in reply to Creating tuples based on sets

Update: I missed the part about needing arbitrarily sized tuples. This only groups in sets of two...my bad. :(

Original message follows...

Ah, the joys of dot products. No need for recursion: 

#!/usr/bin/perl -w

@set = qw(A B C);

print join("\n", map(join(',', @$_), @{&make_tuples(\@set, \@set)}));

sub make_tuples {
  my($set1, $set2) = @_;
  my @product;
  foreach my $x (0 .. $#$set1) {
    foreach my $y (0 .. $#$set2) {
      push(@product, [$set1->[$x], $set2->[$y]]);
    }
  }
  \@product;
}

[download]

I daresay I just answered a homework question for you, but so be it.

Matt

Replies are listed 'Best First'.
Question
by waxmop (Beadle) on Apr 16, 2003 at 15:31 UTC

    At the end of the program, you just do 

    \@product;

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    Does this do the same thing as 
    return \@product;

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    ???
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Re: Re: Creating tuples based on sets
by perlguy (Deacon) on Apr 16, 2003 at 15:31 UTC
    There is a definite possibility that I am missing something, but isn't waxmop asking for something that will solve for an arbitrary number of elements (in the solution), not just 2 necessarily?
[reply]

      Yeah, I need the function to fit any of these situations: 

      my @set1 = qw(A B C D);
my @set2 = qw(X Y Z);
mk_tuples(@set1, 2);
mk_tuples(@set2, 5);

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      etc...

      Here's a side question. How come this doesnt' work in perl? 

      mk_tuples(["A", "B"], 3);

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      I can't replace an array with just a list of elements.
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[d/l]
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        but you can... try this: 

        my $arrayref = ["A", "B"];
print join "\n", @$arrayref;

        [download]
[reply]
[d/l]

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