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NFAs and DFAs

NFAs and DFAs. Sipser 1.2 (pages 47-63). Last time…. 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

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NFAs and DFAs

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  1. NFAs and DFAs Sipser 1.2 (pages 47-63)

  2. Last time…

  3. NFA • A nondeterministic finite automaton (NFA) is a 5-tuple (Q, Σ, δ,q0, 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 • q0∈Q is the start state • F⊆Q is a set of accept states • In-class exercise:

  4. NFA computation • Let N=(Q, Σ, δ, q0, F)be a NFA and let wbe a string over the alphabet Σ • Then Nacceptsw if • w can be written as w1w2w3…wm with each wi∈Σε and • There exists a sequence of states s0,s1,s2,…,sm exists in Q with the following conditions: • s0=q0 • si+1∈δ(si,wi+1) fori = 0,…,m-1 • sn∈F

  5. One last operation

  6. Kleenestar operation • Let A be a language. • The Kleene star operation is A* = {x1x2…xk | k ≥ 0 and each xi ∈ A} Exercise • A = {w | wis a string of 0s and 1s containing an odd number of 1s} • B = {w | wis a string of 0s and 1s containing an even number of 1s} • C = {0,1} • What are A*, B*, and C*?

  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

  8. Kleene star • Theorem: The class of languages recognized by NFAs is closed under the Kleene star operation.

  9. If only… … somebody would prove that the class of languages recognized by NFAs and the class of languages recognized by DFAs were equal…

  10. Then… • We’d have: • The class of regular languages is closed under: • Concatenation • Kleene star

  11. And… a cute proof for closure under union!

  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…

  13. Then there exists a DFA M • M is a 5-tuple (Q, Σ, δM, q0, F), where • Q is a finite set called the states • Σ is a finite set called the alphabet • δ: Q × Σ → Qis the transition function • q0∈Q is the start state • F⊆Q is a set of accept states

  14. We need to show there exists an NFA N • N is a 5-tuple (Q, Σ, δN, q0, 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 • q0∈Q is the start state • F⊆Q is a set of accept states • Can we define N from M?

  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, q0, F)

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

  17. Remember…

  18. A simpler example

  19. Removing choice Proof: • Let A be a language accepted by NFA N = (Q, Σ, δ, q0, F) • We construct a DFA M=(Q’, Σ, δ’, q0’, F’) recognizing A • Q’ = P(Q) • For R∈Q' and a∈Σ, define δ'(R,a) = ∪ δ(r,a) • q0’ = {q0} • F’ = { R∈Q’ | R contains an accept state from F} r∈R

  20. Okay, but what about ε arrows?

  21. Modifying our construction Proof: • Let A be a language accepted by NFA N = (Q, Σ, δ, q0, F) • We construct a DFA M=(Q’, Σ, δ’, q0’, 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} • q0’ = E({q0}) • F’ = { R∈Q’ | R contains an accept state from F} r∈R

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