cws/main_cw03.tex
author Christian Urban <christian.urban@kcl.ac.uk>
Tue, 24 Nov 2020 09:04:06 +0000
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% !TEX program = xelatex
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\documentclass{article}
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\usepackage{../style}
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\usepackage{../langs}
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\usepackage{disclaimer}
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\usepackage{tikz}
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\usepackage{pgf}
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\usepackage{pgfplots}
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\usepackage{stackengine}
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%% \usepackage{accents}
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\newcommand\barbelow[1]{\stackunder[1.2pt]{#1}{\raisebox{-4mm}{\boldmath$\uparrow$}}}
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\begin{filecontents}{re-python2.data}
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1 0.033
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19 0.084 
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20 0.141
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21 0.248
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22 0.485
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23 0.878
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24 1.71
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\end{filecontents}
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\begin{filecontents}{re-java.data}
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5  0.00298
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24  1.70251
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25  3.36112
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26  6.63998
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27  13.35120
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\end{filecontents}
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\begin{filecontents}{re-js.data}
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28  1.994
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30  7.648
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31  15.881 
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\end{filecontents}
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\begin{filecontents}{re-java9.data}
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1000  0.01410
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2000  0.04882
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3000  0.10609
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4000  0.17456
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5000  0.27530
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6000  0.41116
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7000  0.53741
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8000  0.70261
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9000  0.93981
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10000 0.97419
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11000 1.28697
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12000 1.51387
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14000 2.07079
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16000 2.69846
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20000 4.41823
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60000 43.0327746
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\end{filecontents}
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\begin{filecontents}{re-swift.data}
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5   0.001
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\end{filecontents}
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\begin{filecontents}{re-dart.data}
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\end{filecontents}
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\begin{document}
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% BF IDE
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% https://www.microsoft.com/en-us/p/brainf-ck/9nblgggzhvq5
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\section*{Main Part 3 (Scala, 7 Marks)}
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%\mbox{}\hfill\textit{``[Google’s MapReduce] abstraction is inspired by the}\\
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%\mbox{}\hfill\textit{map and reduce primitives present in Lisp and many}\\
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%\mbox{}\hfill\textit{other functional language.''}\smallskip\\
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%\mbox{}\hfill\textit{ --- Dean and Ghemawat, who designed this concept at Google}
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%\bigskip\medskip
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\noindent
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This part is about a regular expression matcher described by
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Brzozowski in 1964. This part is due on \cwEIGHTa{} at 5pm.  The
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background is that ``out-of-the-box'' regular expression matching in
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mainstream languages like Java, JavaScript and Python can sometimes be
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excruciatingly slow.  You are supposed to implement a regular
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expression matcher that is much, much faster. \bigskip
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\IMPORTANTNONE{}
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\noindent
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Also note that the running time of each part will be restricted to a
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maximum of 30 seconds on my laptop.  
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\DISCLAIMER{}
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\subsection*{Reference Implementation}
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This Scala assignment comes with a reference implementation in form of
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a \texttt{jar}-file. This allows you to run any test cases on your own
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computer. For example you can call Scala on the command line with the
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option \texttt{-cp re.jar} and then query any function from the
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\texttt{re.scala} template file. As usual you have to prefix the calls
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with \texttt{CW8c} or import this object.  Since some tasks
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are time sensitive, you can check the reference implementation as
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follows: if you want to know, for example, how long it takes to match
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strings of $a$'s using the regular expression $(a^*)^*\cdot b$ you can
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query as follows:
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\begin{lstlisting}[xleftmargin=1mm,numbers=none,basicstyle=\ttfamily\small]
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$ scala -cp re.jar
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scala> import CW8c._  
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scala> for (i <- 0 to 5000000 by 500000) {
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  | println(f"$i: ${time_needed(2, matcher(EVIL, "a" * i))}%.5f secs.")
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  | }
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0: 0.00002 secs.
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500000: 0.10608 secs.
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1000000: 0.22286 secs.
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1500000: 0.35982 secs.
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2000000: 0.45828 secs.
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2500000: 0.59558 secs.
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3000000: 0.73191 secs.
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3500000: 0.83499 secs.
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4000000: 0.99149 secs.
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4500000: 1.15395 secs.
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5000000: 1.29659 secs.
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\end{lstlisting}%$
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\subsection*{Preliminaries}
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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. Most of the functions are defined by
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recursion over regular expressions and can be elegantly implemented
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using Scala's pattern-matching. The implementation should deal with the
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following regular expressions, which have been predefined in the file
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\texttt{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 single 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 a string with zero or more copies of $r$\\
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\end{tabular}
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\end{center}
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\noindent 
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Why? Regular expressions are
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one of the simplest ways to match patterns in text, and
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are endlessly useful for searching, editing and analysing data in all
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sorts of places (for example analysing network traffic in order to
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detect security breaches). However, you need to be fast, otherwise you
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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{https://blog.cloudflare.com/details-of-the-cloudflare-outage-on-july-2-2019}  
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\item[$\bullet$] \url{https://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{https://davidvgalbraith.com/how-i-fixed-atom}  
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\end{itemize}}
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% Knowing how to match regular expressions and strings will let you
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% solve a lot of problems that vex other humans.
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\subsubsection*{Tasks (file re.scala)}
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The file \texttt{re.scala} has already a definition for regular
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expressions and also defines some handy shorthand notation for
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regular expressions. The notation in this document matches up
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with the code in the file as follows:
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\begin{center}
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  \begin{tabular}{rcl@{\hspace{10mm}}l}
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    & & code: & shorthand:\smallskip \\ 
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  $\ZERO$ & $\mapsto$ & \texttt{ZERO}\\
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  $\ONE$  & $\mapsto$ & \texttt{ONE}\\
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  $c$     & $\mapsto$ & \texttt{CHAR(c)}\\
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  $r_1 + r_2$ & $\mapsto$ & \texttt{ALT(r1, r2)} & \texttt{r1 | r2}\\
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  $r_1 \cdot r_2$ & $\mapsto$ & \texttt{SEQ(r1, r2)} & \texttt{r1 $\sim$ r2}\\
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  $r^*$ & $\mapsto$ &  \texttt{STAR(r)} & \texttt{r.\%}
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\end{tabular}    
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\end{center}  
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\begin{itemize}
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\item[(1)] Implement a function, called \textit{nullable}, by
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  recursion over regular expressions. This function tests whether a
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  regular expression can match the empty string. This means given a
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  regular expression it either returns true or false. The function
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  \textit{nullable}
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  is defined as follows:
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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[(2)] 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 of a 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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    $\textit{der}\;a\;r$ & $=$ & $(\ONE \cdot b)\cdot c$ & \quad($= r'$)\\
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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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    $\textit{der}\;b\;r'$ & $=$ & $((\ZERO \cdot b) + \ONE)\cdot c$ & \quad($= r''$)\\
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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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    (is $\textit{nullable}$)                      
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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[(3)] Implement the function \textit{simp}, which recursively
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  traverses a regular expression, and on the way up simplifies every
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  regular expression on the left (see below) to the regular expression
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  on the right, except it does not simplify inside ${}^*$-regular
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  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 $r_1$. \textbf{Hint:} Regular expressions can be
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  seen as trees and there are several methods for traversing
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  trees. One of them corresponds to the inside-out traversal, which is also
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  sometimes called post-order tra\-versal: you traverse inside the
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  tree and on the way up you apply simplification rules.
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  \textbf{Another Hint:}
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  Remember numerical expressions from school times---there you had expressions
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  like $u + \ldots + (1 \cdot x) - \ldots (z + (y \cdot 0)) \ldots$
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  and simplification rules that looked very similar to rules
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  above. You would simplify such numerical expressions by replacing
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  for example the $y \cdot 0$ by $0$, or $1\cdot x$ by $x$, and then
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  look whether more rules are applicable. If you organise the
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  simplification in an inside-out fashion, it is always clear which
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  simplification should be applied next.\hfill[1 Mark]
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\item[(4)] 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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Note that this function is different from \textit{der}, which only
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takes a single character.
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The second function, called \textit{matcher}, takes a string and a
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regular expression as arguments. It builds first the derivatives
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according to \textit{ders} and after that tests whether the resulting
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derivative regular expression can match the empty string (using
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\textit{nullable}).  For example the \textit{matcher} will produce
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true for the regular expression $(a\cdot b)\cdot c$ and the string
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$abc$, but false if you give it the string $ab$. \hfill[1 Mark]
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\item[(5)] Implement a function, called \textit{size}, by recursion
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  over regular expressions. If a regular expression is seen as a tree,
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  then \textit{size} should return the number of nodes in such a
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  tree. Therefore this function is defined as follows:
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\begin{center}
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\begin{tabular}{lcl}
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$\textit{size}(\ZERO)$ & $\dn$ & $1$\\
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$\textit{size}(\ONE)$  & $\dn$ & $1$\\
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$\textit{size}(c)$     & $\dn$ & $1$\\
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$\textit{size}(r_1 + r_2)$ & $\dn$ & $1 + \textit{size}(r_1) + \textit{size}(r_2)$\\
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$\textit{size}(r_1 \cdot r_2)$ & $\dn$ & $1 + \textit{size}(r_1) + \textit{size}(r_2)$\\
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$\textit{size}(r^*)$ & $\dn$ & $1 + \textit{size}(r)$\\
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\end{tabular}
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\end{center}
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You can use \textit{size} in order to test how much the ``evil'' regular
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expression $(a^*)^* \cdot b$ grows when taking successive derivatives
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according the letter $a$ without simplification and then compare it to
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taking the derivative, but simplify the result.  The sizes
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are given in \texttt{re.scala}. \hfill[1 Mark]
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\item[(6)] You do not have to implement anything specific under this
221
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  task.  The purpose here is that you will be marked for some ``power''
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  test cases. For example can your matcher decide within 30 seconds
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  whether the regular expression $(a^*)^*\cdot b$ matches strings of the
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  form $aaa\ldots{}aaaa$, for say 1 Million $a$'s. And does simplification
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  simplify the regular expression
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   402
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  \[
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  \texttt{SEQ(SEQ(SEQ(..., ONE | ONE) , ONE | ONE), ONE | ONE)}
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  \]  
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  \noindent correctly to just \texttt{ONE}, where \texttt{SEQ} is nested
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  50 or more times?\\
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  \mbox{}\hfill[2 Mark]
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\end{itemize}
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\subsection*{Background}
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Although easily implementable in Scala, the idea behind the derivative
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   415
function might not so easy to be seen. To understand its purpose
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   416
better, assume a regular expression $r$ can match strings of the form
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   417
$c\!::\!cs$ (that means strings which start with a character $c$ and have
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diff changeset
   418
some rest, or tail, $cs$). If you take the derivative of $r$ with
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diff changeset
   419
respect to the character $c$, then you obtain a regular expression
22705d22c105 updated
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that can match all the strings $cs$.  In other words, the regular
22705d22c105 updated
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expression $\textit{der}\;c\;r$ can match the same strings $c\!::\!cs$
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   422
that can be matched by $r$, except that the $c$ is chopped off.
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   423
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Assume now $r$ can match the string $abc$. If you take the derivative
22705d22c105 updated
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diff changeset
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according to $a$ then you obtain a regular expression that can match
22705d22c105 updated
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diff changeset
   426
$bc$ (it is $abc$ where the $a$ has been chopped off). If you now
22705d22c105 updated
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diff changeset
   427
build the derivative $\textit{der}\;b\;(\textit{der}\;a\;r)$ you
22705d22c105 updated
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diff changeset
   428
obtain a regular expression that can match the string $c$ (it is $bc$
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diff changeset
   429
where $b$ is chopped off). If you finally build the derivative of this
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diff changeset
   430
according $c$, that is
22705d22c105 updated
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diff changeset
   431
$\textit{der}\;c\;(\textit{der}\;b\;(\textit{der}\;a\;r))$, you obtain
22705d22c105 updated
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diff changeset
   432
a regular expression that can match the empty string. You can test
22705d22c105 updated
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diff changeset
   433
whether this is indeed the case using the function nullable, which is
22705d22c105 updated
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parents: 111
diff changeset
   434
what your matcher is doing.
22705d22c105 updated
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diff changeset
   435
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diff changeset
   436
The purpose of the $\textit{simp}$ function is to keep the regular
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diff changeset
   437
expressions small. Normally the derivative function makes the regular
221
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diff changeset
   438
expression bigger (see the SEQ case and the example in (2)) and the
218
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diff changeset
   439
algorithm would be slower and slower over time. The $\textit{simp}$
22705d22c105 updated
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diff changeset
   440
function counters this increase in size and the result is that the
22705d22c105 updated
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diff changeset
   441
algorithm is fast throughout.  By the way, this algorithm is by Janusz
22705d22c105 updated
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diff changeset
   442
Brzozowski who came up with the idea of derivatives in 1964 in his PhD
22705d22c105 updated
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diff changeset
   443
thesis.
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diff changeset
   444
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   445
\begin{center}\small
22705d22c105 updated
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diff changeset
   446
\url{https://en.wikipedia.org/wiki/Janusz_Brzozowski_(computer_scientist)}
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diff changeset
   447
\end{center}
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diff changeset
   448
105
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If you want to see how badly the regular expression matchers do in
221
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diff changeset
   451
Java\footnote{Version 8 and below; Version 9 and above does not seem to be as
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diff changeset
   452
  catastrophic, but still much worse than the regular expression
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diff changeset
   453
  matcher based on derivatives.}, JavaScript and Python with the
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diff changeset
   454
`evil' regular expression $(a^*)^*\cdot b$, then have a look at the
351
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diff changeset
   455
graphs below (you can try it out for yourself: have a look at the files
591b9005157e updated
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   456
\texttt{catastrophic9.java}, \texttt{catastrophic.js},
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\texttt{catastrophic.py} etc on KEATS). Compare this with the matcher you
221
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diff changeset
   458
have implemented. How long can the string of $a$'s be in your matcher
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diff changeset
   459
and still stay within the 30 seconds time limit?
78
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diff changeset
   460
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\begin{center}
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diff changeset
   462
\begin{tabular}{@{}cc@{}}
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diff changeset
   463
\multicolumn{2}{c}{Graph: $(a^*)^*\cdot b$ and strings 
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diff changeset
   464
           $\underbrace{a\ldots a}_{n}$}\bigskip\\
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diff changeset
   465
  
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diff changeset
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\begin{tikzpicture}
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diff changeset
   467
\begin{axis}[
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   468
    xlabel={$n$},
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diff changeset
   469
    x label style={at={(1.05,0.0)}},
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    ylabel={time in secs},
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diff changeset
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    y label style={at={(0.06,0.5)}},
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    enlargelimits=false,
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diff changeset
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    xtick={0,5,...,30},
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    xmax=33,
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    ymax=45,
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diff changeset
   476
    ytick={0,5,...,40},
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diff changeset
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    scaled ticks=false,
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diff changeset
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    axis lines=left,
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    width=6cm,
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    height=5.5cm, 
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    legend entries={Python, Java 8, JavaScript, Swift, Dart},  
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    legend pos=north west,
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    legend cell align=left]
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\addplot[blue,mark=*, mark options={fill=white}] table {re-python2.data};
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\addplot[cyan,mark=*, mark options={fill=white}] table {re-java.data};
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\addplot[red,mark=*, mark options={fill=white}] table {re-js.data};
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\addplot[magenta,mark=*, mark options={fill=white}] table {re-swift.data};
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\addplot[brown,mark=*, mark options={fill=white}] table {re-dart.data};
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\end{axis}
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\end{tikzpicture}
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  & 
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\begin{tikzpicture}
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\begin{axis}[
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    xlabel={$n$},
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    x label style={at={(1.05,0.0)}},
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    ylabel={time in secs},
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    y label style={at={(0.06,0.5)}},
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    %enlargelimits=false,
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    %xtick={0,5000,...,30000},
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    xmax=65000,
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    ymax=45,
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    ytick={0,5,...,40},
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    scaled ticks=false,
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    axis lines=left,
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    width=6cm,
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    height=5.5cm, 
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    legend entries={Java 9},  
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    legend pos=north west]
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\addplot[cyan,mark=*, mark options={fill=white}] table {re-java9.data};
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\end{axis}
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\end{tikzpicture}
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\end{tabular}  
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\end{center}
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\newpage
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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: