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
w be a string over the alphabet Σ
Then N accepts w 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) for i = 0,…,m-1
sn∈F

5. One last operation

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

7. - From http://visualbasic.about.com/od/usevb6/a/RegExVB6.htm
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

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. a cute proof for closure under union!
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 → Q is 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