\documentclass{article}

\usepackage{chessboard}

\usepackage[LSBC4,T1]{fontenc}

\let\clipbox\relax

\usepackage{../style}

\usepackage{disclaimer}

\begin{document}

\setchessboard{smallboard,

               zero,

               showmover=false,

               boardfontencoding=LSBC4,

               hlabelformat=\arabic{ranklabel},

               vlabelformat=\arabic{filelabel}}

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

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

\IMPORTANT{}

Also note that the running time of each part will be restricted to a

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):

            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},

\noindent

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}.

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

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,

\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 =>

  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

  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

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.

\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]

  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

  \mbox{}\hfill[1 Mark]

\end{itemize}  

\bigskip

\end{document}

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