[PDF] a^nb^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.22 fév. 2010

  • 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.

  • What is an example of a non-regular language?

    For Example:-
    Let us define the language L = {?, ab, aabb, aaabbb, aaaabbbb, aaaaabbbbb …} We shall now show that this language is non-Regular. We must take a note that though, it is a subset of many regular languages, such as a*b*, which however, also includes such strings as aab and bb that {a^nb^n} does not.
  • A regular language can be recognized by a finite automata, which is a simple machine model consisting of states, transitions, and an initial and final state. Conversely, every regular language can be expressed using finite automata. , corresponding to an even number of 0's in the string.
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