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\section*{Homework 4}

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\begin{enumerate}

\item Given the regular expressions

\begin{center}

\begin{tabular}{ll}    

  1) & $(ab + a)\cdot (\ONE + b)$\\

  2) & $(aa + a)^*$\\

\end{tabular}

\end{center}

there are several values for how these regular expressions can

recognise the strings (for 1) $ab$ and (for 2) $aaa$. Give in each case

\emph{all} the values and indicate which one is the POSIX value.

\item If a regular expression $r$ does not contain any occurrence of $\ZERO$,  

is it possible for $L(r)$ to be empty? Explain why, or give a proof.

\item Define the tokens and regular expressions for a language

  consisting of numbers, left-parenthesis $($, right-parenthesis $)$,

  identifiers and the operations $+$, $-$ and $*$. Can the following

  strings in this language be lexed?

  \begin{itemize}

  \item $(a + 3) * b$

  \item $)()++ -33$

  \item $(a / 3) * 3$

  \end{itemize}

  In case they can, can you give the corresponding token

  sequences.

\item Assume $r$ is nullable. Show that

  \[ 1 + r + r\cdot r \;\equiv\; r\cdot r

  \]

  holds.

\item \textbf{(Deleted)} 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}

      $rev(\ZERO)$   & $\dn$ & $\ZERO$\\

      $rev(\ONE)$         & $\dn$ & $\ONE$\\

      $rev(c)$                      & $\dn$ & $c$\\

      $rev(r_1 + r_2)$        & $\dn$ & $rev(r_1) + rev(r_2)$\\

      $rev(r_1 \cdot r_2)$  & $\dn$ & $rev(r_2) \cdot rev(r_1)$\\

      $rev(r^*)$                   & $\dn$ & $rev(r)^*$\\

    \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 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 Simplify the regular expression

\[

(\ZERO \cdot (b \cdot c)) + 

((\ZERO \cdot c) + \ONE)

\]

      Does simplification always preserve the meaning of a

      regular expression? 

\item The Sulzmann \& Lu algorithm contains the function

      $mkeps$ which answers how a regular expression can match

      the empty string. What is the answer of $mkeps$ for the

      regular expressions:    

  \[

  \begin{array}{l}

  (\ZERO \cdot (b \cdot c)) + 

  ((\ZERO \cdot c) + \ONE)\\

    (a + \ONE) \cdot (\ONE + \ONE)\\

    a^*

  \end{array}

  \]

\item What is the purpose of the record regular expression in

      the Sulzmann \& Lu algorithm?

%\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.

\item \POSTSCRIPT  

\end{enumerate}

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