Automata for Real-time Systems

Automata for Real-time Systems

B. Srivathsan

Chennai Mathematical Institute

Lecture 2:

Timed languages and timed

automata

L₅:={(abcd.Σ^∗, τ )|τ₃−τ₁ ≤2 andτ₄−τ₂ ≥5}

Interleaving distances

0 1 2 3 4 5 6 7

a b c d

q₀ a q₁ q₂ q₃ q₄

{x}

b {y}

x≤1,c y≥5,d Σ

Exercise:Prove thatL5cannot be accepted by a one-clock TA.

ninterleavings⇒neednclocks

n+1 clocks more expressive thannclocks

{(a^k, τ)|τ_i+2−τ_i≤1 for alli≤k−2}

q0 q1 q2

a {x}

a,y≤1

a,x≤1 {y} {x}

{(a^k, τ)|τ_i+2−τ_i≤1 for alli≤k−2}

q0 q1 q2

a {x}

a,y≤1

a,x≤1 {y}

{x}

Timed automata

Runs

1 clock<2 clocks< . . .

L6:={(a^k, τ )|τ_iis some integer for eachi}

0 1 2 3 4 5 6 7

a a a

Claim: No timed automatoncan acceptL6

L6:={(a^k, τ )|τ_iis some integer for eachi}

0 1 2 3 4 5 6 7

a a a

Claim: No timed automatoncan acceptL6

Step 1: SupposeL6=L(A)

Letcmaxbe the maximum constant appearing in a guard ofA

Step 2: For a clockx, x=dcmaxe+1 and x=dcmaxe+1.1

satisfy the same guards

Step 3: (a; dcmaxe+1)∈L6and soAhas an accepting run (q0,v0)−−−−−−−−−→^δ=dcmaxe+1 (q0,v0+δ)−→^a (qF,vF)

Step 4: By Step 2, the following is an accepting run (q0,v0) ^δ

0=dcmaxe+1.1

−−−−−−−−−−→(q0,v0+δ⁰)−→^a (qF,v⁰_F) Hence(a; dcmaxe+1.1)∈L(A)6=L6

Thereforeno timed automatoncan acceptL6

Step 1: SupposeL6=L(A)

Letcmaxbe the maximum constant appearing in a guard ofA Step 2: For a clockx,

x=dcmaxe+1 and x=dcmaxe+1.1 satisfy the same guards

Step 3: (a; dcmaxe+1)∈L6and soAhas an accepting run (q0,v0)−−−−−−−−−→^δ=dcmaxe+1 (q0,v0+δ)−→^a (qF,vF)

Step 4: By Step 2, the following is an accepting run (q0,v0) ^δ

0=dcmaxe+1.1

−−−−−−−−−−→(q0,v0+δ⁰)−→^a (qF,v⁰_F) Hence(a; dcmaxe+1.1)∈L(A)6=L6

Thereforeno timed automatoncan acceptL6

Step 1: SupposeL6=L(A)

Letcmaxbe the maximum constant appearing in a guard ofA Step 2: For a clockx,

x=dcmaxe+1 and x=dcmaxe+1.1 satisfy the same guards

Step 3: (a; dcmaxe+1)∈L6and soAhas an accepting run (q0,v0)−−−−−−−−−→^δ=dcmaxe+1 (q0,v0+δ)−→^a (qF,vF)

Step 4: By Step 2, the following is an accepting run (q0,v0) ^δ

0=dcmaxe+1.1

−−−−−−−−−−→(q0,v0+δ⁰)−→^a (qF,v⁰_F) Hence(a; dcmaxe+1.1)∈L(A)6=L6

Thereforeno timed automatoncan acceptL6

Step 1: SupposeL6=L(A)

Letcmaxbe the maximum constant appearing in a guard ofA Step 2: For a clockx,

x=dcmaxe+1 and x=dcmaxe+1.1 satisfy the same guards

Step 3: (a; dcmaxe+1)∈L6and soAhas an accepting run (q0,v0)−−−−−−−−−→^δ=dcmaxe+1 (q0,v0+δ)−→^a (qF,vF)

Step 4: By Step 2, the following is an accepting run (q0,v0) ^δ

0=dcmaxe+1.1

−−−−−−−−−−→(q0,v0+δ⁰)−→^a (qF,v⁰_F)

Hence(a; dcmaxe+1.1)∈L(A)6=L6

Thereforeno timed automatoncan acceptL6

Step 1: SupposeL6=L(A)

Letcmaxbe the maximum constant appearing in a guard ofA Step 2: For a clockx,

x=dcmaxe+1 and x=dcmaxe+1.1 satisfy the same guards

Step 3: (a; dcmaxe+1)∈L6and soAhas an accepting run (q0,v0)−−−−−−−−−→^δ=dcmaxe+1 (q0,v0+δ)−→^a (qF,vF)

Step 4: By Step 2, the following is an accepting run (q0,v0) ^δ

0=dcmaxe+1.1

−−−−−−−−−−→(q0,v0+δ⁰)−→^a (qF,v⁰_F) Hence(a; dcmaxe+1.1)∈L(A)6=L6

Thereforeno timed automatoncan acceptL6

L7={( (ab)^k, τ )|τ_2i+2−τ_2i+1< τ_2i−τ_2i−1for eachi ≥1}

Convergingabdistances

0 1 2 3 4 5 6 7

a b a b a b

Exercise:Prove thatno timed automatoncan acceptL7

L7={( (ab)^k, τ )|τ_2i+2−τ_2i+1< τ_2i−τ_2i−1for eachi ≥1}

Convergingabdistances

0 1 2 3 4 5 6 7

a b a b a b

Exercise:Prove thatno timed automatoncan acceptL7

L7={( (ab)^k, τ )|τ_2i=i and τ_2i+2−τ_2i+1 < τ_2i−τ_2i−1}

Pivoted convergingabdistances

>1

0 1 2 3 4

b b b b

a a a a

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

q₀ a q₁ q₂

{x}

y=1,b {y} x>1,a

{x}

L7={( (ab)^k, τ )|τ_2i=i and τ_2i+2−τ_2i+1 < τ_2i−τ_2i−1}

Pivoted convergingabdistances

>1

0 1 2 3 4

b b b b

a a a a

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

q₀ a q₁ q₂

{x}

y=1,b {y} x>1,a

{x}

L7={( (ab)^k, τ )|τ_2i=i and τ_2i+2−τ_2i+1 < τ_2i−τ_2i−1}

Pivoted convergingabdistances

>1

0 1 2 3 4

b b b b

a a a a

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

q₀ a q₁ q₂

{x}

y=1,b {y} x>1,a

{x}

L7={( (ab)^k, τ )|τ_2i=i and τ_2i+2−τ_2i+1 < τ_2i−τ_2i−1}

Pivoted convergingabdistances

>1

0 1 2 3 4

b b b b

a a a a

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

q₀ a q₁ q₂

{x}

y=1,b {y}

x>1,a {x}

Timed automata

Runs

1 clock<2 clocks< . . . Role of max constant

Timed regular lngs.

Timed automata

Runs

1 clock<2 clocks< . . . Role of max constant

Timed regular lngs.

Timed regular languages

Timed languages

L⁰6=L(A)

Timed regular languages L=L(A)

L⁰

L

Definition

A timed language is calledtimed regularif it can beacceptedby a timed automaton

L=L(A)

L

L⁰ L⁰=L(A⁰)

L∪L⁰

L∪L⁰=L(A∪)

A= (Q,Σ,X,T,Q0,F) A⁰= (Q⁰,Σ,X⁰,T⁰,Q⁰₀,F⁰)

A∪= (Q∪Q⁰, Σ, X∪X⁰, T∪T⁰, Q₀∪Q⁰₀, F∪F⁰) L(A)∪ L(A⁰) =L(A∪)

Timed regular languages areclosedunderunion

L=L(A)

L

L⁰ L⁰=L(A⁰)

L∩L⁰

L∩L⁰=L(A∩)

A= (Q,Σ,X,T,Q₀,F) A⁰= (Q⁰,Σ,X⁰,T⁰,Q⁰₀,F⁰) A∩= (Q×Q⁰, Σ, X∪X⁰, T∩, Q₀×Q⁰₀, F ×F⁰)

T∩: (q1,q⁰₁) ^a,g∧g

0

−−−−−−−→(q2,q₂⁰)if

R∪R⁰

q1 a,g

−−−−→q2∈T and q⁰₁ ^a,g

0

−−−−→q⁰₂∈T⁰

R R⁰

L : a timed language overΣ Untime(L) ≡ {w∈Σ^∗ | ∃τ.(w, τ)∈L}

Untiming construction

Forevery timedautomatonAthere is afinite automatonAus.t.

Untime(L(A) ) =L(Au)

more about this later. . .

Complementation

Σ :{a,b}

L = {(w, τ)| there is anaat some timetand no action occurs at timet+1}

L = {(w, τ)| everyahas an action at a distance 1 from it}

Claim: No timed automatoncan acceptL

Decision problems for timed automata: A survey

Alur, Madhusudhan.SFM’04: RT

Complementation

Σ :{a,b}

L = {(w, τ)| there is anaat some timetand no action occurs at timet+1}

L = {(w, τ)| everyahas an action at a distance 1 from it}

Claim: No timed automatoncan acceptL

Decision problems for timed automata: A survey

Alur, Madhusudhan.SFM’04: RT

Step 1: L = {(w, τ)| everyahas an action at a distance 1 from it} SupposeLis timed regular

Step 2: LetL⁰ = {(a^∗b^∗, τ)| alla’s occur before time 1 and no twoa’s happen at same time} ClearlyL⁰is timed regular

Step 3: Untime(L∩L⁰ )should be a regular language

Step 4: But, Untime(L∩L⁰) ={aⁿb^m|m≥n},not regular!

ThereforeLcannot be timed regular

Step 1: L = {(w, τ)| everyahas an action at a distance 1 from it} SupposeLis timed regular

Step 2: LetL⁰ = {(a^∗b^∗, τ)| alla’s occur before time 1 and no twoa’s happen at same time} ClearlyL⁰is timed regular

Step 3: Untime(L∩L⁰ )should be a regular language

Step 4: But, Untime(L∩L⁰) ={aⁿb^m|m≥n},not regular!

ThereforeLcannot be timed regular

Step 1: L = {(w, τ)| everyahas an action at a distance 1 from it} SupposeLis timed regular

Step 2: LetL⁰ = {(a^∗b^∗, τ)| alla’s occur before time 1 and no twoa’s happen at same time} ClearlyL⁰is timed regular

Step 3: Untime(L∩L⁰ )should be a regular language

Step 4: But, Untime(L∩L⁰) ={aⁿb^m|m≥n},not regular!

ThereforeLcannot be timed regular

Step 1: L = {(w, τ)| everyahas an action at a distance 1 from it} SupposeLis timed regular

Step 2: LetL⁰ = {(a^∗b^∗, τ)| alla’s occur before time 1 and no twoa’s happen at same time} ClearlyL⁰is timed regular

Step 3: Untime(L∩L⁰ )should be a regular language

Step 4: But, Untime(L∩L⁰ ) ={aⁿb^m|m≥n},not regular!

ThereforeLcannot be timed regular

Step 1: L = {(w, τ)| everyahas an action at a distance 1 from it} SupposeLis timed regular

Step 2: LetL⁰ = {(a^∗b^∗, τ)| alla’s occur before time 1 and no twoa’s happen at same time} ClearlyL⁰is timed regular

Step 3: Untime(L∩L⁰ )should be a regular language

Step 4: But, Untime(L∩L⁰ ) ={aⁿb^m|m≥n},not regular!

L

L

Timed regular languages arenot closedundercomplementation

Timed automata

Runs

1 clock<2 clocks< . . . Role of max constant

Timed regular lngs.

Closure under∪,∩

Non-closure under complement

ε -transitions

Timed automata

Runs

1 clock<2 clocks< . . . Role of max constant

Timed regular lngs.

Closure under∪,∩

Non-closure under complement

ε -transitions

L6:={(a^k, τ )|τ_iis some integer for eachi}

ε ε ε ε

0 1 2 3 4 5 6 7

a a a

q0

x=1, ε, {x}

x=1,a,{x}

L6:={(a^k, τ )|τ_iis some integer for eachi}

ε ε ε ε

0 1 2 3 4 5 6 7

a a a

q0

x=1, ε, {x}

x=1,a,{x}

Claim: No timed automatoncan acceptL6

L6:={(a^k, τ )|τ_iis some integer for eachi}

ε ε ε ε

0 1 2 3 4 5 6 7

a a a

q0

x=1, ε, {x}

x=1,a,{x}

Claim: No timed automatoncan acceptL6

ε-transitions

ε-transitionsadd expressive powerto timed automata.

However, they add poweronlywhen a clock isresetin anε-transition.

Characterization of the expressive power of silent transitions in timed automata

Bérard, Diekert, Gastin, Petit.Fundamenta Informaticae’98

ε-transitions

ε-transitionsadd expressive powerto timed automata. However, they add poweronlywhen a clock isresetin anε-transition.

Characterization of the expressive power of silent transitions in timed automata

Bérard, Diekert, Gastin, Petit.Fundamenta Informaticae’98

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

Recall...

Huge system Property

Higher-level description Higher-level description

AutomatonA AutomatonB

translation translation

Model-Checker

L(A)⊆ L(B)?

L(A)⊆ L(B) iff

L(A) ∩ L(B) =∅

non-closure under complement⇒the abovecannot be donefor TA!

L(A)⊆ L(B) iff

L(A) ∩ L(B) =∅

non-closure under complement⇒the abovecannot be donefor TA!

Course plan

Timed automaton

Emptiness L(A) =∅?

Inclusion L(A)⊆ L(B)?

Decidable Complexity Better algos

Undecidable Variations of TA for decidability

One-clock Alternation Event-clock, Integer reset L.1

L.2

L.3 L.4

L.5, L.6 L.7, L.8 L.9, L.10 L.11-12

L.13-18

Special topics

Diagonal constraints Infinite timed words Example from industry L.19

L.20-21 L.22

Course plan

Timed automaton

Emptiness L(A) =∅?

Inclusion L(A)⊆ L(B)?

Decidable Complexity Better algos

Undecidable Variations of TA for decidability

One-clock Alternation Event-clock, Integer reset L.1

L.2

L.3 L.4

L.5, L.6 L.7, L.8 L.9, L.10 L.11-12

L.13-18

Special topics

Diagonal constraints Infinite timed words Example from industry L.19

L.20-21 L.22