--- a/cws/cw02.tex	Tue Nov 14 13:14:47 2017 +0000
+++ b/cws/cw02.tex	Tue Nov 14 21:34:22 2017 +0000
@@ -188,7 +188,7 @@
 \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
-$50 \times 50$).  A \emph{position} (or field) on the chessboard is
+$40 \times 40$).  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
@@ -219,7 +219,7 @@
 \item[(1b)] 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-oclock'' following in clockwise order.  For example on an
+  ``12-o'clock'' 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 in this order:
@@ -292,14 +292,14 @@
 \textbf{Testing} The first tour function will be called with board
 sizes of up to $8 \times 8$. 
 
-
+%%\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
-takes too much time. There is a heuristic, called 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
+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 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.
@@ -342,10 +342,12 @@
   a solution when started in the middle of the board (that is
   position $(dimension / 2, dimension / 2)$). \hfill[1 Mark]
 
-\item[(3c)] Implement a first-tour-heuristic function for boards up to $50\times 50$.
-  It is the same function  as in (3b) but searches for tours. You have
-  to be careful to write a tail-recursive version of the first-tour-heuristic
-  function otherwise you will get problems with stack-overflows. \hfill[1 Mark]
+\item[(3c)] Implement a first-tour-heuristic function for boards up to
+  $40\times 40$.  It is the same function as in (3b) but searches for
+  tours (not just closed tours). You have to be careful to write a
+  tail-recursive version of the first-tour-heuristic function
+  otherwise you will get problems with stack-overflows.\\
+  \mbox{}\hfill[1 Mark]
 \end{itemize}  
 
 \end{document}