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\section*{Resit Exam}

The Scala part of the exam is worth 30\%. It is about `jumps'
within lists.

\IMPORTANTEXAM{}

\DISCLAIMEREXAM{}
 
%%\newpage

\subsection*{Task}

\noindent
Suppose you are given a list of numbers. Each number indicates how many
steps can be taken forward from this element. For example in the
list 

\begin{center}
\begin{tikzpicture}[scale=0.8]
  \draw[line width=1mm,cap=round] (0,0) -- (5,0);
  \draw[line width=1mm,cap=round] (0,1) -- (5,1);

  \draw[line width=1mm,cap=round] (0,0) -- (0,1);
  \node at (0.5,0.5) {\textbf{\Large 3}};

  \draw[line width=1mm,cap=round] (1,0) -- (1,1);
  \node at (1.5,0.5) {\textbf{\Large 4}};

  \draw[line width=1mm,cap=round] (2,0) -- (2,1);
  \node at (2.5,0.5) {\textbf{\Large 2}};

  \draw[line width=1mm,cap=round] (3,0) -- (3,1);
  \node at (3.5,0.5) {\textbf{\Large 0}};
  
  \draw[line width=1mm,cap=round] (4,0) -- (4,1);

  \node at (4.5,0.5) {\textbf{\Large 1}};
  
  \draw[line width=1mm,cap=round] (5,0) -- (5,1);

  \draw[->,line width=0.5mm,cap=round,out=90,in=90,relative] (0.5,1) to (1.5,1);
  \draw[->,line width=0.5mm,cap=round,out=90,in=90,relative] (0.5,1) to (2.5,1);
  \draw[->,line width=0.5mm,cap=round,out=90,in=90,relative] (0.5,1) to (3.5,1);

  \draw[->,line width=0.5mm,cap=round,out=-90,in=-90,relative] (2.5,0) to (3.5,0);
  \draw[->,line width=0.5mm,cap=round,out=-90,in=-90,relative] (2.5,0) to (4.5,0);

  \draw[->,line width=0.5mm,cap=round,out=90,in=90,relative] (4.5,1) to (5.7,1);
  \node at (5.7, 0.8) {End};
\end{tikzpicture}
\end{center}  

\noindent
the first 3 indicates that you can step to the next three elements,
that is 4, 2, and 0. The 2 in the middle indicates that you can step
to elements 0 and 1.  From the final 1 you can step to the End of the
list. You can also do this from element 4, since the end of this list
is reachable from there. A 0 always indicates that you cannot
step any further from this element.\medskip

\noindent
The problem is to calculate a sequence of steps to reach the end of
the list by taking only steps indicated by the integers. For the list
above, possible sequence of steps are 3 - 2 - 1 - End, but also 3 - 4
- End.  This is a recursive problem that can be thought of as a tree
where the root is a list and the children are all the lists that are
reachable by a single step. For example for the list above this gives a
tree like

\begin{center}
\begin{tikzpicture}[grow=right,level distance=30mm,child anchor=north]
  \node {[3,4,2,0,1]}
     child {node {[0,1]}}
     child {node {[2,0,1]}
        child {node {[1]} child [level distance=13mm] {node {End}}}
        child {node {[0,1]}}
     }
     child {node {[4,2,0,1]\ldots}};
\end{tikzpicture}
\end{center}

\subsubsection*{Tasks}

\begin{itemize}
\item[(1)] Write a function, called \texttt{steps}, that calculates
  the children of a list. This function takes an integer as one argument
  indicating how many children should be returned. The other argument is a list
  of integers.  In case of 3 and the list [4,2,0,1], it should produce
  the list

  \begin{center}
  {\large[}\;[4,2,0,1],\; [2,0,1],\; [0,1]\;{\large]}
  \end{center}  

  Be careful to account properly for the end of the list. For example
  for the integer 4 and the list [2,0,1], the function should return the list

  \begin{center}
  {\large[}\;[2,0,1], [0,1],\; [1]\;{\large]}
  \end{center}  


  \mbox{}\hfill[Marks: 8\%]
 
\item[(2)] Write a function \texttt{search} that tests whether there
  is a way to reach the end of a list.  This is not always the
  case, for example for the list

  \begin{center}
    [3,5,1,0,0,0,0,0,0,0,0,1]
  \end{center}

  \noindent
  there is no sequence of steps that can bring you to the end of the list.
  If there is a way, \texttt{search} should return true, otherwise false.
  In case of the empty list, \texttt{search} should return true since the
  end of the list is already reached.
  
  \mbox{}\hfill\mbox{[Marks: 10\%]}

\item[(3)] Write a function \texttt{jumps} that calculates the
  shortest sequence of steps needed to reach the end of a list. One
  way to calculate this is to generate \emph{all} sequences to reach
  the end of a list and then select one that has the shortest length.
  This function needs to return a value of type
  \texttt{Option[List[Int]]} because for some lists there does not
  exists a sequence at all. If there exists such a sequence,
  \texttt{jumps} should return \texttt{Some(\ldots)}; otherwise
  \texttt{None}. In the special case of the empty list, \texttt{jumps}
  should return \texttt{None}

  \mbox{}\hfill\mbox{[Marks: 12\%]}
  
\end{itemize}\bigskip


\noindent
\textbf{Hints:} useful list functions: \texttt{.minBy(..)} searches for
the first element in a list that is the minimum according to
a given measure; \texttt{.length} calculates the length of a list;
\texttt{.exists(..)} returns true when an element in a list
satisfies a given predicate, otherwise returns false;
\texttt{.map(..)} applies a given function to each element
in a list; \texttt{.flatten} turns a list of
lists into just a list; \texttt{\_::\_} puts an element on the head of
the list.


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