\documentclass{article}+
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\usepackage{charter}+
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\usepackage{hyperref}+
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\usepackage{amssymb}+
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\usepackage{amsmath}+
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+
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\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions+
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+
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\begin{document}+
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+
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\section*{Homework 4}+
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+
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\begin{enumerate}+
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\item Why is every finite set of strings a regular language?+
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+
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\item What is the language recognised by the regular expressions $(\varnothing^*)^*$.+
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+
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\item If a regular expression $r$ does not contain any occurrence of $\varnothing$ +
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is it possible for $L(r)$ to be empty?+
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+
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\item Assume that $s^{-1}$ stands for the operation of reversing a+
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string $s$. Given the following \emph{reversing} function on regular +
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expressions+
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+
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\begin{center}+
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\begin{tabular}{r@{\hspace{1mm}}c@{\hspace{1mm}}l}+
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$rev(\varnothing)$   & $\dn$ & $\varnothing$\\+
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$rev(\epsilon)$         & $\dn$ & $\epsilon$\\+
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$rev(c)$                      & $\dn$ & $c$\\+
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$rev(r_1 + r_2)$        & $\dn$ & $rev(r_1) + rev(r_2)$\\+
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$rev(r_1 \cdot r_2)$  & $\dn$ & $rev(r_2) \cdot rev(r_1)$\\+
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$rev(r^*)$                   & $\dn$ & $rev(r)^*$\\+
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\end{tabular}+
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\end{center}+
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+
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+
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and the set+
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+
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\begin{center}+
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$Rev\,A \dn \{s^{-1} \;|\; s \in A\}$+
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\end{center}+
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+
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prove whether+
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+
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\begin{center}+
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$L(rev(r)) = Rev (L(r))$+
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\end{center}+
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+
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holds.+
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+
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\item Give a regular expression over the alphabet $\{a,b\}$ recognising all strings +
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that do not contain any substring $bb$ and end in $a$.+
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+
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\item Assume the delimiters for comments are \texttt{$\slash$*} and \texttt{*$\slash$}.+
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Give a regular expression that can recognise comments+
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of the form +
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+
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\begin{center}+
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\texttt{$\slash$*~\ldots{}~*$\slash$} +
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\end{center}+
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+
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where the three dots stand for arbitrary characters, but not comment delimiters.+
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(Hint: You can assume you are already given a regular expression written \texttt{ALL},+
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that can recognise any character, and a regular expression \texttt{NOT} that recognises+
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the complement of a regular expression.)+
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+
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\item Given the alphabet $\{a,b\}$. Draw the automaton that has two states, say  $q_0$ and $q_1$.+
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The starting state is $q_0$ and the final state is $q_1$. The transition+
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function is given by+
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+
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\begin{center}+
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\begin{tabular}{l}+
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$(q_0, a) \rightarrow q_0$\\+
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$(q_0, b) \rightarrow q_1$\\+
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$(q_1, b) \rightarrow q_1$+
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\end{tabular}+
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\end{center}+
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+
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What is the languages recognised by this automaton?+
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+
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\item Give a deterministic finite automaton that can recognise +
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the language $L(a^*\cdot b\cdot b^*)$. +
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+
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+
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\item (Optional) The tokenizer in \texttt{regexp3.scala} takes as+
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argument a string and a list of rules. The result is a list of tokens. Improve this tokenizer so +
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that it filters out all comments and whitespace from the result.+
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+
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\item (Optional) Modify the tokenizer in \texttt{regexp2.scala} so that it+
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implements the \texttt{findAll} function. This function takes a regular+
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expressions and a string, and returns all substrings in this string that +
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match the regular expression.+
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\end{enumerate}+
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+
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% explain what is a context-free grammar and the language it generates +
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%+
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%+
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% Define the language L(M) accepted by a deterministic finite automaton M.+
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%+
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%+
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% does (a + b)*b+ and (a*b+) + (b*b+) define the same language+
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+
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+
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\end{document}+
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+
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%%% Local Variables: +
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%%% mode: latex+
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%%% TeX-master: t+
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%%% End: +
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