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\section*{Coursework 6 (Scala)}

This coursework is about Scala and is worth 10\%. The first and second
part is due on 16 November at 11:00, and the second part on 23
November at 11:00. You are asked to implement three programs about
list processing and recursion.

\subsubsection*{Disclaimer}

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


\subsubsection*{Part 1 (3 Marks)}

This part is about recursion. You are asked to implement a Scala
program that tests examples of the \emph{$3n + 1$-conjecture}. This
conjecture can be described as follows: Start with any positive number
$n$ greater than $0$. If $n$ is even, divide it by $2$ to obtain $n /
2$. If $n$ is odd, multiply it by $3$ and add $1$ to obtain $3n +
1$. Repeat this process and you will always end up with $1$.  For
example if you start with $6$, respectively $9$, you obtain the series

\[
\begin{array}{@{}l@{\hspace{5mm}}l@{}}
6, 3, 10, 5, 16, 8, 4, 2, 1 & \text{(=9 steps)}\\
9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1  & \text{(=20 steps)}\\
\end{array}
\]

\noindent
As you can see, the numbers go up and down like a roller-coaster, but
curiously they seem to always terminate in $1$.  While it is
relatively easy to test this conjecture with particular numbers, it is
an interesting open problem to \emph{prove} that the conjecture is
true for all numbers ($> 0$). Paul Erd\"o{}s, a famous mathematician
you might have hard about, said about this conjecture: ``Mathematics
may not be ready for such problems.'' and also offered \$500 cash
prize for its solution. Jeffrey Lagarias, another mathematician,
claimed that based only on known information about this problem,
``this is an extraordinarily difficult problem, completely out of
reach of present day mathematics.'' There is also a
\href{https://xkcd.com/710/}{xkcd} cartoon about this
conjecture.\medskip

\noindent
\textbf{Tasks:} (1) You are asked to implement a recursive function that
calculates the number of steps needed number until a series ends with
$1$. In case of starting with $6$, it takes $9$ steps and in case of
starting with $9$, it takes $20$ (see above). In order to try out this
function with large numbers, you should use \texttt{Long} as argument
type instead of \texttt{Int}.  You can assume this function will be
called with numbers between $1$ and $10$ million.

(2) Then write a second function that takes an upper bound as argument
and calculates the steps for all numbers in the range from 1 upto this
bound, and returns the maximum of steps needed by a number in that
range. This function should produce for ranges

\begin{itemize}
\item $1 - 10$ where $9$ takes 20 steps
\item $1 - 100$ where $97$ takes 119 steps,
\item $1 - 1,000$ where $871$ takes 179 steps,
\item $1 - 10,000$ where $6,171$ takes 262 steps,
\item $1 - 100,000$ where $77,031$ takes 351 steps,
\item $1 - 1$ million where $837,799$ takes 525 steps, and
\item $1 - 10$ million where $8,400,511$ takes 686 steps
\end{itemize}  


\subsubsection*{Part 2 (4 Marks)}

This part is about list processing---it's a variant of
Buy-low-sell-high in Scala. (1) Given a list of prices for a commodity,
for example

\[
\texttt{List(28.0, 18.0, 20.0, 26.0, 24.0)}
\]

\noindent
you need to write a function that returns a pair for when to buy and
when to sell this commodity. In the example above it should return
$\texttt{(1, 3)}$ because at index $1$ the price is lowest and then at
index $3$ the price is highest. Note the prices are given as lists of
\texttt{Float}s.

(2) Write a function that requests a comma-separated value list from a
Yahoo websevice that provides historical data for stock indices. For
example if you query the URL

\begin{center}
\url{http://ichart.yahoo.com/table.csv?s=GOOG}
\end{center}

\noindent where \texttt{GOOG} stands for Google's stock market symbol
then you will receive a comma-separated value list of the daily stock
prices since Google was floated. You can also try this with other
strock market symbols, for instance AAPL, MSFT, IBM, FB, YHOO, AMZN,
BIDU and so on. This function should return a List of strings, where
each string is one line in this comma-separated value list
(representing one days data). Note that Yahoo generates its answer
such that the newest data is at the front of this list, and the oldest
data is at the end.

(3) As you can see, the financial data from Yahoo is organised in 7 columns,
for example

{\small\begin{verbatim}
Date,Open,High,Low,Close,Volume,Adj Close
2016-11-04,750.659973,770.359985,750.560974,762.02002,2126900,762.02002
2016-11-03,767.25,769.950012,759.030029,762.130005,1914000,762.130005
2016-11-02,778.200012,781.650024,763.450012,768.700012,1872400,768.700012
2016-11-01,782.890015,789.48999,775.539978,783.609985,2404500,783.609985
....
\end{verbatim}}

\noindent
Write a function that ignores the first line (the header) and then
extracts from each line the date (first column) and the Adjusted Close
price (last column). The Adjusted Close price should be converted into
a float. So the result of this function is a list of pairs where the
first components are strings (the dates) and the second are floats
(the adjusted close prices).

(4) Write a function that takes a stock market symbol as argument (you
can assume it is a valid one, like GOOG, AAPL, MSFT, IBM, FB, YHOO,
AMZN, BIDU). The function calculates the dates when you should have
bought Google shares (lowest price) and when you should have sold them
(highest price).\medskip

\noindent
\textbf{Test Data:}
In case of Google, the financial data records 3077 entries starting
from 2004-08-19 until 2016-11-04 (which is the last entry on the day
when I prepared the course work...on 6 November; remember stock
markets are typically closed on weekends and no financial data is
produced then; also I did not count the header line). The lowest
for Goole was on 2004-09-03 with \$49.95513 per share and the
highest on 2016-10-24 with \$813.109985 per share.



\subsubsection*{Part 3 (3 Marks)}

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