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\documentclass{article}
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\usepackage{../style}
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%%\usepackage{../langs}
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\begin{document}
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\section*{Coursework 8 (Scala, Regular Expressions}
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This coursework is worth 10\%. It is about regular expressions and
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pattern matching. The first part is due on 30 November at 11pm; the
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second, more advanced part, is due on 7 December at 11pm. The
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second part is not yet included. For the first part you are
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asked to implement a regular expression matcher. Make sure the files
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you submit can be processed by just calling \texttt{scala
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<<filename.scala>>}.\bigskip
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\noindent
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\textbf{Important:} Do not use any mutable data structures in your
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submission! They are not needed. This excludes the use of
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\texttt{ListBuffer}s, for example. Do not use \texttt{return} in your
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code! It has a different meaning in Scala, than in Java. Do not use
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\texttt{var}! This declares a mutable variable. Make sure the
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functions you submit are defined on the ``top-level'' of Scala, not
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inside a class or object.
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\subsection*{Disclaimer!!!!!!!!}
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It should be understood that the work you submit represents
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your own effort! You have not copied from anyone else. An
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exception is the Scala code I showed during the lectures or
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uploaded to KEATS, which you can freely use.\bigskip
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\subsection*{Part 1 (6 Marks)}
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The task is to implement a regular expression matcher that is based on
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derivatives of regular expressions. The implementation can deal
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with the following regular expressions, which have been predefined
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in the file re.scala:
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\begin{center}
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\begin{tabular}{lcll}
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$r$ & $::=$ & $\ZERO$ & cannot match anything\\
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& $|$ & $\ONE$ & can only match the empty string\\
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& $|$ & $c$ & can match a character (in this case $c$)\\
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& $|$ & $r_1 + r_2$ & can match a string either with $r_1$ or with $r_2$\\
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& $|$ & $r_1\cdot r_2$ & can match the first part of a string with $r_1$ and\\
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& & & then the second part with $r_2$\\
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& $|$ & $r^*$ & can match zero or more times $r$\\
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\end{tabular}
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\end{center}
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\noindent
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Why? Knowing how to match regular expressions and strings fast will
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let you solve a lot of problems that vex other humans. Regular
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expressions are one of the fastest and simplest ways to match patterns
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in text, and are endlessly useful for searching, editing and
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analysing text in all sorts of places. However, you need to be
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fast, otherwise you will stumble over problems such as recently reported at
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{\small
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\begin{itemize}
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\item[$\bullet$] \url{http://stackstatus.net/post/147710624694/outage-postmortem-july-20-2016}
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\item[$\bullet$] \url{https://vimeo.com/112065252}
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\item[$\bullet$] \url{http://davidvgalbraith.com/how-i-fixed-atom/}
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\end{itemize}}
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\subsection*{Tasks (file re.scala)}
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\begin{itemize}
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\item[(1a)] Implement a function, called \textit{nullable}, by recursion over
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regular expressions. This function tests whether a regular expression can match
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the empty string.
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\begin{center}
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\begin{tabular}{lcl}
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$\textit{nullable}(\ZERO)$ & $\dn$ & $\textit{false}$\\
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$\textit{nullable}(\ONE)$ & $\dn$ & $\textit{true}$\\
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$\textit{nullable}(c)$ & $\dn$ & $\textit{false}$\\
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$\textit{nullable}(r_1 + r_2)$ & $\dn$ & $\textit{nullable}(r_1) \vee \textit{nullable}(r_2)$\\
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$\textit{nullable}(r_1 \cdot r_2)$ & $\dn$ & $\textit{nullable}(r_1) \wedge \textit{nullable}(r_2)$\\
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$\textit{nullable}(r^*)$ & $\dn$ & $\textit{true}$\\
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\end{tabular}
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\end{center}\hfill[1 Mark]
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\item[(1b)] Implement a function, called \textit{der}, by recursion over
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regular expressions. It takes a character and a regular expression
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as arguments and calculates the derivative regular expression according
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to the rules:
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\begin{center}
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\begin{tabular}{lcl}
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$\textit{der}\;c\;(\ZERO)$ & $\dn$ & $\ZERO$\\
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$\textit{der}\;c\;(\ONE)$ & $\dn$ & $\ZERO$\\
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$\textit{der}\;c\;(d)$ & $\dn$ & $\textit{if}\; c = d\;\textit{then} \;\ONE \; \textit{else} \;\ZERO$\\
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$\textit{der}\;c\;(r_1 + r_2)$ & $\dn$ & $(\textit{der}\;c\;r_1) + (\textit{der}\;c\;r_2)$\\
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$\textit{der}\;c\;(r_1 \cdot r_2)$ & $\dn$ & $\textit{if}\;\textit{nullable}(r_1)$\\
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& & $\textit{then}\;((\textit{der}\;c\;r_1)\cdot r_2) + (\textit{der}\;c\;r_2)$\\
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& & $\textit{else}\;(\textit{der}\;c\;r_1)\cdot r_2$\\
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$\textit{der}\;c\;(r^*)$ & $\dn$ & $(\textit{der}\;c\;r)\cdot (r^*)$\\
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\end{tabular}
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\end{center}
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For example given the regular expression $r = (a \cdot b) \cdot c$, the derivatives
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w.r.t.~the characters $a$, $b$ and $c$ are
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\begin{center}
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\begin{tabular}{lcll}
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$\textit{der}\;a\;r$ & $=$ & $(\ONE \cdot b)\cdot c$ & ($= r'$)\\
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$\textit{der}\;b\;r$ & $=$ & $(\ZERO \cdot b)\cdot c$\\
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$\textit{der}\;c\;r$ & $=$ & $(\ZERO \cdot b)\cdot c$
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\end{tabular}
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\end{center}
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Let $r'$ stand for the first derivative, then taking the derivatives of $r'$
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w.r.t.~the characters $a$, $b$ and $c$ gives
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\begin{center}
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\begin{tabular}{lcll}
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$\textit{der}\;a\;r'$ & $=$ & $((\ZERO \cdot b) + \ZERO)\cdot c$ \\
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$\textit{der}\;b\;r'$ & $=$ & $((\ZERO \cdot b) + \ONE)\cdot c$ & ($= r''$)\\
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$\textit{der}\;c\;r'$ & $=$ & $((\ZERO \cdot b) + \ZERO)\cdot c$
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\end{tabular}
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\end{center}
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One more example: Let $r''$ stand for the second derivative above,
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then taking the derivatives of $r''$ w.r.t.~the characters $a$, $b$
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and $c$ gives
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\begin{center}
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\begin{tabular}{lcll}
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$\textit{der}\;a\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ZERO$ \\
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$\textit{der}\;b\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ZERO$\\
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$\textit{der}\;c\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ONE$
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\end{tabular}
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\end{center}
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Note, the last derivative can match the empty string, that is it is \textit{nullable}.\\
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\mbox{}\hfill\mbox{[1 Mark]}
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\item[(1c)] Implement the function \textit{simp}, which recursively
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traverses a regular expression from the inside to the outside, and
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simplifies every sub-regular-expression on the left (see below) to
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the regular expression on the right, except it does not simplify inside
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${}^*$-regular expressions.
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\begin{center}
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\begin{tabular}{l@{\hspace{4mm}}c@{\hspace{4mm}}ll}
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$r \cdot \ZERO$ & $\mapsto$ & $\ZERO$\\
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$\ZERO \cdot r$ & $\mapsto$ & $\ZERO$\\
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$r \cdot \ONE$ & $\mapsto$ & $r$\\
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$\ONE \cdot r$ & $\mapsto$ & $r$\\
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$r + \ZERO$ & $\mapsto$ & $r$\\
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$\ZERO + r$ & $\mapsto$ & $r$\\
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$r + r$ & $\mapsto$ & $r$\\
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\end{tabular}
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\end{center}
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For example the regular expression
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\[(r_1 + \ZERO) \cdot \ONE + ((\ONE + r_2) + r_3) \cdot (r_4 \cdot \ZERO)\]
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simplifies to just $r_1$.
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\hfill[1 Mark]
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\item[(1d)] Implement two functions: The first, called \textit{ders},
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takes a list of characters and a regular expression as arguments, and
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builds the derivative w.r.t.~the list as follows:
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\begin{center}
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\begin{tabular}{lcl}
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$\textit{ders}\;(Nil)\;r$ & $\dn$ & $r$\\
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$\textit{ders}\;(c::cs)\;r$ & $\dn$ &
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$\textit{ders}\;cs\;(\textit{simp}(\textit{der}\;c\;r))$\\
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\end{tabular}
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\end{center}
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The second, called \textit{matcher}, takes a string and a regular expression
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as arguments. It builds first the derivatives according to \textit{ders}
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and after that tests whether the resulting derivative regular expression can match
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the empty string (using \textit{nullable}).
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For example the \textit{matcher} will produce true given the
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regular expression $(a\cdot b)\cdot c$ and the string $abc$.
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\hfill[1 Mark]
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\item[(1e)] Implement the function $\textit{replace}\;r\;s_1\;s_2$: it searches
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(from the left to
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right) in the string $s_1$ all the non-empty substrings that match the
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regular expression $r$---these substrings are assumed to be
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the longest substrings matched by the regular expression and
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assumed to be non-overlapping. All these substrings in $s_1$ matched by $r$
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are replaced by $s_2$. For example given the regular expression
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\[(a \cdot a)^* + (b \cdot b)\]
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\noindent the string $s_1 = aabbbaaaaaaabaaaaabbaaaabb$ and
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replacement string $s_2 = c$ yields the string
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\[
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ccbcabcaccc
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\]
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\hfill[2 Mark]
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\end{itemize}
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\textbf{Background} Although easily implementable in Scala, the idea
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behind the derivative function might not so easy to be seen. To
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understand its purpose better, assume a regular expression $r$ can
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match strings of the form $c::cs$ (that means strings which start with
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a character $c$ and have some rest $cs$). If you now take the
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derivative of $r$ with respect to the character $c$, then you obtain a
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regular expressions that can match all the strings $cs$. In other
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words the regular expression $\textit{der}\;c\;r$ can match the same
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strings $c::cs$ that can be matched by $r$, except that the $c$ is
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chopped off.
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Assume now $r$ can match the string $abc$. If you take the derivative
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according to $a$ then you obtain a regular expression that can match
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$bc$ (it is $abc$ where the $a$ has been chopped off). If you now
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build the derivative $\textit{der}\;b\;(\textit{der}\;a\;r))$ you obtain a regular
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expression that can match the string "c" (it is "bc" where 'b' is
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chopped off). If you finally build the derivative of this according
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'c', that is der('c', der('b', der('a', r))), you obtain a regular
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expression that can match the empty string. You can test this using
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the function nullable, which is what your matcher is doing.
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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.
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By the way this whole idea is by Janusz Brzozowski who came up with this in 1964 in his PhD thesis.
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https://en.wikipedia.org/wiki/Janusz_Brzozowski_(computer_scientist)
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\subsection*{Part 2 (4 Marks)}
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Coming soon.
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\end{document}
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: t
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%%% End:
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