cws/cw05.tex
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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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\end{filecontents}
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\begin{filecontents}{re-java.data}
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\end{filecontents}
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\begin{filecontents}{re-js.data}
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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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\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*{Coursework 9 (Scala)}
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This coursework is worth 10\%. It is about a regular expression
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matcher and the shunting yard algorithm by Dijkstra. The first part is
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due on 6 December at 11pm; the second, more advanced part, is due on
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21 December at 11pm. In the first part, you are asked to implement a
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regular expression matcher based on derivatives of regular
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expressions. The reason is that regular expression matching in
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languages like Java, JavaScipt and Python can sometimes be extremely
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slow. The advanced part is about the shunting yard algorithm that
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transforms the usual infix notation of arithmetic expressions into the
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postfix notation, which is for example used in compilers.\bigskip
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\IMPORTANT{}
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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*{Part 1 (6 Marks, Regular Expression Matcher)}
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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? Knowing how to match regular expressions and strings will let you
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solve a lot of problems that vex other humans. Regular expressions are
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one of the fastest and 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{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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\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 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 from the inside to the outside, and
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  on the way simplifies every regular expression on the left (see
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  below) to the regular expression on the right, except it does not
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  simplify inside ${}^*$-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 $r_1$. \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
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  sometimes also called post-order traversal'' you traverse inside the
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  tree and on the way up, you apply simplification rules.
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  Furthermore,
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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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  rule 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
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  task.  The purpose is that you will be marked for some ``power''
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  test cases. For example can your matcher decide withing 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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  \[
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  \texttt{SEQ(SEQ(SEQ(..., ONE | ONE) , ONE | ONE), ONE | ONE)}
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  \]  
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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[1 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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function might not so easy to be seen. To understand its purpose
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better, assume a regular expression $r$ can match strings of the form
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$c\!::\!cs$ (that means strings which start with a character $c$ and have
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some rest, or tail, $cs$). If you take the derivative of $r$ with
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respect to the character $c$, then you obtain a regular expression
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that can match all the strings $cs$.  In other words, the regular
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expression $\textit{der}\;c\;r$ can match the same strings $c\!::\!cs$
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that can be matched by $r$, except that the $c$ is 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
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obtain a regular expression that can match the string $c$ (it is $bc$
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where $b$ is chopped off). If you finally build the derivative of this
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according $c$, that is
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$\textit{der}\;c\;(\textit{der}\;b\;(\textit{der}\;a\;r))$, you obtain
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a regular expression that can match the empty string. You can test
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whether this is indeed the case using the function nullable, which is
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what your matcher is doing.
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The purpose of the $\textit{simp}$ function is to keep the regular
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expressions small. Normally the derivative function makes the regular
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expression bigger (see the SEQ case and the example in (2)) and the
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algorithm would be slower and slower over time. The $\textit{simp}$
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function counters this increase in size and the result is that the
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algorithm is fast throughout.  By the way, this algorithm is by Janusz
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Brzozowski who came up with the idea of derivatives in 1964 in his PhD
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thesis.
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\begin{center}\small
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\url{https://en.wikipedia.org/wiki/Janusz_Brzozowski_(computer_scientist)}
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\end{center}
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If you want to see how badly the regular expression matchers do in
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Java\footnote{Version 8 and below; Version 9 and above does not seem to be as
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  catastrophic, but still much worse than the regular expression
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  matcher based on derivatives.}, JavaScript and in Python with the
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`evil' regular expression $(a^*)^*\cdot b$, then have a look at the
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graphs below (you can try it out for yourself: have a look at the file
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\texttt{catastrophic9.java}, \texttt{catastrophic.js} and
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\texttt{catastrophic.py} on KEATS). Compare this with the matcher you
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have implemented. How long can the string of $a$'s be in your matcher
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and still stay within the 30 seconds time limit?
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\begin{center}
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\begin{tabular}{@{}cc@{}}
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\multicolumn{2}{c}{Graph: $(a^*)^*\cdot b$ and strings 
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           $\underbrace{a\ldots a}_{n}$}\bigskip\\
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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={0,5,...,30},
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    xmax=33,
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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={Python, Java 8, JavaScript},  
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    legend pos=north west]
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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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\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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\subsection*{Part 2 (4 Marks)}
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Coming from Java or C++, you might think Scala is a quite esoteric
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programming language.  But remember, some serious companies have built
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their business on
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Scala.\footnote{\url{https://en.wikipedia.org/wiki/Scala_(programming_language)\#Companies}}
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And there are far, far more esoteric languages out there. One is
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called \emph{brainf***}. You are asked in this part to implement an
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interpreter for this language.
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Urban M\"uller developed brainf*** in 1993.  A close relative of this
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language was already introduced in 1964 by Corado B\"ohm, an Italian
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computer pioneer, who unfortunately died a few months ago. The main
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feature of brainf*** is its minimalistic set of instructions---just 8
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instructions in total and all of which are single characters. Despite
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the minimalism, this language has been shown to be Turing
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complete\ldots{}if this doesn't ring any bell with you: it roughly
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means that every algorithm we know can, in principle, be implemented in
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brainf***. It just takes a lot of determination and quite a lot of
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memory resources. Some relatively sophisticated sample programs in
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brainf*** are given in the file \texttt{bf.scala}.\bigskip
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\noindent
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As mentioned above, brainf*** has 8 single-character commands, namely
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\texttt{'>'}, \texttt{'<'}, \texttt{'+'}, \texttt{'-'}, \texttt{'.'},
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\texttt{','}, \texttt{'['} and \texttt{']'}. Every other character is
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considered a comment.  Brainf*** operates on memory cells containing
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integers. For this it uses a single memory pointer that points at each
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stage to one memory cell. This pointer can be moved forward by one
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memory cell by using the command \texttt{'>'}, and backward by using
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\texttt{'<'}. The commands \texttt{'+'} and \texttt{'-'} increase,
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respectively decrease, by 1 the content of the memory cell to which
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   480
the memory pointer currently points to. The commands for input/output
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   481
are \texttt{','} and \texttt{'.'}. Output works by reading the content
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diff changeset
   482
of the memory cell to which the memory pointer points to and printing
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   483
it out as an ASCII character. Input works the other way, taking some
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   484
user input and storing it in the cell to which the memory pointer
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   485
points to. The commands \texttt{'['} and \texttt{']'} are looping
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   486
constructs. Everything in between \texttt{'['} and \texttt{']'} is
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   487
repeated until a counter (memory cell) reaches zero.  A typical
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diff changeset
   488
program in brainf*** looks as follows:
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diff changeset
   489
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diff changeset
   490
\begin{center}
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diff changeset
   491
\begin{verbatim}
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   492
 ++++++++[>++++[>++>+++>+++>+<<<<-]>+>+>->>+[<]<-]>>.>---.+++++++
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   493
 ..+++.>>.<-.<.+++.------.--------.>>+.>++.
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   494
\end{verbatim}
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   495
\end{center}  
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diff changeset
   496
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   497
\noindent
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   498
This one prints out Hello World\ldots{}obviously. 
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   499
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   500
\subsubsection*{Tasks (file bf.scala)}
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   502
\begin{itemize}
218
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   503
\item[(2a)] Brainf*** memory is represented by a \texttt{Map} from
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  integers to integers. The empty memory is represented by
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  \texttt{Map()}, that is nothing is stored in the
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  memory. \texttt{Map(0 -> 1, 2 -> 3)} clearly stores \texttt{1} at
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   507
  memory location \texttt{0}; at \texttt{2} it stores \texttt{3}. The
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   508
  convention is that if we query the memory at a location that is
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   509
  \emph{not} defined in the \texttt{Map}, we return \texttt{0}. Write
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diff changeset
   510
  a function, \texttt{sread}, that takes a memory (a \texttt{Map}) and
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   511
  a memory pointer (an \texttt{Int}) as argument, and safely reads the
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diff changeset
   512
  corresponding memory location. If the \texttt{Map} is not defined at
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parents: 111
diff changeset
   513
  the memory pointer, \texttt{sread} returns \texttt{0}.
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parents: 111
diff changeset
   514
22705d22c105 updated
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parents: 111
diff changeset
   515
  Write another function \texttt{write}, which takes a memory, a
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diff changeset
   516
  memory pointer and an integer value as argument and updates the
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parents: 111
diff changeset
   517
  \texttt{Map} with the value at the given memory location. As usual
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parents: 111
diff changeset
   518
  the \texttt{Map} is not updated `in-place' but a new map is created
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diff changeset
   519
  with the same data, except the value is stored at the given memory
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diff changeset
   520
  pointer.\hfill[1 Mark]
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parents: 111
diff changeset
   521
22705d22c105 updated
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diff changeset
   522
\item[(2b)] Write two functions, \texttt{jumpRight} and
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parents: 111
diff changeset
   523
  \texttt{jumpLeft} that are needed to implement the loop constructs
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parents: 111
diff changeset
   524
  of brainf***. They take a program (a \texttt{String}) and a program
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parents: 111
diff changeset
   525
  counter (an \texttt{Int}) as argument and move right (respectively
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parents: 111
diff changeset
   526
  left) in the string in order to find the \textbf{matching}
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parents: 111
diff changeset
   527
  opening/closing bracket. For example, given the following program
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parents: 111
diff changeset
   528
  with the program counter indicated by an arrow:
22705d22c105 updated
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parents: 111
diff changeset
   529
22705d22c105 updated
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parents: 111
diff changeset
   530
  \begin{center}
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parents: 111
diff changeset
   531
  \texttt{--[\barbelow{.}.+>--],>,++}
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parents: 111
diff changeset
   532
  \end{center}
22705d22c105 updated
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parents: 111
diff changeset
   533
22705d22c105 updated
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diff changeset
   534
  then the matching closing bracket is in 9th position (counting from 0) and
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parents: 111
diff changeset
   535
  \texttt{jumpRight} is supposed to return the position just after this
109
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diff changeset
   536
  
218
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diff changeset
   537
  \begin{center}
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parents: 111
diff changeset
   538
  \texttt{--[..+>--]\barbelow{,}>,++}
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parents: 111
diff changeset
   539
  \end{center}
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parents: 111
diff changeset
   540
22705d22c105 updated
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diff changeset
   541
  meaning it jumps to after the loop. Similarly, if you are in 8th position
22705d22c105 updated
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parents: 111
diff changeset
   542
  then \texttt{jumpLeft} is supposed to jump to just after the opening
22705d22c105 updated
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parents: 111
diff changeset
   543
  bracket (that is jumping to the beginning of the loop):
109
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parents: 105
diff changeset
   544
218
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diff changeset
   545
  \begin{center}
22705d22c105 updated
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parents: 111
diff changeset
   546
    \texttt{--[..+>-\barbelow{-}],>,++}
22705d22c105 updated
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parents: 111
diff changeset
   547
    \qquad$\stackrel{\texttt{jumpLeft}}{\longrightarrow}$\qquad
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parents: 111
diff changeset
   548
    \texttt{--[\barbelow{.}.+>--],>,++}
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parents: 111
diff changeset
   549
  \end{center}
22705d22c105 updated
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parents: 111
diff changeset
   550
22705d22c105 updated
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parents: 111
diff changeset
   551
  Unfortunately we have to take into account that there might be
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parents: 111
diff changeset
   552
  other opening and closing brackets on the `way' to find the
22705d22c105 updated
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parents: 111
diff changeset
   553
  matching bracket. For example in the brainf*** program
22705d22c105 updated
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parents: 111
diff changeset
   554
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   555
  \begin{center}
22705d22c105 updated
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parents: 111
diff changeset
   556
  \texttt{--[\barbelow{.}.[+>]--],>,++}
22705d22c105 updated
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parents: 111
diff changeset
   557
  \end{center}
109
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parents: 105
diff changeset
   558
218
22705d22c105 updated
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parents: 111
diff changeset
   559
  we do not want to return the index for the \texttt{'-'} in the 9th
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parents: 111
diff changeset
   560
  position, but the program counter for \texttt{','} in 12th
22705d22c105 updated
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parents: 111
diff changeset
   561
  position. The easiest to find out whether a bracket is matched is by
22705d22c105 updated
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parents: 111
diff changeset
   562
  using levels (which are the third argument in \texttt{jumpLeft} and
22705d22c105 updated
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parents: 111
diff changeset
   563
  \texttt{jumpLeft}). In case of \texttt{jumpRight} you increase the
22705d22c105 updated
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parents: 111
diff changeset
   564
  level by one whenever you find an opening bracket and decrease by
22705d22c105 updated
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parents: 111
diff changeset
   565
  one for a closing bracket. Then in \texttt{jumpRight} you are looking
22705d22c105 updated
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parents: 111
diff changeset
   566
  for the closing bracket on level \texttt{0}. For \texttt{jumpLeft} you
22705d22c105 updated
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parents: 111
diff changeset
   567
  do the opposite. In this way you can find \textbf{matching} brackets
22705d22c105 updated
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parents: 111
diff changeset
   568
  in strings such as
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   569
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   570
  \begin{center}
22705d22c105 updated
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parents: 111
diff changeset
   571
  \texttt{--[\barbelow{.}.[[-]+>[.]]--],>,++}
22705d22c105 updated
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parents: 111
diff changeset
   572
  \end{center}
109
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parents: 105
diff changeset
   573
218
22705d22c105 updated
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parents: 111
diff changeset
   574
  for which \texttt{jumpRight} should produce the position:
22705d22c105 updated
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parents: 111
diff changeset
   575
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   576
  \begin{center}
22705d22c105 updated
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parents: 111
diff changeset
   577
  \texttt{--[..[[-]+>[.]]--]\barbelow{,}>,++}
22705d22c105 updated
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parents: 111
diff changeset
   578
  \end{center}
22705d22c105 updated
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parents: 111
diff changeset
   579
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   580
  It is also possible that the position returned by \texttt{jumpRight} or
22705d22c105 updated
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parents: 111
diff changeset
   581
  \texttt{jumpLeft} is outside the string in cases where there are
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   582
  no matching brackets. For example
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   583
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   584
  \begin{center}
22705d22c105 updated
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parents: 111
diff changeset
   585
  \texttt{--[\barbelow{.}.[[-]+>[.]]--,>,++}
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   586
  \qquad$\stackrel{\texttt{jumpRight}}{\longrightarrow}$\qquad
22705d22c105 updated
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parents: 111
diff changeset
   587
  \texttt{--[..[[-]+>[.]]-->,++\barbelow{\;\phantom{+}}}
22705d22c105 updated
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parents: 111
diff changeset
   588
  \end{center}
22705d22c105 updated
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parents: 111
diff changeset
   589
  \hfill[1 Mark]
109
293ea84d82ca updated
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parents: 105
diff changeset
   590
293ea84d82ca updated
Christian Urban <urbanc@in.tum.de>
parents: 105
diff changeset
   591
218
22705d22c105 updated
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parents: 111
diff changeset
   592
\item[(2c)] Write a recursive function \texttt{run} that executes a
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   593
  brainf*** program. It takes a program, a program counter, a memory
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   594
  pointer and a memory as arguments. If the program counter is outside
22705d22c105 updated
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parents: 111
diff changeset
   595
  the program string, the execution stops and \texttt{run} returns the
22705d22c105 updated
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parents: 111
diff changeset
   596
  memory. If the program counter is inside the string, it reads the
22705d22c105 updated
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parents: 111
diff changeset
   597
  corresponding character and updates the program counter \texttt{pc},
22705d22c105 updated
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parents: 111
diff changeset
   598
  memory pointer \texttt{mp} and memory \texttt{mem} according to the
22705d22c105 updated
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parents: 111
diff changeset
   599
  rules shown in Figure~\ref{comms}. It then calls recursively
22705d22c105 updated
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parents: 111
diff changeset
   600
  \texttt{run} with the updated data.
22705d22c105 updated
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parents: 111
diff changeset
   601
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   602
  Write another function \texttt{start} that calls \texttt{run} with a
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   603
  given brainfu** program and memory, and the program counter and memory pointer
22705d22c105 updated
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parents: 111
diff changeset
   604
  set to~$0$. Like \texttt{run} it returns the memory after the execution
22705d22c105 updated
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parents: 111
diff changeset
   605
  of the program finishes. You can test your brainf**k interpreter with the
22705d22c105 updated
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parents: 111
diff changeset
   606
  Sierpinski triangle or the Hello world programs or have a look at
109
293ea84d82ca updated
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parents: 105
diff changeset
   607
218
22705d22c105 updated
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parents: 111
diff changeset
   608
  \begin{center}
22705d22c105 updated
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parents: 111
diff changeset
   609
  \url{https://esolangs.org/wiki/Brainfuck}
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   610
  \end{center}\hfill[2 Marks]
109
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parents: 105
diff changeset
   611
  
218
22705d22c105 updated
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parents: 111
diff changeset
   612
  \begin{figure}[p]
22705d22c105 updated
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parents: 111
diff changeset
   613
  \begin{center}
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   614
    \begin{tabular}{|@{}p{0.8cm}|l|}
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   615
      \hline
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   616
      \hfill\texttt{'>'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   617
                       $\bullet$ & $\texttt{pc} + 1$\\
22705d22c105 updated
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parents: 111
diff changeset
   618
                       $\bullet$ & $\texttt{mp} + 1$\\
22705d22c105 updated
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parents: 111
diff changeset
   619
                       $\bullet$ & \texttt{mem} unchanged
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   620
                     \end{tabular}\\\hline   
22705d22c105 updated
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parents: 111
diff changeset
   621
      \hfill\texttt{'<'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
22705d22c105 updated
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parents: 111
diff changeset
   622
                       $\bullet$ & $\texttt{pc} + 1$\\
22705d22c105 updated
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parents: 111
diff changeset
   623
                       $\bullet$ & $\texttt{mp} - 1$\\
22705d22c105 updated
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parents: 111
diff changeset
   624
                       $\bullet$ & \texttt{mem} unchanged
22705d22c105 updated
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parents: 111
diff changeset
   625
                     \end{tabular}\\\hline   
22705d22c105 updated
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parents: 111
diff changeset
   626
      \hfill\texttt{'+'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
22705d22c105 updated
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parents: 111
diff changeset
   627
                       $\bullet$ & $\texttt{pc} + 1$\\
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   628
                       $\bullet$ & $\texttt{mp}$ unchanged\\
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   629
                       $\bullet$ & \texttt{mem} updated with \texttt{mp -> mem(mp) + 1}\\
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   630
                     \end{tabular}\\\hline   
22705d22c105 updated
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parents: 111
diff changeset
   631
      \hfill\texttt{'-'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
22705d22c105 updated
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parents: 111
diff changeset
   632
                       $\bullet$ & $\texttt{pc} + 1$\\
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   633
                       $\bullet$ & $\texttt{mp}$ unchanged\\
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   634
                       $\bullet$ & \texttt{mem} updated with \texttt{mp -> mem(mp) - 1}\\
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   635
                     \end{tabular}\\\hline   
22705d22c105 updated
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parents: 111
diff changeset
   636
      \hfill\texttt{'.'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
22705d22c105 updated
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parents: 111
diff changeset
   637
                       $\bullet$ & $\texttt{pc} + 1$\\
22705d22c105 updated
Christian Urban <urbanc@in.tum.de>
parents: 111
diff changeset
   638
                       $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\\
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   639
                       $\bullet$ & print out \,\texttt{mem(mp)} as a character\\
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                     \end{tabular}\\\hline   
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      \hfill\texttt{','} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
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   642
                       $\bullet$ & $\texttt{pc} + 1$\\
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   643
                       $\bullet$ & $\texttt{mp}$ unchanged\\
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   644
                       $\bullet$ & \texttt{mem} updated with \texttt{mp -> \textrm{input}}\\
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   645
                       \multicolumn{2}{@{}l}{the input is given by \texttt{Console.in.read().toByte}}
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                     \end{tabular}\\\hline   
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   647
      \hfill\texttt{'['} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
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   648
                       \multicolumn{2}{@{}l}{if \texttt{mem(mp) == 0} then}\\
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   649
                       $\bullet$ & $\texttt{pc = jumpRight(prog, pc + 1, 0)}$\\
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   650
                       $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\medskip\\
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   651
                       \multicolumn{2}{@{}l}{otherwise if \texttt{mem(mp) != 0} then}\\
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   652
                       $\bullet$ & $\texttt{pc} + 1$\\
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   653
                       $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\\
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   654
                     \end{tabular}
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                     \\\hline   
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      \hfill\texttt{']'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
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   657
                       \multicolumn{2}{@{}l}{if \texttt{mem(mp) != 0} then}\\
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   658
                       $\bullet$ & $\texttt{pc = jumpLeft(prog, pc - 1, 0)}$\\
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   659
                       $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\medskip\\
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diff changeset
   660
                       \multicolumn{2}{@{}l}{otherwise if \texttt{mem(mp) == 0} then}\\
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   661
                       $\bullet$ & $\texttt{pc} + 1$\\
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diff changeset
   662
                       $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\\
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   663
                     \end{tabular}\\\hline   
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   664
      any other char & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}}
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diff changeset
   665
                         $\bullet$ & $\texttt{pc} + 1$\\
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diff changeset
   666
                         $\bullet$ & \texttt{mp} and \texttt{mem} unchanged
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   667
                       \end{tabular}\\
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      \hline                 
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   669
    \end{tabular}
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   670
  \end{center}
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  \caption{The rules for how commands in the brainf*** language update the program counter \texttt{pc},
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   672
    memory pointer \texttt{mp} and memory \texttt{mem}.\label{comms}}
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  \end{figure}
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\end{itemize}\bigskip  
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   676
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6
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\end{document}
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68
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%%% Local Variables: 
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%%% mode: latex
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%%% TeX-master: t
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%%% End: