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\section*{Coursework 6 (Scala)}

This coursework is about Scala and is worth 10\%. The first and second

part are due on 16 November at 11pm, and the third part on 21 December

at 11pm. You are asked to implement three programs about list

processing and recursion. The third part is more advanced and might

include material you have not yet seen in the first lecture.

\bigskip

\IMPORTANT{}

\noindent

Also note that the running time of each part will be restricted to a

maximum of 360 seconds on my laptop.

\DISCLAIMER{}

\subsection*{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}, also

called \emph{Collatz conjecture}. This conjecture can be described as

follows: Start with any positive number $n$ greater than $0$:

\begin{itemize}

\item If $n$ is even, divide it by $2$ to obtain $n / 2$.

\item If $n$ is odd, multiply it by $3$ and add $1$ to obtain $3n +

  1$.

\item Repeat this process and you will always end up with $1$.

\end{itemize}

\noindent

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 $1$. The conjecture is that

this will \emph{always} happen for every number greater than

0.\footnote{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 \emph{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 a \$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

  (click \href{https://xkcd.com/710/}{here}). If you are able to solve

  this conjecture, you will definitely get famous.}\bigskip

\noindent

\textbf{Tasks (file collatz.scala):}

\begin{itemize}

\item[(1)] You are asked to implement a recursive function that

  calculates the number of steps needed 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

  $1$ Million. \hfill[2 Marks]

\item[(2)] Write a second function that takes an upper bound as

  argument and calculates the steps for all numbers in the range from

  1 up to this bound. It returns the maximum number of steps and the

  corresponding number that needs that many steps.  More precisely

  it returns a pair where the first

  component is the number of steps and the second is the

  corresponding number. \hfill\mbox{[1 Mark]}

\end{itemize}

\noindent

\textbf{Test Data:} Some test ranges are:

\begin{itemize}

\item 1 to 10 where $9$ takes 20 steps 

\item 1 to 100 where $97$ takes 119 steps,

\item 1 to 1,000 where $871$ takes 179 steps,

\item 1 to 10,000 where $6,171$ takes 262 steps,

\item 1 to 100,000 where $77,031$ takes 351 steps, 

\item 1 to 1 Million where $837,799$ takes 525 steps

  %%\item[$\bullet$] $1 - 10$ million where $8,400,511$ takes 686 steps

\end{itemize}

\noindent

\textbf{Hints:} useful math operators: \texttt{\%} for modulo; useful

functions: \mbox{\texttt{(1\,to\,10)}} for ranges, \texttt{.toInt},

\texttt{.toList} for conversions, \texttt{List(...).max} for the

maximum of a list, \texttt{List(...).indexOf(...)} for the first index of

a value in a list.

\subsection*{Part 2 (3 Marks)}

This part is about web-scraping and list-processing in Scala. It uses

online data about the per-capita alcohol consumption for each country

(per year?), and a file containing the data about the population size of

each country.  From this data you are supposed to estimate how many

litres of pure alcohol are consumed worldwide.\bigskip

\noindent

\textbf{Tasks (file alcohol.scala):}

\begin{itemize}

\item[(1)] Write a function that given an URL requests a

  comma-separated value (CSV) list.  We are interested in the list

  from the following URL

\begin{center}

  \url{https://raw.githubusercontent.com/fivethirtyeight/data/master/alcohol-consumption/drinks.csv}

\end{center}

\noindent Your function should take a string (the URL) as input, and

produce a list of strings as output, where each string is one line in

the corresponding CSV-list.  This list from the URL above should

contain 194 lines.\medskip

\noindent

Write another function that can read the file \texttt{population.csv}

from disk (the file is distributed with the coursework). This

function should take a string as argument, the file name, and again

return a list of strings corresponding to each entry in the

CSV-list. For \texttt{population.csv}, this list should contain 216

lines.\hfill[1 Mark]

\item[(2)] Unfortunately, the CSV-lists contain a lot of ``junk'' and we

  need to extract the data that interests us.  From the header of the

  alcohol list, you can see there are 5 columns

  \begin{center}

    \begin{tabular}{l}

      \texttt{country (name),}\\

      \texttt{beer\_servings,}\\

      \texttt{spirit\_servings,}\\

      \texttt{wine\_servings,}\\

      \texttt{total\_litres\_of\_pure\_alcohol}

    \end{tabular}  

  \end{center}

  \noindent

  Write a function that extracts the data from the first column,

  the country name, and the data from the fifth column (converted into

  a \texttt{Double}). For this go through each line of the CSV-list

  (except the first line), use the \texttt{split(",")} function to

  divide each line into an array of 5 elements. Keep the data from the

  first and fifth element in these arrays.\medskip

  \noindent

  Write another function that processes the population size list. This

  is already of the form country name and population size.\footnote{Your

    friendly lecturer already did the messy processing for you from the

  Worldbank database, see \url{https://github.com/datasets/population/tree/master/data} for the original.} Again, split the

  strings according to the commas. However, this time generate a

  \texttt{Map} from country names to population sizes.\hfill[1 Mark]

\item[(3)] In (2) you generated the data about the alcohol consumption

  per capita for each country, and also the population size for each

  country. From this generate next a sorted(!) list of the overall

  alcohol consumption for each country. The list should be sorted from

  highest alcohol consumption to lowest. The difficulty is that the

  data is scraped off from ``random'' sources on the Internet and

  annoyingly the spelling of some country names does not always agree in both

  lists. For example the alcohol list contains

  \texttt{Bosnia-Herzegovina}, while the population writes this country as

  \texttt{Bosnia and Herzegovina}. In your sorted

  overall list include only countries from the alcohol list, whose

  exact country name is also in the population size list. This means

  you can ignore countries like Bosnia-Herzegovina from the overall

  alcohol consumption. There are 177 countries where the names

  agree. The UK is ranked 10th on this list by

  consuming 671,976,864 Litres of pure alcohol each year.\medskip

  \noindent

  Finally, write another function that takes an integer, say

  \texttt{n}, as argument. You can assume this integer is between 0

  and 177 (the number of countries in the sorted list above).  The

  function should return a triple, where the first component is the

  sum of the alcohol consumption in all countries (on the list); the

  second component is the sum of the \texttt{n}-highest alcohol

  consumers on the list; and the third component is the percentage the

  \texttt{n}-highest alcohol consumers drink with respect to the

  the world consumption. You will see that according to our data, 164

  countries (out of 177) gobble up 100\% of the World alcohol

  consumption.\hfill\mbox{[1 Mark]}

\end{itemize}

\noindent

\textbf{Hints:} useful list functions: \texttt{.drop(n)},

\texttt{.take(n)} for dropping or taking some elements in a list,

\texttt{.getLines} for separating lines in a string;

\texttt{.sortBy(\_.\_2)} sorts a list of pairs according to the second

elements in the pairs---the sorting is done from smallest to highest;

useful \texttt{Map} functions: \texttt{.toMap} converts a list of

pairs into a \texttt{Map}, \texttt{.isDefinedAt(k)} tests whether the

map is defined at that key, that is would produce a result when

called with this key; useful data functions: \texttt{Source.fromURL},

\texttt{Source.fromFile} for obtaining a webpage and reading a file.

\newpage

\subsection*{Advanced Part 3 (4 Marks)}

A purely fictional character named Mr T.~Drumb inherited in 1978

approximately 200 Million Dollar from his father. Mr Drumb prides

himself to be a brilliant business man because nowadays it is

estimated he is 3 Billion Dollar worth (one is not sure, of course,

because Mr Drumb refuses to make his tax records public).

Since the question about Mr Drumb's business acumen remains open,

let's do a quick back-of-the-envelope calculation in Scala whether his

claim has any merit. Let's suppose we are given \$100 in 1978 and we

follow a really dumb investment strategy, namely:

\begin{itemize}

\item We blindly choose a portfolio of stocks, say some Blue-Chip stocks

  or some Real Estate stocks.

\item If some of the stocks in our portfolio are traded in January of

  a year, we invest our money in equal amounts in each of these

  stocks.  For example if we have \$100 and there are four stocks that

  are traded in our portfolio, we buy \$25 worth of stocks

  from each.

\item Next year in January, we look how our stocks did, liquidate

  everything, and re-invest our (hopefully) increased money in again

  the stocks from our portfolio (there might be more stocks available,

  if companies from our portfolio got listed in that year, or less if

  some companies went bust or were de-listed).

\item We do this for 39 years until January 2017 and check what would

  have become out of our \$100.

\end{itemize}

\noindent

Until Yahoo was bought by Altaba this summer, historical stock market

data for such back-of-the-envelope calculations was freely available

online. Unfortuantely nowadays this kind of data is difficult to

obtain, unless you are prepared to pay extortionate prices or be

severely rate-limited.  Therefore this coursework comes with a number

of files containing CSV-lists with the historical stock prices for the

companies in our portfolios. Use these files for the following

tasks.\bigskip

\noindent

\textbf{Tasks (file drumb.scala):}

\begin{itemize}

\item[(1.a)] Write a function \texttt{get\_january\_data} that takes a

  stock symbol and a year as arguments. The function reads the

  corresponding CSV-file and returns the list of strings that start

  with the given year (each line in the CSV-list is of the form

  \texttt{year-01-someday,someprice}).

\item[(1.b)] Write a function \texttt{get\_first\_price} that takes

  again a stock symbol and a year as arguments. It should return the

  first January price for the stock symbol in given the year. For this

  it uses the list of strings generated by

  \texttt{get\_january\_data}.  A problem is that normally a stock

  exchange is not open on 1st of January, but depending on the day of

  the week on a later day (maybe 3rd or 4th). The easiest way to solve

  this problem is to obtain the whole January data for a stock symbol

  and then select the earliest, or first, entry in this list. The

  stock price of this entry should be converted into a double.  Such a

  price might not exist, in case the company does not exist in the given

  year. For example, if you query for Google in January of 1980, then

  clearly Google did not exist yet.  Therefore you are asked to

  return a trade price as \texttt{Option[Double]}\ldots\texttt{None}

  will be the value for when no price exists.

\item[(1.c)] Write a function \texttt{get\_prices} that takes a

  portfolio (a list of stock symbols), a years range and gets all the

  first trading prices for each year in the range. You should organise

  this as a list of lists of \texttt{Option[Double]}'s. The inner

  lists are for all stock symbols from the portfolio and the outer

  list for the years.  For example for Google and Apple in years 2010

  (first line), 2011 (second line) and 2012 (third line) you obtain:

\begin{verbatim}

  List(List(Some(311.349976), Some(27.505054)), 

       List(Some(300.222351), Some(42.357094)),

       List(Some(330.555054), Some(52.852215)))

\end{verbatim}\hfill[2 Marks]

\item[(2.a)] Write a function that calculates the \emph{change factor} (delta)

  for how a stock price has changed from one year to the next. This is

  only well-defined, if the corresponding company has been traded in both

  years. In this case you can calculate

  \[

  \frac{price_{new} - price_{old}}{price_{old}}

  \]

  If the change factor is defined, you should return it

  as \texttt{Some(change factor)}; if not, you should return

  \texttt{None}.

\item[(2.b)] Write a function that calculates all change factors

  (deltas) for the prices we obtained under Task 1. For the running

  example of Google and Apple for the years 2010 to 2012 you should

  obtain 4 change factors:

\begin{verbatim}  

  List(List(Some(-0.03573992567129673), Some(0.5399749442411563))

        List(Some(0.10103412653643493), Some(0.2477771728154912)))

\end{verbatim}

  That means Google did a bit badly in 2010, while Apple did very well.

  Both did OK in 2011. Make sure you handle the cases where a company is

  not listed in a year. In such cases the change factor should be \texttt{None}

  (see 2.a).\\

  \mbox{}\hfill\mbox{[1 Mark]}

\item[(3.a)] Write a function that calculates the ``yield'', or

  balance, for one year for our portfolio.  This function takes the

  change factors, the starting balance and the year as arguments. If

  no company from our portfolio existed in that year, the balance is

  unchanged. Otherwise we invest in each existing company an equal

  amount of our balance. Using the change factors computed under Task

  2, calculate the new balance. Say we had \$100 in 2010, we would have

  received in our running example involving Google and Apple:

  \begin{verbatim}

  $50 * -0.03573992567129673 + $50 * 0.5399749442411563

                                       = $25.21175092849298

  \end{verbatim}

  as profit for that year, and our new balance for 2011 is \$125 when

  converted to a \texttt{Long}.

\item[(3.b)] Write a function that calculates the overall balance

  for a range of years where each year the yearly profit is compounded to

  the new balances and then re-invested into our portfolio.\\

  \mbox{}\hfill\mbox{[1 Mark]}

\end{itemize}\medskip  

\noindent

\textbf{Test Data:} File \texttt{drumb.scala} contains two portfolios

collected from the S\&P 500, one for blue-chip companies, including

Facebook, Amazon and Baidu; and another for listed real-estate

companies, whose names I have never heard of. Following the dumb

investment strategy from 1978 until 2017 would have turned a starting

balance of \$100 into roughly \$30,895 for real estate and a whopping

\$349,597 for blue chips.  Note when comparing these results with your

own calculations: there might be some small rounding errors, which

when compounded lead to moderately different values.\bigskip

\noindent

\textbf{Hints:} useful string functions: \texttt{.startsWith(...)} for

checking whether a string has a given prefix, \texttt{\_ ++ \_} for

concatenating two strings; useful option functions: \texttt{.flatten}

flattens a list of options such that it filters way all

\texttt{None}'s, \texttt{Try(...) getOrElse ...} runs some code that

might raise an exception---if yes, then a default value can be given;

useful list functions: \texttt{.head} for obtaining the first element

in a non-empty list, \texttt{.length} for the length of a

list.\bigskip

\noindent

\textbf{Moral:} Reflecting on our assumptions, we are over-estimating

our yield in many ways: first, who can know in 1978 about what will

turn out to be a blue chip company.  Also, since the portfolios are

chosen from the current S\&P 500, they do not include the myriad

of companies that went bust or were de-listed over the years.

So where does this leave our fictional character Mr T.~Drumb? Well, given

his inheritance, a really dumb investment strategy would have done

equally well, if not much better.\medskip

\end{document}

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