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Lecture 8 From NFA to Regular Language

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Induction on k= # of states other than initial and final states K=0 a a* b a c d c*a(d+bc*a)*

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k > 1 c a b d e ac*e bc*d bc*e ac*d

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Example 1 0 1 0 1 0 1 0 0+101 10 (0+10)*0(1+10(0+10)*0)* 0+10 0 (0+10)*0 0+10 0 10 1 10(1+10)*0 1 How to get in the final state? Cycles at the final state. Why?

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Example 2 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 ε ε (0+10)*0(1+10(0+10)*0)* + 0*(1+01*1)(00*(1+01*1))* Solution 1

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0 0 1 1 0 1 ε ε 0 0 ε 01* 1+01*1 01*+(1+01*1) 0+(1+01*1)0 (0+(1+01*1)0)*(01*+1+01*1) Solution 2

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0 0 1 1 0 1 ε ε 0+10 (0+10+01*10)*(1+01*(ε+1)) Solution 3 1 ε+1 10 1 0+10+01*10 1+01*(ε+1) Different ways may give different regular expressions for the same language.

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Theorem A language is regular if and only if it can be accepted by an NFA if and only if it can be accepted by a DFA.

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Closure Properties If A is regular, so is its complement A. If A and B are regular, then A∩B, A\B, A U B, AB are regular.

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Quotient L 1 /L 2 = {x | there exists y in L 2 s.t. xy in L 1 } If L 1 is regular, so is L 1 /L 2. L 1 =L(M), M=(Q,Σ,δ,s,F) L 1 /L 2 = L(M’), M’ = (Q,Σ,δ,s, F’) where F’={q in Q | there exists y in L 2 s.t. δ(q,y) in F}

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Example L 1 = L(M) 0 1 0 1 0 1 0 1 L 2 = 0*1+0*11

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