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% BF IDE

% https://www.microsoft.com/en-us/p/brainf-ck/9nblgggzhvq5

\section*{Main Part 3 (Scala, 7 Marks)}

\mbox{}\hfill\textit{``Java is the most distressing thing to happen to computing since MS-DOS.''}\smallskip\\

\mbox{}\hfill\textit{ --- Alan Kay, the inventor of object-oriented programming}\bigskip\medskip

\noindent

This part is about a regular expression matcher described by

Brzozowski in 1964.  The

background is that ``out-of-the-box'' regular expression matching in

mainstream languages like Java, JavaScript and Python can sometimes be

excruciatingly slow.  You are supposed to implement a regular

expression matcher that is much, much faster. \bigskip

\IMPORTANTNONE{}

\noindent

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

maximum of 30 seconds on my laptop.  

\DISCLAIMER{}

\subsection*{Reference Implementation}

This Scala assignment comes with a reference implementation in form of

a \texttt{jar}-file. This allows you to run any test cases on your own

computer. For example you can call Scala on the command line with the

option \texttt{-cp re.jar} and then query any function from the

\texttt{re.scala} template file. As usual you have to prefix the calls

with \texttt{M3} or import this object.  Since some tasks

are time sensitive, you can check the reference implementation as

follows: if you want to know, for example, how long it takes to match

strings of $a$'s using the regular expression $(a^*)^*\cdot b$ you can

query as follows:

\begin{lstlisting}[xleftmargin=1mm,numbers=none,basicstyle=\ttfamily\small]

$ scala -cp re.jar

scala> import M3._  

scala> for (i <- 0 to 5000000 by 500000) {

  | println(f"$i: ${time_needed(2, matcher(EVIL, "a" * i))}%.5f secs.")

  | }

0: 0.00002 secs.

500000: 0.10608 secs.

1000000: 0.22286 secs.

1500000: 0.35982 secs.

2000000: 0.45828 secs.

2500000: 0.59558 secs.

3000000: 0.73191 secs.

3500000: 0.83499 secs.

4000000: 0.99149 secs.

4500000: 1.15395 secs.

5000000: 1.29659 secs.

\end{lstlisting}%$

\subsection*{Preliminaries}

The task is to implement a regular expression matcher that is based on

derivatives of regular expressions. Most of the functions are defined by

recursion over regular expressions and can be elegantly implemented

using Scala's pattern-matching. The implementation should deal with the

following regular expressions, which have been predefined in the file

\texttt{re.scala}:

\begin{center}

\begin{tabular}{lcll}

  $r$ & $::=$ & $\ZERO$     & cannot match anything\\

      &   $|$ & $\ONE$      & can only match the empty string\\

      &   $|$ & $c$         & can match a single 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 a string with zero or more copies of $r$\\

\end{tabular}

\end{center}

\noindent 

Why? Regular expressions are

one of the 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{https://blog.cloudflare.com/details-of-the-cloudflare-outage-on-july-2-2019}  

\item[$\bullet$] \texttt{\href{https://web.archive.org/web/20160801163029/https://www.stackstatus.net/post/147710624694/outage-postmortem-july-20-2016}{https://stackstatus.net/post/147710624694/outage-postmortem-july-20-2016}}

\item[$\bullet$] \url{https://vimeo.com/112065252}

\item[$\bullet$] \url{https://davidvgalbraith.com/how-i-fixed-atom}  

\end{itemize}}

% Knowing how to match regular expressions and strings will let you

% solve a lot of problems that vex other humans.

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

The file \texttt{re.scala} has already a definition for regular

expressions and also defines some handy shorthand notation for regular

expressions. The notation in this coursework description matches up

with the code as follows:

\begin{center}

  \begin{tabular}{rcl@{\hspace{10mm}}l}

    & & code: & shorthand:\smallskip \\ 

  $\ZERO$ & $\mapsto$ & \texttt{ZERO}\\

  $\ONE$  & $\mapsto$ & \texttt{ONE}\\

  $c$     & $\mapsto$ & \texttt{CHAR(c)}\\

  $\sum rs$ & $\mapsto$ & \texttt{ALTs(rs)}\\  

  $r_1 + r_2$ & $\mapsto$ & \texttt{ALT(r1, r2)} & \texttt{r1 | r2}\\

  $\prod rs$ & $\mapsto$ & \texttt{SEQs(rs)}\\  

  $r_1 \cdot r_2$ & $\mapsto$ & \texttt{SEQ(r1, r2)} & \texttt{r1 $\sim$ r2}\\

  $r^*$ & $\mapsto$ &  \texttt{STAR(r)} & \texttt{r.\%}

\end{tabular}    

\end{center}  

\noindent

binary (+ / \texttt{ALT}) and the other is n-ary ($\sum$ /

\texttt{ALTs}). The latter takes a list of regular expressions as

argument.  In what follows we shall use $rs$ to stand for lists of

regular expressions. When the list is empty, we shall write $\sum\,[]$;

if it is non-empty, we sometimes write $\sum\,[r_1,..., r_n]$.

The binary alternative can be seen as an abbreviation,

that is $r_1 + r_2 \dn \sum\,[r_1, r_2]$. As a result we can ignore the

binary version and only implement the n-ary alternative. Similarly

the sequence regular expression is only implemented with lists and the

binary version can be obtained by defining $r_1 \cdot r_2 \dn \prod\,[r_1, r_2]$.

\begin{itemize}

\item[(1)] Implement a function, called \textit{nullable}, by

  recursion over regular expressions. This function tests whether a

  regular expression can match the empty string. This means given a

  regular expression, it either returns true or false. The function

  \textit{nullable}

  is defined as follows:

\begin{center}

\begin{tabular}{lcl}

$\textit{nullable}(\ZERO)$ & $\dn$ & $\textit{false}$\\

$\textit{nullable}(\ONE)$  & $\dn$ & $\textit{true}$\\

$\textit{nullable}(c)$     & $\dn$ & $\textit{false}$\\

$\textit{nullable}(\sum rs)$ & $\dn$ & $\exists r \in rs.\;\textit{nullable}(r)$\\

$\textit{nullable}(\prod rs)$ & $\dn$ & $\forall r\in rs.\;\textit{nullable}(r)$\\

$\textit{nullable}(r^*)$ & $\dn$ & $\textit{true}$\\

\end{tabular}

\end{center}~\hfill[0.5 Marks]

\item[(2)] Implement a function, called \textit{der}, by recursion over

  regular expressions. It takes a character and a regular expression

  as arguments and calculates the \emph{derivative} of a regular expression according

  to the rules:

\begin{center}

\begin{tabular}{lcl}

$\textit{der}\;c\;(\ZERO)$ & $\dn$ & $\ZERO$\\

$\textit{der}\;c\;(\ONE)$  & $\dn$ & $\ZERO$\\

$\textit{der}\;c\;(d)$     & $\dn$ & $\textit{if}\; c = d\;\textit{then} \;\ONE \; \textit{else} \;\ZERO$\\

$\textit{der}\;c\;(\sum\;[r_1,..,r_n])$ & $\dn$ & $\sum\;[\textit{der}\;c\;r_1,..,\textit{der}\;c\;r_n]$\\

$\textit{der}\;c\;(\prod\;[])$ & $\dn$ & $\ZERO$\\  

$\textit{der}\;c\;(\prod\;r\!::\!rs)$ & $\dn$ & $\textit{if}\;\textit{nullable}(r)$\\

      & & $\textit{then}\;(\prod\;(\textit{der}\;c\;r)\!::\!rs) + (\textit{der}\;c\;(\prod rs))$\\

      & & $\textit{else}\;(\prod\;(\textit{der}\;c\;r)\!::\! rs)$\\

$\textit{der}\;c\;(r^*)$ & $\dn$ & $(\textit{der}\;c\;r)\cdot (r^*)$\\

\end{tabular}

\end{center}

\mbox{}\hfill\mbox{[1 Mark]}

\item[(3)] We next want to simplify regular expressions: essentially

  we want to remove $\ZERO$ in regular expressions like $r + \ZERO$

  and $\ZERO + r$.  However, our n-ary alternative takes a list of

  regular expressions as argument, and we therefore need a more general

  ``denesting'' function, which deletes $\ZERO$s and ``spills out'' nested

  $\sum$s.  This function, called \texttt{denest}, should analyse a

  list of regular expressions, say $rs$, as follows:

  \begin{center}

    \begin{tabular}{lllll}

      1) &$rs = []$                & $\dn$ & $[]$ & (empty list)\\

      2) &$rs = \ZERO :: rest$     & $\dn$ & $\texttt{denest}\;rest$ & (throw away $\ZERO$)\\

      3) &$rs = (\sum rs) :: rest$ & $\dn$ & $rs ::: \texttt{denest}\;rest$ & (spill out $\sum$)\\

      4) &$rs = r :: rest$         & $\dn$ & $r :: \texttt{denest}\;rest$ & (otherwise)\\

    \end{tabular}  

  \end{center}  

  The first clause states that empty lists cannot be further

  denested. The second removes the first $\ZERO$ from the list and recurses.

  The third is when the first regular expression is an \texttt{ALTs}, then

  the content of this alternative should be spilled out and appended

  with the denested rest of the list. The last case is for all other

  cases where the head of the list is not $\ZERO$ and not an \texttt{ALTs},

  then we just keep the head of the list and denest the rest.\\

  \mbox{}\hfill\mbox{[1 Mark]}

\item[(4)] Implement the function \texttt{flts} which flattens our

  n-ary sequence regular expression $\prod$. Like \texttt{denest}, this

  function is intended to delete $\ONE$s and spill out nested $\prod$s.

  Unfortunately, there is a special case to do with $\ZERO$: If this function encounters a $\ZERO$, then

  the whole ``product'' should be $\ZERO$. The problem is that the $\ZERO$ can be anywhere

  inside the list. The easiest way to implement this function is therefore by using an

  accumulator, which when called is set to \texttt{Nil}. This means \textit{flts} takes

  two arguments (which are both lists of regular expressions)

  \[

  \texttt{flts}\;rs\;acc  

  \]

  This function analyses the list $rs$ as follows

  \begin{center}

    \begin{tabular}{@{}lllll@{}}

      1) &$rs = []$                   & $\dn$ & $acc$ & (empty list)\\

      2) &$rs = \ZERO :: rest$        & $\dn$ & $[\ZERO]$ & (special case for $\ZERO$)\\

      3) &$rs = \ONE :: rest$         & $\dn$ & $\texttt{flts}\,rest\,acc$ & (throw away $\ONE$)\\

      4) &$rs = (\prod rs) :: rest$ & $\dn$ & $\texttt{flts}\;rest\,(acc ::: rs)$ & (spill out $\prod$)\\

      5) &$rs = r :: rest$            & $\dn$ & $\texttt{flts}\;rest\,(acc ::: [r])$& (otherwise)\\

    \end{tabular}  

  \end{center}

  In the first case we just return whatever has accumulated in

  $acc$. In the fourth case we spill out the $rs$ by appending the

  $rs$ to the end of the accumulator. Similarly in the last case we

  append the single regular expression $r$ to the end of the

  accumulator. I let you think why the ``end'' is needed. \mbox{}\hfill\mbox{[1 Mark]}

\item[(5)] Before we can simplify regular expressions, we need what is often called

  \emph{smart constructors} for $\sum$ and $\prod$. While the ``normal'' constructors

  \texttt{ALTs} and \texttt{SEQs} give us alternatives and sequences, respectively, \emph{smart}

  constructors might return something different depending on what list of regular expressions

  they are given as argument.

  \begin{center}

  \begin{tabular}{@{}c@{\hspace{9mm}}c@{}}

    \begin{tabular}{l@{\hspace{2mm}}l@{\hspace{1mm}}ll}

      & $\sum^{smart}$\smallskip\\

      1) & $rs = []$  & $\dn$ & $\ZERO$\\ 

      2) & $rs = [r]$ & $\dn$ & $r$\\ \\

      3) & otherwise  & $\dn$ & $\sum\,rs$

    \end{tabular} &

    \begin{tabular}{l@{\hspace{2mm}}l@{\hspace{1mm}}ll}

        & $\prod^{smart}$\smallskip\\              

        1) & $rs = []$ & $\dn$ & $\ONE$\\

        2a) & $rs = [\ZERO]$ & $\dn$ & $\ZERO$\\

        2b) & $rs = [r]$ & $\dn$ & $r$\\

        3) & otherwise  & $\dn$ & $\prod\,rs$

    \end{tabular}              

  \end{tabular}    

  \end{center}

  \mbox{}\hfill\mbox{[0.5 Marks]}

\item[(6)] Implement the function \textit{simp}, which recursively

  traverses a regular expression, and on the way up simplifies every

  regular expression on the left (see below) to the regular expression

  on the right, except it does not simplify inside ${}^*$-regular

  expressions and also does not simplify $\ZERO$, $\ONE$ and characters.

  \begin{center}

    \begin{tabular}{@{}l@{\hspace{3mm}}c@{\hspace{3mm}}ll@{}}

      LHS: & & RHS:\smallskip\\

$\sum\;[r_1,..,r_n]$ & $\mapsto$ & $\sum^{smart}\;(\texttt{(denest + distinct)}[simp(r_1),..,simp(r_n)])$\\

      $\prod\;[r_1,..,r_n]$ & $\mapsto$ & $\prod^{smart}\;(\texttt{(flts)}[simp(r_1),..,simp(r_n)])$\\

      $r$ & $\mapsto$ & $r$ \quad (all other cases)

\end{tabular}

\end{center}

The first case is as follows: first apply $simp$ to all regular

expressions $r_1,.. ,r_n$; then denest the resulting list using

\texttt{denest}; after that remove all duplicates in this list (this can be

done in Scala using the function

\texttt{\_.distinct}). Finally, you end up with a list of (simplified)

regular expressions; apply the smart constructor $\sum^{smart}$ to this list.

Similarly in the $\prod$ case: simplify first all regular

expressions $r_1,.. ,r_n$; then flatten the resulting list using \texttt{flts} and apply the

smart constructor $\prod^{smart}$ to the result. In all other cases, just return the

input $r$ as is.

  For example the regular expression

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

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

\item[(7)] 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

true for the regular expression $(a\cdot b)\cdot c$ and the string

$abc$, but false if you give it the string $ab$. \hfill[0.5 Mark]

\item[(8)] Implement a function, called \textit{size}, by recursion

  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}

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

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

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

$\textit{size}(\sum\,[r_1,..,r_n]$ & $\dn$ & $1 + \textit{size}(r_1) + ... + \textit{size}(r_n)$\\

$\textit{size}(\prod\,[r_1,..,r_n]$ & $\dn$ & $1 + \textit{size}(r_1) + ... + \textit{size}(r_n)$\\

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

\end{tabular}

\end{center}

You can use \textit{size} in order to test how much the ``evil'' regular

expression $(a^*)^* \cdot b$ grows when taking successive derivatives

according the letter $a$ without simplification and then compare it to

taking the derivative, but simplify the result.  The sizes

are given in \texttt{re.scala}. \hfill[0.5 Marks]

\item[(9)] You do not have to implement anything specific under this

  task.  The purpose here is that you will be marked for some ``power''

  test cases. For example can your matcher decide within 30 seconds

  whether the regular expression $(a^*)^*\cdot b$ matches strings of the

  form $aaa\ldots{}aaaa$, for say 1 Million $a$'s. And does simplification

  simplify the regular expression

  \[

  \texttt{SEQ(SEQ(SEQ(..., ONE | ONE) , ONE | ONE), ONE | ONE)}

  \]  

  \noindent correctly to just \texttt{ONE}, where \texttt{SEQ} is nested

  50 or more times?\\

  \mbox{}\hfill[1 Mark]

\end{itemize}

\subsection*{Background}

Although easily implementable in Scala (ok maybe the \texttt{simp} functions and

the constructors \texttt{ALTs}/\texttt{SEQs} needs a bit more thinking), 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, or tail, $cs$). If you take the derivative of

$r$ with respect to the character $c$, then you obtain a regular

expression 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

$\textit{der}\;c\;(\textit{der}\;b\;(\textit{der}\;a\;r))$, you obtain