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SAT Solver as coNP Solver Beyond Resolution Norbert Manthey - - PowerPoint PPT Presentation

Apr 06, 2023 137 likes •642 views

SAT Solver as coNP Solver Beyond Resolution Norbert Manthey International Center for Computational Logic Technische Universit at Dresden Germany Motivation Propositional Logic and Satisfiability Proof Theory and SAT Solving

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◮ Motivation ◮ Propositional Logic and Satisfiability ◮ Proof Theory and SAT Solving ◮ Beyond Resolution ◮ Conclusion

Norbert Manthey SAT Solver as coNP Solver – Beyond Resolution 1

SAT Solver as coNP Solver – Beyond Resolution

Norbert Manthey

International Center for Computational Logic Technische Universit¨ at Dresden Germany

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Motivation

Why is solving SAT interesting? Who benefits from improved SAT solvers?

Norbert Manthey SAT Solver as coNP Solver – Beyond Resolution 2

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Applications of SAT Solving

◮ Many industrial applications have the complexity NP ◮ Satisfiability Testing (SAT) is NP-complete ◮ There are many industrial problems that are successfully solved with modern

solvers

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Applications of SAT Solving

◮ Many industrial applications have the complexity NP ◮ Satisfiability Testing (SAT) is NP-complete ◮ There are many industrial problems that are successfully solved with modern

solvers

◮ Translating applications directly into SAT

⊲ (Bounded) Model Checking ⊲ Planning ⊲ Periodic Event Scheduling ⊲ Termination Analysis ⊲ Bioinformatics ⊲ Hardware and Software Verification ⊲ Information Flow Quantification

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Applications of SAT Solving

◮ Many industrial applications have the complexity NP ◮ Satisfiability Testing (SAT) is NP-complete ◮ There are many industrial problems that are successfully solved with modern

solvers

◮ Using (incremental) SAT solvers as back-end

⊲ Optimization variant of SAT: MaxSAT ⊲ SAT Modulo Theories ⊲ Constraint Satisfaction Problem ⊲ Minimum Unsatisfiable Subformula Extraction ⊲ Minimum Correction Subset Extraction

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Applications of SAT Solving

◮ Many industrial applications have the complexity NP ◮ Satisfiability Testing (SAT) is NP-complete ◮ There are many industrial problems that are successfully solved with modern

solvers

◮ SAT solvers usually build the backbone . . .

⊲ ISABELLE uses NITPICK ⊲ NITPICK uses KODKOD ⊲ KODKOD uses MINISAT (or another SAT solver)

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Propositional Logic and Satisfiability

Preliminaries and definitions. Properties of propositional logic proofs.

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Propositional Logic

◮ Let V be the set of variables

V = {a, b, ...} which can be mapped to the truth values {⊤, ⊥}

◮ Literals are positive or negative variables:

the complement of a literal x, or ¬x is the literal ¬x, or x, respectively.

◮ A clause

is a disjunction of literals C1 = (a ∨ ¬b ∨ c)

◮ A formula in conjunctive normal form (CNF)

is a conjunction of clauses F = (C1 ∧ C2 ∧ . . . )

◮ Clauses and formulas can be represented as sets ◮ A resolvent C = C1 ⊗ C2 of two clauses C1 and C2 is defined as:

C := {(C1 \ {x}) ∪ (C2 \ {¬x})}

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Propositional Logic

◮ An interpretation J maps variables truth values. ◮ The value of a formula F under an interpretation J is evaluated according to

propositional logic.

◮ A formula F is satisfied by an interpretation J, if J maps F to ⊤.

Such a satisfying interpretation J is called model.

◮ A formula F is unsatisfiable, if there exists no J that maps F to ⊤. ◮ A formula F is called a tautology, if F is equivalent to ⊤ ◮ Let TAUT be the set of all tautological formulas

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Satisfiability Testing (SAT Problem)

◮ SAT Problem: Given a formula F, is there a model J for the formula? ◮ Complexity: N P complete ◮ Algorithms:

Theoretically, non-deterministic algorithms Practically: Truth table DPLL Algorithm CDCL Algorithm and with learning, restarts and simplification

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Satisfiability Testing (SAT Problem)

◮ SAT Problem: Given a formula F, is there a model J for the formula? ◮ Complexity: N P complete ◮ Algorithms:

Theoretically, non-deterministic algorithms Practically: Truth table DPLL Algorithm CDCL Algorithm and with learning, restarts and simplification

◮ If F is satisfiable, a good heuristic finds the solution in linear time

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Satisfiability Testing (SAT Problem)

◮ SAT Problem: Given a formula F, is there a model J for the formula? ◮ Complexity: N P complete ◮ Algorithms:

Theoretically, non-deterministic algorithms Practically: Truth table DPLL Algorithm CDCL Algorithm and with learning, restarts and simplification

◮ If F is satisfiable, a good heuristic finds the solution in linear time ◮ However, what happens if a search decision leads to an unsatisfiable

sub-formula?

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Satisfiability Testing (SAT Problem)

◮ SAT Problem: Given a formula F, is there a model J for the formula? ◮ Complexity: N P complete ◮ Algorithms:

Theoretically, non-deterministic algorithms Practically: Truth table DPLL Algorithm CDCL Algorithm and with learning, restarts and simplification

◮ If F is satisfiable, a good heuristic finds the solution in linear time ◮ However, what happens if a search decision leads to an unsatisfiable

sub-formula? ⊲ What about complexity here?

◮ . . . and what about the complexity of showing unsatisfiability?

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Proof Theory and SAT Solving

Why do we need proofs and proof theory? How strong are current SAT solvers?

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Proof Theory

◮ Usual question in applications: Is my problem sound? ◮ More formal: F ∈ TAUT ? ◮ More easily: ¬F ∈ ¬TAUT ? ◮ . . . or: ¬F ≡ ⊥ ? ◮ Complexity: coN P

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Proof Theory

◮ Usual question in applications: Is my problem sound? ◮ More formal: F ∈ TAUT ? ◮ More easily: ¬F ∈ ¬TAUT ? ◮ . . . or: ¬F ≡ ⊥ ? ◮ Complexity: coN P ◮ How to show the unsatisfiability of a formula?

⊲ Truth table, DPLL, CDCL, general resolution, . . . ⊲ Basically, show that F | = ( ) ⊲ Can the check be done in a linear number of steps? ⊲ Are there any lower or upper bounds known?

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Proof Theory - Proof Properties

◮ Resolution style proofs P = (C1, C2, C3, . . . ( )) for a formula F:

⊲ P := P ∪ {C0 | C ∈ F}, assign the depth 0 to each clause of F ⊲ Resolution: Cmaxh+1

i

:= ⊗{Ch

j | j < i, Cj ∈ ForCj ∈ P}

⊲ Length: the number of resolvents before ( ) is derived ⊲ Width: the maximum size of intermediate resolvents ⊲ Space: the maximum number of clauses used for a resolution step ⊲ Height: the maximum depth of a clause in a proof

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Proof Theory - Proof Properties

◮ Resolution style proofs P = (C1, C2, C3, . . . ( )) for a formula F:

⊲ P := P ∪ {C0 | C ∈ F}, assign the depth 0 to each clause of F ⊲ Resolution: Cmaxh+1

i

:= ⊗{Ch

j | j < i, Cj ∈ ForCj ∈ P}

⊲ Length: the number of resolvents before ( ) is derived ⊲ Width: the maximum size of intermediate resolvents ⊲ Space: the maximum number of clauses used for a resolution step ⊲ Height: the maximum depth of a clause in a proof

◮ Upper bounds:

⊲ Length: possible number of clauses for the given variables (2n+1 − 1) ⊲ Space: n + 2, where n is the number of variables in the formula F

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Proof Theory - Proof Properties

◮ Resolution style proofs P = (C1, C2, C3, . . . ( )) for a formula F:

⊲ P := P ∪ {C0 | C ∈ F}, assign the depth 0 to each clause of F ⊲ Resolution: Cmaxh+1

i

:= ⊗{Ch

j | j < i, Cj ∈ ForCj ∈ P}

⊲ Length: the number of resolvents before ( ) is derived ⊲ Width: the maximum size of intermediate resolvents ⊲ Space: the maximum number of clauses used for a resolution step ⊲ Height: the maximum depth of a clause in a proof

◮ Upper bounds:

⊲ Length: possible number of clauses for the given variables (2n+1 − 1) ⊲ Space: n + 2, where n is the number of variables in the formula F

◮ Lower bounds:

⊲ This is an open question.

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Proof Theory - Proof Properties

◮ Resolution style proofs P = (C1, C2, C3, . . . ( )) for a formula F:

⊲ P := P ∪ {C0 | C ∈ F}, assign the depth 0 to each clause of F ⊲ Resolution: Cmaxh+1

i

:= ⊗{Ch

j | j < i, Cj ∈ ForCj ∈ P}

⊲ Length: the number of resolvents before ( ) is derived ⊲ Width: the maximum size of intermediate resolvents ⊲ Space: the maximum number of clauses used for a resolution step ⊲ Height: the maximum depth of a clause in a proof

◮ Upper bounds:

⊲ Length: possible number of clauses for the given variables (2n+1 − 1) ⊲ Space: n + 2, where n is the number of variables in the formula F

◮ Lower bounds:

⊲ This is an open question.

◮ In this talk, lets consider length only, because it relates to run time

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Resolution Style Proof Systems

◮ A proof system outputs a unsatisfiability proof for an unsatisfiable formula F ◮ Treelike Resolution is a variant of general resolution

⊲ Derived resolvents Ci can be put into a tree-like dependence scheme

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Proof Systems - the Overall Picture

◮ Hence, CDCL can be considered more powerful than DPLL

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Beyond Resolution

What else does theory provide? Which techniques are currently used in leading SAT solvers?

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Resolution as Proof System

◮ Pigeonhole principle: n + 1 pigeons cannot size in n holes ◮ CNF formula: xi,j represents pigeon i sitting in hole j

⊲ In each hole there is a pigeon (at-least-one)

  • 1≤j≤n xij

for all pigeons 1 ≤ i ≤ n + 1 ⊲ In each hole, there cannot be two pigeons (at-most-hole) ¬xij ∨ ¬xi′j for all pigeon pairs 1 ≤ i < i′ ≤ n + 1, and 1 ≤ j ≤ n

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Resolution as Proof System

◮ Pigeonhole principle: n + 1 pigeons cannot size in n holes ◮ CNF formula: xi,j represents pigeon i sitting in hole j

⊲ In each hole there is a pigeon (at-least-one)

  • 1≤j≤n xij

for all pigeons 1 ≤ i ≤ n + 1 ⊲ In each hole, there cannot be two pigeons (at-most-hole) ¬xij ∨ ¬xi′j for all pigeon pairs 1 ≤ i < i′ ≤ n + 1, and 1 ≤ j ≤ n

◮ Proof, for 3 pigeons and 2 holes (n = 2):

(x22 ∨ x32) ⊗ (¬x31 ∨ ¬x22) = (x32 ∨ ¬x31) (x32 ∨ ¬x31) ⊗ (x31 ∨ x12) = (x32 ∨ x12) (x32 ∨ x12) ⊗ (¬x21 ∨ ¬x32) = (x12 ∨ ¬x21) (x12 ∨ ¬x21) ⊗ (¬x21 ∨ ¬x12) = (¬x21) . . . (propagating the unit in the formula) · · · = ( ), in total, 11 simple resolutions are required

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Automatizability

◮ How to find such a proof? ◮ Does there exist an algorithm to find a short proof, if there exists one?

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Automatizability

◮ How to find such a proof? ◮ Does there exist an algorithm to find a short proof, if there exists one? ◮ . . . so that the generated proof is polynomial in the size of the shortest proof?

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Automatizability

◮ How to find such a proof? ◮ Does there exist an algorithm to find a short proof, if there exists one? ◮ . . . so that the generated proof is polynomial in the size of the shortest proof? ◮ Some results from the literature:

⊲ Truth Tables are automatizable. ⊲ Tree-Like Resolution is automatizable in quasi-polynomial time ( nO(log n) ) ⊲ Resolution is not automatizable, unless FTP = W[P]. ⊲ Extended Resolution is not automatizable, unless RSA is insecure.

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Automatizability

◮ How to find such a proof? ◮ Does there exist an algorithm to find a short proof, if there exists one? ◮ . . . so that the generated proof is polynomial in the size of the shortest proof? ◮ Some results from the literature:

⊲ Truth Tables are automatizable. ⊲ Tree-Like Resolution is automatizable in quasi-polynomial time ( nO(log n) ) ⊲ Resolution is not automatizable, unless FTP = W[P]. ⊲ Extended Resolution is not automatizable, unless RSA is insecure.

◮ In this talk, look at existing SAT algorithms

namely the ones proposed for CDCL solvers

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Variable Elimination - Resolution Inside SAT Solvers

◮ Formula simplifications reduce the number of variables - developed 2005 ◮ (Bounded) Variable Elimination

⊲ S = Sx ⊗ S¬x, the set of all pairwise resolvents ⊲ replaces Sx ∪ S¬x with S

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Variable Elimination - Resolution Inside SAT Solvers

◮ Formula simplifications reduce the number of variables - developed 2005 ◮ (Bounded) Variable Elimination

⊲ S = Sx ⊗ S¬x, the set of all pairwise resolvents ⊲ replaces Sx ∪ S¬x with S if |Sx ∪ S¬x| ≤ |S|, ⊲ Number of clauses and variables decreases ⊲ F := (F \ (Sx ∪ S¬x)) ∪ S ⊲ Implementation tries replacement for all variables x (sometimes multiple times)

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Variable Elimination as Preprocessing

0.1 1 10 100 1000 10000 0.1 1 10 100 1000 10000 bve-benchmark plain-benchmark run time comparison run time

◮ Riss without (638) and with applying variable elimination before search (755)

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Variable Elimination on the Pigeonhole Problem

0.1 1 10 100 1000 10000 0.1 1 10 100 1000 10000 bve-ph plain-ph run time comparison run time 5 10 15 20 25 30 35 40

◮ Riss without (10) and with variable elimination (10)

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The Cutting Planes Proof System

◮ Reason on cardinality constraints

i aixi ≤ c, instead of clauses

◮ Follow two rules:

⊲ Linear combination of already existing constraints ⊲ divide a sum by some integer d:

i⌈ ai d ⌉xi ≤ ⌈ c d ⌉

◮ Pigeonhole for n pigeons:

⊲

1≤j≤n xij ≥ 1

for all pigeons 1 ≤ i ≤ n + 1 ⊲

1≤i≤n+1 xij ≤ 1

for all holes 1 ≤ i ≤ n

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The Cutting Planes Proof System

◮ Reason on cardinality constraints

i aixi ≤ c, instead of clauses

◮ Follow two rules:

⊲ Linear combination of already existing constraints ⊲ divide a sum by some integer d:

i⌈ ai d ⌉xi ≤ ⌈ c d ⌉

◮ Pigeonhole for n pigeons:

⊲

1≤j≤n xij ≥ 1

for all pigeons 1 ≤ i ≤ n + 1 ⊲

1≤i≤n+1 xij ≤ 1

for all holes 1 ≤ i ≤ n

◮ Formula (for n = 2):

⊲ 1 ≤ x11 + x12, 1 ≤ x21 + x22 and 1 ≤ x31 + x32 ⊲ x11 + x21 + x31 ≤ 1 ⊲ x12 + x22 + x32 ≤ 1

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The Cutting Planes Proof System

◮ Reason on cardinality constraints

i aixi ≤ c, instead of clauses

◮ Pigeonhole for n pigeons:

⊲

1≤j≤n xij ≥ 1

for all pigeons 1 ≤ i ≤ n + 1 ⊲

1≤i≤n+1 xij ≤ 1

for all holes 1 ≤ i ≤ n

◮ Formula (for n = 2):

⊲ 1 ≤ x11 + x12, 1 ≤ x21 + x22 and 1 ≤ x31 + x32 ⊲ x11 + x21 + x31 ≤ 1 ⊲ x12 + x22 + x32 ≤ 1

◮ Proof:

⊲ x21 + x31 ≤ x12 (from 1 ≤ x11 + x12 and (x11 + x21 + x31 ≤ 1) ⊲ 1 + x31 ≤ x12 + x22 ( with 1 ≤ x21 + x22) ⊲ 2 ≤ x12 + x22 + x32 ( with 1 ≤ x31 + x32) ⊲ 1 ≤ 0 ( with x12 + x22 + x32 ≤ 1)

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Cutting Planes Inside SAT Solvers

◮ Used as formula simplification (also during search again) in Lingeling and Riss ◮ Needs to find cardinality constraints within the formula

⊲ Pattern matching ⊲ Limited to naive encodings of at-most-one and at-most-two, at-least-one

◮ Given the constraints C, perform elimination, as BVE on the constraints

⊲ S := Sx + S¬x, pairwise linear combination ⊲ C := (C \ (Sx ∪ S¬x)) ∪ S if |S| ≤ |Sx ∪ S¬x|

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Cutting Planes as Preprocessing

0.1 1 10 100 1000 10000 0.1 1 10 100 1000 10000 fm-benchmark plain-benchmark run time comparison run time

◮ Riss without (638) and with using cutting planes as preprocessing (623)

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Cutting Planes on the Pigeonhole Problem

0.1 1 10 100 1000 10000 0.1 1 10 100 1000 10000 fm-ph plain-ph run time comparison run time 5 10 15 20 25 30 35 40

◮ Riss without (10) and with cutting planes (38)

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Extended Resolution Inside SAT Solvers

◮ Use extended resolution to improve learned clauses (resolvents in the proof) ◮ Published as extended clause learning

⊲ Given a long learned clause C = (l1 ∨ l′

1 ∨ l2 · · · ∨ ln)

introduce a fresh variable x with the clauses for x ↔ (l1 ∨ l′

1)

and learn C′ = (x ∨ l2 ∨ · · · ∨ ln) instead

◮ Or restricted extended resolution

⊲ Given two learned clauses C = (¬l1 ∨ l2 ∨ · · · ∨ ln), C′ = (¬l′

1 ∨ l2 ∨ · · · ∨ ln)

introduce a fresh variable x with the clauses for ¬x ↔ (¬l1 ∧ ¬l′

1) (which is equal to x ↔ (l1 ∨ l′ 1))

and replace C and C′ with C′′ = (¬x ∨ l2 ∨ · · · ∨ ln)

◮ The disjunction (l1 ∨ l′

1) could be replaced with x in the whole formula

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Extended Resolution As Formula Simplification

◮ Use extended resolution to improve encoded instance ◮ Published as Bounded Variable Addition ◮ Opposite operation of variable elimination ◮ Initially, used to decrease the number of clauses inside the formula

⊲ let x be a fresh variable ⊲ S = Sx ⊗ S¬x, the set of all pairwise resolvents ⊲ replaces S with Sx ∪ S¬x if |S| ≥ |Sx ∪ S¬x| ⊲ Number of clauses decreases, and number of variables increases ⊲ F := (F \ S) ∪ (Sx ∪ S¬x)

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Extended Resolution As Formula Simplification

◮ How to find S, such that there exists an Sx and S¬x?

⊲ Pattern matching for x ↔

i li – AND-gates

⊲ Pattern matching for x ↔ l1 ⊕ l2 – XOR-gates ⊲ Pattern matching for x ↔ ITE(s, t, f) – If-then-else-gates

◮ Let F = (a ∨ c) ∧ (b ∨ c) ∧ (a ∨ d) ∧ (b ∨ d) ∧ (a ∨ e) ∧ (b ∨ e) ◮ Then, let x be fresh, and lets have the clauses (x ∨ a) and (x ∨ b) ◮ BVA generates the formula

F ′ = (¬x ∨ c) ∧ (¬x ∨ d) ∧ (¬x ∨ e) ∧ (x ∨ a) ∧ (x ∨ b)

◮ With respect to the common variables, F ′ and F are equivalent ◮ F can be generated from F ′ by variable elimination

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Extended Resolution in SAT Solving

0.1 1 10 100 1000 10000 0.1 1 10 100 1000 10000 rer-benchmark plain-benchmark run time comparison run time

◮ Riss without (638) and with using extended resolution (638, ecl: 559)

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Variable Addition as Preprocessing

0.1 1 10 100 1000 10000 0.1 1 10 100 1000 10000 bvaA-benchmark plain-benchmark run time comparison run time

◮ Riss (638) vs. variable addition (AND-gate: 748, ITE-gate: 698, XOR-gate: 664)

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Variable Addition on the Pigeonhole Problem

0.1 1 10 100 1000 10000 0.1 1 10 100 1000 10000 bvaA-ph plain-ph run time comparison run time 5 10 15 20 25 30 35 40

◮ Riss (10) vs. variable addition (AND-gate: 11, ITE-gate: 10, XOR-gate: 10)

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Conclusion

What is it all about? What is expected for the next solver generation?

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Conclusion

◮ There are reasoning techniques that are stronger than resolution ◮ Theoretical results are not yet well integrated ◮ Their practical relevance is not well understood ◮ Integrating more powerful reasoning into CDCL SAT solvers is a hot research

topic

◮ Basically, the tools are available — however, understanding and good

heuristics are missing

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Future Work

◮ All presented techniques are implemented in our SAT solver Riss

⊲ Left out here, but also there (with similar picture)

◮ ◮ Parity reasoning via Gaussian elimination ◮ ◮ Reencoding parts of the formula ◮ How to deal with all these techniques?

⊲ When to use which technique, how long, how often, on which variables? ⊲ The system gives a provides a good portfolio because of its diversity

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Thank you for your attention

Many thanks to Jakob Nordstr ˜ A¶m and Olaf Beyersdorff for all the questions about proof theory.

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SAT Solver as coNP Solver – Beyond Resolution

Norbert Manthey

International Center for Computational Logic Technische Universit¨ at Dresden Germany