Matching a^n A^n with regular expression

Question

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.

Answer 1

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*$)\1(?i)\1$

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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}

Related questions

• Case sensitive and insensitive in the same pattern
• How does the regular expression ‘(?<=#)[^#]+(?=#)’ work?

References

• regular-expressions.info/Lookaround, Grouping and backreferences
• Modifiers (esp. Turning Modes On and Off for Only Part of The Regular Expression)

See also

• Why is {a^nb^n | n >= 0} not regular?
• Wikipedia/Regular expression: Patterns for non-regular languages