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-\documentclass{article}[11pt]
-\usepackage{../style}
-\usepackage{../graphics}
-\usepackage{disclaimer}
-
-\begin{document}
-
-\section*{Resit Exam}
-
-The Scala part of the exam is worth 50\%. 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 sequences 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,line width=0.5mm]
-  \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: 12\%]
- 
-\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: 18\%]}
-
-\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: 20\%]}
-  
-\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.
-
-
-\end{document}
-
-
-%%% Local Variables: 
-%%% mode: latex
-%%% TeX-master: t
-%%% End: 




