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Module 19
• LNFA subset of LFSA
– Theorem 4.1 on page 105 of Martin textb ook
– Compare with set closure proofs
• Main idea
– A state in FSA represents a set of states in
original NFA
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LNFA subset LFSA
• Let L be an arbitrary language in LNFA
• Let M be the NFA such that L(M) = L
– M exists by definition of L in LNFA
• Construct an FSA M’ such that L(M’ ) = L
• Argue L(M’) = L
• There exists an FSA M’ such that L(M’) = L
• L is in LFSA
– By definition of L in LFSA
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Visualization
LNFA LFSA
NFA’s FSA’s
LL
MM’
•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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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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• An NFA can be in
several states after
processing an input
string x
Difficulty *
a a ab
a,b a,b
Input string aaaaba
(1, aaaaba)
(1, aaaba) (2, aaaba)
(1, aaba) (2, aaba) (3, aaba)
(1, aba) (2, aba) (3, aba) crash
(1, ba) (2, ba) (3, ba)
(1, a) (4, a)
(1, λ) (2, λ) (5, λ)
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• All strings which end up
in the set of states {1,2,3}
are indistinguishable with
respect to L(M)
Observation *
a a ab
a,b a,b
Input string aaaaba
(1, aaaaba)
(1, aaaba) (2, aaaba)
(1, aaba) (2, aaba) (3, aaba)
(1, aba) (2, aba) (3,aba) crash
(1, ba) (2,ba) (3, ba)
(1, a) (4, a)
(1, λ) (2, λ) (5, λ)
