\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}\noindentSuppose you are given a list of numbers. Each number indicates how manysteps can be taken forward from this element. For example in thelist \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} \noindentthe 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 stepto elements 0 and 1. From the final 1 you can step to the End of thelist. You can also do this from element 4, since the end of this listis reachable from there. A 0 always indicates that you cannotstep any further from this element.\medskip\noindentThe problem is to calculate a sequence of steps to reach the end ofthe list by taking only steps indicated by the integers. For the listabove, 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 treewhere the root is a list and the children are all the lists that arereachable by a single step. For example for the list above this gives atree 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 forthe first element in a list that is the minimum according toa given measure; \texttt{.length} calculates the length of a list;\texttt{.exists(..)} returns true when an element in a listsatisfies a given predicate, otherwise returns false;\texttt{.map(..)} applies a given function to each elementin a list; \texttt{.flatten} turns a list oflists into just a list; \texttt{\_::\_} puts an element on the head ofthe list.\end{document}%%% Local Variables: %%% mode: latex%%% TeX-master: t%%% End: