The Complexity of Satisfiability for Fragments of CTL ...

Introduction

Results

Conclusion

The Complexity of Satisfiability for Fragments of CTL and CTL? Arne Meier1

Martin Mundhenk2 Michael Thomas1 Heribert Vollmer1 1 Institut

f¨ ur Theoretische Informatik Leibniz Universit¨ at Hannover

2 Institut f¨ ur Informatik Friedrich-Schiller-Universit¨ at Jena

09. September 2008

1/25

Arne Meier, Martin Mundhenk, Michael Thomas, Heribert Vollmer The Complexity of Satisfiability for Fragments of CTL and CTL?

Introduction

Results

Conclusion

Introduction

Results Restricting the CTL-Operators Restricting the Boolean functions

Conclusion

2/25


Introduction

Results

Conclusion

CTL-Syntax and -Semantics

Path Quantifier P ∈ {A, E} Temporal Operators T ∈ {X, F, G, U}

(all, exists) (next, future, global, until)

CTL-formulae p, p ∧ q, p ∨ q, PT ϕ, P[ϕUψ] CTL? -formulae all CTL-formulae, Pχ, T χ where p, q propositions, ϕ, ψ CTL-formulae, χ CTL? -formula.

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Introduction

Results

Conclusion

Satisfiability in CTL

Input encoded CTL-formulae hϕi. Question Is there a M = (S, R, l), s ∈ S, s.t. M, s |= ϕ? Notion ϕ ∈ CTL-SAT(T , B)

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Introduction

Results

Conclusion

Outline

Introduction


Conclusion

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Introduction

Results

Conclusion

Complexity of CTL-SAT w.r.t. CTL-operators AX, AF, EU

AF, EU

AG, AU

AX, AU

AX, EU AX, AF, AG

AX, AF

AF, AG

AX, AG

AU

EU

AF

AX

AG

EXPTIME-complete ∅

PSPACE-complete NP-complete

6/25


Introduction

Results

Conclusion

NP-complete Fragments Proof Ideas

CTL-SAT(∅): SAT NP-complete CTL-SAT({AF}): formulae are satisfiable in polynomial-size models (follows from induction over |ϕ|), hence guess model and verify it

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Introduction

Results

Conclusion


AF, EU

AG, AU

AX, AU

AX, EU AX, AF, AG

AX, AF

AF, AG

AX, AG

AU

EU

AF

AX

AG



8/25


Introduction

Results

Conclusion

PSPACE-complete Fragments Proof Idea for CTL-SAT({AX}) (1)

Hardness-Proof Reduction from QBF-3SAT. Idea: Force a model that encodes an assignment-tree. Di = (AX)(n) (qi → (q0 ∧ · · · ∧ qi−1 ∧ qi+1 ∧ · · · ∧ qn ))

(Idea from T. Schneider, 2002) 9/25


Introduction

Results

Conclusion


Hardness-Proof Reduction from QBF-3SAT. Idea: Force a model that encodes an assignment-tree. Di = (AX)(n) (qi → (q0 ∧ · · · ∧ qi−1 ∧ qi+1 ∧ · · · ∧ qn )) Bi = qi → EX qi+1 ∧ xi+1 ∧ C(xi+1 ) ∧ EX qi+1 ∧ xi+1 ∧ C(xi+1 )



Introduction

Results

Conclusion


Hardness-Proof Reduction from QBF-3SAT. Idea: Force a model that encodes an assignment-tree. Di = (AX)(n) (qi → (q0 ∧ · · · ∧ qi−1 ∧ qi+1 ∧ · · · ∧ qn )) Bi = qi → EX qi+1 ∧ xi+1 ∧ C(xi+1 ) ∧ EX qi+1 ∧ xi+1 ∧ C(xi+1 ) Si = xi ∧ C(xi ) → AX xi ∧ C(xi ) ∧ ∧ xi ∧ C(xi ) → AX xi ∧ C(xi )



Introduction

Results

Conclusion


reduction function for QBF-3SAT ≤Pm CTL-SAT

Q1 x1 Q2 x2 . . . Qn xn F 7→

∧

∧

∧

∧

∧

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Introduction

Results

Conclusion



Q1 x1 Q2 x2 . . . Qn xn F 7→ q0 ∧

∧

∧

∧

∧

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Introduction

Results

Conclusion



Q1 x1 Q2 x2 . . . Qn xn F 7→ q0 ∧

n−1 ^

Di ∧

∧

∧

i=0

∧ Di : Level subformula

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Introduction

Results

Conclusion



Q1 x1 Q2 x2 . . . Qn xn F 7→ q0 ∧

n−1 ^

Di ∧

i=0

n ^

(AX)i Bi ∧

∧

i=0

∧ Di : Level subformula Bi : Branching subformula

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Introduction

Results

Conclusion



Q1 x1 Q2 x2 . . . Qn xn F 7→ q0 ∧

n−1 ^

Di ∧

i=0

n ^ i=0

(AX)i Bi ∧

n−1 ^ ^ n−1

(AX)j Si ∧

i=1 j=i

∧ Di : Level subformula Bi : Branching subformula Si : Invariant subformula

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Introduction

Results

Conclusion



Q1 x1 Q2 x2 . . . Qn xn F 7→ q0 ∧

n−1 ^

Di ∧

i=0

n ^ i=0

(AX)i Bi ∧

n−1 ^ ^ n−1

(AX)j Si ∧

i=1 j=i

∧ Q1 Q2 . . . Qn (C1 ∧ · · · ∧ Cm ) Di : Level subformula Bi : Branching subformula Si : Invariant subformula Qi : EX if Qi = ∃, else AX. 10/25


Introduction

Results

Conclusion


Membership in PSPACE classify four sets of propositions and search for contradictions

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Introduction

Results

Conclusion


Membership in PSPACE classify four sets of propositions and search for contradictions • satisfied prop. in a state,

(p)

(Idea from Ladner, 1977)

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Introduction

Results

Conclusion



(p)

• falsified prop. in a state,

(¬p)


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Introduction

Results

Conclusion



(p)


(¬p)

• satisfied prop. in all successive states,

(AXp)


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Introduction

Results

Conclusion



(p)


(¬p)

• satisfied prop. in all successive states,

(AXp)

• satisfied prop. in one successive state.

(EXp)

Depth of recursion in the algorithm is ≤ O(|ϕ|). (Idea from Ladner, 1977)

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Introduction

Results

Conclusion

PSPACE-complete Fragments Proof Ideas

CTL-SAT({AG}): almost a similar proof as for the AX-case

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Introduction

Results

Conclusion


CTL-SAT({AG}): almost a similar proof as for the AX-case CTL-SAT({AF, AG}): argumentation over so-called quasi models, main idea: guess minimal quasi model and verify it; if there is a contradiction, then it can be found within linear depth; hardness follows from CTL-SAT({AG})

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Introduction

Results

Conclusion


CTL-SAT({AG}): almost a similar proof as for the AX-case CTL-SAT({AF, AG}): argumentation over so-called quasi models, main idea: guess minimal quasi model and verify it; if there is a contradiction, then it can be found within linear depth; hardness follows from CTL-SAT({AG}) CTL-SAT({AX, AF}): problem exp. modelsize; two-steps algorithm with generalization of quasi models: 1. replace EGψ ∈ ϕ with ψ, and verify obtained formula ϕ0 2. for the EG-case use the fixpoint-characterisation of EGπ and test each subformula seperately 12/25


Introduction

Results

Conclusion


AF, EU

AG, AU

AX, AU

AX, EU AX, AF, AG

AX, AF

AF, AG

AX, AG

AU

EU

AF

AX

AG



13/25


Introduction

Results

Conclusion

EXPTIME-complete Fragments Proof Ideas

CTL-SAT({AX, AG}) generic reduction from PSPACE-ATMs

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Introduction

Results

Conclusion

EXPTIME-complete Fragments Proof Ideas

CTL-SAT({AX, AG}) generic reduction from PSPACE-ATMs

CTL-SAT({EU}),CTL-SAT({AU}) adapted the reduction for CTL-SAT({AX, AG}) to work only with EU, resp., AU

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Introduction

Results

Conclusion

Complexity of CTL-SAT w.r.t. Boolean functions BF R1

R0 R2

M M1

M0 M2

S20 S30

S21 S202

S201

S302

S301

S02

S01

S200

S210

S300

S0

D2

V V1

L1

V2

L3

S311

S312

S11

S12

S31

S1

S10

L V0

S212

S310

D D1

S00

S211

E L0

L2

E1

E0 E2

N N2

Complexity results for CTL: EXPTIME-complete in EXPTIME NC1 -complete TC0 -complete

I I1

I0 I2

Complexity results for CTL? : 2EXPTIME-complete in 2EXPTIME NC1 -complete TC0 -complete

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Introduction

Results

Conclusion

Complexity of CTL-SAT w.r.t. Boolean functions Lower Bound in CTL-SAT(S1 )

Theorem All results for restricting the operators hold for each set of Boolean functions B with S1 ⊆ [B].

Proof. 1. [B ∪ {1}] = [S1 ∪ {1}] = BF.

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Introduction

Results

Conclusion



Proof. 1. [B ∪ {1}] = [S1 ∪ {1}] = BF. 2. Replace all connectives with short [B ∪ {1}] representations.

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Introduction

Results

Conclusion



Proof. 1. [B ∪ {1}] = [S1 ∪ {1}] = BF. 2. Replace all connectives with short [B ∪ {1}] representations. 3. Replace all 1s with a new introduced variable t.

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Introduction

Results

Conclusion



Proof. 1. [B ∪ {1}] = [S1 ∪ {1}] = BF. 2. Replace all connectives with short [B ∪ {1}] representations. 3. Replace all 1s with a new introduced variable t. 4. For each ψ ∈ SF(ϕ) write ψ ∧ t.

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Introduction

Results

Conclusion

Complexity of CTL-SAT w.r.t. Boolean functions Testing the Syntactical Correctness

Theorem Determining for given w ∈ Σ? and set of Boolean functions B, if 0 w ∈ CTL(B) is TC0 -complete under ≤AC m -reductions.

Proof of Hardness.

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Introduction

Results

Conclusion



Proof of Hardness. 1. Reduction from MAJ = {w ∈ {0, 1}? | |w |1 > |w |0 }

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Introduction

Results

Conclusion



Proof of Hardness. 1. Reduction from MAJ = {w ∈ {0, 1}? | |w |1 > |w |0 } 2. |w |1 > |w |0 iff there is a 0 ≤ k ≤ n, s.t. |w |1 = |w 0k |0

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Introduction

Results

Conclusion



Proof of Hardness. 1. Reduction from MAJ = {w ∈ {0, 1}? | |w |1 > |w |0 } 2. |w |1 > |w |0 iff there is a 0 ≤ k ≤ n, s.t. |w |1 = |w 0k |0 ( 1 7→ A[>U, 3. Homomorphism h : 0 7→ >].

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Introduction

Results

Conclusion



Proof of Hardness. 1. Reduction from MAJ = {w ∈ {0, 1}? | |w |1 > |w |0 } 2. |w |1 > |w |0 iff there is a 0 ≤ k ≤ n, s.t. |w |1 = |w 0k |0 ( 1 7→ A[>U, 3. Homomorphism h : 0 7→ >]. W 4. w ∈ MAJ iff 0≤k≤n h(1n w 0n+k ) ∈ CTL(B)

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Introduction

Results

Conclusion

TC0 -complete Fragments Proof Ideas

In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ⊆ R1 : always satisfiable Φ s

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Introduction

Results

Conclusion


In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ⊆ R1 : always satisfiable [B] ⊆ D: always satisfiable Φ s

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Introduction

Results

Conclusion


In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ⊆ R1 : always satisfiable [B] ⊆ D: always satisfiable [B] = N: w.l.o.g. considerh the form i O1 O2 · · · Ok P1 ψUP2 · · · UPl [· · · Uψ 0 ] · · · ,

Φ s

0 λ where ψ 0 ≡ O10 O20 · · · Om

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Introduction

Results

Conclusion


In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ⊆ R1 : always satisfiable [B] ⊆ D: always satisfiable [B] = N: w.l.o.g. considerh the form i O1 O2 · · · Ok P1 ψUP2 · · · UPl [· · · Uψ 0 ] · · · ,

Φ s

0 λ where ψ 0 ≡ O10 O20 · · · Om

[B] ∈ {V, V0 , E, E0 }: substitute propositions with >

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Introduction

Results

Conclusion

NC1 -complete Fragments Proof Ideas

In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). S11 ⊆ [B] ⊆ M: 1. substitute propositions with > 2. evaluate (almost) like in propositional logic (H. Schnoor, 2005)

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Introduction

Results

Conclusion

Fragments in P Proof Ideas

In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ∈ {L, L0 }:

(modified Algorithm from Bauland et al, 2005) 20/25


Introduction

Results

Conclusion


In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ∈ {L, L0 }: 1. Delete preceeding operators in front of > and ⊥



Introduction

Results

Conclusion


In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ∈ {L, L0 }: 1. Delete preceeding operators in front of > and ⊥ 2. Remove ⊥s.



Introduction

Results

Conclusion


In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ∈ {L, L0 }: 1. Delete preceeding operators in front of > and ⊥ 2. Remove ⊥s. 3. Order formula lexicographically to π1 ⊕ . . . ⊕ πn .



Introduction

Results

Conclusion


In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ∈ {L, L0 }: 1. Delete preceeding operators in front of > and ⊥ 2. Remove ⊥s. 3. Order formula lexicographically to π1 ⊕ . . . ⊕ πn . 4. Replace πi ⊕ πi+1 with ⊥, where πi ≡ πi+1 syntactically. (modified Algorithm from Bauland et al, 2005) 20/25


Introduction

Results

Conclusion


In the following: B is a set of Boolean functions and we will investigate CTL-SAT(B). [B] ∈ {L, L0 }: 1. Delete preceeding operators in front of > and ⊥ 2. Remove ⊥s. 3. Order formula lexicographically to π1 ⊕ . . . ⊕ πn . 4. Replace πi ⊕ πi+1 with ⊥, where πi ≡ πi+1 syntactically. 5. Goto 1. until formula becomes empty or does not change. (modified Algorithm from Bauland et al, 2005) 20/25


0

1

S200

Introduction S302

S210

Results

S301 S300

S0

Conclusion S311

S312

S310

D Promise Problem CTL-SAT P without Until S02

Adjusting Results forD Boolean Restrictions

S01

S11

1

S1

S12

S D S Promise: The given formula ϕ is syntactically correct. 00

2

V V1

10

L V0

L1

V2

L3

E L0

L2

E1

E0 E2

N N2 EXPTIME-complete solvable in P NC1 -complete TC0 -complete NLOGTIME-complete

I I1

I0

coNLOGTIME-complete trivial

I2

(H. Schnoor, 2005)

21/25 Figure 4.3: The complexity of CTL-SATP with no restrictions on the CTL-operators, depicted Post’s lattice. Michael Thomas, Heribert Vollmer The Complexity of Satisfiability for Fragments of CTL and CTL? Arne Meier, in Martin Mundhenk,

0

1

S200

Introduction S302

S210

Results

S301 S300

S0

Conclusion S311

S312

S310



S01

S11

1

S1

S12


2

V V1

10

L V0

L1

V2

L3

E L0

L2

E1

E0 E2

N N2 EXPTIME-complete

[V, V under ≤dlt 0 ]: NLOGTIME-complete solvable in P proj -reductions NC1 -complete TC0 -complete NLOGTIME-complete

I I1

I0


I2

(H. Schnoor, 2005)


0

1

S200

Introduction S302

S210

Results

S301 S300

S0

Conclusion S311

S312

S310



S01

S11

1

S1

S12


2

V V1

10

L V0

L1

V2

L3

E L0

L2

E1

E0 E2


[V, V under ≤dlt 0 ]: NLOGTIME-complete solvable in P proj -reductions NC1 -complete

[E, E under ≤dlt 0 ]: coNLOGTIME-complete proj -reductions I TC -complete 0

NLOGTIME-complete

I1

I0


I2

(H. Schnoor, 2005)


0

1

S200

Introduction S302

S210

Results

S301 S300

S0

Conclusion S311

S312

S310



S01

S11

1

S1

S12


2

V V1

10

L V0

L1

V2

L3

E L0

L2

E1

E0 E2




NLOGTIME-complete [N]: AC0 [2]-complete under ≤I dlt I proj -reductions 1

0


I2

(H. Schnoor, 2005)


0

1

S200

Introduction S302

S210

Results

S301 S300

S0

Conclusion S311

S312

S310



S01

S11

1

S1

S12


2

V V1

10

L V0

L1

V2

L3

E L0

L2

E1

E0 E2




NLOGTIME-complete [N]: AC0 [2]-complete under ≤I dlt I proj -reductions 1

0

coNLOGTIME-complete

and manytrivial trivial cases...

I2

(H. Schnoor, 2005)


Introduction

Results

Conclusion

Complexity of CTL? -SAT w.r.t. Temporal Operators A, X, U

(Bauland et al, 2007)

LTL

X, U

U

X, F

F

X

A, U

A, X, F

A, F

A, X

A

2EXPTIME-complete ∅

PSPACE-complete

22/25 NP-complete Arne Meier, Martin Mundhenk, Michael Thomas, Heribert Vollmer The Complexity of Satisfiability for Fragments of CTL and CTL?

Introduction

Results

Conclusion

Complexity of CTL? -SAT Proof Ideas

∅, {A} Equals SAT.

23/25


Introduction

Results

Conclusion



{A, F}, {A, X} Generalized proofs of the CTL-cases.

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Introduction

Results

Conclusion




{A, U}, {A, X, F} Modified version of Vardi’s EXPSPACE-ATM reduction.

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Introduction

Results

Conclusion




{A, U}, {A, X, F} Modified version of Vardi’s EXPSPACE-ATM reduction.

Clones (except BF down to S1 ) All CTL-SAT-results were already results for CTL? -SAT.

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Introduction

Results

Conclusion

Outline

Introduction


Conclusion

24/25


Introduction

Results

Conclusion

Conclusion Complexity of CTL-SAT • Restricting operators leads to a trichotomy (EXPTIME-,

PSPACE-, and NP-complete). • Restricting Boolean functions leads to a quadotomy

(EXPTIME-, NC1 -, TC0 -complete, and membership in P). • Promise Problem leads to a much finer classification with

NLOGTIME- and coNLOGTIME-completeness.

Complexity of CTL? -SAT • Restricting operators leads to a trichotomy (2EXPTIME-,

PSPACE-, and NP-complete). • Restricting Boolean functions leads to a similar hierarchy as

for CTL. 25/25


The Complexity of Satisfiability for Fragments of CTL ...

The Complexity of Satisfiability for Fragments of CTL ...

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