For amusement, here is the recursive regexp modified to match the case where there are as many 1's as 0's (in any order): 
  my $re;
  $re = qr{
          (?:
             0            # 0
             (??{$re})    # Balanced 1's and 0's
             1            # 1
           |
             1            # 1
             (??{$re})    # Balanced 1's and 0's
             0            # 0
          )*
         }x;
   /^$re\z/;

[download]
A random note. For amusement I tried to optimize this, but that failed. I believe the following should be fast, but it isn't, and that's probably a bug (in my understanding, or in the RE engine): 
  my $re;
  $re = qr{
          (?:
             0            # 0
             (?>          # If I get here, do not backtrack!
               (??{$re})  # Balanced 1's and 0's
             )
             1            # 1
           |
             1            # 1
             (?>          # If I get here, do not backtrack!
               (??{$re})  # Balanced 1's and 0's
             )
             0            # 0
          )*
         }x;
   for (qw(1001 11100 000111 00111), "101"x50) {
     if (/^$re\z/) {
       print "$_ matched\n";
     }
     else {
       print "$_ did not match\n";
     }
   }

[download]
Given the performance problems, I would not recommend using this regular expression in real life.

UPDATE: It was a bug in my understanding. The following works and is fast: 

  my $re;
  $re = qr{
          (?:
             0            # 0
             (??{$re})    # Balanced 1's and 0's
             1            # 1
           |
             1            # 1
             (?>          # If I get here, do not backtrack!
               (??{$re})  # Balanced 1's and 0's
             )
             0            # 0
          )*?
         }x;
   for (qw(1001 11100 000111 00111), "101"x50) {
     if (/^$re\z/) {
       print "$_ matched\n";
     }
     else {
       print "$_ did not match\n";
     }
   }

[download]
And so is this: 
  my $re;
  $re = qr{
          (?>             # Avoid backtracking
             0            # 0
             (??{$re})    # Balanced 1's and 0's
             1            # 1
           |
             1            # 1
             (??{$re})    # Balanced 1's and 0's
             0            # 0
          )*?
         }x;
   for (qw(1001 11100 000111 00111), "101"x50) {
     if (/^$re\z/) {
       print "$_ matched\n";
     }
     else {
       print "$_ did not match\n";
     }
   }

[download]
The key is that you need to avoid having to backtrack on success.

In reply to Re^2: A (non) reg-ex question by tilly
in thread A (non) reg-ex question by Ido

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