in reply to One Zero variants_without_repetition

Do you mean like this? 

print unpack 'B8', chr for 0 .. 255;;

00000000
00000001
00000010
00000011
00000100
00000101
...
11111011
11111100
11111101
11111110
11111111

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Re^2: One Zero variants_without_repetition
by thenetfreaker (Friar) on Aug 07, 2007 at 11:03 UTC
    no, i need to print all the variations with a spesific number of ones and zeroes without repeats, for example all the variants with 6 zeroes and 14 ones.
    as i gave the example earlier with 2 ones and 3 zeroes.
[reply]

      Okay, sorry! Try this iterator then. It will handle upto 32 0s + 1s.

      If you uncomment the second example it runs on a bit.

      Update: Had to tweak the termination condition. It works now but I'm not happy with it.

      Update2: D'oh! No need to count both 1s and 0s. 

      #! perl -slw
use strict;

sub combs {
    my( $ones, $zeros ) = @_;
    my $n = $ones+$zeros;
    my $max = 2**$n;
    my $p = 0;
    return sub {
        my $x = '';
        $x = unpack "b$n", pack 'V', $p++
            until $x =~ tr[1][] == $ones
            or $p > $max and return;
        return $x;
    }
}

my $iter = combs( 2, 3 );
print while $_ = $iter->();

#my $iter = combs( 14, 6 );
#print while $_ = $iter->();
__END__
C:\test>junk7
11000
10100
01100
10010
01010
00110
10001
01001
00101
00011

      [download]
      Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
      "Science is about questioning the status quo. Questioning authority".
      In the absence of evidence, opinion is indistinguishable from prejudice.
[reply]
[d/l]
        That's a very nice algorythm, but it's a bit slow, and 2 questions:
        1. how can i work with (much)more that 32 1's+0's ?
        2. why do i need $max to be 2**$n ? when combs(2,3) gives 10 reasults not 32 ???

        i need this sub{} to play with the strings that contain that number of 1's and 0's, not the numbers that contain them.
[reply]

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