Module21 - 1 Module 21 Closure Properties for LFSA using...

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Unformatted text preview: 1 Module 21 Closure Properties for LFSA using NFAs From now on, when I say NFA, I mean any NFA including an NFA- unless I add a specific restriction union (second proof) concatenation Kleene closure 2 LFSA closed under set union (again) 3 LFSA closed under set union Let L 1 and L 2 be arbitrary languages in LFSA Let M 1 and M 2 be NFA s s.t. L(M 1 ) = L 1 , L(M 2 ) = L 2 M 1 and M 2 exist by definition of L 1 and L 2 in LFSA and the fact that every FSA is an NFA Construct NFA M 3 from NFA s M 1 and M 2 Argue L(M 3 ) = L 1 union L 2 There exists NFA M 3 s.t. L(M 3 ) = L 1 union L 2 L 1 union L 2 is in LFSA 4 Visualization L 1 union L 2 L 1 L 2 LFSA M 3 M 1 M 2 NFA s Let L 1 and L 2 be arbitrary languages in LFSA Let M 1 and M 2 be NFA s s.t. L(M 1 ) = L 1 , L(M 2 ) = L 2 M 1 and M 2 exist by definition of L 1 and L 2 in LFSA and the fact that every FSA is an NFA Construct NFA M 3 from NFA s M 1 and M 2 Argue L(M 3 ) = L 1 union L 2 There exists NFA M 3 s.t. L(M 3 ) = L 1 union L 2 L 1 union L 2 is in LFSA 5 Algorithm Specification Input Two NFA s M 1 and M 2 Output NFA M 3 such that L(M 3 ) = ? NFA M 1 NFA M 2 NFA M 3 A 6 Use -transition NFA M 1 NFA M 2 NFA M 3 A a M 1 a,b a,b a,b M 2 a a,b a,b a,b M 3 7 General Case * NFA M 1 NFA M 2 NFA M 3 A M 1 M 2 M 3 8 Construction * Input NFA M 1 = (Q 1 , 1 , q 1 , 1 , A 1 ) NFA M 2 = (Q 2 , 2 , q 2 , 2 , A 2 ) Output NFA M 3 = (Q 3 , 3 , q...
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This note was uploaded on 07/25/2008 for the course CSE 460 taught by Professor Torng during the Fall '07 term at Michigan State University.

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Module21 - 1 Module 21 Closure Properties for LFSA using...

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