sub recurse {
  if (grep /A/, @a > 1){ my @c = splice @a, 0, 2; splice @c, 1, 0, (@a
+, @b); }
  if (grep /B/, @b > 1){ my @c = splice @b, 0, 2; splice @c, 1, 0, (@a
+, @b); }
  if (grep /A/, @a == 1){ shift @a; }
  if (grep /B/, @b == 1){ shift @b; }
}
[download]

Here's a simple implementation of the suggested recursive algorithm. Note that you don't really need the arrays, so I'm just using a count of how many of each letter are left:

#!/usr/bin/perl

use strict;

print_interleave(1,1);
print_interleave(2,1);
print_interleave(1,2);
print_interleave(2,2);
print_interleave(3,3);
print_interleave(4,14);
print_interleave(7,4);

sub print_interleave {
  my ($total_a, $total_b) = @_;
  print '(', $total_a, ', ', $total_b, ")\t";
  print interleave($total_a, $total_b), "\n";
}

sub interleave {
  my ($remaining_a, $remaining_b) = @_;
  if ($remaining_a > 1) {
    return 'A' . interleave($remaining_a - 2, $remaining_b) . 'A';
  } elsif ($remaining_b > 1) {
    return 'B' . interleave($remaining_a, $remaining_b - 2) . 'B';
  } elsif ($remaining_a == 1) {
    return 'A';
  } elsif ($remaining_b == 1) {
    return 'B';
  } else {
    return ''; # Just for completeness
  }
}
[download]

Unfortunately, it doesn't produce the correct results:

(1, 1)  A
(2, 1)  ABA
(1, 2)  BAB
(2, 2)  ABBA
(3, 3)  ABABA
(4, 14) AABBBBBBBBBBBBBBAA
(7, 4)  AAABBABBAAA
[download]

As for how to terminate it, every time interleave() calls itself, it reduces either $remaining_a or $remaining_b. One of them will eventually reach 0 and terminate that branch of the recursion. Since nothing in interleave() ever increases them, all branches will terminate.

UPDATE: Replacing the above interleave() with this interleave2()

sub interleave2 {
  my ($remaining_a, $remaining_b) = @_;
  if ($remaining_a > 1 && $remaining_b) {
    my $b_range = int($remaining_b/($remaining_a - 1));
    return 'A' . 'B' x $b_range . interleave2($remaining_a - 1, $remai
+ning_b - $b_range);
  } elsif ($remaining_a) {
    return 'A' . interleave2($remaining_a - 1, $remaining_b);
  } elsif ($remaining_b) {
    return 'B' . interleave2($remaining_a, $remaining_b - 1);
  } else {
    return ''; # Just for completeness
  }
}
[download]

appears to produce correct results:

(1, 1)  AB
(2, 1)  ABA
(1, 2)  ABB
(2, 2)  ABBA
(3, 3)  ABABBA
(4, 14) ABBBBABBBBBABBBBBA
(7, 4)  AAABABABABA
[download]

esper:

++ Nice implementation, test & improvement!

I was actually not proposing my algorithm as being a correct one or not, just suggesting a mode of attack. (I like recursion for tasks like this.) But judging from your results, I wasn't terribly far off.

...roboticus