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This coursework is worth 10\%. It is about regular expressions,

November at 11pm; the second, more advanced part, is due on 21

\noindent

\textbf{Important:}

\begin{itemize}

\item Make sure the files you submit can be processed by just calling\\

  \mbox{\texttt{scala <<filename.scala>>}} on the commandline. Use the

  template files provided and do not make any changes to arguments of

  functions or to any types. You are free to implement any auxiliary

  function you might need.

\item Do not use any mutable data structures in your

submissions! They are not needed. This means you cannot create new 

\texttt{Array}s or \texttt{ListBuffer}s, for example. 

\item Do not use \texttt{return} in your code! It has a different

  meaning in Scala, than in Java.

\item Do not use \texttt{var}! This declares a mutable variable. Only

  use \texttt{val}!

\item Do not use any parallel collections! No \texttt{.par} therefore!

  Our testing and marking infrastructure is not set up for it.

\end{itemize}

\noindent

Also note that the running time of each part will be restricted to a

maximum of 360 seconds on my laptop

\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

\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 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 data in all

sorts of places (for example analysing network traffic in order to

detect security breaches). 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}}

\subsubsection*{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 it either returns true or false.

\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}

    $\textit{der}\;a\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ZERO$ \\

    $\textit{der}\;b\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ZERO$\\

    $\textit{der}\;c\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ONE$ &

    (is $\textit{nullable}$)                      

  \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

  \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

  \[(r_1 + \ZERO) \cdot \ONE + ((\ONE + r_2) + r_3) \cdot (r_4 \cdot \ZERO)\]

  seen as trees and there are several methods for traversing

  like $u + \ldots + (1 \cdot x) - \ldots (z + (y \cdot 0)) \ldots$

  and simplification rules that looked very similar to rules

  above. You would simplify such numerical expressions by replacing

  for example the $y \cdot 0$ by $0$, or $1\cdot x$ by $x$, and then

  look whether more rules are applicable. If you organise the

  simplification in an inside-out fashion, it is always clear which

\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}

Note that this function is different from \textit{der}, which only

takes a single character.

The second function, 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

  over regular expressions. If a regular expression is seen as a tree,

  then \textit{size} should return the number of nodes in such a

  tree. Therefore this function is defined as follows:

\begin{center}

\begin{tabular}{lcl}

$\textit{size}(\ZERO)$ & $\dn$ & $1$\\

$\textit{size}(\ONE)$  & $\dn$ & $1$\\

$\textit{size}(c)$     & $\dn$ & $1$\\

$\textit{size}(r_1 + r_2)$ & $\dn$ & $1 + \textit{size}(r_1) + \textit{size}(r_2)$\\

$\textit{size}(r_1 \cdot r_2)$ & $\dn$ & $1 + \textit{size}(r_1) + \textit{size}(r_2)$\\

$\textit{size}(r^*)$ & $\dn$ & $1 + \textit{size}(r)$\\

\end{tabular}

\end{center}

\subsection*{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

that can match all the strings $cs$.  In other words, the regular

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

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

\begin{center}\small

\url{https://en.wikipedia.org/wiki/Janusz_Brzozowski_(computer_scientist)}

\end{center}

\end{document}

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