cws/cw01.tex
author Christian Urban <urbanc@in.tum.de>
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\documentclass{article}
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\usepackage{../style}
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%%\usepackage{../langs}
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\begin{document}
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\section*{Coursework 6 (Scala)}
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This coursework is about Scala and is worth 10\%. The first and second
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part are due on 16 November at 11pm, and the third part on 23 November
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at 11pm. You are asked to implement three programs about list
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processing and recursion. The third part is more advanced and might
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include material you have not yet seen in the first lecture.
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Make sure the files you submit can be processed by just calling
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\texttt{scala <<filename.scala>>}.\bigskip
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\noindent
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\textbf{Important:} Do not use any mutable data structures in your
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submissions! They are not needed. This means you cannot use 
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\texttt{ListBuffer}s, for example. Do not use \texttt{return} in your
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code! It has a different meaning in Scala, than in Java.
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Do not use \texttt{var}! This declares a mutable variable. Make sure the
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functions you submit are defined on the ``top-level'' of Scala, not
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inside a class or object. Also note that the running time of
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each part will be restricted to a maximum of 360 seconds on my laptop.
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\subsection*{Disclaimer}
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It should be understood that the work you submit represents
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your own effort. You have not copied from anyone else. An
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exception is the Scala code I showed during the lectures or
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uploaded to KEATS, which you can freely use.\bigskip
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\subsection*{Part 1 (3 Marks)}
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This part is about recursion. You are asked to implement a Scala
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program that tests examples of the \emph{$3n + 1$-conjecture}, also
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called \emph{Collatz conjecture}. This conjecture can be described as
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follows: Start with any positive number $n$ greater than $0$:
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\begin{itemize}
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\item If $n$ is even, divide it by $2$ to obtain $n / 2$.
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\item If $n$ is odd, multiply it by $3$ and add $1$ to obtain $3n +
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  1$.
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\item Repeat this process and you will always end up with $1$.
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\end{itemize}
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\noindent
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For example if you start with $6$, respectively $9$, you obtain the
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series
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\[
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\begin{array}{@{}l@{\hspace{5mm}}l@{}}
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6, 3, 10, 5, 16, 8, 4, 2, 1 & \text{(= 9 steps)}\\
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9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1  & \text{(= 20 steps)}\\
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\end{array}
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\]
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\noindent
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As you can see, the numbers go up and down like a roller-coaster, but
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curiously they seem to always terminate in $1$. The conjecture is that
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this will \emph{always} happen for every number greater than
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0.\footnote{While it is relatively easy to test this conjecture with
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  particular numbers, it is an interesting open problem to
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  \emph{prove} that the conjecture is true for \emph{all} numbers ($>
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  0$). Paul Erd\"o{}s, a famous mathematician you might have hard
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  about, said about this conjecture: ``Mathematics may not be ready
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  for such problems.'' and also offered a \$500 cash prize for its
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  solution. Jeffrey Lagarias, another mathematician, claimed that
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  based only on known information about this problem, ``this is an
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  extraordinarily difficult problem, completely out of reach of
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  present day mathematics.'' There is also a
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  \href{https://xkcd.com/710/}{xkcd} cartoon about this conjecture
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  (click \href{https://xkcd.com/710/}{here}). If you are able to solve
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  this conjecture, you will definitely get famous.}\bigskip
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\newpage
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\noindent
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\textbf{Tasks (file collatz.scala):}
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\begin{itemize}
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\item[(1)] You are asked to implement a recursive function that
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  calculates the number of steps needed until a series ends
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  with $1$. In case of starting with $6$, it takes $9$ steps and in
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  case of starting with $9$, it takes $20$ (see above). In order to
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  try out this function with large numbers, you should use
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  \texttt{Long} as argument type, instead of \texttt{Int}.  You can
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  assume this function will be called with numbers between $1$ and
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  $1$ million. \hfill[2 Marks]
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\item[(2)] Write a second function that takes an upper bound as
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  argument and calculates the steps for all numbers in the range from
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  1 up to this bound. It returns the maximum number of steps and the
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  corresponding number that needs that many steps.  More precisely
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  it returns a pair where the first
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  component is the number of steps and the second is the
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  corresponding number. \hfill\mbox{[1 Mark]}
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\end{itemize}
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\noindent
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\textbf{Test Data:} Some test ranges are:
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\begin{itemize}
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\item 1 to 10 where $9$ takes 20 steps 
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\item 1 to 100 where $97$ takes 119 steps,
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\item 1 to 1,000 where $871$ takes 179 steps,
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\item 1 to 10,000 where $6,171$ takes 262 steps,
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\item 1 to 100,000 where $77,031$ takes 351 steps, 
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\item 1 to 1 million where $837,799$ takes 525 steps
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  %%\item[$\bullet$] $1 - 10$ million where $8,400,511$ takes 686 steps
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\end{itemize}\bigskip
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\subsection*{Part 2 (4 Marks)}
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This part is about list processing---it's a variant of
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``buy-low-sell-high'' in Scala. It uses the online financial data
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service from Yahoo.\bigskip 
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\noindent
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\textbf{Tasks (file trade.scala):}
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\begin{itemize}
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\item[(1)] Given a list of prices for a commodity, for example
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\[
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\texttt{List(28.0, 18.0, 20.0, 26.0, 24.0)}
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\]
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\noindent
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you need to write a function that returns a pair of indices for when
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to buy and when to sell this commodity. In the example above it should
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return the pair $\texttt{(1, 3)}$ because at index $1$ the price is lowest and
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then at index $3$ the price is highest. Note the prices are given as
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lists of \texttt{Double}s.\newline \mbox{} \hfill[1 Mark]
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\item[(2)] Write a function that requests a comma-separated value (CSV) list
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  from the Yahoo websevice that provides historical data for stock
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  indices. For example if you query the URL
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\begin{center}
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\url{http://ichart.yahoo.com/table.csv?s=GOOG}
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\end{center}
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\noindent where \texttt{GOOG} stands for Google's stock market symbol,
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then you will receive a CSV-list of the daily stock prices since
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Google was listed. You can also try this with other stock market
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symbols, for instance AAPL, MSFT, IBM, FB, YHOO, AMZN, BIDU and so
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on. 
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This function should return a List of strings, where each string
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is one line in this CVS-list (representing one day's
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data). Note that Yahoo generates its answer such that the newest data
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is at the front of this list, and the oldest data is at the end.
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\hfill[1 Mark]
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\item[(3)] As you can see, the financial data from Yahoo is organised in 7 columns,
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for example
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{\small\begin{verbatim}
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Date,Open,High,Low,Close,Volume,Adj Close
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2016-11-04,750.659973,770.359985,750.560974,762.02002,2126900,762.02002
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2016-11-03,767.25,769.950012,759.030029,762.130005,1914000,762.130005
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2016-11-02,778.200012,781.650024,763.450012,768.700012,1872400,768.700012
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2016-11-01,782.890015,789.48999,775.539978,783.609985,2404500,783.609985
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....
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\end{verbatim}}
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\noindent
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Write a function that ignores the first line (the header) and then
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extracts from each line the date (first column) and the Adjusted Close
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price (last column). The Adjusted Close price should be converted into
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a \texttt{Double}. So the result of this function is a list of pairs where the
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first components are strings (the dates) and the second are doubles
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(the adjusted close prices).\newline\mbox{}\hfill\mbox{[1 Mark]}
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\item[(4)] Write a function that takes a stock market symbol as
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  argument (you can assume it is a valid one, like GOOG or AAPL). The
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  function calculates the \underline{dates} when you should have
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  bought the corresponding shares (lowest price) and when you should
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  have sold them (highest price).\hfill\mbox{[1 Mark]}
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\end{itemize}
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\noindent
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\textbf{Test Data:}
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In case of Google, the financial data records 3077 entries starting
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from 2004-08-19 until 2016-11-04 (which is the last entry on the day
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when I prepared the course work...namely on 6 November; remember stock
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markets are typically closed on weekends and no financial data is
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produced then; also I did not count the header line). The lowest
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shareprice for Google was on 2004-09-03 with \$49.95513 per share and the
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highest on 2016-10-24 with \$813.109985 per share.\bigskip
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\subsection*{Advanced Part 3 (3 Marks)}
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A purely fictional character named Mr T.~Drumb inherited in 1978
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approximately 200 Million Dollar from his father. Mr Drumb prides
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himself to be a brilliant business man because nowadays it is
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estimated he is 3 Billion Dollar worth (one is not sure, of course,
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because Mr Drumb refuses to make his tax records public).
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Since the question about Mr Drumb's business acumen remains, let's do a
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quick back-of-the-envelope calculation in Scala whether his claim has
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any merit. Let's suppose we are given \$100 in 1978 and we follow a
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really dumb investment strategy, namely:
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\begin{itemize}
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\item We blindly choose a portfolio of stocks, say some Blue-Chip stocks
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  or some Real Estate stocks.
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\item If some of the stocks in our portfolio are traded in January of
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  a year, we invest our money in equal amounts in each of these
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  stocks.  For example if we have \$100 and there are four stocks that
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  are traded in our portfolio, we buy \$25 worth of stocks
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  from each.
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\item Next year in January, we look how our stocks did, liquidate
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  everything, and re-invest our (hopefully) increased money in again
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  the stocks from our portfolio (there might be more stocks available,
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  if companies from our portfolio got listed in that year, or less if
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  some companies went bust or de-listed).
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\item We do this for 38 years until January 2016 and check what would
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  have become out of our \$100.
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\end{itemize}\medskip  
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\noindent
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\textbf{Tasks (file drumb.scala):}
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\begin{itemize}
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\item[(1.a)] Write a function that queries the Yahoo financial data
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  service and obtains the first trade (adjusted close price) of a
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  stock symbol and a year. A problem is that normally a stock exchange
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  is not open on 1st of January, but depending on the day of the week
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  on a later day (maybe 3rd or 4th). The easiest way to solve this
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  problem is to obtain the whole January data for a stock symbol as
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  CSV-list and then select the earliest entry in this list. For this
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  you can specify a date range with the Yahoo service. For example if
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  you want to obtain all January data for Google in 2000, you can form
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  the query:\mbox{}\\[-8mm]
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  \begin{center}\small
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    \mbox{\url{http://ichart.yahoo.com/table.csv?s=GOOG&a=0&b=1&c=2000&d=1&e=1&f=2000}}
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  \end{center}
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  For other companies and years, you need to change the stock symbol
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  (\texttt{GOOG}) and the year \texttt{2000} (in the \texttt{c} and
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  \texttt{f} argument of the query). Such a request might fail, if the
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  company does not exist during this period. For example, if you query
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  for Google in January of 1980, then clearly Google did not exists yet.
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  Therefore you are asked to return a trade price as
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  \texttt{Option[Double]}.
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\item[(1.b)] Write a function that takes a portfolio (a list of stock symbols),
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  a years range and gets all the first trading prices for each year. You should
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  organise this as a list of lists of \texttt{Option[Double]}'s. The inner lists
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  are for all stock symbols from the portfolio and the outer list for the years.
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  For example for Google and Apple in years 2010 (first line), 2011
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  (second line) and 2012 (third line) you obtain:
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\begin{verbatim}
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  List(List(Some(313.062468), Some(27.847252)), 
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       List(Some(301.873641), Some(42.884065)),
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       List(Some(332.373186), Some(53.509768)))
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\end{verbatim}\hfill[1 Mark]
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\item[(2.a)] Write a function that calculates the \emph{change factor} (delta)
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  for how a stock price has changed from one year to the next. This is
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  only well-defined, if the corresponding company has been traded in both
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  years. In this case you can calculate
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  \[
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  \frac{price_{new} - price_{old}}{price_{old}}
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  \]
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\item[(2.b)] Write a function that calculates all change factors
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  (deltas) for the prices we obtained under Task 1. For the running
24
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  example of Google and Apple for the years 2010 to 2012 you should
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  obtain 4 change factors:
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\begin{verbatim}  
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  List(List(Some(-0.03573991820699504), Some(0.5399747522663995))
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          List(Some(0.10103414428290529), Some(0.24777742035415723)))
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\end{verbatim}
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  That means Google did a bit badly in 2010, while Apple did very well.
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  Both did OK in 2011.\hfill\mbox{[1 Mark]}
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\item[(3.a)] Write a function that calculates the ``yield'', or
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  balance, for one year for our portfolio.  This function takes the
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  change factors, the starting balance and the year as arguments. If
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  no company from our portfolio existed in that year, the balance is
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  unchanged. Otherwise we invest in each existing company an equal
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  amount of our balance. Using the change factors computed under Task
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  2, calculate the new balance. Say we had \$100 in 2010, we would have
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  received in our running example
6
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parents:
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  \begin{verbatim}
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  $50 * -0.03573991820699504 + $50 * 0.5399747522663995
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                                         = $25.211741702970222
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  \end{verbatim}
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  as profit for that year, and our new balance for 2011 is \$125 when
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  converted to a \texttt{Long}.
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\item[(3.b)] Write a function that calculates the overall balance
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  for a range of years where each year the yearly profit is compounded to
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  the new balances and then re-invested into our portfolio.\mbox{}\hfill\mbox{[1 Mark]}
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\end{itemize}\medskip  
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\noindent
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\textbf{Test Data:} File \texttt{drumb.scala} contains two portfolios
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collected from the S\&P 500, one for blue-chip companies, including
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Facebook, Amazon and Baidu; and another for listed real-estate companies, whose
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names I have never heard of. Following the dumb investment strategy
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from 1978 until 2016 would have turned a starting balance of \$100
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into \$23,794 for real estate and a whopping \$524,609 for blue chips.\medskip
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\noindent
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\textbf{Moral:} Reflecting on our assumptions, we are over-estimating
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our yield in many ways: first, who can know in 1978 about what will
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turn out to be a blue chip company.  Also, since the portfolios are
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chosen from the current S\&P 500, they do not include the myriad
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of companies that went bust or were de-listed over the years.
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So where does this leave our fictional character Mr T.~Drumb? Well, given
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his inheritance, a really dumb investment strategy would have done
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equally well, if not much better.
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\end{document}
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%%% Local Variables: 
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%%% mode: latex
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%%% TeX-master: t
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%%% End: