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\section*{Homework 4}

\HEADER

\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.

  \solution{

    The property to prove is

    \begin{center}

    $P(r)$: If $r$ does not contain $\ZERO$, then $L(r) \not= \emptyset$.  

  \end{center}

  For this you have to now go through all cases.

  Case $r = 0$: $P(\ZERO)$ says: If $\ZERO$ does not contain $\ZERO$

  then \ldots. The premise is obviously false, so everything follows,

  in particular $L(r) \not= \emptyset$.\medskip

  Case $r = \ONE$ and $r = c$ are similar, just that the premise is

  true, but also $L(\ONE)$ and $L(c)$ are not empty. So we shown

  $L(r) \not= \emptyset$.\medskip

  Case $r = r_1 + r_2$: We know $P(r_1)$ and $P(r_2)$ as IHs. We need to show

  $P(r_1 + r_2)$: If $r_1 + r_2$ does not contain $\ZERO$, then $L(r_1 + r_2) \not= \emptyset$.

  If $r_1 + r_2$ does not contain $\ZERO$, then also $r_1$ does not contain $\ZERO$

  and $r_2$ does not contain $\ZERO$. So we can apply the two IHs $P(r_1)$ and $P(r_2)$,

  which allow us to infer that $L(r_1) \not= \emptyset$ and $L(r_2) \not= \emptyset$.

  But if this is the case, then also $L(r_1 + r_2) \not= \emptyset$, which is what we needed

  to show in this case.\medskip

  The other cases are similar.\bigskip

  This lemma essentially says that for basic regular expressions, if

  they do not match anything at all, they must contain $\ZERO$(s)

  somewhere.

}

\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.

  \solution{

    If $r$ is nullable, then $1 + r \equiv r$. With this you can replace

    \begin{align}

      (1 + r) + r\cdot r & \equiv  r + r\cdot r\\

                         & \equiv  r \cdot (1 + r)\\

                         & \equiv  r \cdot r

    \end{align}  

    where in (2) you pull out the ``factor'' $r$ (because $r_1 \cdot (r_2 + r_3) \equiv r_1 \cdot r_2 + r_1 \cdot r_3$).

  }

\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 Recall the functions \textit{nullable} and

      \textit{zeroable}.  Define recursive functions

      \textit{atmostempty} (for regular expressions that match no

      string or only the empty string), \textit{somechars} (for

      regular expressions that match some non-empty string),

      \textit{infinitestrings} (for regular expressions that can match

      infinitely many strings).

      \solution{

        \textbf{zeroable}: The property is $z(r) \;\text{iff}\; L(r) = \emptyset$:

        \begin{align}

          z(\ZERO) &\dn true\\

          z(\ONE)  &\dn false\\

          z(c)     &\dn false\\

          z(r_1 + r_2) &\dn z(r_1) \wedge z(r_2)\\

          z(r_1 \cdot r_2) &\dn z(r_1) \vee z(r_2)\\

          z(r^*)  &\dn false

        \end{align}\bigskip

        \textbf{atmostempty}: The property is ``either $L(r) = \emptyset$ or $L(r) = \{[]\}$'', which

        is more formally $a(r) \;\text{iff}\; L(r) \subseteq \{[]\}$:

        \begin{align}

          a(\ZERO) &\dn true\\

          a(\ONE)  &\dn true\\

          a(c)     &\dn false\\

          a(r_1 + r_2) &\dn a(r_1) \wedge a(r_2)\\

          a(r_1 \cdot r_2) &\dn z(r_1) \vee z(r_2) \vee (a(r_1) \wedge a(r_2))\\

          a(r^*)  &\dn a(r)

        \end{align}

        For this it is good to remember the regex should either not

        match anything at all, or just the empty string.\bigskip

        \textbf{somechars}: The property is ``$L(r)$ must contain a string which is not the empty string'', which

        is more formally $s(r) \;\text{iff}\; \exists\,s. s \not= [] \wedge s \in L(r)$:

        \begin{align}

          s(\ZERO) &\dn false\\

          s(\ONE)  &\dn false\\

          s(c)     &\dn true\\

          s(r_1 + r_2) &\dn s(r_1) \vee s(r_2)\\

          s(r_1 \cdot r_2) &\dn (\neg z(r_1) \wedge s(r_2)) \;\vee\; (\neg z(r_2) \wedge s(r_1))\\

          s(r^*)  &\dn s(r)

        \end{align}

        Here the interesting case is $r_1 \cdot r_2$ where one essentially has to make sure

        that one of the regexes is not zeroable, because then the resulting regex cannot match any

        string.\bigskip

        \textbf{infinitestrings}: The property is

        $i(r) \;\text{iff}\; L(r)\;\text{is infinite}$:

        \begin{align}

          i(\ZERO) &\dn false\\

          i(\ONE)  &\dn false\\

          i(c)     &\dn false\\

          i(r_1 + r_2) &\dn i(r_1) \vee i(r_2)\\

          i(r_1 \cdot r_2) &\dn (\neg z(r_1) \wedge i(r_2)) \;\vee\; (\neg z(r_2) \wedge i(r_1))\\

          i(r^*)  &\dn \neg a(r)

        \end{align}

        Here the interesting bit is that as soon $r$ can match at least a single string, then $r^*$

        will match infinitely many strings.

        }

\item There are two kinds of automata that are generated for 

  regular expression matching---DFAs and NFAs. (1) Regular expression engines like

  the one in Python generate NFAs.  Explain what is the problem with such

  NFAs and what is the reason why they use NFAs. (2) Regular expression

  engines like the one in Rust generate DFAs. Explain what is the

  problem with these regex engines and also what is the problem with $a^{\{1000\}}$

  in these engines.

%\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}

\end{document}

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