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Let L = {a^f(m) | m >= 1 } where f: Z^+ -> Z^+ is monotone increasing and complies that for all element n in Z^+ there is an m belonging to Z^+ such that f(m+1) - f(m) >= n.

Is it possible to prove that L is a regular language?

icktoofay
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Alejandro Barreiro
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  • OK, I have let those things happen. Now what? – KevinDTimm Aug 15 '11 at 21:29
  • My guess is no, since I don't think L so defined is regular. Doesn't f(x) = 2^x work, and doesn't an elementary argument based on equivalence classes under indistinguishability suffice? – Patrick87 Aug 16 '11 at 01:12

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Let f(x) = 2^x. For any positive n, f(n+1) - f(n) >= n.

L = {a^f(m)} is not regular. Consider the strings a^(2^x + 1). After an FA processes such a string, the smallest string which leads to an accepting state is a^(2^x - 1), having length 2^x - 1. Therefore, a separate state will be needed for every value of x. Since there are infinitely many values of x (positive integers), no FA exists to recognize L; ergo, L is not a regular language.

Patrick87
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