php - Match a^n b^n c^n (e.g. "aaabbbccc") using regular expressions (PCRE)

It is a well known fact that modern regular expression implementations (most notably PCRE) have little in common with the original notion of regular grammars. For example you can parse the classical example of a context-free grammar {aⁿbⁿ; n>0} (e.g. aaabbb) using this regex (demo):

~^(a(?1)?b)$~

My question is: How far can you go? Is it also possible to parse the context-sensitive grammar {aⁿbⁿcⁿ;n>0} (e.g. aaabbbccc) using PCRE?

Answer

Inspired by NullUserExceptions answer (which he already deleted as it failed for one case) I think I have found a solution myself:

$regex = '~^
    (?=(a(?-1)?b)c)
     a+(b(?-1)?c)
$~x';


var_dump(preg_match($regex, 'aabbcc'));    // 1
var_dump(preg_match($regex, 'aaabbbccc')); // 1
var_dump(preg_match($regex, 'aaabbbcc'));  // 0
var_dump(preg_match($regex, 'aaaccc'));    // 0
var_dump(preg_match($regex, 'aabcc'));     // 0
var_dump(preg_match($regex, 'abbcc'));     // 0

Try it yourself: http://codepad.viper-7.com/1erq9v

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Explanation

If you consider the regex without the positive lookahead assertion (the (?=...) part), you have this:

~^a+(b(?-1)?c)$~

This does nothing more than check that there's an arbitrary number of as, followed by an equal number of bs and cs.

This doesn't yet satisfy our grammar, because the number of as must be the same, too. We can ensure that by checking that the number of as equals the number of bs. And this is what the expression in the lookahead assertion does: (a(?-1)?b)c. The c is necessary so we don't only match a part of the bs.

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Conclusion

I think this impressively shows that modern regex is not only capable of parsing non-regular grammars, but can even parse non-context-free grammars. Hopefully this will lay to rest the endless parroting of "you can't do X with regex because X isn't regular"