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\begin{document}

% BF IDE

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

\section*{Coursework 9 (Scala)}

This coursework is worth 10\%. It is about a regular expression

matcher and the shunting yard algorithm by Dijkstra. The first part is

due on 6 December at 11pm; the second, more advanced part, is due on

20 December at 11pm. In the first part, you are asked to implement a

regular expression matcher based on derivatives of regular

expressions. The background is that regular expression matching in

languages like Java, JavaScript and Python can sometimes be excruciatingly

slow. The advanced part is about the shunting yard algorithm that

transforms the usual infix notation of arithmetic expressions into the

postfix notation, which is for example used in compilers.\bigskip

\IMPORTANT{}

\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 three reference implementations in form of

\texttt{jar}-files you can download from KEATS. 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{CW9a}, \texttt{CW9b} and \texttt{CW9c}.

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 $(a^*)^*\cdot b$

$ scala -cp re.jar

scala> import CW9a._  

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

  | println(i + " " + "%.5f".format(time_needed(2, matcher(EVIL, "a" * i))) + "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*{Part 1 (6 Marks)}

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{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}}

% 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 document matches up

with the code in the file 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)}\\

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

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

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

\end{tabular}    

\end{center}  

\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}

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

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

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

$\textit{nullable}(r_1 + r_2)$ & $\dn$ & $\textit{nullable}(r_1) \vee \textit{nullable}(r_2)$\\

$\textit{nullable}(r_1 \cdot r_2)$ & $\dn$ & $\textit{nullable}(r_1) \wedge \textit{nullable}(r_2)$\\

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

\end{tabular}

\end{center}~\hfill[1 Mark]

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

  regular expressions. It takes a character and a regular expression

  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}

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

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

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

  \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}

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

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

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

  \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[(3)] 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.

  \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 $r_1$. \textbf{Hint:} Regular expressions can be

  seen as trees and there are several methods for traversing

  tree and on the way up you apply simplification rules.

  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

  simplification should be applied next.\hfill[1 Mark]

\item[(4)] 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[1 Mark]

\item[(5)] 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}

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

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

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

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

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

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

\end{tabular}

\end{center}

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[1 Mark]

\item[(6)] 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

  \mbox{}\hfill[1 Mark]

\end{itemize}

\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

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

a regular expression that can match the empty string. You can test

whether this is indeed the case using the function nullable, which is

what your matcher is doing.

The purpose of the $\textit{simp}$ function is to keep the regular

expressions small. Normally the derivative function makes the regular

expression bigger (see the SEQ case and the example in (2)) and the

algorithm would be slower and slower over time. The $\textit{simp}$

function counters this increase in size and the result is that the

algorithm is fast throughout.  By the way, this algorithm is by Janusz

Brzozowski who came up with the idea of derivatives in 1964 in his PhD

thesis.

\begin{center}\small

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

\end{center}

If you want to see how badly the regular expression matchers do in

Java\footnote{Version 8 and below; Version 9 and above does not seem to be as

  catastrophic, but still much worse than the regular expression

  matcher based on derivatives.}, JavaScript and Python with the

`evil' regular expression $(a^*)^*\cdot b$, then have a look at the

graphs below (you can try it out for yourself: have a look at the file

\texttt{catastrophic9.java}, \texttt{catastrophic.js} and

\texttt{catastrophic.py} on KEATS). Compare this with the matcher you

have implemented. How long can the string of $a$'s be in your matcher

and still stay within the 30 seconds time limit?

\begin{center}

\begin{tabular}{@{}cc@{}}

\multicolumn{2}{c}{Graph: $(a^*)^*\cdot b$ and strings 

           $\underbrace{a\ldots a}_{n}$}\bigskip\\

\begin{tikzpicture}

\begin{axis}[

    xlabel={$n$},

    x label style={at={(1.05,0.0)}},

    ylabel={time in secs},

    y label style={at={(0.06,0.5)}},

    enlargelimits=false,

    xtick={0,5,...,30},

    xmax=33,

    ymax=45,

    ytick={0,5,...,40},

    scaled ticks=false,

    axis lines=left,

    width=6cm,

    height=5.5cm, 

    legend entries={Python, Java 8, JavaScript},  

    legend pos=north west,

    legend cell align=left]

\addplot[blue,mark=*, mark options={fill=white}] table {re-python2.data};

\addplot[cyan,mark=*, mark options={fill=white}] table {re-java.data};

\addplot[red,mark=*, mark options={fill=white}] table {re-js.data};

\end{axis}

\end{tikzpicture}

  & 

\begin{tikzpicture}

\begin{axis}[

    xlabel={$n$},

    x label style={at={(1.05,0.0)}},

    ylabel={time in secs},

    y label style={at={(0.06,0.5)}},

    %enlargelimits=false,

    %xtick={0,5000,...,30000},

    xmax=65000,

    ymax=45,

    ytick={0,5,...,40},

    scaled ticks=false,

    axis lines=left,

    width=6cm,

    height=5.5cm, 

    legend entries={Java 9},  

    legend pos=north west]

\addplot[cyan,mark=*, mark options={fill=white}] table {re-java9.data};

\end{axis}

\end{tikzpicture}

\end{tabular}  

\end{center}

\newpage

\subsection*{Part 2 (4 Marks)}

The \emph{Shunting Yard Algorithm} has been developed by Edsger Dijkstra,

an influential computer scientist who developed many well-known