\documentclass{article}
\usepackage{chessboard}
\usepackage[LSBC4,T1]{fontenc}
\usepackage{../style}

\begin{document}

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

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

\section*{Coursework 7 (Scala, Knight's Tour)}

This coursework is about searching and backtracking, and worth
10\%. The first part is due on 23 November at 11pm; the second, more
advanced part, is due on 30 November at 11pm. You are asked to
implement Scala programs that solve various versions of the
\textit{Knight's Tour Problem} on a chessboard. Make sure the files
you submit can be processed by just calling \texttt{scala
  <<filename.scala>>}.
 
\subsection*{Disclaimer}

It should be understood that the work you submit represents
your own effort. You have not copied from anyone else. An
exception is the Scala code I showed during the lectures or
uploaded to KEATS, which you can freely use.\medskip

\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
The 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 (it is open) 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 bigger, 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}]

\subsection*{Part 1 (6 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 as an argument.  The fun with this problem is that
even for small chessbord dimensions it has already an incredably large
search space---finding a tour is like finding a needle in a
haystack. In the first task we want to see 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 boyond board sizes of
$100 \times 100$).  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[(1a)] Implement a is-legal-move function that takes a dimension, a path
and a position as argument and tests whether the position is inside
the board and not yet element in the path. \hfill[1 Mark]

\item[(1b)] Implement a 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-oclock'' following in clockwise order.  For example on an
  $8\times 8$ board for a knight on position $(2, 2)$ and otherwise
  empty board, the legal-moves function should produce the onward
  positions

  \begin{center}
  \texttt{List((3,4), (4,3), (4,1), (3,0), (1,0), (0,1), (0,3), (1,4))}
  \end{center}

  in this order.  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[(1c)] Implement two recursive functions (count-tours and
  enum-tours). They each take a dimension and a path as
  arguments. They exhaustively search for \underline{\bf open} tours
  starting from the given path. The first function counts all possible
  open tours (there can be none for certain board sizes) and the second
  collects all open tours in a list of paths.\hfill[2 Marks]
\end{itemize}

\noindent \textbf{Test data:} For the marking, the functions in (1c)
will be called with board sizes up to $5 \times 5$. If you only search
for open tours on $5 \times 5$ board starting from field $(0, 0)$,
there are 304 of them. If you try out every field of a $5 \times
5$-board as a starting field and add up all open tours, you obtain
1728. A $6\times 6$ board is already too large to be searched
exhaustively.\footnote{For your interest, the number of open tours on
  $6\times 6$, $7\times 7$ and $8\times 8$ are 6637920, 165575218320,
  19591828170979904, respectively.}

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

\begin{itemize}
\item[(2a)] Implement a first-function. This function takes a list of
  positions and a function $f$ as arguments. The function $f$ takes a
  position as argument and produces an optional path. The idea behind
  the 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}.\newline\mbox{}\hfill[1 Mark]
  
\item[(2b)] Implement a first-tour function. Using the first-function
  from (2a), search recursively for an open tour.  As there might not
  be such a tour at all, the first-tour function needs to return an
  \texttt{Option[Path]}.\hfill[2 Marks]
\end{itemize}

\noindent
\textbf{Testing} The first tour function will be called with board
sizes of up to $8 \times 8$. 


\subsection*{Part 2 (4 Marks)}

As you should have seen in Part 1, a naive search for open tours
beyond $8 \times 8$ boards and also for searching for closed tours
takes too much time. There is a heuristic (called Warnsdorf's rule)
that can speed up finding a tour. This heuristice states that a knight
is moved so that it always proceeds to the square 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 sould be
tried out in the order

\[
(0, 1), (0, 5), (2, 1), (2, 5), (3, 4), (3, 2)
\]

\noindent
Whenever there are ties, the correspoding 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[(3a)] orderered-moves

\item[(3b)] first-closed tour heuristics; up to $6\times 6$

\item[(3c)] first tour heuristics; up to $50\times 50$
\end{itemize}  

\end{document}

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