--- a/cws/main_cw04.tex Thu Nov 02 23:34:53 2023 +0000
+++ b/cws/main_cw04.tex Sat Nov 04 18:53:37 2023 +0000
@@ -7,6 +7,35 @@
\usepackage{../styles/langs}
\usepackage{disclaimer}
\usepackage{ulem}
+%\usepackage{tipauni}
+
+
+
+\tikzset
+{%
+ pics/piece/.style n args={1}{
+ code={%
+ \fill[rounded corners] (-0.1,-0.1) rectangle (-0.9, -0.9);
+ \fill[left color=white,rounded corners,
+ right color=gray,
+ opacity=0.7] (-0.1,-0.1) rectangle (-0.9, -0.9);
+ \draw[line width=0.4mm,rounded corners] (-0.1,-0.1) rectangle (-0.9, -0.9);
+ \draw[line width=0.2mm,rounded corners] (-0.2,-0.2) rectangle (-0.8, -0.8);
+ \draw[anchor=mid] (-0.5,-0.6) node {#1};
+ }},
+ pics/king/.style n args={1}{
+ code={%
+ \fill[rounded corners] (-0.1,-0.1) rectangle (-0.9, -0.9);
+ \fill[left color=white,rounded corners,
+ right color=gray,
+ opacity=0.7] (-0.1,-0.1) rectangle (-0.9, -0.9);
+ \draw[line width=0.4mm,rounded corners] (-0.1,-0.1) rectangle (-0.9, -0.9);
+ \draw[line width=0.2mm,rounded corners] (-0.2,-0.2) rectangle (-0.8, -0.8);
+ \draw[anchor=mid] (-0.5,-0.6) node {#1};
+ \draw[anchor=center] (-0.5,-0.25) node {\includegraphics[scale=0.015]{crown.png}};
+ }}
+}
+
\begin{document}
@@ -19,472 +48,515 @@
\mbox{}\\[-18mm]\mbox{}
-\section*{Main Part 4 (Scala, 11 Marks)}
+\section*{Main Part 4:\\ Implementing the Shogun Board Game (7 Marks)}
\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.
-\medskip
+You are asked to implement a Scala program for playing the Shogun
+board game.\medskip
-% Note the core, more advanced, part might include material you have not
-%yet seen in the first three lectures. \bigskip
+%The deadline for your submission is on 26th July at
+%16:00. There will be no automated tests for the resit, but there are
+%many testcases in the template and the task description. Make sure
+%you use Scala \textbf{2.13.XX} for the resit---the same version as
+%during the lectures. \medskip
\IMPORTANTNONE{}
\noindent
-Also note that the running time of each part will be restricted to a
+Also note that the running time of each task 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}.
+try to avoid to calculate the result again.
\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:
+Shogun
+(\faVolumeUp\,[shōgoon]) is a game played by two players on a chess board and is somewhat
+similar to chess and checkers. A real Shogun board looks
+like in the pictures on the left.
-\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
- ]
+\begin{center}
+\begin{tabular}{@{}ccc@{}}
+\raisebox{2mm}{\includegraphics[scale=0.1]{shogun2.jpeg}}
+&
+\raisebox{2mm}{\includegraphics[scale=0.14]{shogun.jpeg}}
+&
+\begin{tikzpicture}[scale=0.45,every node/.style={scale=0.45}]
+% chessboard
+\draw[very thick,gray] (0,0) rectangle (8,8);
+\foreach\x in {0,...,7}\foreach\y in {7,...,0}
+{
+ \pgfmathsetmacro\blend{Mod(\x+\y,2)==0?75:25}
+ \fill[gray!\blend] (\x,\y) rectangle ++ (1,1);
+}
+% black pieces
+\foreach\x/\y/\e in {1/1/1,2/1/3,3/1/2,4/1/3,6/1/3,7/1/1,8/1/2}
+ \pic[fill=white] at (\x,\y) {piece={\e}};
+% white pieces
+\foreach\x/\y/\e in {1/8/4,2/8/2,3/8/4,5/8/4,6/8/2,7/8/3,8/8/1}
+ \pic[fill=red] at (\x,\y) {piece={\e}};
+\pic[fill=white] at (5.0,1.0) {king={1}};
+\pic[fill=red] at (4.0,8.0) {king={2}};
+% numbers
+\foreach\x in {1,...,8}
+{\draw (\x - 0.5, -0.4) node {\x};
+}
+\foreach\y in {1,...,8}
+{\draw (-0.4, \y - 0.6, -0.4) node {\y};
+}
+\end{tikzpicture}
+\end{tabular}
+\end{center}
\noindent
-where the 35th move can join up again with the 0th move.
+In what follows we shall use board illustrations as shown on the right. As
+can be seen there are two colours in Shogun for the pieces, red and white. Each
+player has 8 pieces, one of which is a king (the piece with the crown)
+and seven are pawns. At the beginning the pieces are lined up as shown
+above. What sets Shogun apart from chess and checkers is that each
+piece has, what I call, a kind of \textit{energy}---which for pawns is
+a number between 1 and 4, and for kings between 1 and 2. The energy
+determines how far a piece has to move. In the physical version of
+Shogun, the pieces and the board have magnets that can change the
+energy of a piece from move to move---so a piece on one field can have
+energy 2 and on a different field the same piece might have energy
+3. There are some further constraints on legal moves, which are
+explained below. The point of this part is to implement functions
+about moving pieces on the Shogun board.\medskip\medskip
-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):
+%and testing for when a
+%checkmate occurs---i.e.~the king is attacked and cannot move
+%anymore to an ``unattacked'' field (to simplify matters for
+%the resit we leave out the case where the checkmate can be averted by capturing
+%the attacking piece).\medskip
-{\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]}
+\noindent
+Like in chess, in Shogun the players take turns of moving and
+possibly capturing opposing pieces.
+There are the following rules on how pieces can move:
-\subsection*{Reference Implementation}
-
-%\mbox{}\alert{}\textcolor{red}{You need to download \texttt{knight1.jar} from K%EATS. The one
-%supplied with github does not contain the correct code. See Scala coursework
-%section on KEATS.}\medskip
+\begin{itemize}
+\item The energy of a piece determines how far, that is how many
+ fields, a piece has to move (remember pawns have an energy between 1 --
+ 4, kings have an energy of only 1 -- 2). The energy of a piece might
+ change when the piece moves to new field.
+\item Pieces can move in straight lines (up, down, left, right), or in
+ L-shape moves, meaning a move can make a single
+ 90$^{\circ}$-turn. S-shape moves with more than one turn are not
+ allowed. Also in a single move a piece cannot go forward and then
+ go backward---for example with energy 3 you cannot move 2 fields up and
+ then 1 field down. A piece can never move diagonally.
+\item A piece cannot jump over another piece and cannot stack up on top of your own pieces.
+ But you can capture an opponent's piece if you move to an occupied field. A captured
+ piece is removed from the board.
+\end{itemize}
\noindent
-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{M4a}, \texttt{M4b}, \texttt{M4c} and \texttt{M4d}.
-Since some of the calls are time sensitive, I included some timing
-information. For example
+Like in chess, checkmate is determined when the king of a player cannot
+move anymore to a field that is not attacked, or a player cannot
+capture or block the attacking piece, or the king is the only
+piece left for a player. A board that is checkmate is the following:
+
+\begin{center}
+\begin{tikzpicture}[scale=0.5,every node/.style={scale=0.5}]
+% chessboard
+\draw[very thick,gray] (0,0) rectangle (8,8);
+\foreach\x in {0,...,7}\foreach\y in {7,...,0}
+{
+ \pgfmathsetmacro\blend{Mod(\x+\y,2)==0?75:25}
+ \fill[gray!\blend] (\x,\y) rectangle ++ (1,1);
+}
+% redpieces
+\pic[fill=red] at (4,2) {king={2}};
+\pic[fill=red] at (6,1) {piece={3}};
+\pic[fill=red] at (4,4) {piece={4}};
+\pic[fill=red] at (5,3) {piece={4}};
+% white pieces
+\pic[fill=white] at (7,1) {king={2}};
+\pic[fill=white] at (8,5) {piece={2}};
+\pic[fill=white] at (4,1) {piece={2}};
+% numbers
+\foreach\x in {1,...,8}
+{\draw (\x - 0.5, -0.4) node {\x};
+}
+\foreach\y in {1,...,8}
+{\draw (-0.4, \y - 0.6, -0.4) node {\y};
+}
+\end{tikzpicture}
+\end{center}
+
+\noindent
+The reason for the checkmate is that the white king on field (7, 1) is
+attacked by the red pawn on \mbox{(5, 3)}. There is nowhere for the
+white king to go, and no white pawn can be moved into the way of this
+red pawn and white can also not capture it. When determining a possible
+move, you need to be careful with pieces that might be in the
+way. Consider the following position:
-\begin{lstlisting}[language={},numbers=none,basicstyle=\ttfamily\small]
-$ scala -cp knight1.jar
-scala> M4a.enum_tours(5, List((0, 0))).length
-Time needed: 1.722 secs.
-res0: Int = 304
-
-scala> M4a.print_board(8, M4a.first_tour(8, List((0, 0))).get)
-Time needed: 15.411 secs.
+\begin{equation}\label{moves}
+\begin{tikzpicture}[scale=0.5,every node/.style={scale=0.5}]
+% chessboard
+\draw[very thick,gray] (0,0) rectangle (8,8);
+\foreach\x in {0,...,7}\foreach\y in {7,...,0}
+{
+ \pgfmathsetmacro\blend{Mod(\x+\y,2)==0?75:25}
+ \fill[gray!\blend] (\x,\y) rectangle ++ (1,1);
+}
+% redpieces
+\fill[blue!50] (0,2) rectangle ++ (1,1);
+\fill[blue!50] (1,1) rectangle ++ (1,1);
+\fill[blue!50] (0,4) rectangle ++ (1,1);
+\fill[blue!50] (1,5) rectangle ++ (1,1);
+\fill[blue!50] (2,6) rectangle ++ (1,1);
+%%\fill[blue!50] (3,7) rectangle ++ (1,1);
+\fill[blue!50] (4,6) rectangle ++ (1,1);
+\fill[blue!50] (5,5) rectangle ++ (1,1);
+\fill[blue!50] (6,4) rectangle ++ (1,1);
+\fill[blue!50] (6,2) rectangle ++ (1,1);
+\fill[blue!50] (7,3) rectangle ++ (1,1);
+\fill[blue!50] (4,0) rectangle ++ (1,1);
+\fill[blue!50] (2,0) rectangle ++ (1,1);
+\pic[fill=red] at (4,4) {piece={4}};
+\pic[fill=red] at (4,8) {piece={4}};
+\pic[fill=white] at (2,5) {piece={3}};
+\pic[fill=white] at (4,3) {piece={2}};
+\pic[fill=white] at (6,3) {piece={1}};
+\pic[fill=white] at (8,4) {piece={1}};
+% numbers
+\foreach\x in {1,...,8}
+{\draw (\x - 0.5, -0.4) node {\x};
+}
+\foreach\y in {1,...,8}
+{\draw (-0.4, \y - 0.6, -0.4) node {\y};
+}
+\end{tikzpicture}
+\end{equation}
- 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}%$
+\noindent
+The red piece in the centre on field (4, 4) can move to all the blue fields.
+In particular it can move to (2, 6), because it can move 2 fields up
+and 2 fields to the left---it cannot reach this field by moving two
+fields to the left and then two up, because jumping over the white
+piece at (2, 5) is not allowed. Similarly, the field at (6, 2) is
+unreachable for the red piece because of the two white pieces at (4,
+3) and (6, 3) are in the way and no S-shape move is allowed in
+Shogun. The red piece on (4, 4) cannot move to the field (4, 8) at the
+top, because a red piece is already there; but it can move to (8, 4)
+and capture the white piece there. The moral is we always have to
+explore all possible ways in order to determine whether a piece can be
+moved to a field or not: in general there might be several ways and some of
+them might be blocked.
\subsection*{Hints}
+Useful functions about pieces and boards are defined at the beginning
+of the template file. The function \texttt{.map} applies a function to
+each element of a list or set; \texttt{.flatMap} works like
+\texttt{map} followed by a \texttt{.flatten}---this is useful if a
+function returns a set of sets, which need to be ``unioned up''. Sets
+can be partitioned according to a predicate with the function
+\texttt{.partition}. For example
+
+\begin{lstlisting}
+val (even, odd) = Set(1,2,3,4,5).partition(_ % 2 == 0)
+// --> even = Set(2,4)
+// odd = Set(1,3,5)
+\end{lstlisting}
+
\noindent
-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;
-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
+The function \texttt{.toList} transforms a set into a list. The function
+\texttt{.count} counts elements according to a predicate. For example
+\begin{lstlisting}
+Set(1,2,3,4,5).count(_ % 2 == 0)
+// --> 2
+\end{lstlisting}
-%%\newpage
+%% \newpage
\subsection*{Tasks}
-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
+You are asked to implement how pieces can move on a Shogun board. Let
+us first fix the basic datastructures for the implementation. A
+\emph{position} (or field) is a pair of integers, like $(3, 2)$. The
+board's dimension is always 8 $\times$ 8. A \emph{colour} is either
+red (\texttt{Red}) or white (\texttt{Wht}). A \emph{piece} is either
+a pawn or a king, and has a position, a colour and an energy (an
+integer). In the template file there are functions \texttt{incx},
+\texttt{decx}, \texttt{incy} and \texttt{decy} for incrementing and
+decrementing the x- and y-coordinates of positions of pieces.
-\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*{Task 1 (file knight1.scala)}
+A \emph{board} consists of a set of pieces. We always assume that we
+start with a consistent board and every move generates another
+consistent board. In this way we do not need to check, for example,
+whether pieces are stacked on top of each other or located outside the
+board, or have an energy outside the permitted range. There are
+functions \texttt{-} and \texttt{+} for removing, respectively adding,
+single pieces to a board. The function \texttt{occupied} takes a
+position and a board as arguments, and returns an \texttt{Option} of a
+piece when this position is occupied, otherwise \texttt{None}. The
+function \texttt{occupied\_by} returns the colour of a potential piece
+on that position. The function \texttt{is\_occupied} returns a boolean
+for whether a position is occupied or not; \texttt{print\_board} is a
+rough function that prints out a board on the console. This function
+is meant for testing purposes.
\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[(1)] You need to calculate all possible moves for a piece on a Shogun board. In order to
+ make sure no piece moves forwards and backwards at the same time,
+ and also exclude all S-shape moves, the data-structure \texttt{Move}
+ is introduced. A \texttt{Move} encodes all simple moves (up, down, left,
+ right) and L-shape moves (first right, then up and so on). This is defined
+ as follows:
+
+{\small\begin{lstlisting}
+abstract class Move
+case object U extends Move // up
+case object D extends Move // down
+case object R extends Move // right
+case object L extends Move // left
+case object RU extends Move // ...
+case object LU extends Move
+case object RD extends Move
+case object LD extends Move
+case object UR extends Move
+case object UL extends Move
+case object DR extends Move
+case object DL extends Move
+\end{lstlisting}}
-\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:
+You need to implement an \texttt{eval} function that takes a piece
+\texttt{pc}, a move \texttt{m}, an energy \texttt{en} and a board
+\texttt{b} as arguments. The idea is to recursively calculate all
+fields that can be reached by the move \texttt{m} (there might be more than
+one). The energy acts as a counter and decreases in each recursive
+call until 0 is reached (the final field). The function \texttt{eval} for a piece \texttt{pc}
+should behave as follows:
- \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{itemize}
+\item If the position of a piece is outside the board, then no field can be reached (represented by
+ the empty set \texttt{Set()}).
+\item If the energy is 0 and the position of the piece is \textit{not} occupied, then the field can be reached
+ and the set \texttt{Set(pc)} is returned whereby \texttt{pc} is the piece given as argument.
+\item If the energy is 0 and the position of the piece \textit{is} occupied, but occupied by a piece
+ of the opposite colour, then also the set \texttt{Set(pc)} is returned.
+\item In case the energy is > 0 and the position of the piece
+ \texttt{pc} is occupied, then this move is blocked and the set
+ \texttt{Set()} is returned.
+\item In all other cases we have to analyse the move
+ \texttt{m}. First, the simple moves (that is \texttt{U}, \texttt{D},
+ \texttt{L} and \texttt{R}) we only have to increment / decrement the
+ x- or y-position of the piece, decrease the energy and call eval
+ recursively with the updated arguments. For example for \texttt{U}
+ you need to increase the y-coordinate:
\begin{center}
- \texttt{List((6,5), (5,6))}
+ \texttt{U} $\quad\Rightarrow\quad$ new arguments: \texttt{incy(pc)}, \texttt{U}, energy - 1, same board
\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[1 Mark]
-\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
-\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:
-
- \[
- \begin{array}{lcl}
- \textit{first}(\texttt{Nil}, f) & \dn & \texttt{None}\\
- \textit{first}(x\!::\!xs, f) & \dn & \begin{cases}
- f(x) & \textit{if}\;f(x) \not=\texttt{None}\\
- \textit{first}(xs, f) & \textit{otherwise}\\
- \end{cases}
- \end{array}
- \]
+ The move \texttt{U} here acts like a ``mode'', meaning if you move
+ up, you can only move up; the mode never changes. Similarly for the other simple moves: if
+ you move right, you can only move right and so on. In this way it is
+ prevented to go first to the right, and then change direction in order to go
+ left (same with up and down).
+
+ For the L-shape moves (\texttt{RU}, \texttt{LU}, \texttt{RD} and so on) you need to calculate two
+ sets of reachable fields. Say we analyse \texttt{RU}, then we first have to calculate all fields
+ reachable by moving to the right; then we have to calculate all moves by changing the mode to \texttt{U}.
+ That means there are two recursive calls to \texttt{eval}:
- \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]
+ \begin{center}
+ \begin{tabular}{@{}lll@{}}
+ \texttt{RU} & $\Rightarrow$ & new args for call 1: \texttt{incx(pc)}, \texttt{RU}, energy - 1, same board\\
+ & & new args for call 2: \texttt{pc}, \texttt{U}, same energy, same board
+ \end{tabular}
+ \end{center}
+
+ In each case we receive some new piece(s) on reachable fields and therefore we return the set
+ containing all these fields. Similarly in the other cases.
\end{itemize}
-\noindent
-\textbf{Testing:} The \texttt{first\_tour} function will be called with board
-sizes of up to $8 \times 8$.
-\bigskip
-
-%%\newpage
-\subsubsection*{Task 2 (file knight2.scala)}
+For example on the left board below, \texttt{eval} is called with the white
+piece in the centre and the move \texttt{RU} generates then a set of
+new pieces corresponding to the blue fileds. The difference on the
+right board is that \texttt{eval} is called with a red piece and therefore the
+field (4, 8) is not reachable anymore because it is already occupied by
+another red piece.
-\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.
+\begin{center}
+\begin{tabular}{cc}
+\begin{tikzpicture}[scale=0.45,every node/.style={scale=0.45}]
+% chessboard
+\draw[very thick,gray] (0,0) rectangle (8,8);
+\foreach\x in {0,...,7}\foreach\y in {7,...,0}
+{
+ \pgfmathsetmacro\blend{Mod(\x+\y,2)==0?75:25}
+ \fill[gray!\blend] (\x,\y) rectangle ++ (1,1);
+}
+\fill[blue!50] (5,5) rectangle ++ (1,1);
+\fill[blue!50] (3,7) rectangle ++ (1,1);
+\fill[blue!50] (4,6) rectangle ++ (1,1);
+\fill[blue!50] (6,4) rectangle ++ (1,1);
+\fill[blue!50] (7,3) rectangle ++ (1,1);
+
+% black pieces
+\foreach\x/\y/\e in {1/1/1,2/1/3,3/1/2,4/1/3,6/1/3,7/1/1,8/1/2}
+ \pic[fill=white] at (\x,\y) {piece={\e}};
+% white pieces
+\foreach\x/\y/\e in {1/8/4,2/8/2,3/8/4,5/8/4,6/8/2,7/8/3,8/8/1}
+ \pic[fill=red] at (\x,\y) {piece={\e}};
+\pic[fill=white] at (5.0,1.0) {king={1}};
+\pic[fill=red] at (4.0,8.0) {king={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}]
+\pic[fill=white] at (4,4) {piece={4}};
+% numbers
+\foreach\x in {1,...,8}
+{\draw (\x - 0.5, -0.4) node {\x};
+}
+\foreach\y in {1,...,8}
+{\draw (-0.4, \y - 0.6, -0.4) node {\y};
+}
+\end{tikzpicture}
+&
+\begin{tikzpicture}[scale=0.45,every node/.style={scale=0.45}]
+% chessboard
+\draw[very thick,gray] (0,0) rectangle (8,8);
+\foreach\x in {0,...,7}\foreach\y in {7,...,0}
+{
+ \pgfmathsetmacro\blend{Mod(\x+\y,2)==0?75:25}
+ \fill[gray!\blend] (\x,\y) rectangle ++ (1,1);
+}
+\fill[blue!50] (5,5) rectangle ++ (1,1);
+\fill[blue!50] (4,6) rectangle ++ (1,1);
+\fill[blue!50] (6,4) rectangle ++ (1,1);
+\fill[blue!50] (7,3) rectangle ++ (1,1);
-\noindent
-Warnsdorf's Rule states that the moves on the board above should be
-tried in the order
+% black pieces
+\foreach\x/\y/\e in {1/1/1,2/1/3,3/1/2,4/1/3,6/1/3,7/1/1,8/1/2}
+ \pic[fill=white] at (\x,\y) {piece={\e}};
+% white pieces
+\foreach\x/\y/\e in {1/8/4,2/8/2,3/8/4,5/8/4,6/8/2,7/8/3,8/8/1}
+ \pic[fill=red] at (\x,\y) {piece={\e}};
+\pic[fill=white] at (5.0,1.0) {king={1}};
+\pic[fill=red] at (4.0,8.0) {king={2}};
-\[
-(0, 1), (0, 5), (2, 1), (2, 5), (3, 4), (3, 2)
-\]
-
-\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.
+\pic[fill=red] at (4,4) {piece={4}};
+% numbers
+\foreach\x in {1,...,8}
+{\draw (\x - 0.5, -0.4) node {\x};
+}
+\foreach\y in {1,...,8}
+{\draw (-0.4, \y - 0.6, -0.4) node {\y};
+}
+\end{tikzpicture}
+\\[-5mm]
+\end{tabular}
+\end{center}\hfill[3 Marks]
-\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[(2)] Implement an \texttt{all\_moves} function that calculates for a
+ piece and a board, \textit{all} pieces on legal onward positions. For this
+ you have to call \texttt{eval} for all possible moves \texttt{m} (that is \texttt{U},
+ \texttt{D}, \ldots, \texttt{DL}). An example for all moves for the red piece on (4, 4) are
+ shown in \eqref{moves} on page \pageref{moves}.\\
+ \mbox{}\hfill[1 Mark]
+
+\item[(3)] Implement a function \texttt{attacked} that takes a colour and a board
+ and calculates all pieces of the opposite side that are attacked. For example
+ below on the left are all the attacked pieces by red, and on the right for white:
-\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}
+\begin{center}
+\begin{tabular}{cc}
+\begin{tikzpicture}[scale=0.45,every node/.style={scale=0.45}]
+% chessboard
+\draw[very thick,gray] (0,0) rectangle (8,8);
+\foreach\x in {0,...,7}\foreach\y in {7,...,0}
+{
+ \pgfmathsetmacro\blend{Mod(\x+\y,2)==0?75:25}
+ \fill[gray!\blend] (\x,\y) rectangle ++ (1,1);
+}
+\fill[blue!50] (7,3) rectangle ++ (1,1);
+\fill[blue!50] (6,0) rectangle ++ (1,1);
+
+
+% black pieces
+\foreach\x/\y/\e in {6/1/3,4/4/4,5/3/4,6/5/3}
+ \pic[fill=red] at (\x,\y) {piece={\e}};
+% white pieces
+\foreach\x/\y/\e in {8/4/2,4/1/2,8/7/3}
+ \pic[fill=white] at (\x,\y) {piece={\e}};
+
+\pic[fill=red] at (4,2) {king={2}};
+\pic[fill=white] at (7,1) {king={2}};
-\subsubsection*{Task 3 (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
+% numbers
+\foreach\x in {1,...,8}
+{\draw (\x - 0.5, -0.4) node {\x};
+}
+\foreach\y in {1,...,8}
+{\draw (-0.4, \y - 0.6, -0.4) node {\y};
+}
+\end{tikzpicture}
+&
+\begin{tikzpicture}[scale=0.45,every node/.style={scale=0.45}]
+% chessboard
+\draw[very thick,gray] (0,0) rectangle (8,8);
+\foreach\x in {0,...,7}\foreach\y in {7,...,0}
+{
+ \pgfmathsetmacro\blend{Mod(\x+\y,2)==0?75:25}
+ \fill[gray!\blend] (\x,\y) rectangle ++ (1,1);
+}
+\fill[blue!50] (5,0) rectangle ++ (1,1);
+
+
+% black pieces
+\foreach\x/\y/\e in {6/1/3,4/4/4,5/3/4,6/5/3}
+ \pic[fill=red] at (\x,\y) {piece={\e}};
+% white pieces
+\foreach\x/\y/\e in {8/4/2,4/1/2,8/7/3}
+ \pic[fill=white] at (\x,\y) {piece={\e}};
- 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> M4c.tour_on_mega_board(70, List((0, 0)))
-Time needed: 9.484 secs.
-...<<long_list>>...
-\end{lstlisting}%$
+\pic[fill=red] at (4,2) {king={2}};
+\pic[fill=white] at (7,1) {king={2}};
+% numbers
+\foreach\x in {1,...,8}
+{\draw (\x - 0.5, -0.4) node {\x};
+}
+\foreach\y in {1,...,8}
+{\draw (-0.4, \y - 0.6, -0.4) node {\y};
+}
+\end{tikzpicture}
+\\[-5mm]
+\end{tabular}
+\end{center}\mbox{}\hfill[1 Mark]
+
+\item[(4)] Implement a function \texttt{attackedN} that takes a piece and a board
+ and calculates the number of times this pieces is attacked by pieces of the opposite colour.
+ For example the piece on field (8, 4) above is attacked by 3 red pieces, and
+ the piece on (6, 1) by 1 white piece.
+ \\
+ \mbox{}\hfill[1 Mark]
+
+\item[(5)] Implement a function \texttt{protectedN} that takes a piece and a board
+ and calculates the number of times this pieces is protected by pieces of the same colour.
+ For example the piece on field (8, 4) above is protected by 1 white pieces (the one on (8, 7)),
+ and the piece on (5, 3) is protected by three red pieces ((6, 1), (4, 2), and (6, 5)).
+ \\
\mbox{}\hfill[1 Mark]
\end{itemize}
-\subsubsection*{Task 4 (file knight4.scala)}
-\begin{itemize}
-\item[(10)] In this task we want to solve the problem of finding a single
- tour (if there exists one) on what is sometimes called ``mutilated''
- chessboards, for example rectangular chessbords or chessboards where
- fields are missing. For this implement a search function
-
- \begin{center}
- \begin{tabular}{@{}l@{}}
- \texttt{def one\_tour\_pred(dim: Int, path: Path,}\\
- \texttt{\phantom{def one\_tour\_pred(}n: Int, f: Pos => Boolean): Option[Path]}
- \end{tabular}
- \end{center}
-
- which has, like before, the dimension and current path as arguments,
- and in addtion it takes an integer, which specifies the length of
- the longest path (or length of the path after which the search
- should backtrack), and a function from positions to Booleans. This
- function acts as a predicate in order to restrict which onward legal
- moves should be considered in the search. The function
- \texttt{one\_tour\_pred} should return a single tour
- (\texttt{Some}-case), if one or more tours exist, and \texttt{None}, if no
- tour exists. For example when called with
-
- \begin{center}
- \texttt{one\_tour\_pred(7, List((0, 0)), 35, x => x.\_1 < 5)}
- \end{center}
-
- we are looking for a tour starting from position \texttt{(0,0)} on a
- 7 $\times$ 5 board, where the maximum length of the path is 35. The
- predicate \texttt{x => x.\_1 < 5} ensures that no position with an
- x-coordinate bigger than 4 is considered. One possible solution is
-
- \begin{lstlisting}[language={},numbers=none,basicstyle=\ttfamily\small]
- 0 13 22 33 28 11 20
- 23 32 1 12 21 34 27
- 14 7 16 29 2 19 10
- 31 24 5 8 17 26 3
- 6 15 30 25 4 9 18
- -1 -1 -1 -1 -1 -1 -1
- -1 -1 -1 -1 -1 -1 -1
-\end{lstlisting}%$
-
-where \texttt{print\_board} prints a \texttt{-1} for all fields that
-have not been visited.
-
- \mbox{}\hfill[2 Marks]
-\end{itemize}
-
-
-
\end{document}
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