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% BF IDE

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

\section*{Part 9 (Scala)}

\mbox{}\hfill\textit{``[Google’s MapReduce] abstraction is inspired by the}\\

\mbox{}\hfill\textit{map and reduce primitives present in Lisp and many}\\

\mbox{}\hfill\textit{other functional language.''}\smallskip\\

\mbox{}\hfill\textit{ --- Dean and Ghemawat, who designed this concept at Google}

\bigskip\medskip

\noindent

This part is about the shunting yard algorithm by Dijkstra and a

regular expression matcher by Brzozowski. The preliminary part is due on

\cwNINE{} at 4pm; the core, more advanced part, is due on \cwNINEa{}

at 4pm. The preliminary 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. In the core

part, you are asked to implement a regular expression matcher based on

derivatives of regular expressions. 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 an regular expression matcher that is

much, much faster. \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 $(a^*)^*\cdot b$

you can query as follows:

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

$ scala -cp re.jar

scala> import CW9c._  

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*{Preliminary Part (4 Marks)}

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

an influential computer scientist who developed many well-known

algorithms. This algorithm transforms the usual infix notation of

arithmetic expressions into the postfix notation, sometimes also

called reverse Polish notation.

Why on Earth do people use the postfix notation? It is much more

convenient to work with the usual infix notation for arithmetic

expressions. Most modern calculators (as opposed to the ones used 20

years ago) understand infix notation. So why on Earth? \ldots{}Well,

many computers under the hood, even nowadays, use postfix notation

extensively. For example if you give to the Java compiler the

expression $1 + ((2 * 3) + (4 - 3))$, it will generate the Java Byte

code

\begin{lstlisting}[language=JVMIS,numbers=none]

ldc 1 

ldc 2 

ldc 3 

imul 

ldc 4 

ldc 3 

isub 

iadd 

iadd

\end{lstlisting}

\noindent

where the command \texttt{ldc} loads a constant onto the stack, and \texttt{imul},

\texttt{isub} and \texttt{iadd} are commands acting on the stack. Clearly this

is the arithmetic expression in postfix notation.\bigskip

\noindent

The shunting yard algorithm processes an input token list using an

operator stack and an output list. The input consists of numbers,

operators ($+$, $-$, $*$, $/$) and parentheses, and for the purpose of

the assignment we assume the input is always a well-formed expression

in infix notation.  The calculation in the shunting yard algorithm uses

information about the

precedences of the operators (given in the template file). The

algorithm processes the input token list as follows:

\begin{itemize}

\item If there is a number as input token, then this token is

  transferred directly to the output list. Then the rest of the input is

  processed.

\item If there is an operator as input token, then you need to check

  what is on top of the operator stack. If there are operators with

  a higher or equal precedence, these operators are first popped off

  from the stack and moved to the output list. Then the operator from the input

  is pushed onto the stack and the rest of the input is processed.

\item If the input is a left-parenthesis, you push it on to the stack

  and continue processing the input.

\item If the input is a right-parenthesis, then you pop off all operators

  from the stack to the output list until you reach the left-parenthesis.

  Then you discharge the $($ and $)$ from the input and stack, and continue

  processing the input list.

\item If the input is empty, then you move all remaining operators

  from the stack to the output list.  

\end{itemize}  

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

\begin{itemize}

\item[(1)] Implement the shunting yard algorithm described above. The

  function, called \texttt{syard}, takes a list of tokens as first

  argument. The second and third arguments are the stack and output

  list represented as Scala lists. The most convenient way to

  implement this algorithm is to analyse what the input list, stack

  and output list look like in each step using pattern-matching. The

  algorithm transforms for example the input

  \[

  \texttt{List(3, +, 4, *, (, 2, -, 1, ))}

  \]

  into the postfix output

  \[

  \texttt{List(3, 4, 2, 1, -, *, +)}

  \]  

  You can assume the input list is always a  list representing

  a well-formed infix arithmetic expression.\hfill[1 Mark]

\item[(2)] Implement a compute function that takes a postfix expression

  as argument and evaluates it generating an integer as result. It uses a

  stack to evaluate the postfix expression. The operators $+$, $-$, $*$

  are as usual; $/$ is division on integers, for example $7 / 3 = 2$.

  \hfill[1 Mark]

\end{itemize}

\subsubsection*{Task (file postfix2.scala)}

\begin{itemize}

\item[(3/4)] Extend the code in (7) and (8) to include the power

  operator.  This requires proper account of associativity of

  the operators. The power operator is right-associative, whereas the

  other operators are left-associative.  Left-associative operators

  are popped off if the precedence is bigger or equal, while

  right-associative operators are only popped off if the precedence is

  bigger. The compute function in this task should use

  \texttt{Long}s, rather than \texttt{Int}s.\hfill[2 Marks]

\end{itemize}

\subsection*{Core Part (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{https://blog.cloudflare.com/details-of-the-cloudflare-outage-on-july-2-2019}  

\item[$\bullet$] \url{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 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[(5)] 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[(6)] 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 of a regular expression according

  to the rules:

\begin{center}

\begin{tabular}{lcl}

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

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

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

$\textit{der}\;c\;(r_1 + r_2)$ & $\dn$ & $(\textit{der}\;c\;r_1) + (\textit{der}\;c\;r_2)$\\

$\textit{der}\;c\;(r_1 \cdot r_2)$ & $\dn$ & $\textit{if}\;\textit{nullable}(r_1)$\\

      & & $\textit{then}\;((\textit{der}\;c\;r_1)\cdot r_2) + (\textit{der}\;c\;r_2)$\\

      & & $\textit{else}\;(\textit{der}\;c\;r_1)\cdot r_2$\\

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

\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[(7)] 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

  trees. One of them corresponds to the inside-out traversal, which is also

  sometimes called post-order tra\-versal: you traverse inside the

  tree and on the way up you apply simplification rules.

  \textbf{Another Hint:}

  Remember numerical expressions from school times---there you had expressions

  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[(8)] 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[(9)] 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[(10)] You do not have to implement anything specific under this

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