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\section*{Coursework 6 (Scala)}+
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+
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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>>}. +
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+
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+
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\subsection*{Disclaimer}+
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+
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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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+
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+
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\subsection*{Part 1 (3 Marks)}+
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+
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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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+
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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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+
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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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\[+
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\begin{array}{@{}l@{\hspace{5mm}}l@{}}+
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6, 3, 10, 5, 16, 8, 4, 2, 1 & \text{(= 9 steps)}\\+
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9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1  & \text{(= 20 steps)}\\+
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\end{array}+
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\]+
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+
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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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+
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\newpage+
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\noindent+
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\textbf{Tasks (file collatz.scala):}+
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+
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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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+
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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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+
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\noindent+
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\textbf{Test Data:} Some test ranges are:+
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+
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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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  +
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+
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+
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\subsection*{Part 2 (4 Marks)}+
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+
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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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+
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\noindent+
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\textbf{Tasks (file trade.scala):}+
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+
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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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\[+
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\texttt{List(28.0, 18.0, 20.0, 26.0, 24.0)}+
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\]+
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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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+
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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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+
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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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+
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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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+
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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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+
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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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+
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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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+
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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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+
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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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+
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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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+
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\subsection*{Advanced Part 3 (3 Marks)}+
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+
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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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+
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Since the question about Mr Drumb's business acumen remains open, 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 dump investment strategy, namely:+
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+
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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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+
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\noindent+
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\textbf{Tasks (file drumb.scala):}+
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+
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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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+
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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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+
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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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+
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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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+
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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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 +
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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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  \[+
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  \frac{price_{new} - price_{old}}{price_{old}}+
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  \]+
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+
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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+
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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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+
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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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+
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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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+
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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+
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+
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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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+
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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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  +
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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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+
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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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+
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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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