diff -r e689375abcc1 -r 22705d22c105 cws/cw04.tex --- a/cws/cw04.tex Fri Nov 23 01:52:37 2018 +0000 +++ b/cws/cw04.tex Tue Nov 27 21:41:59 2018 +0000 @@ -1,184 +1,646 @@ \documentclass{article} \usepackage{../style} \usepackage{../langs} +\usepackage{disclaimer} +\usepackage{tikz} +\usepackage{pgf} +\usepackage{pgfplots} +\usepackage{stackengine} +%% \usepackage{accents} +\newcommand\barbelow[1]{\stackunder[1.2pt]{#1}{\raisebox{-4mm}{\boldmath$\uparrow$}}} + +\begin{filecontents}{re-python2.data} +1 0.033 +5 0.036 +10 0.034 +15 0.036 +18 0.059 +19 0.084 +20 0.141 +21 0.248 +22 0.485 +23 0.878 +24 1.71 +25 3.40 +26 7.08 +27 14.12 +28 26.69 +\end{filecontents} + +\begin{filecontents}{re-java.data} +5 0.00298 +10 0.00418 +15 0.00996 +16 0.01710 +17 0.03492 +18 0.03303 +19 0.05084 +20 0.10177 +21 0.19960 +22 0.41159 +23 0.82234 +24 1.70251 +25 3.36112 +26 6.63998 +27 13.35120 +28 29.81185 +\end{filecontents} + +\begin{filecontents}{re-java9.data} +1000 0.01410 +2000 0.04882 +3000 0.10609 +4000 0.17456 +5000 0.27530 +6000 0.41116 +7000 0.53741 +8000 0.70261 +9000 0.93981 +10000 0.97419 +11000 1.28697 +12000 1.51387 +14000 2.07079 +16000 2.69846 +20000 4.41823 +24000 6.46077 +26000 7.64373 +30000 9.99446 +34000 12.966885 +38000 16.281621 +42000 19.180228 +46000 21.984721 +50000 26.950203 +60000 43.0327746 +\end{filecontents} + \begin{document} -\section*{Replacement Coursework 1 (Roman Numerals)} +% BF IDE +% https://www.microsoft.com/en-us/p/brainf-ck/9nblgggzhvq5 + +\section*{Coursework 8 (Regular Expressions and Brainf***)} -This coursework is worth 10\%. It is about translating roman numerals -into integers and also about validating roman numerals. The coursework -is due on 2 February at 5pm. Make sure the files you submit can be -processed by just calling \texttt{scala <>}.\bigskip +This coursework is worth 10\%. It is about regular expressions, +pattern matching and an interpreter. The first part is due on 30 +November at 11pm; the second, more advanced part, is due on 21 +December at 11pm. In the first part, you are asked to implement a +regular expression matcher based on derivatives of regular +expressions. The reason is that regular expression matching in Java +and Python can sometimes be extremely slow. The advanced part is about +an interpreter for a very simple programming language.\bigskip + +\IMPORTANT{} \noindent -\textbf{Important:} Do not use any mutable data structures in your -submission! They are not needed. This menas you cannot use -\texttt{ListBuffer}s, for example. Do not use \texttt{return} in your -code! It has a different meaning in Scala, than in Java. Do not use -\texttt{var}! This declares a mutable variable. Make sure the -functions you submit are defined on the ``top-level'' of Scala, not -inside a class or object. Also note that the running time will be -restricted to a maximum of 360 seconds on my laptop. +Also note that the running time of each part will be restricted to a +maximum of 30 seconds on my laptop. + +\DISCLAIMER{} -\subsection*{Disclaimer} +\subsection*{Part 1 (6 Marks)} + +The task is to implement a regular expression matcher that is based on +derivatives of regular expressions. Most of the functions are defined by +recursion over regular expressions and can be elegantly implemented +using Scala's pattern-matching. The implementation should deal with the +following regular expressions, which have been predefined in the file +\texttt{re.scala}: -It should be understood that the work you submit represents your own -effort! You have not copied from anyone else. An exception is the -Scala code I showed during the lectures or uploaded to KEATS, which -you can freely use.\bigskip - +\begin{center} +\begin{tabular}{lcll} + $r$ & $::=$ & $\ZERO$ & cannot match anything\\ + & $|$ & $\ONE$ & can only match the empty string\\ + & $|$ & $c$ & can match a single character (in this case $c$)\\ + & $|$ & $r_1 + r_2$ & can match a string either with $r_1$ or with $r_2$\\ + & $|$ & $r_1\cdot r_2$ & can match the first part of a string with $r_1$ and\\ + & & & then the second part with $r_2$\\ + & $|$ & $r^*$ & can match zero or more times $r$\\ +\end{tabular} +\end{center} -\subsection*{Part 1 (Translation)} +\noindent +Why? Knowing how to match regular expressions and strings will let you +solve a lot of problems that vex other humans. Regular expressions are +one of the fastest and simplest ways to match patterns in text, and +are endlessly useful for searching, editing and analysing data in all +sorts of places (for example analysing network traffic in order to +detect security breaches). However, you need to be fast, otherwise you +will stumble over problems such as recently reported at + +{\small +\begin{itemize} +\item[$\bullet$] \url{http://stackstatus.net/post/147710624694/outage-postmortem-july-20-2016} +\item[$\bullet$] \url{https://vimeo.com/112065252} +\item[$\bullet$] \url{http://davidvgalbraith.com/how-i-fixed-atom/} +\end{itemize}} + +\subsubsection*{Tasks (file re.scala)} -\noindent -Roman numerals are strings consisting of the letters $I$, $V$, $X$, -$L$, $C$, $D$, and $M$. Such strings should be transformed into an -internal representation using the datatypes \texttt{RomanDigit} and -\texttt{RomanNumeral} (defined in \texttt{roman.scala}), and then from -this internal representation converted into Integers. +The file \texttt{re.scala} has already a definition for regular +expressions and also defines some handy shorthand notation for +regular expressions. The notation in this document matches up +with the code in the file as follows: + +\begin{center} + \begin{tabular}{rcl@{\hspace{10mm}}l} + & & code: & shorthand:\smallskip \\ + $\ZERO$ & $\mapsto$ & \texttt{ZERO}\\ + $\ONE$ & $\mapsto$ & \texttt{ONE}\\ + $c$ & $\mapsto$ & \texttt{CHAR(c)}\\ + $r_1 + r_2$ & $\mapsto$ & \texttt{ALT(r1, r2)} & \texttt{r1 | r2}\\ + $r_1 \cdot r_2$ & $\mapsto$ & \texttt{SEQ(r1, r2)} & \texttt{r1 $\sim$ r2}\\ + $r^*$ & $\mapsto$ & \texttt{STAR(r)} & \texttt{r.\%} +\end{tabular} +\end{center} + \begin{itemize} -\item[(1)] First write a polymorphic function that recursively - transforms a list of options into an option of a list. For example, - if you have the lists on the left-hand side, they should be transformed into - the options on the right-hand side: +\item[(1a)] Implement a function, called \textit{nullable}, by + recursion over regular expressions. This function tests whether a + regular expression can match the empty string. This means given a + regular expression it either returns true or false. The function + \textit{nullable} + is defined as follows: + +\begin{center} +\begin{tabular}{lcl} +$\textit{nullable}(\ZERO)$ & $\dn$ & $\textit{false}$\\ +$\textit{nullable}(\ONE)$ & $\dn$ & $\textit{true}$\\ +$\textit{nullable}(c)$ & $\dn$ & $\textit{false}$\\ +$\textit{nullable}(r_1 + r_2)$ & $\dn$ & $\textit{nullable}(r_1) \vee \textit{nullable}(r_2)$\\ +$\textit{nullable}(r_1 \cdot r_2)$ & $\dn$ & $\textit{nullable}(r_1) \wedge \textit{nullable}(r_2)$\\ +$\textit{nullable}(r^*)$ & $\dn$ & $\textit{true}$\\ +\end{tabular} +\end{center}~\hfill[1 Mark] - \begin{center} - \begin{tabular}{lcl} - \texttt{List(Some(1), Some(2), Some(3))} & $\Rightarrow$ & - \texttt{Some(List(1, 2, 3))} \\ - \texttt{List(Some(1), None, Some(3))} & $\Rightarrow$ & - \texttt{None} \\ - \texttt{List()} & $\Rightarrow$ & \texttt{Some(List())} - \end{tabular} - \end{center} +\item[(1b)] Implement a function, called \textit{der}, by recursion over + regular expressions. It takes a character and a regular expression + as arguments and calculates the derivative regular expression according + to the rules: - This means the function should produce \texttt{None} as soon - as a \texttt{None} is inside the list. Otherwise it produces - a list of all \texttt{Some}s. In case the list is empty, it - produces \texttt{Some} of the empty list. \hfill[1 Mark] +\begin{center} +\begin{tabular}{lcl} +$\textit{der}\;c\;(\ZERO)$ & $\dn$ & $\ZERO$\\ +$\textit{der}\;c\;(\ONE)$ & $\dn$ & $\ZERO$\\ +$\textit{der}\;c\;(d)$ & $\dn$ & $\textit{if}\; c = d\;\textit{then} \;\ONE \; \textit{else} \;\ZERO$\\ +$\textit{der}\;c\;(r_1 + r_2)$ & $\dn$ & $(\textit{der}\;c\;r_1) + (\textit{der}\;c\;r_2)$\\ +$\textit{der}\;c\;(r_1 \cdot r_2)$ & $\dn$ & $\textit{if}\;\textit{nullable}(r_1)$\\ + & & $\textit{then}\;((\textit{der}\;c\;r_1)\cdot r_2) + (\textit{der}\;c\;r_2)$\\ + & & $\textit{else}\;(\textit{der}\;c\;r_1)\cdot r_2$\\ +$\textit{der}\;c\;(r^*)$ & $\dn$ & $(\textit{der}\;c\;r)\cdot (r^*)$\\ +\end{tabular} +\end{center} + +For example given the regular expression $r = (a \cdot b) \cdot c$, the derivatives +w.r.t.~the characters $a$, $b$ and $c$ are - -\item[(2)] Write first a function that converts the characters $I$, $V$, - $X$, $L$, $C$, $D$, and $M$ into an option of a \texttt{RomanDigit}. - If it is one of the roman digits, it should produce \texttt{Some}; - otherwise \texttt{None}. - - Next write a function that converts a string into a - \texttt{RomanNumeral}. Again, this function should return an - \texttt{Option}: If the string consists of $I$, $V$, $X$, $L$, $C$, - $D$, and $M$ only, then it produces \texttt{Some}; otherwise if - there is any other character in the string, it should produce - \texttt{None}. The empty string is just the empty - \texttt{RomanNumeral}, that is the empty list of - \texttt{RomanDigit}'s. You should use the function under Task (1) - to produce the result. \hfill[2 Marks] +\begin{center} + \begin{tabular}{lcll} + $\textit{der}\;a\;r$ & $=$ & $(\ONE \cdot b)\cdot c$ & ($= r'$)\\ + $\textit{der}\;b\;r$ & $=$ & $(\ZERO \cdot b)\cdot c$\\ + $\textit{der}\;c\;r$ & $=$ & $(\ZERO \cdot b)\cdot c$ + \end{tabular} +\end{center} + +Let $r'$ stand for the first derivative, then taking the derivatives of $r'$ +w.r.t.~the characters $a$, $b$ and $c$ gives + +\begin{center} + \begin{tabular}{lcll} + $\textit{der}\;a\;r'$ & $=$ & $((\ZERO \cdot b) + \ZERO)\cdot c$ \\ + $\textit{der}\;b\;r'$ & $=$ & $((\ZERO \cdot b) + \ONE)\cdot c$ & ($= r''$)\\ + $\textit{der}\;c\;r'$ & $=$ & $((\ZERO \cdot b) + \ZERO)\cdot c$ + \end{tabular} +\end{center} -\item[(3)] Write a recursive function \texttt{RomanNumral2Int} that - converts a \texttt{RomanNumeral} into an integer. You can assume the - generated integer will be between 0 and 3999. The argument of the - function is a list of roman digits. It should look how this list - starts and then calculate what the corresponding integer is for this - ``start'' and add it with the integer for the rest of the list. That - means if the argument is of the form shown on the left-hand side, it - should do the calculation on the right-hand side. +One more example: Let $r''$ stand for the second derivative above, +then taking the derivatives of $r''$ w.r.t.~the characters $a$, $b$ +and $c$ gives + +\begin{center} + \begin{tabular}{lcll} + $\textit{der}\;a\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ZERO$ \\ + $\textit{der}\;b\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ZERO$\\ + $\textit{der}\;c\;r''$ & $=$ & $((\ZERO \cdot b) + \ZERO) \cdot c + \ONE$ & + (is $\textit{nullable}$) + \end{tabular} +\end{center} + +Note, the last derivative can match the empty string, that is it is \textit{nullable}.\\ +\mbox{}\hfill\mbox{[1 Mark]} + +\item[(1c)] Implement the function \textit{simp}, which recursively + traverses a regular expression from the inside to the outside, and + on the way simplifies every regular expression on the left (see + below) to the regular expression on the right, except it does not + simplify inside ${}^*$-regular expressions. \begin{center} - \begin{tabular}{lcl} - $M::r$ & $\Rightarrow$ & $1000 + \text{roman numeral of rest}\; r$\\ - $C::M::r$ & $\Rightarrow$ & $900 + \text{roman numeral of rest}\; r$\\ - $D::r$ & $\Rightarrow$ & $500 + \text{roman numeral of rest}\; r$\\ - $C::D::r$ & $\Rightarrow$ & $400 + \text{roman numeral of rest}\; r$\\ - $C::r$ & $\Rightarrow$ & $100 + \text{roman numeral of rest}\; r$\\ - $X::C::r$ & $\Rightarrow$ & $90 + \text{roman numeral of rest}\; r$\\ - $L::r$ & $\Rightarrow$ & $50 + \text{roman numeral of rest}\; r$\\ - $X::L::r$ & $\Rightarrow$ & $40 + \text{roman numeral of rest}\; r$\\ - $X::r$ & $\Rightarrow$ & $10 + \text{roman numeral of rest}\; r$\\ - $I::X::r$ & $\Rightarrow$ & $9 + \text{roman numeral of rest}\; r$\\ - $V::r$ & $\Rightarrow$ & $5 + \text{roman numeral of rest}\; r$\\ - $I::V::r$ & $\Rightarrow$ & $4 + \text{roman numeral of rest}\; r$\\ - $I::r$ & $\Rightarrow$ & $1 + \text{roman numeral of rest}\; r$ - \end{tabular} - \end{center} - - The empty list will be converted to integer $0$.\hfill[1 Mark] - -\item[(4)] Write a function that takes a string and if possible - converts it into the internal representation. If successful, it then - calculates the integer (an option of an integer) according to the - function in (3). If this is not possible, then return - \texttt{None}.\hfill[1 Mark] - - -\item[(5)] The file \texttt{roman.txt} contains a list of roman numerals. - Read in these numerals, convert them into integers and then add them all - up. The Scala function for reading a file is - - \begin{center} - \texttt{Source.fromFile("filename")("ISO-8859-9")} +\begin{tabular}{l@{\hspace{4mm}}c@{\hspace{4mm}}ll} +$r \cdot \ZERO$ & $\mapsto$ & $\ZERO$\\ +$\ZERO \cdot r$ & $\mapsto$ & $\ZERO$\\ +$r \cdot \ONE$ & $\mapsto$ & $r$\\ +$\ONE \cdot r$ & $\mapsto$ & $r$\\ +$r + \ZERO$ & $\mapsto$ & $r$\\ +$\ZERO + r$ & $\mapsto$ & $r$\\ +$r + r$ & $\mapsto$ & $r$\\ +\end{tabular} \end{center} - Make sure you process the strings correctly by ignoring whitespaces - where needed.\\ \mbox{}\hfill[1 Mark] + For example the regular expression + \[(r_1 + \ZERO) \cdot \ONE + ((\ONE + r_2) + r_3) \cdot (r_4 \cdot \ZERO)\] + + simplifies to just $r_1$. \textbf{Hint:} Regular expressions can be + seen as trees and there are several methods for traversing + trees. One of them corresponds to the inside-out traversal, which is + sometimes also called post-order traversal. Furthermore, + remember numerical expressions from school times: there you had expressions + like $u + \ldots + (1 \cdot x) - \ldots (z + (y \cdot 0)) \ldots$ + and simplification rules that looked very similar to rules + above. You would simplify such numerical expressions by replacing + for example the $y \cdot 0$ by $0$, or $1\cdot x$ by $x$, and then + look whether more rules are applicable. If you organise the + simplification in an inside-out fashion, it is always clear which + rule should be applied next.\hfill[2 Marks] + +\item[(1d)] Implement two functions: The first, called \textit{ders}, + takes a list of characters and a regular expression as arguments, and + builds the derivative w.r.t.~the list as follows: + +\begin{center} +\begin{tabular}{lcl} +$\textit{ders}\;(Nil)\;r$ & $\dn$ & $r$\\ + $\textit{ders}\;(c::cs)\;r$ & $\dn$ & + $\textit{ders}\;cs\;(\textit{simp}(\textit{der}\;c\;r))$\\ +\end{tabular} +\end{center} + +Note that this function is different from \textit{der}, which only +takes a single character. + +The second function, called \textit{matcher}, takes a string and a +regular expression as arguments. It builds first the derivatives +according to \textit{ders} and after that tests whether the resulting +derivative regular expression can match the empty string (using +\textit{nullable}). For example the \textit{matcher} will produce +true for the regular expression $(a\cdot b)\cdot c$ and the string +$abc$, but false if you give it the string $ab$. \hfill[1 Mark] + +\item[(1e)] Implement a function, called \textit{size}, by recursion + over regular expressions. If a regular expression is seen as a tree, + then \textit{size} should return the number of nodes in such a + tree. Therefore this function is defined as follows: + +\begin{center} +\begin{tabular}{lcl} +$\textit{size}(\ZERO)$ & $\dn$ & $1$\\ +$\textit{size}(\ONE)$ & $\dn$ & $1$\\ +$\textit{size}(c)$ & $\dn$ & $1$\\ +$\textit{size}(r_1 + r_2)$ & $\dn$ & $1 + \textit{size}(r_1) + \textit{size}(r_2)$\\ +$\textit{size}(r_1 \cdot r_2)$ & $\dn$ & $1 + \textit{size}(r_1) + \textit{size}(r_2)$\\ +$\textit{size}(r^*)$ & $\dn$ & $1 + \textit{size}(r)$\\ +\end{tabular} +\end{center} + +You can use \textit{size} in order to test how much the `evil' regular +expression $(a^*)^* \cdot b$ grows when taking successive derivatives +according the letter $a$ without simplification and then compare it to +taking the derivative, but simplify the result. The sizes +are given in \texttt{re.scala}. \hfill[1 Mark] \end{itemize} +\subsection*{Background} + +Although easily implementable in Scala, the idea behind the derivative +function might not so easy to be seen. To understand its purpose +better, assume a regular expression $r$ can match strings of the form +$c\!::\!cs$ (that means strings which start with a character $c$ and have +some rest, or tail, $cs$). If you take the derivative of $r$ with +respect to the character $c$, then you obtain a regular expression +that can match all the strings $cs$. In other words, the regular +expression $\textit{der}\;c\;r$ can match the same strings $c\!::\!cs$ +that can be matched by $r$, except that the $c$ is chopped off. + +Assume now $r$ can match the string $abc$. If you take the derivative +according to $a$ then you obtain a regular expression that can match +$bc$ (it is $abc$ where the $a$ has been chopped off). If you now +build the derivative $\textit{der}\;b\;(\textit{der}\;a\;r)$ you +obtain a regular expression that can match the string $c$ (it is $bc$ +where $b$ is chopped off). If you finally build the derivative of this +according $c$, that is +$\textit{der}\;c\;(\textit{der}\;b\;(\textit{der}\;a\;r))$, you obtain +a regular expression that can match the empty string. You can test +whether this is indeed the case using the function nullable, which is +what your matcher is doing. + +The purpose of the $\textit{simp}$ function is to keep the regular +expressions small. Normally the derivative function makes the regular +expression bigger (see the SEQ case and the example in (1b)) and the +algorithm would be slower and slower over time. The $\textit{simp}$ +function counters this increase in size and the result is that the +algorithm is fast throughout. By the way, this algorithm is by Janusz +Brzozowski who came up with the idea of derivatives in 1964 in his PhD +thesis. + +\begin{center}\small +\url{https://en.wikipedia.org/wiki/Janusz_Brzozowski_(computer_scientist)} +\end{center} + -\subsection*{Part 2 (Validation)} +If you want to see how badly the regular expression matchers do in +Java\footnote{Version 8 and below; Version 9 does not seem to be as + catastrophic, but still worse than the regular expression matcher +based on derivatives.} and in Python with the `evil' regular +expression $(a^*)^*\cdot b$, then have a look at the graphs below (you +can try it out for yourself: have a look at the file +\texttt{catastrophic.java} and \texttt{catastrophic.py} on +KEATS). Compare this with the matcher you have implemented. How long +can the string of $a$'s be in your matcher and still stay within the +30 seconds time limit? -As you can see the function under Task (3) can produce some unexpected -results. For example for $XXCIII$ it produces 103. The reason for this -unexpected result is that $XXCIII$ is actually not a valid roman -number, neither is $IIII$ for 4 nor $MIM$ for 1999. Although actual -Romans were not so fussy about this,\footnote{They happily used - numbers like $XIIX$ or $IIXX$ for 18.} but modern times declared -that there are precise rules for what a valid roman number is, namely: +\begin{center} +\begin{tabular}{@{}cc@{}} +\multicolumn{2}{c}{Graph: $(a^*)^*\cdot b$ and strings + $\underbrace{a\ldots a}_{n}$}\bigskip\\ + +\begin{tikzpicture} +\begin{axis}[ + xlabel={$n$}, + x label style={at={(1.05,0.0)}}, + ylabel={time in secs}, + y label style={at={(0.06,0.5)}}, + enlargelimits=false, + xtick={0,5,...,30}, + xmax=33, + ymax=45, + ytick={0,5,...,40}, + scaled ticks=false, + axis lines=left, + width=6cm, + height=5.5cm, + legend entries={Python, Java 8}, + legend pos=north west] +\addplot[blue,mark=*, mark options={fill=white}] table {re-python2.data}; +\addplot[cyan,mark=*, mark options={fill=white}] table {re-java.data}; +\end{axis} +\end{tikzpicture} + & +\begin{tikzpicture} +\begin{axis}[ + xlabel={$n$}, + x label style={at={(1.05,0.0)}}, + ylabel={time in secs}, + y label style={at={(0.06,0.5)}}, + %enlargelimits=false, + %xtick={0,5000,...,30000}, + xmax=65000, + ymax=45, + ytick={0,5,...,40}, + scaled ticks=false, + axis lines=left, + width=6cm, + height=5.5cm, + legend entries={Java 9}, + legend pos=north west] +\addplot[cyan,mark=*, mark options={fill=white}] table {re-java9.data}; +\end{axis} +\end{tikzpicture} +\end{tabular} +\end{center} +\newpage + +\subsection*{Part 2 (4 Marks)} + +Coming from Java or C++, you might think Scala is a quite esoteric +programming language. But remember, some serious companies have built +their business on +Scala.\footnote{\url{https://en.wikipedia.org/wiki/Scala_(programming_language)\#Companies}} +And there are far, far more esoteric languages out there. One is +called \emph{brainf***}. You are asked in this part to implement an +interpreter for this language. + +Urban M\"uller developed brainf*** in 1993. A close relative of this +language was already introduced in 1964 by Corado B\"ohm, an Italian +computer pioneer, who unfortunately died a few months ago. The main +feature of brainf*** is its minimalistic set of instructions---just 8 +instructions in total and all of which are single characters. Despite +the minimalism, this language has been shown to be Turing +complete\ldots{}if this doesn't ring any bell with you: it roughly +means that every algorithm we know can, in principle, be implemented in +brainf***. It just takes a lot of determination and quite a lot of +memory resources. Some relatively sophisticated sample programs in +brainf*** are given in the file \texttt{bf.scala}.\bigskip + +\noindent +As mentioned above, brainf*** has 8 single-character commands, namely +\texttt{'>'}, \texttt{'<'}, \texttt{'+'}, \texttt{'-'}, \texttt{'.'}, +\texttt{','}, \texttt{'['} and \texttt{']'}. Every other character is +considered a comment. Brainf*** operates on memory cells containing +integers. For this it uses a single memory pointer that points at each +stage to one memory cell. This pointer can be moved forward by one +memory cell by using the command \texttt{'>'}, and backward by using +\texttt{'<'}. The commands \texttt{'+'} and \texttt{'-'} increase, +respectively decrease, by 1 the content of the memory cell to which +the memory pointer currently points to. The commands for input/output +are \texttt{','} and \texttt{'.'}. Output works by reading the content +of the memory cell to which the memory pointer points to and printing +it out as an ASCII character. Input works the other way, taking some +user input and storing it in the cell to which the memory pointer +points to. The commands \texttt{'['} and \texttt{']'} are looping +constructs. Everything in between \texttt{'['} and \texttt{']'} is +repeated until a counter (memory cell) reaches zero. A typical +program in brainf*** looks as follows: + +\begin{center} +\begin{verbatim} + ++++++++[>++++[>++>+++>+++>+<<<<-]>+>+>->>+[<]<-]>>.>---.+++++++ + ..+++.>>.<-.<.+++.------.--------.>>+.>++. +\end{verbatim} +\end{center} + +\noindent +This one prints out Hello World\ldots{}obviously. + +\subsubsection*{Tasks (file bf.scala)} \begin{itemize} -\item Repeatable roman digits are $I$, $X$, $C$ and $M$. The other ones - are non-repeatable. Repeatable digits can be repeated upto 3 times in a - number (for example $MMM$ is OK); non-repeatable digits cannot be - repeated at all (for example $VV$ is excluded). +\item[(2a)] Brainf*** memory is represented by a \texttt{Map} from + integers to integers. The empty memory is represented by + \texttt{Map()}, that is nothing is stored in the + memory. \texttt{Map(0 -> 1, 2 -> 3)} clearly stores \texttt{1} at + memory location \texttt{0}; at \texttt{2} it stores \texttt{3}. The + convention is that if we query the memory at a location that is + \emph{not} defined in the \texttt{Map}, we return \texttt{0}. Write + a function, \texttt{sread}, that takes a memory (a \texttt{Map}) and + a memory pointer (an \texttt{Int}) as argument, and safely reads the + corresponding memory location. If the \texttt{Map} is not defined at + the memory pointer, \texttt{sread} returns \texttt{0}. + + Write another function \texttt{write}, which takes a memory, a + memory pointer and an integer value as argument and updates the + \texttt{Map} with the value at the given memory location. As usual + the \texttt{Map} is not updated `in-place' but a new map is created + with the same data, except the value is stored at the given memory + pointer.\hfill[1 Mark] + +\item[(2b)] Write two functions, \texttt{jumpRight} and + \texttt{jumpLeft} that are needed to implement the loop constructs + of brainf***. They take a program (a \texttt{String}) and a program + counter (an \texttt{Int}) as argument and move right (respectively + left) in the string in order to find the \textbf{matching} + opening/closing bracket. For example, given the following program + with the program counter indicated by an arrow: + + \begin{center} + \texttt{--[\barbelow{.}.+>--],>,++} + \end{center} + + then the matching closing bracket is in 9th position (counting from 0) and + \texttt{jumpRight} is supposed to return the position just after this -\item If a smaller digits precedes a bigger digit, then $I$ can precede $V$ and $X$; $X$ can preced - $L$ and $C$; and $C$ can preced $D$ and $M$. No other combination is permitted in this case. + \begin{center} + \texttt{--[..+>--]\barbelow{,}>,++} + \end{center} + + meaning it jumps to after the loop. Similarly, if you are in 8th position + then \texttt{jumpLeft} is supposed to jump to just after the opening + bracket (that is jumping to the beginning of the loop): -\item If a smaller digit precedes a bigger digit (for example $IV$), then the smaller number - must be either the first digit in the number, or follow a digit which is at least 10 times its value. - So $VIV$ is excluded, because $I$ follows $V$ and $I * 10$ is bigger than $V$; but $XIV$ is - allowed, because $I$ follows $X$ and $I * 10$ is equal to $X$. + \begin{center} + \texttt{--[..+>-\barbelow{-}],>,++} + \qquad$\stackrel{\texttt{jumpLeft}}{\longrightarrow}$\qquad + \texttt{--[\barbelow{.}.+>--],>,++} + \end{center} + + Unfortunately we have to take into account that there might be + other opening and closing brackets on the `way' to find the + matching bracket. For example in the brainf*** program + + \begin{center} + \texttt{--[\barbelow{.}.[+>]--],>,++} + \end{center} -\item Let us say two digits are called a \emph{compound} roman digit - when a smaller digit precedes a bigger digit (so $IV$, $XL$, $CM$ - for example). If a compound digit is followed by another digit, then - this digit must be smaller than the first digit in the compound - digit. For example $IXI$ is excluded, but $XLI$ is not. + we do not want to return the index for the \texttt{'-'} in the 9th + position, but the program counter for \texttt{','} in 12th + position. The easiest to find out whether a bracket is matched is by + using levels (which are the third argument in \texttt{jumpLeft} and + \texttt{jumpLeft}). In case of \texttt{jumpRight} you increase the + level by one whenever you find an opening bracket and decrease by + one for a closing bracket. Then in \texttt{jumpRight} you are looking + for the closing bracket on level \texttt{0}. For \texttt{jumpLeft} you + do the opposite. In this way you can find \textbf{matching} brackets + in strings such as + + \begin{center} + \texttt{--[\barbelow{.}.[[-]+>[.]]--],>,++} + \end{center} -\item The empty roman numeral is valid. -\end{itemize} + for which \texttt{jumpRight} should produce the position: + + \begin{center} + \texttt{--[..[[-]+>[.]]--]\barbelow{,}>,++} + \end{center} + + It is also possible that the position returned by \texttt{jumpRight} or + \texttt{jumpLeft} is outside the string in cases where there are + no matching brackets. For example + + \begin{center} + \texttt{--[\barbelow{.}.[[-]+>[.]]--,>,++} + \qquad$\stackrel{\texttt{jumpRight}}{\longrightarrow}$\qquad + \texttt{--[..[[-]+>[.]]-->,++\barbelow{\;\phantom{+}}} + \end{center} + \hfill[1 Mark] -\noindent -The tasks in this part are as follows: -\begin{itemize} -\item[(6)] Implement a recursive function \texttt{isValidNumeral} that - takes a \texttt{RomanNumeral} as argument and produces true if \textbf{all} - the rules above are satisfied, and otherwise false. +\item[(2c)] Write a recursive function \texttt{run} that executes a + brainf*** program. It takes a program, a program counter, a memory + pointer and a memory as arguments. If the program counter is outside + the program string, the execution stops and \texttt{run} returns the + memory. If the program counter is inside the string, it reads the + corresponding character and updates the program counter \texttt{pc}, + memory pointer \texttt{mp} and memory \texttt{mem} according to the + rules shown in Figure~\ref{comms}. It then calls recursively + \texttt{run} with the updated data. + + Write another function \texttt{start} that calls \texttt{run} with a + given brainfu** program and memory, and the program counter and memory pointer + set to~$0$. Like \texttt{run} it returns the memory after the execution + of the program finishes. You can test your brainf**k interpreter with the + Sierpinski triangle or the Hello world programs or have a look at - Hint: It might be more convenient to test when the rules fail and then return false; - return true in all other cases. - \mbox{}\hfill[2 Marks] - -\item[(7)] Write a recursive function that converts an Integer into a \texttt{RomanNumeral}. - You can assume the function will only be called for integers between 0 and 3999.\mbox{}\hfill[1 Mark] + \begin{center} + \url{https://esolangs.org/wiki/Brainfuck} + \end{center}\hfill[2 Marks] -\item[(8)] Write a function that reads a text file (for example \texttt{roman2.txt}) - containing valid and invalid roman numerals. Convert all valid roman numerals into - integers, add them up and produce the result as a \texttt{RomanNumeral} (using the function - from (7)). \hfill[1 Mark] -\end{itemize} - + \begin{figure}[p] + \begin{center} + \begin{tabular}{|@{}p{0.8cm}|l|} + \hline + \hfill\texttt{'>'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & $\texttt{mp} + 1$\\ + $\bullet$ & \texttt{mem} unchanged + \end{tabular}\\\hline + \hfill\texttt{'<'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & $\texttt{mp} - 1$\\ + $\bullet$ & \texttt{mem} unchanged + \end{tabular}\\\hline + \hfill\texttt{'+'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & $\texttt{mp}$ unchanged\\ + $\bullet$ & \texttt{mem} updated with \texttt{mp -> mem(mp) + 1}\\ + \end{tabular}\\\hline + \hfill\texttt{'-'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & $\texttt{mp}$ unchanged\\ + $\bullet$ & \texttt{mem} updated with \texttt{mp -> mem(mp) - 1}\\ + \end{tabular}\\\hline + \hfill\texttt{'.'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\\ + $\bullet$ & print out \,\texttt{mem(mp)} as a character\\ + \end{tabular}\\\hline + \hfill\texttt{','} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & $\texttt{mp}$ unchanged\\ + $\bullet$ & \texttt{mem} updated with \texttt{mp -> \textrm{input}}\\ + \multicolumn{2}{@{}l}{the input is given by \texttt{Console.in.read().toByte}} + \end{tabular}\\\hline + \hfill\texttt{'['} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + \multicolumn{2}{@{}l}{if \texttt{mem(mp) == 0} then}\\ + $\bullet$ & $\texttt{pc = jumpRight(prog, pc + 1, 0)}$\\ + $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\medskip\\ + \multicolumn{2}{@{}l}{otherwise if \texttt{mem(mp) != 0} then}\\ + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\\ + \end{tabular} + \\\hline + \hfill\texttt{']'} & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + \multicolumn{2}{@{}l}{if \texttt{mem(mp) != 0} then}\\ + $\bullet$ & $\texttt{pc = jumpLeft(prog, pc - 1, 0)}$\\ + $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\medskip\\ + \multicolumn{2}{@{}l}{otherwise if \texttt{mem(mp) == 0} then}\\ + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & $\texttt{mp}$ and \texttt{mem} unchanged\\ + \end{tabular}\\\hline + any other char & \begin{tabular}[t]{@{}l@{\hspace{2mm}}l@{}} + $\bullet$ & $\texttt{pc} + 1$\\ + $\bullet$ & \texttt{mp} and \texttt{mem} unchanged + \end{tabular}\\ + \hline + \end{tabular} + \end{center} + \caption{The rules for how commands in the brainf*** language update the program counter \texttt{pc}, + memory pointer \texttt{mp} and memory \texttt{mem}.\label{comms}} + \end{figure} +\end{itemize}\bigskip + + + \end{document}