1. CS 350 — Fall 2018 gilray.org/classes/fall2018/cs350/
Deterministic finite automata (DFA) &
Non-deterministic finite automata (NFA)
CS 350 — Fall 2018 gilray.org/classes/fall2018/cs350/

2. DFA transition diagram
𝐿={hello} h e l o q0 q1 q2 q3 l q4 q5
DFA transition diagram

3. Formally, each DFA is a 5-tuple:
A set of final states
Formally, each DFA is a 5-tuple:
(𝑄,Σ,𝛿, 𝑞 0 ,𝐹)
A starting state
A finite set of states {q…}
A transition function
An alphabet

4. Formally, each DFA is a 5-tuple:
(𝑄,Σ,𝛿, 𝑞 0 ,𝐹)
𝛿⊆(𝑄×Σ⇀𝑄)
𝐹⊆𝑄
𝑞 0 ∈𝑄

5. (𝑄,Σ,𝛿, 𝑞 0 ,𝐹),where 𝑄={ 𝑞 0 ,…, 𝑞 5 } Σ={a,b,…,z} 𝐹={ 𝑞 5 }l l o q0 q1 q2 q3 q4 q5
(𝑄,Σ,𝛿, 𝑞 0 ,𝐹),where
𝑄={ 𝑞 0 ,…, 𝑞 5 }
Σ={a,b,…,z}
𝛿={( 𝑞 0 ,h, 𝑞 1 ),( 𝑞 1 ,e, 𝑞 2 ),
( 𝑞 2 ,l, 𝑞 3 ),( 𝑞 3 ,l, 𝑞 4 ),( 𝑞 4 ,o, 𝑞 5 )}
𝐹={ 𝑞 5 }

6. h e l o q0 q1 q2 q3 l q4 q5 𝛿 h l e o … …
𝑞 0
𝑞 1
DFA transition table
𝑞 1
𝑞 2
𝑞 2
𝑞 3
𝑞 3
𝑞 4
𝑞 4
𝑞 5
𝑞 5
𝛿={( 𝑞 0 ,h, 𝑞 1 ),( 𝑞 1 ,e, 𝑞 2 ),
( 𝑞 2 ,l, 𝑞 3 ),( 𝑞 3 ,l, 𝑞 4 ),( 𝑞 4 ,o, 𝑞 5 )}

7. Extended transition functionh e l q3 l o q0 q1 q2 q4 q5
Extended transition function
𝛿 ̂ ={( 𝑞 0 ,h, 𝑞 1 ),( 𝑞 1 ,e, 𝑞 2 ),
( 𝑞 2 ,l, 𝑞 3 ),( 𝑞 3 ,l, 𝑞 4 ),( 𝑞 4 ,o, 𝑞 5 ),
( 𝑞 0 ,he, 𝑞 2 ),( 𝑞 1 ,el, 𝑞 3 ),( 𝑞 2 ,ll, 𝑞 4 ),( 𝑞 3 ,lo, 𝑞 5 ),
( 𝑞 0 ,hel, 𝑞 3 ),( 𝑞 1 ,ell, 𝑞 4 ),( 𝑞 2 ,llo, 𝑞 5 ),
( 𝑞 0 ,hell, 𝑞 4 ),( 𝑞 1 ,ello, 𝑞 5 ),( 𝑞 0 ,hello, 𝑞 5 )}
𝛿={( 𝑞 0 ,h, 𝑞 1 ),( 𝑞 1 ,e, 𝑞 2 ),
( 𝑞 2 ,l, 𝑞 3 ),( 𝑞 3 ,l, 𝑞 4 ),( 𝑞 4 ,o, 𝑞 5 )}

8. Computing the extended transition function as a least fixed point
𝛿 ̂ ,where 𝑓 𝛿,𝑄 ( 𝛿 ̂ )= 𝛿 ̂
𝑓 𝛿,𝑄 ( 𝛿 ̂ )={( 𝑞 𝑖 , 𝑠 0 𝑠 1 , 𝑞 𝑗 )|( 𝑞 𝑖 , 𝑠 0 , 𝑞 𝑘 )∈𝛿∧( 𝑞 𝑘 , 𝑠 1 , 𝑞 𝑗 )∈ 𝛿 ̂ }
∪{( 𝑞 𝑖 ,𝜖, 𝑞 𝑖 )| 𝑞 𝑖 ∈𝑄}

9. Computing the extended transition function as a least fixed point
𝑓 𝛿,𝑄 ( 𝛿 ̂ )={( 𝑞 𝑖 , 𝑠 0 𝑠 1 , 𝑞 𝑗 )|( 𝑞 𝑖 , 𝑠 0 , 𝑞 𝑘 )∈𝛿∧( 𝑞 𝑘 , 𝑠 1 , 𝑞 𝑗 )∈ 𝛿 ̂ }
∪{( 𝑞 𝑖 ,𝜖, 𝑞 𝑖 )| 𝑞 𝑖 ∈𝑄}
ℒ( a ∗ ) 𝛿 a
𝑞 0
𝑞 0 a q0

10. Computing the extended transition function as a least fixed point
𝑓 𝛿,𝑄 ( 𝛿 ̂ )={( 𝑞 𝑖 , 𝑠 0 𝑠 1 , 𝑞 𝑗 )|( 𝑞 𝑖 , 𝑠 0 , 𝑞 𝑘 )∈𝛿∧( 𝑞 𝑘 , 𝑠 1 , 𝑞 𝑗 )∈ 𝛿 ̂ }
∪{( 𝑞 𝑖 ,𝜖, 𝑞 𝑖 )| 𝑞 𝑖 ∈𝑄}
ℒ( a ∗ )
𝛿 ̂ 𝜖 a
𝑞 0
𝑞 0
𝑞 0 a q0

11. Computing the extended transition function as a least fixed point
𝑓 𝛿,𝑄 ( 𝛿 ̂ )={( 𝑞 𝑖 , 𝑠 0 𝑠 1 , 𝑞 𝑗 )|( 𝑞 𝑖 , 𝑠 0 , 𝑞 𝑘 )∈𝛿∧( 𝑞 𝑘 , 𝑠 1 , 𝑞 𝑗 )∈ 𝛿 ̂ }
∪{( 𝑞 𝑖 ,𝜖, 𝑞 𝑖 )| 𝑞 𝑖 ∈𝑄}
ℒ( a ∗ )
𝛿 ̂ 𝜖 a aa
aaa
𝑞 0
𝑞 0
𝑞 0
𝑞 0
𝑞 0 a q0

12. ℒ(𝑄,Σ,𝛿, 𝑞 0 ,𝐹)={𝑠|( 𝑞 0 ,𝑠, 𝑞 𝑓 )∈ 𝛿 ̂ ∧ 𝑞 𝑓 ∈𝐹}
The Language L of a DFA:
and
𝛿 ̂
(𝑄,Σ,𝛿, 𝑞 0 ,𝐹)
ℒ(𝑄,Σ,𝛿, 𝑞 0 ,𝐹)={𝑠|( 𝑞 0 ,𝑠, 𝑞 𝑓 )∈ 𝛿 ̂ ∧ 𝑞 𝑓 ∈𝐹}
where 𝛿 ̂ is the least fixed point 𝛿 ̂ = 𝑓 𝛿,𝑄 ( 𝛿 ̂ )

13. L is the language L((a|b)*cc(a|b)*).
Try an example:
L is the language L((a|b)*cc(a|b)*).

14. L is the language L((a|b)*cc(a|b)*).
Try an example:
L is the language L((a|b)*cc(a|b)*). c c q0 q1 q2
a,b
a,b

15. L is the language L((a|b)*bab).
Try an example:
L is the language L((a|b)*bab).

16. L is the language L((a|b)*bab).
Try an example:
L is the language L((a|b)*bab). b b b a b q0 q1 q2 q3 a a a

17. Try an example: L is the language, over {a,b}, of strings
with an even number of a’s and of b’s.

18. Try an example: L is the language, over {a,b}, of strings
with an even number of a’s and of b’s. a q0 q2 a b b b b q1 a q3 a

19. There are an even number of a’s There are an odd number of a’sq0 q2 a b b b b q1 a q3 a

20. a a b b b b a a There are an even number of b’sq0 q2 a b b b b q1 a q3 a
There are an odd number of bs

21. So this is just the DFAs for (aa). and for (bb). composed
So this is just the DFAs for (aa)* and for (bb)* composed. Can this idea be generalized? a
(E,E)
(O,E) a b b b b
(E,O) a
(O,O) a

22. 𝐹 ∩ ={( 𝑞 0𝐹 , 𝑞 1𝐹 )| 𝑞 0𝐹 ∈ 𝐹 0 ∧ 𝑞 1𝐹 ∈ 𝐹 1 }
ℒ( 𝑄 0 ,Σ, 𝛿 0 , 𝑞 00 , 𝐹 0 )∩ℒ( 𝑄 1 ,Σ, 𝛿 1 , 𝑞 10 , 𝐹 1 )
=ℒ( 𝑄 ∩ ,Σ, 𝛿 ∩ , 𝑞 ∩ , 𝐹 ∩ )
𝑄 ∩ = 𝑄 0 × 𝑄 1
𝛿 ∩ ={(( 𝑞 0 , 𝑞 1 ),𝑠,( 𝑞 0 ′ , 𝑞 1 ′ ))
|( 𝑞 0 ,𝑠, 𝑞 0 ′ )∈ 𝛿 0
𝑞 ∩ =( 𝑞 00 , 𝑞 10 )
∧( 𝑞 1 ,𝑠, 𝑞 1 ′ )∈ 𝛿 1 }
𝐹 ∩ ={( 𝑞 0𝐹 , 𝑞 1𝐹 )| 𝑞 0𝐹 ∈ 𝐹 0 ∧ 𝑞 1𝐹 ∈ 𝐹 1 }

23. Try an example: L is the language, over {0,1}, of strings
with an even number of 0s and two 1s.

24. Try an example: L is the language, over {0,1}, of strings
with an even number of 0s and two 1s. 1 1 q0 q1 q2 q0 q1 1 1

25. Try an example: L is the language, over {0,1}, of strings
with an even number of 0s and two 1s. 1 1
q0,q0
q0,q1
q0,q2 1 1
q1,q0
q1,q1
q1,q2

26. is the complement of language (a|b)*bab
Try an example:
is the complement of language (a|b)*bab 𝐿

27. (Recall that this is how we defined L as a DFA.)
Try an example:
is the complement of language (a|b)*bab 𝐿 b b b a b q0 q1 q2 q3 a a a
(Recall that this is how we defined L as a DFA.)

28. is the complement of language (a|b)*bab
Try an example:
is the complement of language (a|b)*bab 𝐿 b b b a b q0 q1 q2 q3 a a a

29. (Recall that this is how we defined {a} as a DFA.)
Try an example:
is the complement of language a 𝐿 a q0 q1
(Recall that this is how we defined {a} as a DFA.)

30. is the complement of language a
Try an example:
is the complement of language a 𝐿 a a q0 q1 q2 a

31. 𝑄 ′ =𝑄∪{ 𝑞 dead } 𝐹 ′ = 𝑄 ′ −𝐹 𝛿 ′ =𝛿∪{( 𝑞 dead ,𝑠, 𝑞 dead )|𝑠∈Σ}
The complement ofℒ(𝑄,Σ,𝛿, 𝑞 0 ,𝐹)isℒ( 𝑄 ′ ,Σ, 𝛿 ′ , 𝑞 0 , 𝐹 ′ )
𝑄 ′ =𝑄∪{ 𝑞 dead }
𝐹 ′ = 𝑄 ′ −𝐹
𝛿 ′ =𝛿∪{( 𝑞 dead ,𝑠, 𝑞 dead )|𝑠∈Σ}
∪{(𝑞,𝑠, 𝑞 dead )|𝑠∈Σ∧𝑞∈𝑄∧(𝑞,𝑠, 𝑞 ′ )∉𝛿}

32. Non-deterministic finite automata (NFA)

33. Models of regular languages
“Converts to”
GNFA
“Converts to” RE
DFA
“Minimizes to”
NFA
“Converts to”
“Converts to”

34. L is the language L((a|b)*bab).
Try an example:
L is the language L((a|b)*bab). b a b q0 q1 q2 q3
a,b

35. b a b a,b Let’s run this nondeterministically on the string babab and
track what paths we can take: b a b q0 q1 q2 q3
a,b

36. /
Let’s run this nondeterministically on the string babab and
track what paths we can take: b a b q0 q1 q2 q3
a,b

37. / /
Let’s run this nondeterministically on the string babab and
track what paths we can take: b a b q0 q1 q2 q3
a,b

38. / / /
Let’s run this nondeterministically on the string babab and
track what paths we can take: b a b q0 q1 q2 q3
a,b

39. / / / /
Let’s run this nondeterministically on the string babab and
track what paths we can take: b a b q0 q1 q2 q3
a,b

40. If any possible path would accept, then the NFA accepts./ / / / /
Let’s run this nondeterministically on the string babab and
track what paths we can take:
If any possible path would accept, then the NFA accepts. b a b q0 q1 q2 q3
a,b

41. Formally, each NFA is also a 5-tuple:
A set of final states
Formally, each NFA is also a 5-tuple:
(𝑄,Σ,𝛿, 𝑞 0 ,𝐹)
A starting state
A finite set of states {q0,…}
A transition relation
An alphabet

42. 𝛿:𝑄×Σ×𝑄
A transition relation

43. 𝛿:𝑄×Σ×𝑄
𝛿:𝑄×Σ→𝒫(𝑄)
A transition relation

44. b 𝛿 a 𝑞 0 𝑞 1 𝑞 2 𝑞 3 NFA transition table { 𝑞 0 } { 𝑞 0 , 𝑞 1 }q0 q1 q2 q3
a,b
NFA transition table b 𝛿 a
𝑞 0
{ 𝑞 0 }
{ 𝑞 0 , 𝑞 1 }
𝑞 1
{ 𝑞 2 }
𝑞 2
{ 𝑞 3 }
𝑞 3

45. NFA extended transition tableq0 q1 q2 q3
NFA extended transition table
a,b
𝛿 ̂ b ab ba
bab a aa
𝑞 0
{ 𝑞 0 }
{ 𝑞 0 , 𝑞 1 }
{ 𝑞 0 }
{ 𝑞 0 , 𝑞 1 }
{ 𝑞 0 , 𝑞 2 }
{ 𝑞 0 , 𝑞 3 }
𝑞 1
{ 𝑞 2 }
{ 𝑞 3 } …
𝑞 2
{ 𝑞 3 }
𝑞 3

46. This is exactly the same as for a DFA, except now a relation.
Computing an NFA’s extended transition relation as a least fixed point
This is exactly the same as for a DFA, except now a relation.
𝛿 ̂ ,where 𝑓 𝛿,𝑄 ( 𝛿 ̂ )= 𝛿 ̂
𝑓 𝛿,𝑄 ( 𝛿 ̂ )={( 𝑞 𝑖 , 𝑠 0 𝑠 1 , 𝑞 𝑗 )|( 𝑞 𝑖 , 𝑠 0 , 𝑞 𝑘 )∈𝛿∧( 𝑞 𝑘 , 𝑠 1 , 𝑞 𝑗 )∈ 𝛿 ̂ }
∪{( 𝑞 𝑖 ,𝜖, 𝑞 𝑖 )| 𝑞 𝑖 ∈𝑄}

47. The Language L of an NFA:
and
𝛿 ̂
(𝑄,Σ,𝛿, 𝑞 0 ,𝐹)
𝐿={𝑠|( 𝛿 ̂ ( 𝑞 0 ,𝑠)∩𝐹)⊃Ø}