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3 Pages

### Pandurengan, Vignesh_14

Course: CS 530, Fall 2008
School: Illinois Tech
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Word Count: 370

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Notes Lecture - Lecture Dated 10/07/08 Lecture Number 14 Vignesh Pandurengan CWID : 10428060 October 29, 2008 Turing Machine Its a acceptor that is it basically accepts the language Its a function computator Its a generator ( enumerator ) , the strings will be printed on the tape is the language generated by the turing machine. G(M1) = Set of strings printed on tape. There is a language L(M1) that will accept (...

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Notes Lecture - Lecture Dated 10/07/08 Lecture Number 14 Vignesh Pandurengan CWID : 10428060 October 29, 2008 Turing Machine Its a acceptor that is it basically accepts the language Its a function computator Its a generator ( enumerator ) , the strings will be printed on the tape is the language generated by the turing machine. G(M1) = Set of strings printed on tape. There is a language L(M1) that will accept ( ie., it will end in accept state) Sample Problem L(M) = G(M1 ) Given M ,Construct M1 We have the strings w1,w2, ... .wn ( Countable set ) ie.., if its 0,1) ,then the sorted set will be like ,0,1,00,01,.... We run turing machine on w1 and if its accepted it will be printed.If its not printed , it means its rejected. NFA and DFA will accept regular languages PDA will accept Context Free Every regular language is context free The language can be recursive or recursive enumerable Recursive language contains the CFL Size of L = Size of R A turing machine is a 7-tuple , ( Q , ,,,q0,qaccept,qreject) , where Q , , are all finite sets and 1.Q is the set of states, is 2. the input alphabet not containing the blank symbol , 3. is the tape alphabet 4. q0 is the start state , qaccept is the final accepted state, qreject is the 1 rejected state is 0,1 Assume it has k states, that q0,q1 .. .qk-1 Transition function ( qi,a) - (qr,b,- ) When it enters accept state ,it will stop If L1 is regular complement of L1 is regular If L1 is recursive ,then complement of L1 is also recursive? Proof: For any string w ,we have a turing machine M1 ,it will give yes / no ( ie.., ACCEPT / REJECT ) Usually its ( qi,a) = ( qf,a) where qi is the initial state and qf is the final state Now its ( qi,a) = (ACCEPT / REJECT ) So , now L1 L2 is recursive similarly L1 L2 is also recursive L1 , L2 is Recursive enumerable then L1 L2 is recursive enumerable too. If a Language L is recursive, its complement is recursive but not recursive enumerable If a language L is recursive enumerable its complement is not recursive and is actually recursive enumerable If a language L is recursive , then its complement is recursive enumerable but not recursive. 2
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Illinois Tech - CS - 530
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%!PS-Adobe-2.0 %Creator: dvips(k) 5.96dev Copyright 2007 Radical Eye Software %Title: lecture_10_14.dvi %CreationDate: Wed Oct 22 16:40:22 2008 %Pages: 4 %PageOrder: Ascend %BoundingBox: 0 0 595 842 %DocumentFonts: SFRM1728 SFRM1200 SFTT1440 SFTT1000
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Caltech - CNS - 120
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Caltech - CNS - 120
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Caltech - CNS - 120
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