47 Slides--Turing Machines

47 Slides--Turing Machines - CS103 HO#47 Slides-Turing...

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CS103 HO#47 Slides--Turing Machines May 14, 2010 1 T R R u v x y z Context-Free Grammars Long strings force repeated variables in their derivations. ... ... y z x DFA’s Long strings force repeated states. The idea behind the pumping lemmas Strings that are long enough satisfy the pumping lemmas. If we can find one that doesn’t, the language must not be regular or context-free. Theorem 2.34. Pumping Lemma for Context-Free Languages If A is a context-free 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 five pieces s = uvxyz satisfying the conditions 1. for each i 0, uv i xy i z A, 2. |vy| > 0, and 3. |vxy| p. T R R u v x y z T * u R z * u v R y z * u v x y z T R R u v R y z v x y uv 2 xy 2 z T R u z x uv 0 xy 0 z Example 2.37. B = { a n b n c n | n 0 } is not context-free. Assume B is CF. Let s = a p b p c p , where p is the pumping length. a p b p c p We don’t know exactly where vxy falls, but |vxy| p and |vy| 0. At least one of v, y is not empty, suppose it is v. v might contain just one symbol, in which case pumping s unbalances the number of a's, b's, and c's, or v might contain 2 symbols, in which case pumping s gets symbols out of order. So s cannot be pumped, and B is not context-free. Suppose s = uvxyz Typo in Sipser: Page 125, third and fourth lines of the third paragraph.
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47 Slides--Turing Machines - CS103 HO#47 Slides-Turing...

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