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NFAs

NFAs. Sipser 1.2 (pages 47–54). Recall…. Last time we showed that the class of regular languages is closed under: Complement Union Intersection. Concatenation operation. Let A and B be languages. The concatenation of A and B is A ° B = { xy | x ∈ A and y ∈ B} .

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NFAs

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  1. NFAs Sipser 1.2 (pages 47–54)

  2. Recall… • Last time we showed that the class of regular languages is closed under: • Complement • Union • Intersection

  3. Concatenation operation • Let A and B be languages. • The concatenation of A and B is A°B = {xy | x ∈A and y ∈ B}. Example • Let A = {w | wis a string of 0s and 1s containing an odd number of 1s} • Let B = {w | wis a string of 0s and 1s containing an even number of 1s} • What are A°B; B°A; A°A; B°B? • Are any of these languages regular?

  4. Concatenation • Conjecture: The class of regular languages is closed under the concatenation operation. • Proof idea:

  5. Hmm… • 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} • Create a machine by gluing… how?

  6. We need another approach ?

  7. The empty symbol ε • Seems like it might work • Let’s try running it…

  8. Relaxing the rules Deterministic (DFA) Nondeterministic (NFA)

  9. Running DFA vs NFA DFA computation (path) NFA computation (tree)

  10. An example computation • What about 1011?

  11. Another example • What language is being recognized? • Hint: can you start listing strings accepted?

  12. Formally… • 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:

  13. 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 in 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

  14. Nondeterminism makes life easier • Let’s build an NFA that recognizes B = {w | wis a string over {a,b} that starts and ends with the same symbol} C = {w | wis a string over {0,1} that contains at least three 1’s} D = {w | wis a string over {0,1} that contains at least three consecutive 1’s}

  15. If at first you don’t succeed… … adjust your goal! We wanted to prove: The class of regular languages is closed under the concatenation operation.

  16. If at first you don’t succeed… … adjust your goal! Instead, let’s prove the theorem: The class of languages recognized by NFAs is closed under the concatenation operation.

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