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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.
\end{document}
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