SAT Solver as coNP Solver – Beyond Resolution · Norbert Manthey SAT Solver as coNP Solver – Beyond Resolution 24. Resolution as Proof System I Pigeonhole principle: n+ 1 pigeons

I Motivation

I Propositional Logic and Satisfiability

I Proof Theory and SAT Solving

I Beyond Resolution

I Conclusion

Norbert MantheySAT Solver as coNP Solver – Beyond Resolution 1

SAT Solver as coNP Solver – Beyond Resolution

Norbert MantheyInternational Center for Computational LogicTechnische Universitat DresdenGermany

Motivation

Why is solving SAT interesting?

Who benefits from improved SAT solvers?


Applications of SAT Solving

I Many industrial applications have the complexityNP

I Satisfiability Testing (SAT) isNP-complete

I There are many industrial problems that are successfully solved with modernsolvers






I Translating applications directly into SAT

. (Bounded) Model Checking

. Planning

. Periodic Event Scheduling

. Termination Analysis

. Bioinformatics

. Hardware and Software Verification

. Information Flow Quantification






I 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






I SAT solvers usually build the backbone . . .

. ISABELLE uses NITPICK

. NITPICK uses KODKOD

. KODKOD uses MINISAT (or another SAT solver)


Propositional Logic and Satisfiability

Preliminaries and definitions.

Properties of propositional logic proofs.


Propositional Logic

I Let V be the set of variables

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

I Literals are positive or negative variables:

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

I A clause

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

I A formula in conjunctive normal form (CNF)

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

I Clauses and formulas can be represented as sets

I A resolvent C = C1 ⊗ C2 of two clauses C1 and C2 is defined as:C := {(C1 \ {x}) ∪ (C2 \ {¬x})}


Propositional Logic

I An interpretation J maps variables truth values.

I The value of a formula F under an interpretation J is evaluated according topropositional logic.

I A formula F is satisfied by an interpretation J, if J maps F to>.

Such a satisfying interpretation J is called model.

I A formula F is unsatisfiable, if there exists no J that maps F to>.

I A formula F is called a tautology, if F is equivalent to>

I Let TAUT be the set of all tautological formulas


Satisfiability Testing (SAT Problem)

I SAT Problem: Given a formula F , is there a model J for the formula?

I Complexity: NP complete

I Algorithms:

Theoretically, non-deterministic algorithms

Practically:

Truth table

DPLL Algorithm

CDCL Algorithm and with learning, restarts and simplification

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

I However, what happens if a search decision leads to an unsatisfiablesub-formula?

. What about complexity here?

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





I Algorithms:


Practically:

Truth table

DPLL Algorithm










I Algorithms:


Practically:

Truth table

DPLL Algorithm










I Algorithms:


Practically:

Truth table

DPLL Algorithm







Proof Theory and SAT Solving

Why do we need proofs and proof theory?

How strong are current SAT solvers?


Proof Theory

I Usual question in applications: Is my problem sound?

I More formal: F ∈ TAUT ?

I More easily: ¬F ∈ ¬TAUT ?

I . . . or: ¬F ≡ ⊥ ?

I Complexity: coNP

I 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?


Proof Theory

I Usual question in applications: Is my problem sound?

I More formal: F ∈ TAUT ?

I More easily: ¬F ∈ ¬TAUT ?

I . . . or: ¬F ≡ ⊥ ?

I Complexity: coNP

I 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?


Proof Theory - Proof Properties

I 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+1i := ⊗{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

I 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

I Lower bounds:

. This is an open question.

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











I Upper bounds:



I Lower bounds:













I Upper bounds:



I Lower bounds:













I Upper bounds:



I Lower bounds:




Resolution Style Proof Systems

I A proof system outputs a unsatisfiability proof for an unsatisfiable formula F

I Treelike Resolution is a variant of general resolution

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


Proof Systems - the Overall Picture

I Hence, CDCL can be considered more powerful than DPLL


Beyond Resolution

What else does theory provide?

Which techniques are currently used in leading SAT solvers?


Resolution as Proof SystemI Pigeonhole principle: n + 1 pigeons cannot size in n holes

I 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

I 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


Resolution as Proof SystemI Pigeonhole principle: n + 1 pigeons cannot size in n holes

I 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

I 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


Automatizability

I How to find such a proof?

I Does there exist an algorithm to find a short proof, if there exists one?

I . . . so that the generated proof is polynomial in the size of the shortest proof?

I 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.

I In this talk, look at existing SAT algorithms

namely the ones proposed for CDCL solvers


Automatizability












Automatizability












Automatizability












Variable Elimination - Resolution Inside SAT Solvers

I Formula simplifications reduce the number of variables - developed 2005

I (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 multipletimes)


Variable Elimination - Resolution Inside SAT Solvers

I Formula simplifications reduce the number of variables - developed 2005

I (Bounded) Variable Elimination


. 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 multipletimes)


Variable Elimination as Preprocessing

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Variable Elimination on the Pigeonhole Problem

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The Cutting Planes Proof System

I Reason on cardinality constraints∑

i ai xi ≤ c, instead of clauses

I Follow two rules:

. Linear combination of already existing constraints

. divide a sum by some integer d:∑

idaid exi ≤ d c

d e

I 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





I Follow two rules:

. Linear combination of already existing constraints

. divide a sum by some integer d:∑

idaid exi ≤ d c

d e


.∑


.∑


I Formula (for n = 2):

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

. x11 + x21 + x31 ≤ 1

. x12 + x22 + x32 ≤ 1






.∑


.∑


I Formula (for n = 2):

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

. x11 + x21 + x31 ≤ 1

. x12 + x22 + x32 ≤ 1

I 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)


Cutting Planes Inside SAT Solvers

I Used as formula simplification (also during search again) in Lingeling and Riss

I 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

I 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 |


Cutting Planes as Preprocessing

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Cutting Planes on the Pigeonhole Problem

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Extended Resolution Inside SAT Solvers

I Use extended resolution to improve learned clauses (resolvents in the proof)

I 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

I 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)

I The disjunction (l1 ∨ l′1) could be replaced with x in the whole formula


Extended Resolution As Formula Simplification

I Use extended resolution to improve encoded instance

I Published as Bounded Variable Addition

I Opposite operation of variable elimination

I Initially, used to decrease the number of clauses inside the formula

. let x be a fresh variable


. 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 )


Extended Resolution As Formula Simplification

I 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

I Let F = (a ∨ c) ∧ (b ∨ c) ∧ (a ∨ d) ∧ (b ∨ d) ∧ (a ∨ e) ∧ (b ∨ e)

I Then, let x be fresh, and lets have the clauses (x ∨ a) and (x ∨ b)

I BVA generates the formulaF ′ = (¬x ∨ c) ∧ (¬x ∨ d) ∧ (¬x ∨ e) ∧ (x ∨ a) ∧ (x ∨ b)

I With respect to the common variables, F ′ and F are equivalent

I F can be generated from F ′ by variable elimination


Extended Resolution in SAT Solving

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Variable Addition as Preprocessing

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


Variable Addition on the Pigeonhole Problem

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Conclusion

What is it all about?

What is expected for the next solver generation?


Conclusion

I There are reasoning techniques that are stronger than resolution

I Theoretical results are not yet well integrated

I Their practical relevance is not well understood

I Integrating more powerful reasoning into CDCL SAT solvers is a hot researchtopic

I Basically, the tools are available — however, understanding and goodheuristics are missing


Future Work

I All presented techniques are implemented in our SAT solver Riss

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

II Parity reasoning via Gaussian elimination

II Reencoding parts of the formula

I 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


Thank you for your attention

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


SAT Solver as coNP Solver – Beyond Resolution

Norbert MantheyInternational Center for Computational LogicTechnische Universitat DresdenGermany

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SAT Solver as coNP Solver – Beyond Resolution · Norbert Manthey SAT Solver as coNP Solver – Beyond Resolution 24. Resolution as Proof System I Pigeonhole principle: n+ 1 pigeons