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Asked: 2026-06-07T02:00:39+00:00

Here on SO people sometimes say something like you cannot parse X with regular

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Here on SO people sometimes say something like “you cannot parse X with regular expressions, because X is not a regular language”. From my understanding however, modern regular expressions engines can match more than just regular languages in Chomsky’s sense. My questions:

given a regular expression engine that supports

  • backreferences
  • lookaround assertions of unlimited width
  • recursion, like (?R)

what kind of languages can it parse? Can it parse any context-free language, and if not, what would be the counterexample?

(To be precise, by “parse” I mean “build a single regular expression that would accept all strings generated by the grammar X and reject all other strings”).

Add.: I’m particularly interested to see an example of a context-free language that modern regex engines (Perl, Net, python regex module) would be unable to parse.

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1 Answer

  1. Editorial Team

    I recently wrote a rather long article on this topic: The true power of regular expressions.

    To summarize:

    • Regular expressions with support for recursive subpattern references can match all context-free languages (e.g a^n b^n).
    • Regular expressions with lookaround assertions and subpattern references can match at least some context-sensitive languages (e.g. ww and a^n b^n c^n).
    • If the assertions have unlimited width (as you say), then all context-sensitive grammars can be matched. I don’t know any regex flavor though that does not have fixed-width restrictions on lookbehind (and at the same time supports subpattern references).
    • Regular expressions with backreferences are NP-complete, so any other NP problem can be solved using regular expressions (after applying a polynomial-time transformation).

    Some examples:

    • Matching the context-free language {a^n b^n, n>0}:

      /^(a(?1)?b)$/
# or
/^ (?: a (?= a* (\1?+ b) ) )+ \1 $/x

    • Matching the context-sensitive language {a^n b^n c^n, n>0}:

      /^
    (?=(a(?-1)?b)c)
    a+(b(?-1)?c)
$/x
# or
/^ (?: a (?= a* (\1?+ b) b* (\2?+ c) ) )+ \1 \2 $/x
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