Module19 - Module 19 LNFA subset of LFSA Theorem 4.1 on...

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1 Module 19 LNFA subset of LFSA Theorem 4.1 on page 131 of Martin textbook Compare with set closure proofs Main idea A state in FSA represents a set of states in original NFA
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2 LNFA subset LFSA Let L be an arbitrary language in Let M be M exists by definition of Construct an M’ such that L(M’) Argue L(M’) = There exists an M’ such that L(M’) = L is in By definition of
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3 Visualization LNFA LFSA NFA’s FSA’s L L M M’ Let L be an arbitrary language in LNFA Let M be an NFA such that L(M) = L M exists by definition of L in LNFA Construct FSA M’ from NFA M Argue L(M’) = L There exists an FSA M’ such that L(M’) = L L is in LFSA
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4 Construction Specification We need an algorithm which does the following Input: NFA M Output: FSA M’ such that L(M’) = L(M)
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5 An NFA can be in several states after processing an input string x Difficulty * a a a b a,b a,b Input string aaaaba (1, a aaaba) (1, a aaba) (2, a aaba) (1, a aba) (2, a aba) (3, a aba) (1, a ba) (2, a ba) (3, a ba) crash (1, b a) (2, b a) (3, b a) (1, a ) (4, a ) (1, λ ) (2, λ ) (5, λ )
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6 All strings which end up in the set of states {1,2,3} are indistinguishable with respect to L(M) Observation * a a a b a,b a,b Input string aaaaba (1, a aaaba) (1, a aaba) (2, a aaba) (1, a aba) (2, a aba) (3, a aba) (1, a ba) (2, a ba) (3, a ba) crash (1, b a) (2, b a) (3, b a) (1, a ) (4, a ) (1, λ ) (2, λ ) (5, λ )
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7 Given an NFA M = (Q, Σ ,q 0 , δ ,A), the equivalent FSA M’ will have state set 2 Q (one state for each subset of Q) Example In this case there are 5 states in 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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Module19 - Module 19 LNFA subset of LFSA Theorem 4.1 on...

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