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NFAs and DFAs Sipser 1.2 (pages 47-63)

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Last time…

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CS 311 Fall 2008 3 NFA A nondeterministic finite automaton (NFA) is a 5-tuple (Q, Σ, δ, q 0, F), where – Q is a finite set called the states – Σ is a finite set called the alphabet – δ: Q × Σε → P(Q) is the transition function – q 0 ∈ Q is the start state – F ⊆ Q is a set of accept states In-class exercise:

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CS 311 Fall 2008 4 NFA computation Let N=(Q, Σ, δ, q 0, F) be a NFA and let w be a string over the alphabet Σ Then N accepts w if – w can be written as w 1 w 2 w 3 …w m with each w i ∈ Σε and –There exists a sequence of states s 0,s 1,s 2,…,s m exists in Q with the following conditions: 1.s 0 =q 0 2.s i+1 ∈ δ(s i,w i+1 ) for i = 0,…,m-1 3.s n ∈ F

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One last operation

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CS 311 Fall 2008 6 Kleene star operation Let A be a language. The Kleene star operation is A* = {x 1 x 2 …x k | k ≥ 0 and each x i ∈ A} Exercise A = {w | w is a string of 0 s and 1 s containing an odd number of 1 s } B = {w | w is a string of 0 s and 1 s containing an even number of 1 s } C = {0,1} What are A*, B*, and C* ?

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CS 311 Fall 2008 7 Clay Knee? Kleene pronounced his last name klay'nee. His son, Ken Kleene, wrote: "As far as I am aware this pronunciation is incorrect in all known languages. I believe that this novel pronunciation was invented by my father." Dr. "Clay Knee" must have been a geek of the first water, to be sure! - From http://visualbasic.about.com/od/usevb6/a/RegExVB6.htm Kleene pronounced his last name klay'nee. His son, Ken Kleene, wrote: "As far as I am aware this pronunciation is incorrect in all known languages. I believe that this novel pronunciation was invented by my father." Dr. "Clay Knee" must have been a geek of the first water, to be sure! - From http://visualbasic.about.com/od/usevb6/a/RegExVB6.htm

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CS 311 Fall 2008 8 Kleene star Theorem: The class of languages recognized by NFAs is closed under the Kleene star operation.

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CS 311 Fall 2008 9 If only… … somebody would prove that the class of languages recognized by NFAs and the class of languages recognized by DFAs were equal…

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CS 311 Fall 2008 10 Then… We’d have: –The class of regular languages is closed under: Concatenation Kleene star

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CS 311 Fall 2008 11 And… a cute proof for closure under union!

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CS 311 Fall 2008 12 We can be that somebody! Theorem: A language is regular if and only if there exists an NFA that recognizes it. Proof: ( ⇒ ) Let A be a regular language…

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CS 311 Fall 2008 13 Then there exists a DFA M M is a 5-tuple (Q, Σ, δ M, q 0, F), where – Q is a finite set called the states – Σ is a finite set called the alphabet – δ: Q × Σ → Q is the transition function – q 0 ∈ Q is the start state – F ⊆ Q is a set of accept states

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CS 311 Fall 2008 14 We need to show there exists an NFA N N is a 5-tuple (Q, Σ, δ N, q 0, F), where – Q is a finite set called the states – Σ is a finite set called the alphabet – δ: Q × Σε → P(Q) is the transition function – q 0 ∈ Q is the start state – F ⊆ Q is a set of accept states Can we define N from M?

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CS 311 Fall 2008 15 Now the other way! Theorem: A language is regular if and only if there exists an NFA that recognizes it. Proof: ( ⇒ ) Let A be a language accepted by NFA N = (Q, Σ, δ, q 0, F)

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CS 311 Fall 2008 16 Equivalent machines Definition: Two machines are equivalent if they recognize the same language Let’s prove: Theorem: Every NFA has an equivalent DFA.

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CS 311 Fall 2008 17 Remember…

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CS 311 Fall 2008 18 A simpler example

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CS 311 Fall 2008 19 Removing choice Proof: Let A be a language accepted by NFA N = (Q, Σ, δ, q 0, F) We construct a DFA M=(Q’, Σ, δ’, q 0 ’, F’) recognizing A – Q’ = P(Q) –For R ∈ Q' and a ∈ Σ, define δ'(R,a) = ∪ δ(r,a) – q 0 ’ = {q 0 } – F’ = { R ∈ Q’ | R contains an accept state from F} r∈Rr∈R

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CS 311 Fall 2008 20 Okay, but what about ε arrows?

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CS 311 Fall 2008 21 Modifying our construction Proof: Let A be a language accepted by NFA N = (Q, Σ, δ, q 0, F) We construct a DFA M=(Q’, Σ, δ’, q 0 ’, F’) recognizing A – Q’ = P(Q) –For R ∈ Q' and a ∈ Σ, define δ'(R,a) = ∪ E(δ(r,a)) where E(R) = {q | q can be reached from R along 0 or more ε arrows } – q 0 ’ = E({q 0 }) – F’ = { R ∈ Q’ | R contains an accept state from F} r∈Rr∈R

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