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csci3255 HW 2-3

# csci3255 HW 2-3 - CSCI3255 Math Foundations of CS Homework...

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CSCI3255: Math Foundations of CS Homework 2, (2.3), page 1 September 2006 1. Use the construction of Theorem 2.2 to convert the nfa in the figure below to a dfa. Can you see a simpler answer more directly? The original machine requires an a in order to accept. Since the λ ’s can always get us back to the initial state, any number of additional a ’s can be accepted. That’s the dfa we’ve drawn. 2. Convert the nfa below into an equivalent dfa. 3. Convert the nfa below into an equivalent dfa. a q 0 { q 0 , q 1 , q 2 } q 1 { q 0 , q 1 , q 2 } q 2 { q 0 , q 1 , q 2 } 0 1 q 0 { q 1 , q 2 } { q 1 , q 2 } q 1 { q 0 , q 2 } { q 1 , q 2 } q 2 Ø { q 1 , q 2 } 0 1 q 0 { q 0 , q 1 , q 2 } { q 1 , q 2 } q 1 { q 0 , q 1 , q 2 } { q 1 , q 2 } q 2 { q 2 } { q 1 } λ q 0 q 1 a q 2 λ a a Q 012 Q 0 0, λ 0 0,1 1 0 q 0 q 1 q 2 1 0 1 1 0 1 0 Q 0 Q 012 Q 12 0,1 0 0, λ 1 q 0 q 1 q 2 1 Q 12 Q 0 Q 0 Q 02 0,1 0 1 0,1

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CSCI3255: Math Foundations of CS Homework 2, (2.3), page 2 September 2006 5. Is it true that for any nfa M = ( Q , Σ , δ , q 0 , F ) that L ( M ) = { w Σ *: δ *( q 0 , w ) F = Φ }? Since L ( M ) is defined as { w Σ *: δ *( q 0 , w ) F Φ }. Yes since either something = or Φ .
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csci3255 HW 2-3 - CSCI3255 Math Foundations of CS Homework...

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