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\begin{document}
\section*{Handout 2}
Having specified what problem our matching algorithm, $match$, is supposed to solve, namely
for a given regular expression $r$ and string $s$ answer $true$ if and only if
\[
s \in L(r)
\]
\noindent
Clearly we cannot use the function $L$ directly in order to solve this problem, because in general
the set of strings $L$ returns is infinite (recall what $L(a^*)$ is). In such cases there is no algorithm
then can test exhaustively, whether a string is member of this set.
The algorithm we define below consists of two parts. One is the function $nullable$ which takes a
regular expression as argument and decides whether it can match the empty string (this means it returns a
boolean). This can be easily defined recursively as follows:
\begin{center}
\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}}
$nullable(\varnothing)$ & $\dn$ & $f\!\/alse$\\
$nullable(\epsilon)$ & $\dn$ & $true$\\
$nullable (c)$ & $\dn$ & $f\!alse$\\
$nullable (r_1 + r_2)$ & $\dn$ & $nullable(r_1) \vee nullable(r_2)$\\
$nullable (r_1 \cdot r_2)$ & $\dn$ & $nullable(r_1) \wedge nullable(r_2)$\\
$nullable (r^*)$ & $\dn$ & $true$ \\
\end{tabular}
\end{center}
\noindent
The idea behind this function is that the following property holds:
\[
nullable(r) \;\;\text{if and only if}\;\; ""\in L(r)
\]
\noindent
On the left-hand side we have a function we can implement; on the right we have its specification.
The other function is calculating a \emph{derivative} of a regular expression. This is a function
which will take a regular expression, say $r$, and a character, say $c$, as argument and return
a new regular expression. Beware that the intuition behind this function is not so easy to grasp on first
reading. Essentially this function solves the following problem: if $r$ can match a string of the form
$c\!::\!s$, what does the regular expression look like that can match just $s$. The definition of this
function is as follows:
\begin{center}
\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {\hspace{-10mm}}l@ {}}
$der\, c\, (\varnothing)$ & $\dn$ & $\varnothing$ & \\
$der\, c\, (\epsilon)$ & $\dn$ & $\varnothing$ & \\
$der\, c\, (d)$ & $\dn$ & if $c = d$ then $\epsilon$ else $\varnothing$ & \\
$der\, c\, (r_1 + r_2)$ & $\dn$ & $der\, c\, r_1 + der\, c\, r_2$ & \\
$der\, c\, (r_1 \cdot r_2)$ & $\dn$ & if $nullable (r_1)$\\
& & then $(der\,c\,r_1) \cdot r_2 + der\, c\, r_2$\\
& & else $(der\, c\, r_1) \cdot r_2$\\
$der\, c\, (r^*)$ & $\dn$ & $(der\,c\,r) \cdot (r^*)$ &
\end{tabular}
\end{center}
\noindent
The first two clauses can be rationalised as follows: recall that $der$ should calculate a regular
expression, if the ``input'' regular expression can match a string of the form $c\!::\!s$. Since neither
$\varnothing$ nor $\epsilon$ can match such a string we return $\varnothing$. In the third case
we have to make a case-distinction: In case the regular expression is $c$, then clearly it can recognise
a string of the form $c\!::\!s$, just that $s$ is the empty string. Therefore we return the $\epsilon$-regular
expression. In the other case we again return $\varnothing$ since no string of the $c\!::\!s$ can be matched.
The $+$-case is relatively straightforward: all strings of the form $c\!::\!s$ are either matched by the
regular expression $r_1$ or $r_2$. So we just have to recursively call $der$ with these two regular
expressions and compose the results again with $+$. The $\cdot$-case is more complicated:
if $r_1\cdot r_2$ matches a string of the form $c\!::\!s$, then the first part must be matched by $r_1$.
Consequently, it makes sense to construct the regular expression for $s$ by calling $der$ with $r_1$ and
``appending'' $r_2$. There is however one exception to this simple rule: if $r_1$ can match the empty
string, then all of $c\!::\!s$ is matched by $r_2$. So in case $r_1$ is nullable (that is can match the
empty string) we have to allow the choice $der\,c\,r_2$ for calculating the regular expression that can match
$s$. The $*$-case is again simple: if $r^*$ matches a string of the form $c\!::\!s$, then the first part must be
``matched'' by a single copy of $r$. Therefore we call recursively $der\,c\,r$ and ``append'' $r^*$ in order to
match the rest of $s$.
Another way to rationalise the definition of $der$ is to consider the following operation on sets:
\[
Der\,c\,A\;\dn\;\{s\,|\,c\!::\!s \in A\}
\]
\noindent
which essentially transforms a set of strings $A$ by filtering out all strings that do not start with $c$ and then
strip off the $c$ from all the remaining strings. For example suppose $A = \{"f\!oo", "bar", "f\!rak"\}$ then
\[
Der\,f\,A = \{"oo", "rak"\}\quad,\quad
Der\,b\,A = \{"ar"\} \quad \text{and} \quad
Der\,a\,A = \varnothing
\]
\noindent
Note that in the last case $Der$ is empty, because no string in $A$ starts with $a$. With this operation we can
state the following property about $der$:
\[
L(der\,c\,r) = Der\,c\,(L(r))
\]
\noindent
This property clarifies what regular expression $der$ calculates, namely take the set of strings
that $r$ can match ($L(r)$), filter out all strings not starting with $c$ and strip off the $c$ from the
remaining strings---this is exactly the language that $der\,c\,r$ can match.
For our matching algorithm we need to lift the notion of derivatives from characters to strings. This can be
done using the following function, taking a string and regular expression as input and a regular expression
as output.
\begin{center}
\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {\hspace{-10mm}}l@ {}}
$der\!s\, []\, r$ & $\dn$ & $r$ & \\
$der\!s\, (c\!::\!s)\, r$ & $\dn$ & $der\!s\,s\,(der\,c\,r)$ & \\
\end{tabular}
\end{center}
\noindent
Having $ders$ in place, we can finally define our matching algorithm:
\[
match\,s\,r = nullable(ders\,s\,r)
\]
\noindent
We claim that
\[
match\,s\,r\quad\text{if and only if}\quad s\in L(r)
\]
\noindent
holds, which means our algorithm satisfies the specification. This algorithm was introduced by
Janus Brzozowski in 1964. Its main attractions are simplicity and being fast, as well as
being easily extendable for other regular expressions such as $r^{\{n\}}$, $r^?$, $\sim{}r$ and so on.
\begin{figure}[p]
{\lstset{language=Scala}\texttt{\lstinputlisting{../progs/app5.scala}}}
{\lstset{language=Scala}\texttt{\lstinputlisting{../progs/app6.scala}}}
\caption{Scala implementation of the nullable and derivatives functions.}
\end{figure}
\end{document}
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