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\section*{Coursework 8 (Scala, Regular Expressions}

This coursework is worth 10\%. It is about regular expressions and
pattern matching. The first part is due on 30 November at 11pm; the
second, more advanced part, is due on 7 December at 11pm. The
second part is not yet included. For the first part you are
asked to implement a regular expression matcher. Make sure the files
you submit can be processed by just calling \texttt{scala
  <<filename.scala>>}.\bigskip

\noindent
\textbf{Important:} Do not use any mutable data structures in your
submission! They are not needed. This excludes the use of
\texttt{ListBuffer}s, for example. Do not use \texttt{return} in your
code! It has a different meaning in Scala, than in Java.  Do not use
\texttt{var}! This declares a mutable variable.  Make sure the
functions you submit are defined on the ``top-level'' of Scala, not
inside a class or object.


\subsection*{Disclaimer!!!!!!!!}

It should be understood that the work you submit represents
your own effort! You have not copied from anyone else. An
exception is the Scala code I showed during the lectures or
uploaded to KEATS, which you can freely use.\bigskip


\subsection*{Part 1 (6 Marks)}

The task is to implement a regular expression matcher that is based on
derivatives of regular expressions. The implementation can deal
with the following regular expressions, which have been predefined
in the file re.scala:

\begin{center}
\begin{tabular}{lcll}
  $r$ & $::=$ & $\ZERO$     & cannot match anything\\
      &   $|$ & $\ONE$      & can only match the empty string\\
      &   $|$ & $c$         & can match a character (in this case $c$)\\
      &   $|$ & $r_1 + r_2$ & can match a string either with $r_1$ or with $r_2$\\
  &   $|$ & $r_1\cdot r_2$ & can match the first part of a string with $r_1$ and\\
          &  & & then the second part with $r_2$\\
      &   $|$ & $r^*$       & can match zero or more times $r$\\
\end{tabular}
\end{center}

\noindent 
Why? Knowing how to match regular expressions and strings fast will
let you solve a lot of problems that vex other humans. Regular
expressions are one of the fastest and simplest ways to match patterns
in text, and are endlessly useful for searching, editing and
analysing text in all sorts of places. However, you need to be
fast, otherwise you will stumble over problems such as recently reported at

{\small
\begin{itemize}
\item[$\bullet$] \url{http://stackstatus.net/post/147710624694/outage-postmortem-july-20-2016}
\item[$\bullet$] \url{https://vimeo.com/112065252}
\item[$\bullet$] \url{http://davidvgalbraith.com/how-i-fixed-atom/}  
\end{itemize}}

\subsection*{Tasks (file re.scala)}

\begin{itemize}
\item[(1a)] Implement a function, called \textit{nullable}, by recursion over
  regular expressions. This function tests whether a regular expression can match
  the empty string.

\begin{center}
\begin{tabular}{lcl}
$\textit{nullable}(\ZERO)$ & $\dn$ & $\textit{false}$\\
$\textit{nullable}(\ONE)$  & $\dn$ & $\textit{true}$\\
$\textit{nullable}(c)$     & $\dn$ & $\textit{false}$\\
$\textit{nullable}(r_1 + r_2)$ & $\dn$ & $\textit{nullable}(r_1) \vee \textit{nullable}(r_2)$\\
$\textit{nullable}(r_1 \cdot r_2)$ & $\dn$ & $\textit{nullable}(r_1) \wedge \textit{nullable}(r_2)$\\
$\textit{nullable}(r^*)$ & $\dn$ & $\textit{true}$\\
\end{tabular}
\end{center}\hfill[1 Mark]

\item[(1b)] Implement a function, called \textit{der}, by recursion over
  regular expressions. It takes a character and a regular expression
  as arguments and calculates the derivative regular expression according
  to the rules:

\begin{center}
\begin{tabular}{lcl}
$\textit{der}\;c\;(\ZERO)$ & $\dn$ & $\ZERO$\\
$\textit{der}\;c\;(\ONE)$  & $\dn$ & $\ZERO$\\
$\textit{der}\;c\;(d)$     & $\dn$ & $\textit{if}\; c = d\;\textit{then} \;\ONE \; \textit{else} \;\ZERO$\\
$\textit{der}\;c\;(r_1 + r_2)$ & $\dn$ & $(\textit{der}\;c\;r_1) + (\textit{der}\;c\;r_2)$\\
$\textit{der}\;c\;(r_1 \cdot r_2)$ & $\dn$ & $\textit{if}\;\textit{nullable}(r_1)$\\
      & & $\textit{then}\;((\textit{der}\;c\;r_1)\cdot r_2) + (\textit{der}\;c\;r_2)$\\
      & & $\textit{else}\;(\textit{der}\;c\;r_1)\cdot r_2$\\
$\textit{der}\;c\;(r^*)$ & $\dn$ & $(\textit{der}\;c\;r)\cdot (r^*)$\\
\end{tabular}
\end{center}

For example given the regular expression $r = (a \cdot b) \cdot c$, the derivatives
w.r.t.~the characters $a$, $b$ and $c$ are

\begin{center}
  \begin{tabular}{lcll}
    $\textit{der}\;a\;r$ & $=$ & $(\ONE \cdot b)\cdot c$ & ($= r'$)\\
    $\textit{der}\;b\;r$ & $=$ & $(\ZERO \cdot b)\cdot c$\\
    $\textit{der}\;c\;r$ & $=$ & $(\ZERO \cdot b)\cdot c$
  \end{tabular}
\end{center}

Let $r'$ stand for the first derivative, then taking the derivatives of $r'$
w.r.t.~the characters $a$, $b$ and $c$ gives

\begin{center}
  \begin{tabular}{lcll}
    $\textit{der}\;a\;r'$ & $=$ & $((\ZERO \cdot b) + \ZERO)\cdot c$ \\
    $\textit{der}\;b\;r'$ & $=$ & $((\ZERO \cdot b) + \ONE)\cdot c$ & ($= r''$)\\
    $\textit{der}\;c\;r'$ & $=$ & $((\ZERO \cdot b) + \ZERO)\cdot c$
  \end{tabular}
\end{center}

One more example: Let $r''$ stand for the second derivative above,
then taking the derivatives of $r''$ w.r.t.~the characters $a$, $b$
and $c$ gives

\begin{center}
  \begin{tabular}{lcll}
    $\textit{der}\;a\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ZERO$ \\
    $\textit{der}\;b\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ZERO$\\
    $\textit{der}\;c\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ONE$
  \end{tabular}
\end{center}

Note, the last derivative can match the empty string, that is it is \textit{nullable}.\\
\mbox{}\hfill\mbox{[1 Mark]}

\item[(1c)] Implement the function \textit{simp}, which recursively
  traverses a regular expression from the inside to the outside, and
  simplifies every sub-regular-expression on the left (see below) to
  the regular expression on the right, except it does not simplify inside
  ${}^*$-regular expressions.

  \begin{center}
\begin{tabular}{l@{\hspace{4mm}}c@{\hspace{4mm}}ll}
$r \cdot \ZERO$ & $\mapsto$ & $\ZERO$\\ 
$\ZERO \cdot r$ & $\mapsto$ & $\ZERO$\\ 
$r \cdot \ONE$ & $\mapsto$ & $r$\\ 
$\ONE \cdot r$ & $\mapsto$ & $r$\\ 
$r + \ZERO$ & $\mapsto$ & $r$\\ 
$\ZERO + r$ & $\mapsto$ & $r$\\ 
$r + r$ & $\mapsto$ & $r$\\ 
\end{tabular}
  \end{center}

  For example the regular expression
  \[(r_1 + \ZERO) \cdot \ONE + ((\ONE + r_2) + r_3) \cdot (r_4 \cdot \ZERO)\]

  simplifies to just $r_1$. 
  \hfill[1 Mark]

\item[(1d)] Implement two functions: The first, called \textit{ders},
  takes a list of characters and a regular expression as arguments, and
  builds the derivative w.r.t.~the list as follows:

\begin{center}
\begin{tabular}{lcl}
$\textit{ders}\;(Nil)\;r$ & $\dn$ & $r$\\
  $\textit{ders}\;(c::cs)\;r$  & $\dn$ &
    $\textit{ders}\;cs\;(\textit{simp}(\textit{der}\;c\;r))$\\
\end{tabular}
\end{center}

The second, called \textit{matcher}, takes a string and a regular expression
as arguments. It builds first the derivatives according to \textit{ders}
and after that tests whether the resulting derivative regular expression can match
the empty string (using \textit{nullable}).
For example the \textit{matcher} will produce true given the
regular expression $(a\cdot b)\cdot c$ and the string $abc$.
\hfill[1 Mark]

\item[(1e)] Implement the function $\textit{replace}\;r\;s_1\;s_2$: it searches
  (from the left to 
right) in the string $s_1$ all the non-empty substrings that match the 
regular expression $r$---these substrings are assumed to be
the longest substrings matched by the regular expression and
assumed to be non-overlapping. All these substrings in $s_1$ matched by $r$
are replaced by $s_2$. For example given the regular expression

\[(a \cdot a)^* + (b \cdot b)\]

\noindent the string $s_1 = aabbbaaaaaaabaaaaabbaaaabb$ and
replacement string $s_2 = c$ yields the string

\[
ccbcabcaccc
\]

\hfill[2 Mark]
\end{itemize}

\textbf{Background} Although easily implementable in Scala, the idea
behind the derivative function might not so easy to be seen. To
understand its purpose better, assume a regular expression $r$ can
match strings of the form $c::cs$ (that means strings which start with
a character $c$ and have some rest $cs$). If you now take the
derivative of $r$ with respect to the character $c$, then you obtain a
regular expressions that can match all the strings $cs$.  In other
words the regular expression $\textit{der}\;c\;r$ can match the same
strings $c::cs$ that can be matched by $r$, except that the $c$ is
chopped off.

Assume now $r$ can match the string $abc$. If you take the derivative
according to $a$ then you obtain a regular expression that can match
$bc$ (it is $abc$ where the $a$ has been chopped off). If you now
build the derivative $\textit{der}\;b\;(\textit{der}\;a\;r))$ you obtain a regular
expression that can match the string "c" (it is "bc" where 'b' is
chopped off). If you finally build the derivative of this according
'c', that is der('c', der('b', der('a', r))), you obtain a regular
expression that can match the empty string. You can test this using
the function nullable, which is what your matcher is doing.

The purpose of the simp function is to keep the regular expression small. Normally the derivative function makes the regular expression bigger (see the SEQ case) and the algorithm would be slower and slower over time. The simp function counters this increase in size and the result is that the algorithm is fast throughout. 
By the way this whole idea is by Janusz Brzozowski who came up with this in 1964 in his PhD thesis. 
https://en.wikipedia.org/wiki/Janusz_Brzozowski_(computer_scientist)



\subsection*{Part 2 (4 Marks)}

Coming soon.

\end{document}


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