[PDF] a^2^n is not regular

  • Why is a NB N >= 0 not regular?

    And to do that you have to count both, the no. of 'a' as well as no. of 'b' but because value of 'n' can reach infinity, it's not possible to count up to infinity using a Finite automata. So that's why {a^n b^n n >= 0} is not regular.

  • Is L ={ a2n n >= 1 regular?

    Example 4 – L = { a2^n n>=1 } is not regular. It forms a G.P., so we cannot have a fixed pattern which repeated would generate all strings of this language.17 mai 2023

  • How do you prove an expression is not regular?

    One can then use the following lemma to prove non regularity: KC-Regularity: Let L??? be a regular language, then there exists a constant c which depends only on L, such that for all x???, If y is the n?th string (relative to the lexicographic ordering) in Lx={y???xy?L}, then K(y)?O(logn)+c.
  • If I let string w be a^mb^m then we know that y will consists of only a 's because of the rule xy <= m . And if I set i=0 , then ww^R will have fewer a 's on the left side than on the right side. Thus, it proves that this language is not regular.
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