Take a binary approach: 
my @n = (-25, 14, 50, 20, -7, -8, -10);
my $max = 2**@n;
for (my $i = 0; $i < $max; ++$i) {
  my $sum = 0;
  $sum += $n[$_] for grep $i & 2**$_, 0 .. $#n;
  if ($sum == 0) {
    my ($bits, @used) = $i;
    while ($bits) {
      my $high_bit = int(log($bits)/log(2));
      push @used, $n[$high_bit];
      $bits &= ~(2**$high_bit);
    }
    print "[@used] = 0\n";
  }
}

[download]
Another approach to the end is thus: 
my ($bits, $j, @used) = ($i, 0);
while ($bits) {
  push @used, $n[$j] if $bits & 1;
  $bits >>= 1, ++$j;
}

[download]
which could also be written as a for loop: 
for (
  my ($bits, $j, @used) = ($i, 0);
  $bits;
  $bits >>= 1, ++$j
) {
  push @used, $n[$j] if $bits & 1;
}

[download]

_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker, who'd like a job (NYC-area)
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

In reply to Re: permuation algorithm by japhy
in thread permutation algorithm by Basilides

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