Basics of automata theory

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

2. Nondeterministic finite automaton (NFA) S0 S1 A A A,B • 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.

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

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

6. Example 2 B A A A 1 0 2 A,B B Complete it yourself  0 1 2 01 02 12 012

7. Example 2 B A B 1 A B A B 01 02 A A A 1 0 2 A,B B A,B B 0 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 thatL(A3) = L(A1) ÅL(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) • A1=h, S1, , I1, F1i and A2= h, S2, 2, I2, F2i • 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. Reminder

10. Example – product of two automata b a s0 s1 b a a t0 t1 b L(A1) = (a+b)*a +  (words ending with ‘a’ + empty word) A1: L(A2) = (ba)* + (ba)*b A2: What should be the language ofA1£ A2 ?

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

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

13. Example 2 – product of two automata a t0 t1 b a s0 L(A1) = (ab)* + (ab)*a (words that alternate between a and b ) s1 A1: b A2: L(A2) = (ba)* + (ba)*b (words that alternate between b and a) What should be the language of A1Å A2 ?

14. Example 2 – product of two automata a s0 s1 A1: b a t0 t1 A2: b • States: (s0,t0), (s0,t1), (s1,t0), (s1,t1). • Initial state: (s1,t0). • Accepting states: (s0,t0), (s0,t1), (s1,t0), (s1,t1). A1Å A2:

15. Example 2 – product of two automata a s0 s1 A1: b a t0 t1 A2: b s0,t0 s1,t0 A1Å A2: s0,t1 b a s1,t1 L(A1Å A2) = e

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 b b a a s0 s0 s1 s1 b b a a • The complementation of A is denoted by

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

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

21. Automata over infinite words (DFW) a S0 S1 b a b • 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 ;.

22. Automata over infinite words (DFW) a S0 S1 b a b • 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.

23. Other computations a S0 S1 b a b • 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…, thecomputation 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 ?

24. The language of a Buchi automaton a S0 S1 b B = a b • 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*)!

25. NonDeterministic Buchi automata (NFW) B = S1 S0 a b a • 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

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

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

28. About the alphabet... a,b b a :a,:b a,b a a,:b a b :a,b b • So far a 2 ∑ meant that a is some atom (a,b...) from a given set AP. • We will also use a 22AP • This is useful for representing states • (cont’d on next slide) (same thing, only write positive literals)

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

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 ∑µ2AP Recall that this is {p,q} p,q  p,q p  p p

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

33. Intersecting two Buchi automata (infinite words) a b s0 s1 b a b a t0 t1 a b • Previous method doesn’t work: Infinite a’s 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 0 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 2 a a a b s0, t1 s1, t0 b

37. s0, t0 b a a 0 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 2 a a a b s0, t1 s1, t0 b simplify by removing unreachable states

38. s0, t0 b a 0 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) a b s0 s1 b a b a t0 t1 a b There are total of 12 states in the product automaton. The reachable part of A1Å A2 is: h s0,t0,0 i b a h s1,t0,0 i h s0,t1,1 i a b a b b a b a a h s1,t0,2 i h s0,t1,0 i