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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 Construct a regular expression that can validate passwords. A
password should be at least 8 characters long and consist of upper-
and lower-case letters and digits. It should contain at least a
single lower-case letter, at least a single upper-case letter and at
least a single digit. If possible ise the intersection regular
expression from CW1, written $\_\&\_$, the bounded regular
expressions; you can also assume a regular expression written
\texttt{ALL} that can match any character.
\solution{
You can build-up the different constraints separately and then
use the intersection operator:
\begin{center}
\begin{tabular}{lll}
$ALL^{\{8..\}}$ & \;\&\; & $(ALL^*\cdot [a-z]\cdot ALL^*)$\\
& \;\&\; & $(ALL^*\cdot [A-Z]\cdot ALL^*)$\\
& \;\&\; & $(ALL^*\cdot [0-9]\cdot ALL^*)$\\
\end{tabular}
\end{center}
}
\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}
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