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43 Slides--Context-Free Languages

# 43 Slides--Context-Free Languages - CS103 HO#43...

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CS103 HO#43 Slides--Context-Free Languages 5/10/10 1 The Pumping Lemma If A is a regular language, then there is a number p (the pumping length) such that, if s is any string in A of length at least p , then s may be divided into three pieces, s = xyz satisfying the following three conditions: 1. for each i 0, xy i z A, 2. |y| > 0, and 3. |xy| p. FOR ANY regular language A, THERE EXISTS a number p such that FOR ANY string s in A with |s| p THERE EXIST strings x, y, z, such that s = xyz and … The Pumping Lemma If A is a regular language, then there is a number p (the pumping length) such that, if s is any string in A of length at least p , then s may be divided into three pieces, s = xyz satisfying the following three conditions: 1. for each i 0, xy i z A, 2. |y| > 0, and 3. |xy| p. To show that language A is NOT REGULAR , assume it is and get a contradiction from this FOR ANY regular language A, THERE EXISTS a number p such that FOR ANY string s in A with |s| p THERE EXIST strings x, y, z, such that s = xyz and ... The Pumping Lemma If A is a regular language, then there is a number p (the pumping length) such that, if s is any string in A of length at least p , then s may be divided into three pieces, s = xyz satisfying the following three conditions: 1. for each i 0, xy i z A, 2. |y| > 0, and 3. |xy| p. To show that language A is NOT REGULAR , assume it is and get a contradiction. If A is regular, there is a pumping length p that works for any s where |s| p. Given p, show how to find an s of sufficient length that cannot be pumped. That is, no matter how we divide s in to xyz there is an i such that xy i z A. (This may involve analyzing several cases.) FOR ANY regular language A, THERE EXISTS a number p such that FOR SOME string s in A with |s| p FOR ALL strings x, y, z, such that s = xyz and ... Show that L = { w | w { ) , ( }* and w has balanced parentheses } is not regular.

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