\documentclass{article}
\usepackage{chessboard}
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\usepackage{../style}
\usepackage{disclaimer}

\begin{document}

\setchessboard{smallboard,
               zero,
               showmover=false,
               boardfontencoding=LSBC4,
               hlabelformat=\arabic{ranklabel},
               vlabelformat=\arabic{filelabel}}

\mbox{}\\[-18mm]\mbox{}

\section*{Coursework 7 (Scala)}

This coursework is worth 10\%. It is about searching and
backtracking. The first part is due on 29 November at 11pm; the
second, more advanced part, is due on 20 December at 11pm. You are
asked to implement Scala programs that solve various versions of the
\textit{Knight's Tour Problem} on a chessboard. Note the second, more
advanced, part might include material you have not yet seen in the
first two lectures. \bigskip

\IMPORTANT{}
Also note that the running time of each part will be restricted to a
maximum of 360 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}]


\noindent
\textbf{Hints:} 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}).

\noindent
\textbf{Hints:} 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}.

            
\subsection*{Part 1 (7 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 (we will never go beyond board sizes of
$40 \times 40$).  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.\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.}\bigskip



\subsubsection*{Tasks (file knight2.scala)}

\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]}. 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 (2a), and searches recursively for a 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

\noindent
\textbf{Hints:} a useful list function: \texttt{.filter(..)} filters a
list according to a boolean function; a useful option function:
\texttt{.isDefined} returns true, if an option is \texttt{Some(..)};
anonymous functions can be constructed using \texttt{(x:Int) => ...},
this functions takes an \texttt{Int} as an argument.


%%\newpage
\subsection*{Part 2 (3 Marks)}

As you should have seen in Part 1, 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 heuristic, called \emph{Warnsdorf's
Rule} that can speed up finding a tour. This heuristic 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 knight3.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 the
  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\_heuristic}
  function that searches for a
  \textbf{closed} tour on a $6\times 6$ board. It should use the
  \texttt{first}-function from (4) and tries out onward moves according to
  the \texttt{ordered\_moves} function from (3a). 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\_heuristic} function
  for boards up to
  $40\times 40$.  It is the same function as in (7) but searches for
  tours (not just closed tours). You have to be careful to write a
  tail-recursive function of the \texttt{first\_tour\_heuristic} function
  otherwise you will get problems with stack-overflows.\\
  \mbox{}\hfill[1 Mark]
\end{itemize}  
\bigskip




\end{document}

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