1. Basics of automata theory
Nondeterministic Finite Automata (NFA)
Nondeterministic Finite Automata on infinite words (NFW)

2. Nondeterministic finite automaton (NFA)
Transitions: (S0,A,S0), (S0,B,S0), (S0,A,S1),(S1,A,S1).
What is the language of this automaton?
All words that have a path to an accepting state.
All words ending with A
(A | B)*A
A,B A S0 S1 9

3. Equivalent deterministic automaton
Every NFA can be transformed to DFA. How ?
The price may be exponential.
A,B A S0 S1 B A S0
S0,S1 10

4. Determinization Let M = (S, Σ, , I, F) be an NFA.
Define a DFA Md = (Sd, Σd, d, Id, Fd), where
Sd = P(S) // power-set of S
Σd = Σ
Id = I
d(q, a) = { (r,a) | r 2 q} for all q2 Sd, a 2 Σ
Fd = {q | q 2 Sd ∧ q ∩ F ≠ ;}

5. Example S0 S1 A A,B Sd = {}{S0}{S1}{S0,S1} d =  = {A,B}
Id = I = {S0}
d = ...
FD = {S1}{S0,S1} B
A,B A S0 S1 B B A
S0,S1 A

6. Example 2 A 1 A 2
A,B B A B
Complete it yourself  1 2 01 02 12
012

7. Example 2 A 1 A 2
A,B B
A,B A B B 1 01 02 A B 2 A B 12 B A
012 A

8. Few important questions (1)
Given two automata A1, A2...
How do we build an automaton A3 such that L(A3) = L(A1) Å L(A2)
A method to build A3: compute the product A1 £ A2
We already saw how to compute a product...

9. Product of two NFA-s (finite words)
Reminder
A1=h , S1, , I1, F1 i and A2= h , S2, 2, I2, F2 i
A1 £ A2 =
Each state is a pair (s,t): s 2 S1 and t 2 S2.
Initial states: pairs (s,t) such that s 2 I1 and t 2 I2.
Accepting states: pairs (s,t) such that s 2 F1 and t 2 F2
((s,t) a (s’,t’)) is a transition if (s,a,s’) 2 1, and (t,a,t’) 2 2. 25

10. Example – product of two automatab a s0 s1
L(A1) = (a+b)*a + 
(words ending with ‘a’
+ empty word)
A1: a b t0 t1
L(A2) = (ba)* + (ba)*b
A2:
The \epsilon in A_2 is not needed because it is included in (ba)*. Put it for clarity of the product automaton
What should be the language of A1 £ A2 ? 26

11. Example – product of two automatab a s0 s1
A1: a b t0 t1
A2:
States: (s0,t0), (s0,t1), (s1,t0), (s1,t1).
Initial state: (s0,t0).
Accepting states: (s0,t0), (s0,t1).
A1 £ A2: 26

12. Example – product of two automatab a s0 s1
A1: a b t0 t1
A2:
s0,t0 a
s1,t0
A1 £ A2:
s0,t1 b a b
s1,t1
L(A1 £ A2) = (ba)* 27

13. Example 2 – product of two automatas0
L(A1) = (ab)* + (ab)*a
(words that alternate
between a and b ) s1
A1: b a b t0 t1
A2:
L(A2) = (ba)* + (ba)*b
(words that alternate
between b and a)
What should be the language of A1 Å A2 ? 26

14. Example 2 – product of two automatab s0 s1 t0 t1
States: (s0,t0), (s0,t1), (s1,t0), (s1,t1).
Initial state: (s1,t0).
Accepting states: (s0,t0), (s0,t1), (s1,t0), (s1,t1).
A1 Å A2: 26

15. Example 2 – product of two automatab s0 s1 t0 t1
s0,t0
s1,t0
A1 Å A2:
s0,t1 b a
s1,t1
L(A1 Å A2) = e 27

16. Few important questions (2)
Given a DFA A:
How do we construct A’ such that L(A’) = * - L(A)
In other words, how do we build an automaton that accepts exactly those words that are rejected by A ?
Answer: compute the complement automaton.
How ? Let F’ = S – F.
(i.e., substitute accepting and non-accepting states.)

17. Example: complementation
The complementation of A is denoted by b a s0 s1

18. Few important questions (3)
Given an automaton A:
Universality: is L(A) = * ?
Emptiness: is L(A) = ; ?
Emptiness: is an accepting state reachable?
Universality: check whether

19. ... Automata on infinite words
And now....
... Automata on infinite words
These are called !-automata

20. Automata over infinite words (DFW / NFW)
Similar definition.
Runs on infinite words over .
Accepts when an accepting state occurs infinitely often in the computation.
This is called a Buchi automaton a b S0 S1 12

21. Automata over infinite words (DFW)
Formally, let F be the set of accepting states.
Let inf() µ S be the set of states appearing infinite number of times in a computation .
 is accepted by the automaton if inf() Å F  ;. a b S0 S1 12

22. Automata over infinite words (DFW)
Consider the word a b a b a b a b…
The computation is S0 S0 S1 S0 S1 S0 S1 …
This computation is accepting, since S0 appears infinitely many times. a b S0 S1 13

23. Other computations
For the word b b b b b… the computation is S0 S1 S1 S1 S1… and is not accepting.
For the word a a a b b b b b …, the computation is S0 S0 S0 S0 S1 S1 S1 S1 …and ... (?)
What is the computation for a b a b b a b b b …? Is it accepting ? a b S0 S1 14

24. The language of a Buchi automaton
The language of a Buchi automaton is the set of infinite strings that are accepted by it.
Such languages are called ! – languages.
Buchi automaton defines ! – regular languages.
L(B) = (ab*)! a b S0 S1
B =

25. NonDeterministic Buchi automata (NFW)
As before, a string is accepted if there exists an accepting run.
L(B) = aa*b!
Surprise: there is no determinization procedure
We will focus on DFW a b S0 S1
B =

26. Why do we need Buchi automata ?
The compilation theorem: every LTL formula  can be translated to a Buchi automaton B that accepts the same language as .
-a a  Fa a Ga
-a a
GFa

27. What about the other direction ?
Can every Buchi automata be translated to an LTL formula ?
No LTL formula can express this property. a
“a holds on every even step”

28. About the alphabet...
So far a 2 ∑ meant that a is some atom (a,b...) from a given set AP.
We will also use a 2 2AP
This is useful for representing states
(cont’d on next slide) b a
a,b
a,:b
:a,:b
:a,b
a,b a b
(same thing, only write positive literals)

29. About the alphabet... ... or even a 2 22AP
This can give us a more compact representaiton
a Æ b
a Ç b
:a Æ :b
:aÆb Ç aÆ:b
a,:b
:a, b
:a, :b
a,b
:a Æ b
aÇ:b

30. About the alphabet...
A Buchi automaton can also be represented with labels on states.
There is 1-1 translation to a Buchi automaton with labels on transitions:
Move labels to outgoing edges.

31. About the alphabet...
From labels on states to labels on transitions: Example.
Let ∑ µ 2AP
Recall that this is {p,q}
p,q 
p,q p p  p

32. Again, important questions...
For Buchi automata B1, B2:
How to compute L(B1) Å L(B2) ?
How to complement ?
Find B’ s.t.
How to check for emptiness ?
Is L(B) = ; ?

33. Intersecting two Buchi automata (infinite words)
Previous method doesn’t work: a b s0 s1
Infinite a’s b a t0 t1
Infinite b’s
s0, t0
s1, t1
Empty language ! b a a a b
s0, t1
s1, t0 b

34. Intersecting two Buchi automata (infinite words)
The reason: a path should be accepted if it fulfills two separate acceptance conditions:
passes infinitely many times through s0
passes infinitely many times through t0
An automaton has such multiple acceptance conditions is called a generalized Buchi automata.
We will learn about this later on.
For now, we will see a reduction of this condition to a standard Buchi automaton.
s0, t0
s1, t1
Empty language ! b a a a b
s0, t1
s1, t0 b

35. Intersecting two Buchi automata (infinite words)
Strategy:
“Multiply” the product automaton by 3 (S = S1 £ S2 £ {0,1,2} )
Start from the ‘0’ copy.
Transition to the ‘1’ copy when entering a state from F1
Transition to the ‘2’ copy if in a ‘1’ state and entering a state from F2, and in the next state back to a ‘0’ state.
Make the ‘2’ copy an accepting set.

36. s0, t0
s1, t1 b a a a b
s0, t1
s1, t0 b
s0, t0
s1, t1 b a 1 a a b
s0, t1
s1, t0 b
s0, t0
s1, t1 b a 2 a a b
s0, t1
s1, t0 b

37. s0, t0 b a a a a b
s0, t1
s1, t0 b a
s0, t0 b a 1 a a b
s0, t1
s1, t0 b b b
s0, t0 b a 2 a a b
s0, t1
s1, t0 b
simplify by removing unreachable states

38. s0, t0 b a a b
s0, t1
s1, t0 b a b a b 1 a a
s0, t1
s1, t0 a a b b 2 a a b
s0, t1
s1, t0 b
simplify by removing unreachable states

39. Intersecting two Buchi automata (infinite words)
There are total of 12 states in the product automaton.
The reachable part of A1 Å A2 is:
h s0,t0,0 i a b a b s0 s1
h s0,t1,1 i
h s1,t0,0 i a b a b a b b b a t0 t1 a a
h s1,t0,2 i
h s0,t1,0 i