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Re(3): Puzzle: need a more general algorithm

by FoxtrotUniform (Prior)
on Jul 09, 2002 at 05:28 UTC ( #180416=note: print w/ replies, xml ) Need Help??


in reply to Re: •Re: Puzzle: need a more general algorithm
in thread Puzzle: need a more general algorithm

    '00111', '01011', '01101', '01110', '10011', '10101', '10110', '11001', '11010', '11100'

So it looks like you're shifting the highest bit up by one, then the next highest bit, then the next, until you've run out of empty bits. How about something like (untested):

use Bit::Vector; my $joins = 2; my $splits = 3; my $length = $joins+$splits; my $start = '0'x$joins . '1'x$splits; my $vector = Bit::Vector->new_Bin($length, $start); my @combinations = (); for my $pos ($joins-1..$length-1) { # 0-based, right? for my $bit ($splits-1..0) { $vector->bit_flip($pos+$bit); $vector->bit_flip($pos+$bit-1); push @combinations, $vector->to_Bin(); } }

--
The hell with paco, vote for Erudil!
:wq


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Re: Re(3): Puzzle: need a more general algorithm
by Ovid (Cardinal) on Jul 09, 2002 at 14:17 UTC

    Tempting, but the this is the "swapping bits" that I was referring to. The solution skips these three combinations:

    01101 01110 10110

    If you look at them closely, you'll see why you can't generate them. That's what got me stuck on this track. If that can be solved, this is a good way to go.

    Oh, and since you mentioned this was untested, I won't comment about for my $bit ($splits-1..0) { :)

    Cheers,
    Ovid

    Join the Perlmonks Setiathome Group or just click on the the link and check out our stats.

      D'oh!

      Fortunately, a man far smarter than I has written an article that you might find useful. ("Pre-Fascicle 2c: Generating all combinations", about halfway down the page. The draft is a .ps.gz file, which is why I didn't link to it directly.)

      --
      The hell with paco, vote for Erudil!
      :wq

        After giving this some thought, I realized that traditional permutation generators are going to have a problem here. Consider the following four permutations:

        11100 11100 11100 11100

        Those can be different permutations. The second could have the first and second ones swapped, the third could have the last fourth and fifth zeroes swapped, etc. Even though, to us, they look the same, the permutators that I've played with don't see that (and how would they calculate it?).

        And it was an interesting coincidence to see the Knuth thread on /. this morning. I'm checking it out now.

        Cheers,
        Ovid

        Join the Perlmonks Setiathome Group or just click on the the link and check out our stats.

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