in reply to
Trying to solve N-Queens

`use strict;
use Data::Dumper;
my $SIZE = shift || 4;
my @BOARD = map {[(1) x $SIZE]} (1..$SIZE);
my @SOLUTION;
# i know that the second argument seems unecessary,
# but this will used to allow different threads to
# tackle different starting rows ...
scan(0,0,[@BOARD],[]);
print Dumper \@SOLUTION;
sub scan {
my ($col,$start_row,$board,$possible) = @_;
my $copy = [map[@$_],@$board];
# no more columns?
if ($col == $SIZE) {
# found our solution!
push @SOLUTION,[@$possible];
return;
}
# find first available row
for my $row ($start_row..$SIZE-1) {
if($board->[$row]->[$col]) {
push @$possible,$row;
print "available row: $row in col $col: (@$possible)\n";
mark_attacks($row,$col,$board);
print_matrix($board);
scan($col+1,0,[@$board],$possible);
# i thought that this copy should not even be necessary
@{$board} = map[@$_],@$copy;
pop @$possible;
}
}
return;
}
sub mark_attacks {
my ($r,$c,$array) = @_;
$array->[$r]->[$c] = 'Q';
# mark horizontal
$array->[$r]->[$_] = 0 for ($c+1..$SIZE-1);
# this line will produce 1 solution for n=4 or 5
#$array->[$r] = [ map {0} @{$array->[$r]} ];
# mark r-c diagonal
$array->[--$r]->[++$c] = 0 while ($r > 0) && ($c < $SIZE-1);
# mark r+c diagonal
($r,$c) = @_;
$array->[++$r]->[++$c] = 0 while ($r < $SIZE-1) && ($c < $SIZE-1);
}
sub print_matrix { print join('',@$_),"\n" for @{+shift} }
`