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

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1 Module 21 Closure Properties for LFSA using NFA’s From now on, when I say NFA, I mean any NFA including an NFA- λ unless I add a specific restriction union (second proof) concatenation Kleene closure

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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

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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

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6 Use λ -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 λ λ
7 General Case * NFA M 1 NFA M 2 NFA M 3 A M 1 M 2 λ λ M 3

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8 Construction * Input NFA M 1 = (Q 1 , Σ 1 , q 1 , δ 1 , A 1 ) NFA M 2 = (Q 2 , Σ 2 , q 2 , δ 2 , A 2 ) Output NFA M 3 = (Q 3 , Σ 3 , q 3 , δ 3 , A 3 ) What is Q 3 ?
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Module21 - Module 21 Closure Properties for LFSA using...

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