* \cap \overline{A}^* \cap B_1, \qquad B_1 = \{(B^*\ a\ B^*\ a_1\ B^*) \vee (B^*\ a_1\ B^*\ a\ B^*) \mid a \in A, a_1 \in \overline{A}\} \\
|
||
|
&\Rightarrow A^* \subseteq B^*
|
||
|
\end{align*}
|
||
|
\qed
|
||
|
|
||
|
|
||
|
\end{subexcs}
|
||
|
|
||
|
|
||
|
|
||
|
\end{excs}
|
||
|
\end{document}
|