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The Pumping Lemma states that for a regular set L, there is a constant n such that any word z in L with length at least n can be split into u, v, and w in a specific way. This lemma is a key concept in formal languages and automata theory, providing a method to analyze regular languages and finite automata.
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Pumping Lemma Let L be a regular set. Then there is a constant n such that if z is any word in L, and |z|>=n, we may write z=uvw in such a way that |uv|<=n, |v|>=1, and for all i>=0, u(v^i)w is in L. Furthermore, n is no greater than the number of states of the smallest FA accepting L.
Exercise The set L={0^(i^2) | i is an integer, i>=1}, which consists of all strings of 0’s whose length is a perfect square, is not regular. Prove that L = {(a^n)(b^k): n > k and n>=0} is not regular.
Which of the following languages are regular sets? {0^(2n) | n>=1} {(0^m)(1^n)(0^(m+n)) | m>=1 and n>=1} {0^n | n is a prime} The set of all strings that do not have three consecutive 0’s
The set of all strings with an equal number of 0’s and 1’s {x | x in (0+1)*, and x=x^R} x^R is x written backward; for example, (011)^R=110. {xwx^R | x,w in (0+1)+}.
