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-% !TEX program = xelatex
-\documentclass{article}
-\usepackage{chessboard}
-\usepackage[LSBC4,T1]{fontenc}
-\let\clipbox\relax
-\usepackage{../style}
-\usepackage{../langs}
-\usepackage{disclaimer}
-
-\begin{document}
-
-\setchessboard{smallboard,
-               zero,
-               showmover=false,
-               boardfontencoding=LSBC4,
-               hlabelformat=\arabic{ranklabel},
-               vlabelformat=\arabic{filelabel}}
-
-\mbox{}\\[-18mm]\mbox{}
-
-\section*{Part 8 (Scala)}
-
-\mbox{}\hfill\textit{``The problem with object-oriented languages is they’ve got all this implicit,}\\
-\mbox{}\hfill\textit{environment that they carry around with them. You wanted a banana but}\\
-\mbox{}\hfill\textit{what you got was a gorilla holding the banana and the entire jungle.''}\smallskip\\
-\mbox{}\hfill\textit{ --- Joe Armstrong (creator of the Erlang programming language)}\medskip\bigskip
-
-\noindent
-This part is about searching and backtracking. You are asked to
-implement Scala programs that solve various versions of the
-\textit{Knight's Tour Problem} on a chessboard. The preliminary part (4\%) is
-due on  \cwEIGHT{} at 4pm; the core part is due on \cwEIGHTa{} at 4pm.
-Note the core, more advanced, part might include material you have not
-yet seen in the first three lectures. \bigskip
-
-\IMPORTANT{}
-Also note that the running time of each part will be restricted to a
-maximum of 30 seconds on my laptop: If you calculate a result once,
-try to avoid to calculate the result again. Feel free to copy any code
-you need from files \texttt{knight1.scala}, \texttt{knight2.scala} and
-\texttt{knight3.scala}.
-
-\DISCLAIMER{}
-
-\subsection*{Background}
-
-The \textit{Knight's Tour Problem} is about finding a tour such that
-the knight visits every field on an $n\times n$ chessboard once. For
-example on a $5\times 5$ chessboard, a knight's tour is:
-
-\chessboard[maxfield=d4, 
-            pgfstyle= {[base,at={\pgfpoint{0pt}{-0.5ex}}]text},
-            text = \small 24, markfield=Z4,
-            text = \small 11, markfield=a4,
-            text = \small  6, markfield=b4,
-            text = \small 17, markfield=c4,
-            text = \small  0, markfield=d4,
-            text = \small 19, markfield=Z3,
-            text = \small 16, markfield=a3,
-            text = \small 23, markfield=b3,
-            text = \small 12, markfield=c3,
-            text = \small  7, markfield=d3,
-            text = \small 10, markfield=Z2,
-            text = \small  5, markfield=a2,
-            text = \small 18, markfield=b2,
-            text = \small  1, markfield=c2,
-            text = \small 22, markfield=d2,
-            text = \small 15, markfield=Z1,
-            text = \small 20, markfield=a1,
-            text = \small  3, markfield=b1,
-            text = \small  8, markfield=c1,
-            text = \small 13, markfield=d1,
-            text = \small  4, markfield=Z0,
-            text = \small  9, markfield=a0,
-            text = \small 14, markfield=b0,
-            text = \small 21, markfield=c0,
-            text = \small  2, markfield=d0
-           ]
-           
-\noindent
-This tour starts in the right-upper corner, then moves to field
-$(3,2)$, then $(4,0)$ and so on. There are no knight's tours on
-$2\times 2$, $3\times 3$ and $4\times 4$ chessboards, but for every
-bigger board there is. 
-
-A knight's tour is called \emph{closed}, if the last step in the tour
-is within a knight's move to the beginning of the tour. So the above
-knight's tour is \underline{not} closed because the last
-step on field $(0, 4)$ is not within the reach of the first step on
-$(4, 4)$. It turns out there is no closed knight's tour on a $5\times
-5$ board. But there are on a $6\times 6$ board and on bigger ones, for
-example
-
-\chessboard[maxfield=e5, 
-            pgfstyle={[base,at={\pgfpoint{0pt}{-0.5ex}}]text},
-            text = \small 10, markfield=Z5,
-            text = \small  5, markfield=a5,
-            text = \small 18, markfield=b5,
-            text = \small 25, markfield=c5,
-            text = \small 16, markfield=d5,
-            text = \small  7, markfield=e5,
-            text = \small 31, markfield=Z4,
-            text = \small 26, markfield=a4,
-            text = \small  9, markfield=b4,
-            text = \small  6, markfield=c4,
-            text = \small 19, markfield=d4,
-            text = \small 24, markfield=e4,
-            % 4  11  30  17   8  15 
-            text = \small  4, markfield=Z3,
-            text = \small 11, markfield=a3,
-            text = \small 30, markfield=b3,
-            text = \small 17, markfield=c3,
-            text = \small  8, markfield=d3,
-            text = \small 15, markfield=e3,
-            %29  32  27   0  23  20 
-            text = \small 29, markfield=Z2,
-            text = \small 32, markfield=a2,
-            text = \small 27, markfield=b2,
-            text = \small  0, markfield=c2,
-            text = \small 23, markfield=d2,
-            text = \small 20, markfield=e2,
-            %12   3  34  21  14   1 
-            text = \small 12, markfield=Z1,
-            text = \small  3, markfield=a1,
-            text = \small 34, markfield=b1,
-            text = \small 21, markfield=c1,
-            text = \small 14, markfield=d1,
-            text = \small  1, markfield=e1,
-            %33  28  13   2  35  22 
-            text = \small 33, markfield=Z0,
-            text = \small 28, markfield=a0,
-            text = \small 13, markfield=b0,
-            text = \small  2, markfield=c0,
-            text = \small 35, markfield=d0,
-            text = \small 22, markfield=e0,
-            vlabel=false,
-            hlabel=false
-           ]
-
-
-\noindent
-where the 35th move can join up again with the 0th move.
-
-If you cannot remember how a knight moves in chess, or never played
-chess, below are all potential moves indicated for two knights, one on
-field $(2, 2)$ (blue moves) and another on $(7, 7)$ (red moves):
-
-{\chessboard[maxfield=g7,
-            color=blue!50,
-            linewidth=0.2em,
-            shortenstart=0.5ex,
-            shortenend=0.5ex,
-            markstyle=cross,
-            markfields={a4, c4, Z3, d3, Z1, d1, a0, c0},
-            color=red!50,
-            markfields={f5, e6},
-            setpieces={Ng7, Nb2},
-            boardfontsize=12pt,labelfontsize=9pt]}
-
-\subsection*{Reference Implementation}
-
-This Scala part comes with three reference implementations in form of
-\texttt{jar}-files. This allows you to run any test cases on your own
-computer. For example you can call Scala on the command line with the
-option \texttt{-cp knight1.jar} and then query any function from the
-\texttt{knight1.scala} template file. As usual you have to
-prefix the calls with \texttt{CW8a}, \texttt{CW8b} and \texttt{CW8c}.
-Since some of the calls are time sensitive, I included some timing
-information. For example
-
-\begin{lstlisting}[language={},numbers=none,basicstyle=\ttfamily\small]
-$ scala -cp knight1.jar
-scala> CW8a.enum_tours(5, List((0, 0))).length
-Time needed: 1.722 secs.
-res0: Int = 304
-
-scala> CW8a.print_board(8, CW8a.first_tour(8, List((0, 0))).get)
-Time needed: 15.411 secs.
-
- 51  46  55  44  53   4  21  12 
- 56  43  52   3  22  13  24   5 
- 47  50  45  54  25  20  11  14 
- 42  57   2  49  40  23   6  19 
- 35  48  41  26  61  10  15  28 
- 58   1  36  39  32  27  18   7 
- 37  34  31  60   9  62  29  16 
-  0  59  38  33  30  17   8  63 
-\end{lstlisting}%$
-
-
-\subsection*{Hints}
-
-\noindent
-\textbf{Preliminary Part} useful list functions: \texttt{.contains(..)} checks
-whether an element is in a list, \texttt{.flatten} turns a list of
-lists into just a list, \texttt{\_::\_} puts an element on the head of
-the list, \texttt{.head} gives you the first element of a list (make
-sure the list is not \texttt{Nil}); a useful option function:
-\texttt{.isDefined} returns true, if an option is \texttt{Some(..)};
-anonymous functions can be constructed using \texttt{(x:Int) => ...},
-this function takes an \texttt{Int} as an argument.\medskip
-
-
-\noindent
-\textbf{Core Part} a useful list function: \texttt{.sortBy} sorts a list
-according to a component given by the function; a function can be
-tested to be tail-recursive by annotation \texttt{@tailrec}, which is
-made available by importing \texttt{scala.annotation.tailrec}.\medskip
-
-
-
-
-\subsection*{Preliminary Part (4 Marks)}
-
-You are asked to implement the knight's tour problem such that the
-dimension of the board can be changed.  Therefore most functions will
-take the dimension of the board as an argument.  The fun with this
-problem is that even for small chessboard dimensions it has already an
-incredibly large search space---finding a tour is like finding a
-needle in a haystack. In the first task we want to see how far we get
-with exhaustively exploring the complete search space for small
-chessboards.\medskip
-
-\noindent
-Let us first fix the basic datastructures for the implementation.  The
-board dimension is an integer.
-A \emph{position} (or field) on the chessboard is
-a pair of integers, like $(0, 0)$. A \emph{path} is a list of
-positions. The first (or 0th move) in a path is the last element in
-this list; and the last move in the path is the first element. For
-example the path for the $5\times 5$ chessboard above is represented
-by
-
-\[
-\texttt{List($\underbrace{\texttt{(0, 4)}}_{24}$,
-  $\underbrace{\texttt{(2, 3)}}_{23}$, ...,
-  $\underbrace{\texttt{(3, 2)}}_1$, $\underbrace{\texttt{(4, 4)}}_0$)}
-\]
-
-\noindent
-Suppose the dimension of a chessboard is $n$, then a path is a
-\emph{tour} if the length of the path is $n \times n$, each element
-occurs only once in the path, and each move follows the rules of how a
-knight moves (see above for the rules).
-
-
-\subsubsection*{Tasks (file knight1.scala)}
-
-\begin{itemize}
-\item[(1)] Implement an \texttt{is\_legal} function that takes a
-  dimension, a path and a position as arguments and tests whether the
-  position is inside the board and not yet element in the
-  path. \hfill[1 Mark]
-
-\item[(2)] Implement a \texttt{legal\_moves} function that calculates for a
-  position all legal onward moves. If the onward moves are
-  placed on a circle, you should produce them starting from
-  ``12-o'clock'' following in clockwise order.  For example on an
-  $8\times 8$ board for a knight at position $(2, 2)$ and otherwise
-  empty board, the legal-moves function should produce the onward
-  positions in this order:
-
-  \begin{center}
-  \texttt{List((3,4), (4,3), (4,1), (3,0), (1,0), (0,1), (0,3), (1,4))}
-  \end{center}
-
-  If the board is not empty, then maybe some of the moves need to be
-  filtered out from this list.  For a knight on field $(7, 7)$ and an
-  empty board, the legal moves are
-
-  \begin{center}
-  \texttt{List((6,5), (5,6))}
-  \end{center}
-  \mbox{}\hfill[1 Mark]
-
-\item[(3)] Implement two recursive functions (\texttt{count\_tours} and
-  \texttt{enum\_tours}). They each take a dimension and a path as
-  arguments. They exhaustively search for tours starting
-  from the given path. The first function counts all possible 
-  tours (there can be none for certain board sizes) and the second
-  collects all tours in a list of paths. These functions will be
-  called with a path containing a single position---the starting field.
-  They are expected to extend this path so as to find all tours starting
-  from the given position.\\
-  \mbox{}\hfill[2 Marks]
-\end{itemize}
-
-\noindent \textbf{Test data:} For the marking, the functions in (3)
-will be called with board sizes up to $5 \times 5$. If you search
-for tours on a $5 \times 5$ board starting only from field $(0, 0)$,
-there are 304 of tours. If you try out every field of a $5 \times
-5$-board as a starting field and add up all tours, you obtain
-1728. A $6\times 6$ board is already too large to be searched
-exhaustively.\footnote{For your interest, the number of tours on
-  $6\times 6$, $7\times 7$ and $8\times 8$ are 6637920, 165575218320,
-  19591828170979904, respectively.}\smallskip
-
-
-\subsection*{Core Part (6 Marks)}
-
-
-\subsubsection*{Tasks (file knight1.scala cont.)}
-
-\begin{itemize}
-\item[(4)] Implement a \texttt{first}-function. This function takes a list of
-  positions and a function $f$ as arguments; $f$ is the name we give to
-  this argument). The function $f$ takes a position as argument and
-  produces an optional path. So $f$'s type is \texttt{Pos =>
-    Option[Path]}. The idea behind the \texttt{first}-function is as follows:
-
-  \[
-  \begin{array}{lcl}
-  \textit{first}(\texttt{Nil}, f) & \dn & \texttt{None}\\  
-  \textit{first}(x\!::\!xs, f) & \dn & \begin{cases}
-    f(x) & \textit{if}\;f(x) \not=\texttt{None}\\
-    \textit{first}(xs, f) & \textit{otherwise}\\
-                              \end{cases}
-  \end{array}
-  \]
-
-  \noindent That is, we want to find the first position where the
-  result of $f$ is not \texttt{None}, if there is one. Note that
-  `inside' \texttt{first}, you do not (need to) know anything about
-  the argument $f$ except its type, namely \texttt{Pos =>
-    Option[Path]}. If you want to find out what the result of $f$ is
-  on a particular argument, say $x$, you can just write $f(x)$. 
-  There is one additional point however you should
-  take into account when implementing \texttt{first}: you will need to
-  calculate what the result of $f(x)$ is; your code should do this
-  only \textbf{once} and for as \textbf{few} elements in the list as
-  possible! Do not calculate $f(x)$ for all elements and then see which 
-  is the first \texttt{Some}.\\\mbox{}\hfill[1 Mark]
-  
-\item[(5)] Implement a \texttt{first\_tour} function that uses the
-  \texttt{first}-function from (4), and searches recursively for single tour.
-  As there might not be such a tour at all, the \texttt{first\_tour} function
-  needs to return a value of type
-  \texttt{Option[Path]}.\\\mbox{}\hfill[1 Mark]
-\end{itemize}
-
-\noindent
-\textbf{Testing:} The \texttt{first\_tour} function will be called with board
-sizes of up to $8 \times 8$.
-\bigskip
-
-%%\newpage
-
-\noindent
-As you should have seen in the earlier parts, a naive search for tours beyond
-$8 \times 8$ boards and also searching for closed tours even on small
-boards takes too much time. There is a heuristics, called \emph{Warnsdorf's
-Rule} that can speed up finding a tour. This heuristics states that a
-knight is moved so that it always proceeds to the field from which the
-knight will have the \underline{fewest} onward moves.  For example for
-a knight on field $(1, 3)$, the field $(0, 1)$ has the fewest possible
-onward moves, namely 2.
-
-\chessboard[maxfield=g7,
-            pgfstyle= {[base,at={\pgfpoint{0pt}{-0.5ex}}]text},
-            text = \small 3, markfield=Z5,
-            text = \small 7, markfield=b5,
-            text = \small 7, markfield=c4,
-            text = \small 7, markfield=c2,
-            text = \small 5, markfield=b1,
-            text = \small 2, markfield=Z1,
-            setpieces={Na3}]
-
-\noindent
-Warnsdorf's Rule states that the moves on the board above should be
-tried in the order
-
-\[
-(0, 1), (0, 5), (2, 1), (2, 5), (3, 4), (3, 2)
-\]
-
-\noindent
-Whenever there are ties, the corresponding onward moves can be in any
-order.  When calculating the number of onward moves for each field, we
-do not count moves that revisit any field already visited.
-
-\subsubsection*{Tasks (file knight2.scala)}
-
-\begin{itemize}
-\item[(6)] Write a function \texttt{ordered\_moves} that calculates a list of
-  onward moves like in (2) but orders them according to 
-  Warnsdorf’s Rule. That means moves with the fewest legal onward moves
-  should come first (in order to be tried out first). \hfill[1 Mark]
-  
-\item[(7)] Implement a \texttt{first\_closed\_tour\_heuristics}
-  function that searches for a single
-  \textbf{closed} tour on a $6\times 6$ board. It should try out
-  onward moves according to
-  the \texttt{ordered\_moves} function from (6). It is more likely to find
-  a solution when started in the middle of the board (that is
-  position $(dimension / 2, dimension / 2)$). \hfill[1 Mark]
-
-\item[(8)] Implement a \texttt{first\_tour\_heuristics} function
-  for boards up to
-  $30\times 30$.  It is the same function as in (7) but searches for
-  tours (not just closed tours). It might be called with any field on the
-  board as starting field.\\
-  %You have to be careful to write a
-  %tail-recursive function of the \texttt{first\_tour\_heuristics} function
-  %otherwise you will get problems with stack-overflows.\\
-  \mbox{}\hfill[1 Mark]
-\end{itemize}    
-
-\subsubsection*{Task (file knight3.scala)}
-\begin{itemize}
-\item[(9)] Implement a function \texttt{tour\_on\_mega\_board} which is
-  the same function as in (8), \textbf{but} should be able to
-  deal with boards up to
-  $70\times 70$ \textbf{within 30 seconds} (on my laptop). This will be tested
-  by starting from field $(0, 0)$. You have to be careful to
-  write a tail-recursive function otherwise you will get problems
-  with stack-overflows. Please observe the requirements about
-  the submissions: no tricks involving \textbf{.par}.\medskip
-
-  The timelimit of 30 seconds is with respect to the laptop on which the
-  marking will happen. You can roughly estimate how well your
-  implementation performs by running \texttt{knight3.jar} on your
-  computer. For example the reference implementation shows
-  on my laptop:
-  
-  \begin{lstlisting}[language={},numbers=none,basicstyle=\ttfamily\small]
-$ scala -cp knight3.jar
-  
-scala> CW8c.tour_on_mega_board(70, List((0, 0)))
-Time needed: 9.484 secs.
-...<<long_list>>...
-\end{lstlisting}%$
-
-  \mbox{}\hfill[1 Mark]
-\end{itemize}  
-\bigskip
-
-
-
-
-\end{document}
-
-%%% Local Variables: 
-%%% mode: latex
-%%% TeX-master: t
-%%% End: 




