Hi,
Really thanks for your answer, yes the second one doesnt compile because of
some error typos like 'if ($elements=>{$elm})', without the hash variable
($elements) it works,
But also, you are right, your code is faster than mine, :D
Again thanks for your answer.
Regards,
On Wed, Aug 13, 2008 at 1:52 AM, Rob Dixon <[EMAIL PROTECTED]> wrote:
> Pablo Zea Aranibar wrote:
> >
> > Hi, realy i need your help with this algorithm.
> > what I need is another algorithm in order to improve my own.
> >
> > example:
> >
> > my $firsSet = [ 'a', 'b', 'c', 'd', 'e' ];
> > my $secondSet = [ 'x', 'y', 'z' ];
> > my $thirdSet = [ 'a', 'x', 'b', 'y' ];
> > my $fourthSet = [ 'b', 'd', 'z' ];
> > my $counter = 0;
> >
> > foreach my $elm1 (@{$firstSet})
> > {
> > foreach my $elm2 (@{$secondSet})
> > {
> > foreach my $elm3 (@{$thirdSet})
> > {
> > foreach my $elm4 (@{$fourthSet})
> > {
> > print "$counter : $elm1, elm2 , elm3 , elm4\n";
> > $counter ++;
> > }
> > }
> > }
> > }
> > # It prints each possible combination taken one element for each set.
> >
> >
> > another (better way):
> >
> > my @Set = ();
> > my $elements = {a=>1, b=>1, c=>1, d=>1, e=>1, x=>1, y=>1, z=>1};
> > $Set[0] = [ 'a', 'b', 'c', 'd', 'e' ];
> > $Set[1] = [ 'x', 'y', 'z' ];
> > $Set[2] = [ 'a', 'x', 'b', 'y' ];
> > $Set[3] = [ 'b', 'd', 'z' ];
> > my $counter = 0;
> >
> > @result = @{$Set[0]};
> > for (1 .. (scalar @Set) - 1) {
> > my $temp = [];
> > foreach my $result (@result) {
> > foreach my $elm (@{$Set[$_]}) {
> > if ($elements=>{$elm}) { #if I dont want repetitions
> > push @{$temp}, $result.', '.$elm;
> > $elements=>{$elm}=0;
> > $counter++;
> > };
> > };
> > };
> > @result = @{$temp};
> > };
> > print "@result\n";
> >
> >
> > But, both are still too slow.
> > So, can you give me an idea to improve it?.
> > someone tell me about mask, and bit operations. but i really dont know
> how
> > to do it. so.. any suggestion?.
> > there was a thread in a forum
> > (here<
> http://forums.devshed.com/software-design-43/algorithm-design-536480.html
> >),
> > but the solution was for python (recursive using "yield").
>
> Looking at your second example (which doesn't compile, so it's not
> surprising
> it's too slow!) I think you want all combinations of the sets that don't
> have
> repeating letters. The program below will do that, and if you want all
> combinations regardless of repeated characters simply remove the line
>
> s/$c//g foreach @sets;
>
> Finally, if you prefer a comma-separated form of output, just change the
> print
> line to read
>
> print join(', ', split ''), "\n" foreach @results;
>
> I hope this helps. I had fun writing it!
>
> Rob
>
>
> use strict;
> use warnings;
>
> my @sets = qw(
> abcde
> xyz
> axby
> bdz
> );
>
> my @results = combine(@sets);
> print "$_\n" foreach @results;
>
> sub combine {
>
> return () if grep length == 0, @_;
>
> my @initials = split '', shift;
> return @initials unless @_;
>
> my @return;
> foreach my $c (@initials) {
> my @sets = @_;
> s/$c//g foreach @sets;
> push @return, map "$c$_", combine(@sets);
> }
>
> @return;
> }
>
>
>
