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Ingy döt Net 2023-07-01 11:58:00 -04:00
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Task/N-queens-problem/XPL0/n-queens-problem.xpl0 Normal file
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def N=8; \board size (NxN)
int R, C; \row and column of board
char B(N,N); \board
include c:\cxpl\codes;
proc Try; \Try adding a queen to the board
int R; \row, for each level of recursion
func Okay;
\Returns 'true' if no row, column, or diagonal from square R,C has a queen
int I;
[for I:= 0 to N-1 do
[if B(I,C) then return false; \row is occupied
if B(R,I) then return false; \column is occupied
if R+I<N & C+I<N then
if B(R+I, C+I) then return false; \diagonal down right
if R-I>=0 & C-I>=0 then
if B(R-I, C-I) then return false; \diagonal up left
if R-I>=0 & C+I<N then
if B(R-I, C+I) then return false; \diagonal up right
if R+I<N & C-I>=0 then
if B(R+I, C-I) then return false; \diagonal down left
];
return true;
]; \Okay
[ \Try
if C>=N then
[for R:= 0 to N-1 do \display solution
[ChOut(0, ^ ); \(avoids scrolling up a color)
for C:= 0 to N-1 do
[Attrib(if (R|C)&1 then $0F else $4F); \checkerboard pattern
ChOut(6, if B(R,C) then $F2 else ^ ); \cute queen symbol
ChOut(6, if B(R,C) then $F3 else ^ );
];
CrLf(0);
];
exit; \one solution is enough
];
for R:= 0 to N-1 do
[if Okay(R,C) then \a queen can be placed here
[B(R,C):= true; \ so do it
C:= C+1; \move to next column
Try; \ and try from there
C:= C-1; \didn't work: backup
B(R,C):= false; \undo queen placement
];
];
]; \Try
[for R:= 0 to N-1 do \clear the board
for C:= 0 to N-1 do
B(R,C):= false;
C:= 0; \start at left column
Try;
]