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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 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.
+
+
+\end{document}
+
+
+%%% Local Variables: 
+%%% mode: latex
+%%% TeX-master: t
+%%% End: 




