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Editorial Team
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Asked: 2026-05-15T13:23:36+00:00

We’re learning about the difference between regular languages and regex, and the teacher explained

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We’re learning about the difference between regular languages and regex, and the teacher explained that the language

a^n b^n

is not regular, but she said that most regex flavors can match

a^n A^n

and she gave this problem for our extra credit homework problem. We’ve been struggling for a couple of days now, and could really use some guidance.

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1 Answer

  1. Editorial Team

    The teacher gave a HUGE hint by limiting the alphabet to {a, A}. The key to solving this problem is realizing that in case-insensitive mode, a matches A and vice versa. The other component of the problem is matching on backreference.

    This pattern will match a{n}A{n} for some n (as seen on rubular.com):

    ^(?=(a*)A*$)\1(?i)\1$

    How it works

    The pattern works as follows:

    • ^(?=(a*)A*$) – Anchored at the beginning of the string, we use positive lookahead to see if we can match (a*)A* until the end of the string
      • If we can succesfully make this assertion, then \1 captures the sequence of a{n}
      • If the assertion fails, then there’s no way the string can be a{n}A{n}, since it’s not even a*A*
    • We then match \1(?i)\1$, that is, the a{n} captured by \1, then in case-insensitive mode (?i), we match a{n} again until the end of the string
    • Thus, since:
      • The string is a*A*
      • \1 is a{n}
      • \1(?i)\1 can only be a{n}A{n}

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