Topics in Timed Automata

Transcription

1 1/32 Topics in Timed Automata B. Srivathsan RWTH-Aachen Software modeling and Verification group

2 2/32 Timed Automata A theory of timed automata R. Alur and D. Dill, TCS 94

3 2/32 Timed Automata Language theory Verification Lectures 1,..., 5 Lectures 6,..., 10 A theory of timed automata R. Alur and D. Dill, TCS 94

4 3/32 Lecture 1: Timed languages and timed automata

5 4/32 Σ : alphabet {a, b} L Σ Σ : words : language {ε, a, b, aa, ab, ba, bb, aab,... } property over words L 1 := {set of words starting with an a } {a, aa, ab, aaa, aab,... } L 2 := {set of words with a non-zero even length } {aa, bb, ab, ba, abab, aaaa,... }

6 4/32 Σ : alphabet {a, b} L Σ Σ : words : language {ε, a, b, aa, ab, ba, bb, aab,... } property over words L 1 := {set of words starting with an a } {a, aa, ab, aaa, aab,... } L 2 := {set of words with a non-zero even length } {aa, bb, ab, ba, abab, aaaa,... } Finite automata, pushdown automata, Turing machines,...

7 5/32 Σ : alphabet {a, b} T Σ : timed words a a a b b π (aa; 0.8, 2.5) (abb; π, 203, 312.3)

8 5/32 Σ : alphabet {a, b} T Σ : timed words a a a b b π (aa; 0.8, 2.5) (abb; π, 203, 312.3) Word w = a 1... a n a i Σ (w, τ) Time sequence τ = τ 1... τ n τ i R 0 τ 1 τ n

9 6/32 L T Σ : Timed language property over timed words L 1 := {( ab(a + b), τ ) τ 2 τ 1 = 1} a b ab b a b a b b L 2 := { (w, τ) τ i+1 τ i 2 for all i < w } a b a a b a

10 6/32 L T Σ : Timed language property over timed words L 1 := {( ab(a + b), τ ) τ 2 τ 1 = 1} a b ab b a b a b b L 2 := { (w, τ) τ i+1 τ i 2 for all i < w } a b a a b a Timed automata

11 7/32

12 Timed automata 7/32

13 Timed automaton: Finite automaton + Finite no. of Clocks 8/32 Clock 0 time

14 Timed automaton: Finite automaton + Finite no. of Clocks 8/32 Clock 0 time {( ab(a + b), τ) τ 2 2} a b q 0 q 1 q 2 a, b

15 Timed automaton: Finite automaton + Finite no. of Clocks 8/32 Clock 0 time {( ab(a + b), τ) τ 2 2} a x 2, b q 0 q 1 q 2 a, b

16 Timed automaton: Finite automaton + Finite no. of Clocks 8/32 Clock 0 time {( ab(a + b), τ) τ 2 2} a x 2, b q 0 q 1 q 2 a, b a b b a b b q 0 q 1 q 2 q 0 q 1 accept reject

17 Timed automaton: Finite automaton + Finite no. of Clocks 8/32 Clock Guards φ := x c x c φ φ φ x Clocks, c Q 0 0 time {( ab(a + b), τ) τ 2 2} a x 2, b q 0 q 1 q 2 a, b a b b a b b q 0 q 1 q 2 q 0 q 1 accept reject

18 Timed automaton: Finite automaton + Finite no. of Clocks 8/32 Clock Guards φ := x c x c φ φ φ x Clocks, c Q 0 0 time {( ab(a + b), τ) τ 2 τ 1 2} a x 2, b q 0 q 1 q 2 a, b

19 Timed automaton: Finite automaton + Finite no. of Clocks 8/32 Clock Guards φ := x c x c φ φ φ 0 time x Clocks, c Q 0 Resets {( ab(a + b), τ) τ 2 τ 1 2} q 0 a q 1 x 2, b q 2 {x} a, b

20 Timed automaton: Finite automaton + Finite no. of Clocks 8/32 Clock Guards φ := x c x c φ φ φ 0 time x Clocks, c Q 0 Resets {( ab(a + b), τ) τ 2 τ 1 2} q 0 a q 1 x 2, b q 2 {x} a, b a b b a bb q 0 q 1 x : 0 accept q 2 x 2 q 0 q 1 x : 0 reject x > 2

21 9/32 L 3 := { ( a k, τ ) k > 0, τ i = i for all i k} An a occurs in every integer from 1,..., k a a a a a

22 9/32 L 3 := { ( a k, τ ) k > 0, τ i = i for all i k} An a occurs in every integer from 1,..., k a a a a a q 0 x = 1, a q 1 {x} x = 1, a {x}

23 10/32 L 4 := { ( a k, τ ) exist i, j s.t. τ j τ i = 1} There are 2 a s which are at distance 1 apart a a a a a a a 0 t t + 1

24 10/32 L 4 := { ( a k, τ ) exist i, j s.t. τ j τ i = 1} There are 2 a s which are at distance 1 apart a a a a a a a 0 t t + 1 a a a q 0 a q 1 x = 1, a q 2 {x}

25 11/32 Three mechanisms to exploit: Reset: to start measuring time Guard: to impose time constraint on action Non-determinism: for existential time constraints

26 12/32 s 2 b, (y = 1) c, (x < 1) c, (x < 1) a, {y} s 0 s 1 s 3 d, (x > 1) A = (Q, Σ, X, T, Q 0, F) T Q Σ guard reset Q a, (y < 1), {y}

27 12/32 (ac; 0.4, 0.9) s 2 b, (y = 1) c, (x < 1) c, (x < 1) a, {y} s 0 s 1 s 3 d, (x > 1) A = (Q, Σ, X, T, Q 0, F) T Q Σ guard reset Q a, (y < 1), {y} x y s s 0 s 1 s a 0.5 c s

28 12/32 (ac; 0.4, 0.9) s 2 b, (y = 1) c, (x < 1) c, (x < 1) a, {y} s 0 s 1 s 3 d, (x > 1) A = (Q, Σ, X, T, Q 0, F) T Q Σ guard reset Q a, (y < 1), {y} x y s s 0 s 1 s a 0.5 c s Run of A over (a 1 a 2... a k ; τ 1 τ 2... τ k ) δ i := τ i τ i 1 ; τ 0 := 0 (q 0, v 0 ) δ 1 (q0, v 0 + δ 1 ) a 1 (q1, v 1 ) δ 2 (q1, v 1 + δ 2 ) (w, τ) L(A) if A has an accepting run over (w, τ) a k (qk, v k )

29 13/32 Timed automata Runs

30 14/32 L 5 := { ( abcd.σ, τ ) τ 3 τ 1 2 and τ 4 τ 2 5} Interleaving distances a b c d

31 14/32 L 5 := { ( abcd.σ, τ ) τ 3 τ 1 2 and τ 4 τ 2 5} Interleaving distances a b c d Σ q 0 a q 1 b q 2 x 1, c q 3 y 5, d q 4 {x} {y}

32 14/32 L 5 := { ( abcd.σ, τ ) τ 3 τ 1 2 and τ 4 τ 2 5} Interleaving distances a b c d Σ q 0 a q 1 b q 2 x 1, c q 3 y 5, d q 4 {x} {y} Exercise: Prove that L 5 cannot be accepted by a one-clock TA.

33 15/32 n interleavings need n clocks n + 1 clocks more expressive than n clocks

34 16/32 Timed automata Runs 1 clock < 2 clocks <...

35 17/32 L 6 := { ( a k, τ ) τ i is some integer for each i} a a a

36 17/32 L 6 := { ( a k, τ ) τ i is some integer for each i} a a a Claim: No timed automaton can accept L 6

37 18/32 Step 1: Suppose L 6 = L(A) Let c max be the maximum constant appearing in a guard of A

38 18/32 Step 1: Suppose L 6 = L(A) Let c max be the maximum constant appearing in a guard of A Step 2: For a clock x, x = c max + 1 and x = c max satisfy the same guards

39 18/32 Step 1: Suppose L 6 = L(A) Let c max be the maximum constant appearing in a guard of A Step 3: Step 2: For a clock x, x = c max + 1 and x = c max satisfy the same guards (a; c max + 1) L 6 and so A has an accepting run (q 0, v 0 ) δ = cmax +1 (q 0, v 0 + δ) a (q F, v F )

40 18/32 Step 1: Suppose L 6 = L(A) Let c max be the maximum constant appearing in a guard of A Step 3: Step 2: For a clock x, x = c max + 1 and x = c max satisfy the same guards (a; c max + 1) L 6 and so A has an accepting run (q 0, v 0 ) δ = cmax +1 (q 0, v 0 + δ) a (q F, v F ) Step 4: By Step 2, the following is an accepting run (q 0, v 0 ) δ = c max +1.1 (q 0, v 0 + δ ) a (q F, v F )

41 Step 1: Suppose L 6 = L(A) Let c max be the maximum constant appearing in a guard of A Step 3: Step 2: For a clock x, x = c max + 1 and x = c max satisfy the same guards (a; c max + 1) L 6 and so A has an accepting run (q 0, v 0 ) δ = cmax +1 (q 0, v 0 + δ) a (q F, v F ) Step 4: By Step 2, the following is an accepting run (q 0, v 0 ) δ = c max +1.1 (q 0, v 0 + δ ) a (q F, v F ) Hence (a; c max + 1.1) L(A) L 6 Therefore no timed automaton can accept L 6 18/32

42 19/32 L 7 = { ( (ab) k, τ ) τ 2i+2 τ 2i+1 < τ 2i τ 2i 1 for each i 1} Converging ab distances a a a b b b

43 19/32 L 7 = { ( (ab) k, τ ) τ 2i+2 τ 2i+1 < τ 2i τ 2i 1 for each i 1} Converging ab distances a a a b b b Exercise: Prove that no timed automaton can accept L 7

44 L 7 = { ( (ab) k, τ ) τ 2i = i and τ 2i+2 τ 2i+1 < τ 2i τ 2i 1 } Pivoted converging ab distances a b a b a b a b

45 L 7 = { ( (ab) k, τ ) τ 2i = i and τ 2i+2 τ 2i+1 < τ 2i τ 2i 1 } Pivoted converging ab distances > 1 a b a b a b a b

46 L 7 = { ( (ab) k, τ ) τ 2i = i and τ 2i+2 τ 2i+1 < τ 2i τ 2i 1 } Pivoted converging ab distances > 1 a b a b a b a b τ 2i+2 τ 2i+1 < τ 2i τ 2i 1 τ 2i+2 τ 2i < τ 2i+1 τ 2i 1 1 < τ 2i+1 τ 2i 1

47 20/32 L 7 = { ( (ab) k, τ ) τ 2i = i and τ 2i+2 τ 2i+1 < τ 2i τ 2i 1 } Pivoted converging ab distances > 1 a b a b a b a b τ 2i+2 τ 2i+1 < τ 2i τ 2i 1 τ 2i+2 τ 2i < τ 2i+1 τ 2i 1 1 < τ 2i+1 τ 2i 1 q 0 a y = 1, b q 1 {x} {y} q 2 x > 1, a {x}

48 21/32 Timed automata Runs 1 clock < 2 clocks <... Role of max constant

49 21/32 Timed automata Timed regular lngs. Runs 1 clock < 2 clocks <... Role of max constant

50 22/32 Timed regular languages Timed languages L L(A) L L Timed regular languages L = L(A) Definition A timed language is called timed regular if it can be accepted by a timed automaton

51 23/32 L L = L(A ) L L L = L(A) L L L = L(A ) A = (Q, Σ, X, T, Q 0, F) A = (Q, Σ, X, T, Q 0, F ) A = ( Q Q, Σ, X X, T T, Q 0 Q 0, F F ) L(A) L(A ) = L(A ) Timed regular languages are closed under union

52 24/32 L L = L(A ) L L L = L(A) L L L = L(A ) A = (Q, Σ, X, T, Q 0, F) A = (Q, Σ, X, T, Q 0, F ) A = ( Q Q, Σ, X X, T, Q 0 Q 0, F F ) T : (q 1, q 1 ) a, g g (q 2, q 2 ) if R R a, g q 1 q 2 T and q 1 a, g q 2 T R R Timed regular languages are closed under intersection

53 25/32 L : a timed language over Σ Untime(L) {w Σ τ. (w, τ) L} Untiming construction For every timed automaton A there is a finite automaton A u s.t. Untime( L(A) ) = L(A u ) more about this in Lecture 6...

54 26/32 Complementation Σ : {a, b} L = { (w, τ) there is an a at some time t and no action occurs at time t + 1 } L = { (w, τ) every a has an action at a distance 1 from it }

55 26/32 Complementation Σ : {a, b} L = { (w, τ) there is an a at some time t and no action occurs at time t + 1 } L = { (w, τ) every a has an action at a distance 1 from it } Claim: No timed automaton can accept L Decision problems for timed automata: A survey Alur, Madhusudhan. SFM 04: RT

56 27/32 Step 1: L = { (w, τ) every a has an action at a distance 1 from it } Suppose L is timed regular

57 27/32 Step 1: L = { (w, τ) every a has an action at a distance 1 from it } Suppose L is timed regular Step 2: Let L = { (a b, τ) all a s occur before time 1 and no two a s happen at same time } Clearly L is timed regular

58 27/32 Step 1: L = { (w, τ) every a has an action at a distance 1 from it } Suppose L is timed regular Step 2: Let L = { (a b, τ) all a s occur before time 1 and no two a s happen at same time } Clearly L is timed regular Step 3: Untime( L L ) should be a regular language

59 27/32 Step 1: L = { (w, τ) every a has an action at a distance 1 from it } Suppose L is timed regular Step 2: Let L = { (a b, τ) all a s occur before time 1 and no two a s happen at same time } Clearly L is timed regular Step 3: Untime( L L ) should be a regular language Step 4: But, Untime( L L ) = {a n b m m n}, not regular!

60 27/32 Step 1: L = { (w, τ) every a has an action at a distance 1 from it } Suppose L is timed regular Step 2: Let L = { (a b, τ) all a s occur before time 1 and no two a s happen at same time } Clearly L is timed regular Step 3: Untime( L L ) should be a regular language Step 4: But, Untime( L L ) = {a n b m m n}, not regular! Therefore L cannot be timed regular

61 28/32 L L Timed regular languages are not closed under complementation

62 29/32 Timed automata Runs 1 clock < 2 clocks <... Role of max constant Timed regular lngs. Closure under, Non-closure under complement

63 29/32 Timed automata Runs 1 clock < 2 clocks <... Role of max constant Timed regular lngs. Closure under, Non-closure under complement ε-transitions

64 30/32 L 6 := { ( a k, τ ) τ i is some integer for each i} a a a Claim: No timed automaton can accept L 6

65 30/32 L 6 := { ( a k, τ ) τ i is some integer for each i} a ε a ε ε ε a

66 30/32 L 6 := { ( a k, τ ) τ i is some integer for each i} a ε a ε ε ε a x = 1, ε, {x} q 0 x = 1, a, {x}

67 31/32 ε-transitions ε-transitions add expressive power to timed automata. Characterization of the expressive power of silent transitions in timed automata Bérard, Diekert, Gastin, Petit. Fundamenta Informaticae 98

68 31/32 ε-transitions ε-transitions add expressive power to timed automata. However, they add power only when a clock is reset in an ε-transition. Characterization of the expressive power of silent transitions in timed automata Bérard, Diekert, Gastin, Petit. Fundamenta Informaticae 98

69 32/32 Timed automata Runs 1 clock < 2 clocks <... Role of max constant Timed regular lngs. Closure under, Non-closure under complement ε-transitions More expressive ε without reset TA

70 32/32 Timed automata Runs 1 clock < 2 clocks <... Role of max constant Timed regular lngs. Closure under, Non-closure under complement ε-transitions More expressive ε without reset TA Next lecture: Determinization