As an exercise related to this problem (not solving the whole problem), I wanted to find an algorithm for 'variations_with_repetitions', (algorithms not being my strong suit), and was able to find a solution. I wouldn't say it is pretty, but it works :-)

It doen't have an iterative solution. Instead it returns all the tuples.

#!/usr/bin/perl
use strict;
use warnings;

my $n = 3;
my @a = "a".."b";

my @b = vw_rep(\@a, $n); # variations with repetition (Algorithm::Comb
+inatorics)

use Data::Dump; dd \@b;

sub vw_rep {
    my ($ref, $n) = @_;
    my @c;
    for my $k (0 .. $n-1) {
        my $L = 0;
        for (1 .. @$ref**$k) {
            for my $i (0 .. $#$ref) {
                for (1 .. @$ref**($n-1 - $k)) {
                    push @{ $c[$L++] }, $ref->[$i];
                }   
            }
        }
    }
    return @c;
}

__END__
C:\Old_Data\perlp>perl var_w_rep.pl
[
  ["a", "a", "a"],
  ["a", "a", "b"],
  ["a", "b", "a"],
  ["a", "b", "b"],
  ["b", "a", "a"],
  ["b", "a", "b"],
  ["b", "b", "a"],
  ["b", "b", "b"],
]
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Update: A better approach using choroba's solution in an iterative fashion could be:

#!/usr/bin/perl
use warnings;
use strict;

# Pm node 1211055

my $n = 3;
my @b = qw( a b );

my $iter = variations_rep_iter(\@b, $n);

while (my $tuple = $iter->()) {
    print "@$tuple\n";  
}


sub variations_rep_iter {
    my ($bases, $n) = @_;
    
    my @indices = (0) x $n;
    my $first = 1;

    my $iter = sub {
        if ($first) {
            $first = 0;
            return [ @$bases[ @indices ] ];
        }
        
        my $r = $#indices;
        while ($r >= 0) {
            if (++$indices[$r] > $#$bases) {
                $indices[$r--] = 0;
            } else {
                last
            }
        }
        return if $r < 0;

        return [ @$bases[ @indices ] ];
    };  
    return $iter;
}
[download]