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\section*{Handout 2}

Having specified what problem our matching algorithm, \pcode{match},

is supposed to solve, namely for a given regular expression $r$ and

string $s$ answer \textit{true} if and only if

\[

s \in L(r)

\]

\noindent we can look at an algorithm to solve this problem.

Clearly we cannot use the function $L$ directly for this,

because in general the set of strings $L$ returns is infinite

(recall what $L(a^*)$ is). In such cases there is no way we

can implement an exhaustive test for whether a string is

member of this set or not. Before we come to the matching 

algorithm, lets have a closer look at what it means when

two regular expressions are equivalent.

\subsection*{Regular Expression Equivalences}

\subsection*{Matching Algorithm}

The algorithm we will 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@ {}}

$nullable(\varnothing)$      & $\dn$ & $f\!\/alse$\\

$nullable(\epsilon)$           & $\dn$ &  $true$\\

\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

Note on the left-hand side we have a function we can implement; on the

right we have its specification.

The other function of our matching algorithm calculates 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. Be careful 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 strips 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 (that is $L(r)$),

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.

If we want to find out whether the string $"abc"$ is matched by the

regular expression $r$ then we can iteratively apply $Der$ as follows

\begin{enumerate}

\item $Der\,a\,(L(r))$

\item $Der\,b\,(Der\,a\,(L(r)))$

\item $Der\,c\,(Der\,b\,(Der\,a\,(L(r))))$

\end{enumerate}

\noindent

In the last step we need to test whether the empty string is in the

set. Our matching algorithm will work similarly, just using regular

expression instead of sets. For this 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.

\subsection*{The Matching Algorithm in Scala}

\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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