\documentclass{article}
\usepackage{charter}
\usepackage{hyperref}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{automata}
\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions
\begin{document}
\section*{Homework 4}
\begin{enumerate}
\item Why is every finite set of strings a regular language?
\item What is the language recognised by the regular expressions $(\varnothing^*)^*$.
\item If a regular expression $r$ does not contain any occurrence of $\varnothing$,
is it possible for $L(r)$ to be empty?
\item Assume that $s^{-1}$ stands for the operation of reversing a
string $s$. Given the following \emph{reversing} function on regular
expressions
\begin{center}
\begin{tabular}{r@{\hspace{1mm}}c@{\hspace{1mm}}l}
$rev(\varnothing)$ & $\dn$ & $\varnothing$\\
$rev(\epsilon)$ & $\dn$ & $\epsilon$\\
$rev(c)$ & $\dn$ & $c$\\
$rev(r_1 + r_2)$ & $\dn$ & $rev(r_1) + rev(r_2)$\\
$rev(r_1 \cdot r_2)$ & $\dn$ & $rev(r_2) \cdot rev(r_1)$\\
$rev(r^*)$ & $\dn$ & $rev(r)^*$\\
\end{tabular}
\end{center}
and the set
\begin{center}
$Rev\,A \dn \{s^{-1} \;|\; s \in A\}$
\end{center}
prove whether
\begin{center}
$L(rev(r)) = Rev (L(r))$
\end{center}
holds.
\item Give a regular expression over the alphabet $\{a,b\}$ recognising all strings
that do not contain any substring $bb$ and end in $a$.
\item Assume the delimiters for comments are \texttt{$\slash$*} and \texttt{*$\slash$}.
Give a regular expression that can recognise comments
of the form
\begin{center}
\texttt{$\slash$*~\ldots{}~*$\slash$}
\end{center}
where the three dots stand for arbitrary characters, but not comment delimiters.
(Hint: You can assume you are already given a regular expression written \texttt{ALL},
that can recognise any character, and a regular expression \texttt{NOT} that recognises
the complement of a regular expression.)
\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 languages 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 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=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] (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 (Optional) The tokenizer in \texttt{regexp3.scala} takes as
%argument a string and a list of rules. The result is a list of tokens. Improve this tokenizer so
%that it filters out all comments and whitespace from the result.
%\item (Optional) Modify the tokenizer in \texttt{regexp2.scala} so that it
%implements the \texttt{findAll} function. This function takes a regular
%expressions and a string, and returns all substrings in this string that
%match the regular expression.
\end{enumerate}
% explain what is a context-free grammar and the language it generates
%
%
% Define the language L(M) accepted by a deterministic finite automaton M.
%
%
% does (a + b)*b+ and (a*b+) + (b*b+) define the same language
\end{document}
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