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\section*{Homework 3}

\begin{enumerate}
\item What is a regular language? Are there alternative ways to define this
  notion? If yes, give an explanation why they define the same notion.

\item Why is every finite set of strings a regular language?

\item Assume you have an alphabet consisting of the letters $a$, $b$
  and $c$ only. (1) Find a regular expression that recognises the two
  strings $ab$ and $ac$. (2) Find a regular expression that matches
  all strings \emph{except} these two strings. Note, you can only use
  regular expressions of the form
      
  \begin{center} $r ::=
    \varnothing \;|\; \epsilon \;|\; c \;|\; r_1 + r_2 \;|\;
    r_1 \cdot r_2 \;|\; r^*$ 
  \end{center}

\item Define the function \textit{zeroable} which takes a regular
  expression as argument and returns a boolean.  The function should
  satisfy the following property:

  \begin{center}
    $\textit{zeroable(r)} \;\text{if and only if}\; L(r) = \varnothing$
  \end{center}

\item Given the alphabet $\{a,b\}$. Draw the automaton that has two
  states, say $q_0$ and $q_1$.  The starting state is $q_0$ and the
  final state is $q_1$. The transition function is given by

  \begin{center}
    \begin{tabular}{l}
      $(q_0, a) \rightarrow q_0$\\
      $(q_0, b) \rightarrow q_1$\\
      $(q_1, b) \rightarrow q_1$
    \end{tabular}
  \end{center}

  What is the language recognised by this automaton?

\item Give a non-deterministic finite automaton that can recognise the
  language $L(a\cdot (a + b)^* \cdot c)$.

\item Given a deterministic finite automata $A(Q, q_0, F, \delta)$,
  define which language is recognised by this automaton. Can you 
  define also the language defined by a non-deterministic automaton?

\item Given the following deterministic finite automata over the
  alphabet $\{a, b\}$, find an automaton that recognises the
  complement language.  (Hint: Recall that for the algorithm from the
  lectures, the automaton needs to be in completed form, that is have
  a transition for every letter from the alphabet.)

  \begin{center}
    \begin{tikzpicture}[scale=2, line width=0.7mm]
      \node[state, initial]        (q0) at ( 0,1) {$q_0$};
      \node[state, accepting]  (q1) at ( 1,1) {$q_1$};
      \path[->] (q0) edge node[above] {$a$} (q1)
                (q1) edge [loop right] node {$b$} ();
    \end{tikzpicture}
  \end{center}

\item Given the following deterministic finite automaton over the
  alphabet $\{0, 1\}$, find the corresponding minimal automaton. In
  case states can be merged, state clearly which states can be merged.

  \begin{center}
    \begin{tikzpicture}[scale=2, line width=0.7mm]
      \node[state, initial]        (q0) at ( 0,1) {$q_0$};
      \node[state]                    (q1) at ( 1,1) {$q_1$};
      \node[state, accepting] (q4) at ( 2,1) {$q_4$};
      \node[state]                    (q2) at (0.5,0) {$q_2$};
      \node[state]                    (q3) at (1.5,0) {$q_3$};
      \path[->] (q0) edge node[above] {$0$} (q1)
                (q0) edge node[right] {$1$} (q2)
                (q1) edge node[above] {$0$} (q4)
                (q1) edge node[right] {$1$} (q2)
                (q2) edge node[above] {$0$} (q3)
                (q2) edge [loop below] node {$1$} ()
                (q3) edge node[left] {$0$} (q4)
                (q3) edge [bend left=95, looseness = 2.2] node [left=2mm] {$1$} (q0)
                (q4) edge [loop right] node {$0, 1$} ();
    \end{tikzpicture}
  \end{center}

%\item Given the following deterministic finite automaton
%
%\begin{center}
%\begin{tikzpicture}[scale=3, line width=0.7mm]
%  \node[state, initial]        (q0) at ( 0,1) {$q_0$};
%  \node[state,accepting]  (q1) at ( 1,1) {$q_1$};
%  \node[state, accepting] (q2) at ( 2,1) {$q_2$};
%  \path[->] (q0) edge node[above] {$b$} (q1)
%                  (q1) edge [loop above] node[above] {$a$} ()
%                  (q2) edge [loop above] node[above] {$a, b$} ()
%                  (q1) edge node[above] {$b$} (q2)
%                  (q0) edge[bend right] node[below] {$a$} (q2)
%                  ;
%\end{tikzpicture}
%\end{center}
%find the corresponding minimal automaton. State clearly which nodes
%can be merged.

\item Given the following non-deterministic finite automaton over the
  alphabet $\{a, b\}$, find a deterministic finite automaton that
  recognises the same language:

  \begin{center}
    \begin{tikzpicture}[scale=3, line width=0.7mm]
      \node[state, initial]        (q0) at ( 0,1) {$q_0$};
      \node[state]                    (q1) at ( 1,1) {$q_1$};
      \node[state, accepting] (q2) at ( 2,1) {$q_2$};
      \path[->] (q0) edge node[above] {$a$} (q1)
                (q0) edge [loop above] node[above] {$b$} ()
                (q0) edge [loop below] node[below] {$a$} ()
                (q1) edge node[above] {$a$} (q2);
    \end{tikzpicture}
  \end{center}

\item Given the following finite deterministic automaton over the alphabet $\{a, b\}$:

  \begin{center}
    \begin{tikzpicture}[scale=2, line width=0.5mm]
      \node[state, initial, accepting]        (q0) at ( 0,1) {$q_0$};
      \node[state, accepting]                    (q1) at ( 1,1) {$q_1$};
      \node[state] (q2) at ( 2,1) {$q_2$};
      \path[->] (q0) edge[bend left] node[above] {$a$} (q1)
                (q1) edge[bend left] node[above] {$b$} (q0)
                (q2) edge[bend left=50] node[below] {$b$} (q0)
                (q1) edge node[above] {$a$} (q2)
                (q2) edge [loop right] node {$a$} ()
                (q0) edge [loop below] node {$b$} ()
            ;
    \end{tikzpicture}
  \end{center}

  Give a regular expression that can recognise the same language as
  this automaton. (Hint: If you use Brzozwski's method, you can assume
  Arden's lemma which states that an equation of the form $q = q\cdot r + s$
  has the unique solution $q = s \cdot r^*$.)
\end{enumerate}

\end{document}

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