Solvers Principles and Architecture (SPA) Lecture 1 SAT Solvers - - PowerPoint PPT Presentation

Solvers Principles and Architecture (SPA)

Lecture 1

SAT Solvers

Master Sciences Informatique (Sif) September 25th, 2017 Rennes

Khalil Ghorbal khalil.ghorbal@inria.fr

• K. Ghorbal (INRIA)

1 SIF M2 1 / 42

Automated Theorem Proving

Specification

Formal specification of the system:

• What the system is expected to do
• What properties the system has to satisfy

Modeling

Mathematical model describing the behavior of the system:

• Finite state or hybrid automata
• Markov chain

Proof

Does the model satisfy the specification? When successful: no corner cases left!

• K. Ghorbal (INRIA)

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Outline

1 Introduction 2 Propositional Logic 3 SAT Solving 4 DPLL-based Algorithms

• K. Ghorbal (INRIA)

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Logic in a Nut Shell

Formally, a logic is a pair of syntax and semantics.

Syntax

• Alphabet: set of symbols
• Expressions: sequences of symbols
• Rules: identifying well-formed expressions

Semantics

• Meaning: what is meant by well-formed expressions
• Rules: infer the meaning from subexpressions
• K. Ghorbal (INRIA)

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Alphabet

Syntax

Alphabet

( left parenthesis ) right parenthesis ¬ Negation ∧ Conjunction ∨ Disjunction (inclusive) ← − Implication ← → Equivalence Propositional symbol“False” 1 Propositional symbol“True” si ith propositional symbol

• K. Ghorbal (INRIA)

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Expressions

Syntax

Expression

Sequence of symbols from the alphabet. (, a1, ∧, a2, ) (a1 ∧ a2) (, ), ∨, a1, ¬, a2 () ∨ a1¬a3 We want to further restrict the allowed combinations.

• K. Ghorbal (INRIA)

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Well-Formed Formulas

Syntax

Well-formed formulas (wff) are defined inductively S: the set of expressions with a single propositional symbol S = {0, 1, s1, s2, . . . } W : the set of wffs is freely generated from S as follows w ::= s | (w) | ¬w | w ∧ w| w ∨ w | w − → w | w ← → w So far we only manipulated symbols or wooden pieces!

• K. Ghorbal (INRIA)

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Semantics with Truth Table

s ¬ 1 1

• K. Ghorbal (INRIA)

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Semantics with Truth Table

s ¬ 1 1 s1 s2 ∧ 1 1 1 1 1 s1 s2 ∨ 1 1 1 1 1 1 1

• K. Ghorbal (INRIA)

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Semantics with Truth Table

s ¬ 1 1 s1 s2 ∧ 1 1 1 1 1 s1 s2 ∨ 1 1 1 1 1 1 1 s1 s2 → 1 1 1 1 1 1 1

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Interpretation Domain

Semantics

Intuition

Given a context, that is a truth value for each propositional symbol, we can determine the truth value of any wff in our context.

Boolean Algebra

• Field structure: Z /2Z = B = {0, 1}
• ”

+” : 0 is the identity, 1 is its own inverse: 1 + 1 = 0

• ”

×” : standard multiplication operator, where 1 is the identity element

• K. Ghorbal (INRIA)

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Transfer Functions

Semantics

• context: σ : S → B. A valuation of all propositional symbols
• σ satisfies σ(0) = 0 and σ(1) = 1
• Define σ : W → B
• σ is well-defined since W is freely generated

Semantics of the Transfer Functions sσ = σ(s) ¬wσ = 1 + wσ w1 ∧ w2σ = w1σ × w2σ w1 ∨ w2σ = w1σ + w2σ + w1σ × w2σ w1 → w2σ = 1 + w1σ + w1σ × w2σ w1 ← → w2σ = 1 + w1σ + w2σ

• K. Ghorbal (INRIA)

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Definitions

• σ: context, valuation, truth assignment
• σ satisfies w if and only if wσ = 1
• w is satisfiable if there exists σ such that σ satisfies w
• w is unsatisfiable if there is no σ such that σ satisfies w

Example:

• (s1 ∨ s2) ∧ (¬s1 ∨ ¬s2) is satisfiable
• (s1 ∨ s2) ∧ (¬s1 ∨ ¬s2) ∧ (s1 ↔ s2) is unsatisfiable
• K. Ghorbal (INRIA)

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Definitions

• σ: context, valuation, truth assignment
• σ satisfies w if and only if wσ = 1
• w is satisfiable if there exists σ such that σ satisfies w
• w is unsatisfiable if there is no σ such that σ satisfies w

Example:

• (s1 ∨ s2) ∧ (¬s1 ∨ ¬s2) is satisfiable
• (s1 ∨ s2) ∧ (¬s1 ∨ ¬s2) ∧ (s1 ↔ s2) is unsatisfiable
• K. Ghorbal (INRIA)

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Definitions

• σ: context, valuation, truth assignment
• σ satisfies w if and only if wσ = 1
• w is satisfiable if there exists σ such that σ satisfies w
• w is unsatisfiable if there is no σ such that σ satisfies w

Example:

• (s1 ∨ s2) ∧ (¬s1 ∨ ¬s2) is satisfiable
• (s1 ∨ s2) ∧ (¬s1 ∨ ¬s2) ∧ (s1 ↔ s2) is unsatisfiable
• K. Ghorbal (INRIA)

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Implications as Satisfiability

Tautological Implication (wi are wffs) w1, . . . , wn | = w if and only if ∀σ.

• i

wiσ = 1 − → wσ = 1

• Every truth assignment that satisfies all wi satisfies necessarily w

Definitions

• |

= w (or 1 | = w): w is a tautology or w is valid

• w1 ∼ w2: w1 |

= w2 and w2 | = w1 (tautological equivalence)

• e.g. s1 → s2 ∼ ¬s1 ∨ s2
• K. Ghorbal (INRIA)

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Implications as Satisfiability

Tautological Implication (wi are wffs) w1, . . . , wn | = w if and only if ∀σ.

• i

wiσ = 1 − → wσ = 1

• Every truth assignment that satisfies all wi satisfies necessarily w

Definitions

• |

= w (or 1 | = w): w is a tautology or w is valid

• w1 ∼ w2: w1 |

= w2 and w2 | = w1 (tautological equivalence)

• e.g. s1 → s2 ∼ ¬s1 ∨ s2
• K. Ghorbal (INRIA)

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Implications as Satisfiability

Tautological Implication (wi are wffs) w1, . . . , wn | = w if and only if ∀σ.

• i

wiσ = 1 − → wσ = 1

• Every truth assignment that satisfies all wi satisfies necessarily w

Definitions

• |

= w (or 1 | = w): w is a tautology or w is valid

• w1 ∼ w2: w1 |

= w2 and w2 | = w1 (tautological equivalence)

• e.g. s1 → s2 ∼ ¬s1 ∨ s2
• K. Ghorbal (INRIA)

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Implications as Satisfiability

Tautological Implication (wi are wffs) w1, . . . , wn | = w if and only if ∀σ.

• i

wiσ = 1 − → wσ = 1

• Every truth assignment that satisfies all wi satisfies necessarily w

Definitions

• |

= w (or 1 | = w): w is a tautology or w is valid

• w1 ∼ w2: w1 |

= w2 and w2 | = w1 (tautological equivalence)

• e.g. s1 → s2 ∼ ¬s1 ∨ s2
• K. Ghorbal (INRIA)

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Proving Theorem with SAT

Tautological Implication as Satisfiability Problem w1, . . . , wn | = w if and only if

• i

wi ∧ ¬w is unsatisfiable Example

• s1, s1 → s2 |

= s2 iff s1 ∧ (s1 → s2) ∧ ¬s2 is unsat.

• s, ¬s |

= (s ∧ ¬s) iff s ∧ ¬s ∧ ¬(s ∧ ¬s) is unsat

• K. Ghorbal (INRIA)

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Proving Theorem with SAT

Tautological Implication as Satisfiability Problem w1, . . . , wn | = w if and only if

• i

wi ∧ ¬w is unsatisfiable Example

• s1, s1 → s2 |

= s2 iff s1 ∧ (s1 → s2) ∧ ¬s2 is unsat.

• s, ¬s |

= (s ∧ ¬s) iff s ∧ ¬s ∧ ¬(s ∧ ¬s) is unsat

• K. Ghorbal (INRIA)

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Proving Theorem with SAT

Tautological Implication as Satisfiability Problem w1, . . . , wn | = w if and only if

• i

wi ∧ ¬w is unsatisfiable Example

• s1, s1 → s2 |

= s2 iff s1 ∧ (s1 → s2) ∧ ¬s2 is unsat.

• s, ¬s |

= (s ∧ ¬s) iff s ∧ ¬s ∧ ¬(s ∧ ¬s) is unsat

• K. Ghorbal (INRIA)

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Outline

1 Introduction 2 Propositional Logic 3 SAT Solving 4 DPLL-based Algorithms

• K. Ghorbal (INRIA)

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SAT Decision Problem

Given a well-formed formula w as an input, if there exists a σ that satisfies w return True, otherwise return False

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2))

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2)) (1) 1 1 1 1 1

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2)) (1) 1 1 1 1 1 (2) 1 1 1 1 1 1

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2)) (1) 1 1 1 1 1 (2) 1 1 1 1 1 1 (3) 1 1 1

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2)) (1) 1 1 1 1 1 (2) 1 1 1 1 1 1 (3) 1 1 1 (4) 1 1 1 1 1 1

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2)) (1) 1 1 1 1 1 (2) 1 1 1 1 1 1 (3) 1 1 1 (4) 1 1 1 1 1 1 (5) 1 1 1

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2)) (1) 1 1 1 1 1 (2) 1 1 1 1 1 1 (3) 1 1 1 (4) 1 1 1 1 1 1 (5) 1 1 1 (6) 1 1 1 1

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2)) (1) 1 1 1 1 1 (2) 1 1 1 1 1 1 (3) 1 1 1 (4) 1 1 1 1 1 1 (5) 1 1 1 (6) 1 1 1 1 (7) 1 1 1

• K. Ghorbal (INRIA)

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Brute Force Algorithm

Example

s1 ∧ (s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2) s1 s2 s3 s1 ∧ ((s2 ∨ ¬s1) ∧ (s3 ∨ ¬s2)) (1) 1 1 1 1 1 (2) 1 1 1 1 1 1 (3) 1 1 1 (4) 1 1 1 1 1 1 (5) 1 1 1 (6) 1 1 1 1 (7) 1 1 1 (8) 1 1 1 1 1 1 1

• K. Ghorbal (INRIA)

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SAT Facts

• Brute force algorithm: exponential complexity:
• 2n for n propositional symbol
• SAT is the first problem to be proven to be NP-complete [Cook 1971]
• SAT solves any decision problem in NP
• Modern SAT Solvers are arguably efficient, why?
• SAT expects an input in Conjunctive Normal Form
• K. Ghorbal (INRIA)

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SAT Facts

• Brute force algorithm: exponential complexity:
• 2n for n propositional symbol
• SAT is the first problem to be proven to be NP-complete [Cook 1971]
• SAT solves any decision problem in NP
• Modern SAT Solvers are arguably efficient, why?
• SAT expects an input in Conjunctive Normal Form
• K. Ghorbal (INRIA)

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SAT Facts

• Brute force algorithm: exponential complexity:
• 2n for n propositional symbol
• SAT is the first problem to be proven to be NP-complete [Cook 1971]
• SAT solves any decision problem in NP
• Modern SAT Solvers are arguably efficient, why?
• SAT expects an input in Conjunctive Normal Form
• K. Ghorbal (INRIA)

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SAT Facts

• Brute force algorithm: exponential complexity:
• 2n for n propositional symbol
• SAT is the first problem to be proven to be NP-complete [Cook 1971]
• SAT solves any decision problem in NP
• Modern SAT Solvers are arguably efficient, why?
• SAT expects an input in Conjunctive Normal Form
• K. Ghorbal (INRIA)

15 SIF M2 15 / 42

SAT Facts

• Brute force algorithm: exponential complexity:
• 2n for n propositional symbol
• SAT is the first problem to be proven to be NP-complete [Cook 1971]
• SAT solves any decision problem in NP
• Modern SAT Solvers are arguably efficient, why?
• SAT expects an input in Conjunctive Normal Form
• K. Ghorbal (INRIA)

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Converting to CNF

Equivalence versus Equisatisfiability

Recall (Tautological) Equivalence w1 ∼ w2 if and only if ∀σ. (w1σ = 1 ← → w2σ = 1) Equisatisfiability w1 ∼SAT w2 if and only if ∃σ. w1σ = 1 ← → ∃σ. w2σ = 1 Equisatisfiability does not imply tautological equivalence!

• w1 := s1 ∧ (s1 ↔ s2) and w2 := s1
• w1 ∼SAT w2 but w1 ∼ w2
• K. Ghorbal (INRIA)

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Converting to CNF

Equivalence versus Equisatisfiability

Recall (Tautological) Equivalence w1 ∼ w2 if and only if ∀σ. (w1σ = 1 ← → w2σ = 1) Equisatisfiability w1 ∼SAT w2 if and only if ∃σ. w1σ = 1 ← → ∃σ. w2σ = 1 Equisatisfiability does not imply tautological equivalence!

• w1 := s1 ∧ (s1 ↔ s2) and w2 := s1
• w1 ∼SAT w2 but w1 ∼ w2
• K. Ghorbal (INRIA)

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Converting to CNF

Equivalence versus Equisatisfiability

Recall (Tautological) Equivalence w1 ∼ w2 if and only if ∀σ. (w1σ = 1 ← → w2σ = 1) Equisatisfiability w1 ∼SAT w2 if and only if ∃σ. w1σ = 1 ← → ∃σ. w2σ = 1 Equisatisfiability does not imply tautological equivalence!

• w1 := s1 ∧ (s1 ↔ s2) and w2 := s1
• w1 ∼SAT w2 but w1 ∼ w2
• K. Ghorbal (INRIA)

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Converting to CNF

Equivalent CNF is Exponential

Converting a wff w to an equivalent formula in CNF using De Morgan’s Laws and distributivity may increase the number of logical operations (Boolean gates) exponentially. Example

• w1 := (s1 ∧ s2) ∨ (s3 ∧ s4), by distributivity
• w2 := (s1 ∨ s3) ∧ (s1 ∨ s4) ∧ (s2 ∨ s3) ∧ (s2 ∨ s4)
• Add extra pairs to w1, (s1 ∧ s2) ∨ (s3 ∧ s4) ∨ (s5 ∧ s6) . . .
• The number of the involved logical operations is exponential
• K. Ghorbal (INRIA)

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Converting to CNF

Equivalent CNF is Exponential

Converting a wff w to an equivalent formula in CNF using De Morgan’s Laws and distributivity may increase the number of logical operations (Boolean gates) exponentially. Example

• w1 := (s1 ∧ s2) ∨ (s3 ∧ s4), by distributivity
• w2 := (s1 ∨ s3) ∧ (s1 ∨ s4) ∧ (s2 ∨ s3) ∧ (s2 ∨ s4)
• Add extra pairs to w1, (s1 ∧ s2) ∨ (s3 ∧ s4) ∨ (s5 ∧ s6) . . .
• The number of the involved logical operations is exponential
• K. Ghorbal (INRIA)

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Converting to CNF

Equivalent CNF is Exponential

Converting a wff w to an equivalent formula in CNF using De Morgan’s Laws and distributivity may increase the number of logical operations (Boolean gates) exponentially. Example

• w1 := (s1 ∧ s2) ∨ (s3 ∧ s4), by distributivity
• w2 := (s1 ∨ s3) ∧ (s1 ∨ s4) ∧ (s2 ∨ s3) ∧ (s2 ∨ s4)
• Add extra pairs to w1, (s1 ∧ s2) ∨ (s3 ∧ s4) ∨ (s5 ∧ s6) . . .
• The number of the involved logical operations is exponential
• K. Ghorbal (INRIA)

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Converting to CNF

Equivalent CNF is Exponential

Converting a wff w to an equivalent formula in CNF using De Morgan’s Laws and distributivity may increase the number of logical operations (Boolean gates) exponentially. Example

• w1 := (s1 ∧ s2) ∨ (s3 ∧ s4), by distributivity
• w2 := (s1 ∨ s3) ∧ (s1 ∨ s4) ∧ (s2 ∨ s3) ∧ (s2 ∨ s4)
• Add extra pairs to w1, (s1 ∧ s2) ∨ (s3 ∧ s4) ∨ (s5 ∧ s6) . . .
• The number of the involved logical operations is exponential
• K. Ghorbal (INRIA)

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Converting to CNF

Tseytin Transformation [Tseytin 1970]

Idea: Converting a w by adding new propositional variables and substitute for nested operations. Example (s1 ∧ s2)

• p1

∨ (s3 ∧ s4)

• p2
• p3
• p1 ↔ (s1 ∧ s2)
• p2 ↔ (s3 ∧ s4)
• p3 ↔ p1 ∨ p2
• CNF: (p1 ↔ (s1 ∧ s2)) ∧ (p2 ↔ (s3 ∧ s4)) ∧ (p3 ↔ p1 ∨ p2) ∧ p3
• K. Ghorbal (INRIA)

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Converting to CNF

Tseytin Transformation [Tseytin 1970]

Idea: Converting a w by adding new propositional variables and substitute for nested operations. Example (s1 ∧ s2)

• p1

∨ (s3 ∧ s4)

• p2
• p3
• p1 ↔ (s1 ∧ s2)
• p2 ↔ (s3 ∧ s4)
• p3 ↔ p1 ∨ p2
• CNF: (p1 ↔ (s1 ∧ s2)) ∧ (p2 ↔ (s3 ∧ s4)) ∧ (p3 ↔ p1 ∨ p2) ∧ p3
• K. Ghorbal (INRIA)

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Converting to CNF

Tseytin Transformation [Tseytin 1970]

• p ↔ (s1 ◦ s2) has at most 3 clauses for any ◦ ∈ {¬, ∧, ∨, ¯

∧, ¯ ∨}

• For n logical operation, the increase is linear O(n)
• Drawback: the number of propositional variables increases (linearly)!
• K. Ghorbal (INRIA)

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Converting to CNF

Tseytin Transformation [Tseytin 1970]

• p ↔ (s1 ◦ s2) has at most 3 clauses for any ◦ ∈ {¬, ∧, ∨, ¯

∧, ¯ ∨}

• For n logical operation, the increase is linear O(n)
• Drawback: the number of propositional variables increases (linearly)!
• K. Ghorbal (INRIA)

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Converting to CNF

Tseytin Transformation [Tseytin 1970]

• p ↔ (s1 ◦ s2) has at most 3 clauses for any ◦ ∈ {¬, ∧, ∨, ¯

∧, ¯ ∨}

• For n logical operation, the increase is linear O(n)
• Drawback: the number of propositional variables increases (linearly)!
• K. Ghorbal (INRIA)

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DiMaCS Standard

• Each propositional variable is represented by a positive integer
• A negative integer refers to negative occurrences
• Clauses are given as sequences of integers separated by spaces
• A 0 terminates the clause

Example:

• (s1 ∨ ¬s3) ∧ (¬s2 ∨ s3 ∨ s4)
• 1 -3 0
• 2 3 4 0
• K. Ghorbal (INRIA)

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DiMaCS Standard

• Each propositional variable is represented by a positive integer
• A negative integer refers to negative occurrences
• Clauses are given as sequences of integers separated by spaces
• A 0 terminates the clause

Example:

• (s1 ∨ ¬s3) ∧ (¬s2 ∨ s3 ∨ s4)
• 1 -3 0
• 2 3 4 0
• K. Ghorbal (INRIA)

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Outline

1 Introduction 2 Propositional Logic 3 SAT Solving 4 DPLL-based Algorithms

• K. Ghorbal (INRIA)

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Definitions

• Literal: propositional symbol (atomic formula) or its negation
• Clause: disjunction of one or more literals
• Conjunctive Normal Form (CNF): conjunction of clauses
• Positive Occurrence: if the symbol occurs unnegated in a clause
• Negative Occurrence: if the symbol occurs negated in a clause

(s1 ∨ ¬s3) ∧ (¬s2 ∨ s3 ∨ s4)

• K. Ghorbal (INRIA)

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Definitions

• Literal: propositional symbol (atomic formula) or its negation
• Clause: disjunction of one or more literals
• Conjunctive Normal Form (CNF): conjunction of clauses
• Positive Occurrence: if the symbol occurs unnegated in a clause
• Negative Occurrence: if the symbol occurs negated in a clause

(s1 ∨ ¬s3) ∧ (¬s2 ∨ s3 ∨ s4)

• K. Ghorbal (INRIA)

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Definitions

• Literal: propositional symbol (atomic formula) or its negation
• Clause: disjunction of one or more literals
• Conjunctive Normal Form (CNF): conjunction of clauses
• Positive Occurrence: if the symbol occurs unnegated in a clause
• Negative Occurrence: if the symbol occurs negated in a clause

(s1 ∨ ¬s3) ∧ (¬s2 ∨ s3 ∨ s4)

• K. Ghorbal (INRIA)

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Definitions

• Literal: propositional symbol (atomic formula) or its negation
• Clause: disjunction of one or more literals
• Conjunctive Normal Form (CNF): conjunction of clauses
• Positive Occurrence: if the symbol occurs unnegated in a clause
• Negative Occurrence: if the symbol occurs negated in a clause

(s1 ∨ ¬s3) ∧ (¬s2 ∨ s3 ∨ s4)

• K. Ghorbal (INRIA)

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Definitions

• Literal: propositional symbol (atomic formula) or its negation
• Clause: disjunction of one or more literals
• Conjunctive Normal Form (CNF): conjunction of clauses
• Positive Occurrence: if the symbol occurs unnegated in a clause
• Negative Occurrence: if the symbol occurs negated in a clause

(s1 ∨ ¬s3) ∧ (¬s2 ∨ s3 ∨ s4)

• K. Ghorbal (INRIA)

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DP Algorithm

Davis, Putnam, 1960

Satisfiability-Preserving Transformations

• Pure literal rule or affirmative-negative rule
• Unit propagation or 1-literal rule
• Resolution rule or rule for eliminating literals (atomic formulas)

DP Algorithm Iteratively apply the rules till reducing the problem to a unique clause

• if the clause has the form s ∧ ¬s the problem is unsat
• otherwise, the problem is sat
• K. Ghorbal (INRIA)

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DP Algorithm

Davis, Putnam, 1960

Satisfiability-Preserving Transformations

• Pure literal rule or affirmative-negative rule
• Unit propagation or 1-literal rule
• Resolution rule or rule for eliminating literals (atomic formulas)

DP Algorithm Iteratively apply the rules till reducing the problem to a unique clause

• if the clause has the form s ∧ ¬s the problem is unsat
• otherwise, the problem is sat
• K. Ghorbal (INRIA)

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Pure Literal Rule

Pure literal i.e. appears only positively or only negatively, ℓ say Delete all clauses containing that literal

• A clause containing ℓ has the form ℓ ∨ w
• (ℓ ↔ 1) ∧ (ℓ ∨ w1) ∧ w2

∼SAT w2

• Repeat till no more clauses contain ℓ

➻ Augment σ such that ℓσ = 1 Nota: not used dynamically while solving as expensive to detect.

• K. Ghorbal (INRIA)

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Example of Preprocessing with Pure Literal Rule

(1 and 7) 1 ∨ 2 1 ∨ 3 ∨ 8 ¯ 2 ∨ ¯ 3 ∨ 4 ¯ 4 ∨ 5 ∨ 7 ¯ 4 ∨ 6 ∨ 8 ¯ 5 ∨ ¯ 6 7 ∨ ¯ 8 7 ∨ ¯ 9 ∨ 10 (¯ 2) 1 ∨ 2 1 ∨ 3 ∨ 8 ¯ 2 ∨ ¯ 3 ∨ 4 ¯ 4 ∨ 5 ∨ 7 ¯ 4 ∨ 6 ∨ 8 ¯ 5 ∨ ¯ 6 7 ∨ ¯ 8 7 ∨ ¯ 9 ∨ 10 (¯ 4 and ¯ 5) 1 ∨ 2 1 ∨ 3 ∨ 8 ¯ 2 ∨ ¯ 3 ∨ 4 ¯ 4 ∨ 5 ∨ 7 ¯ 4 ∨ 6 ∨ 8 ¯ 5 ∨ ¯ 6 7 ∨ ¯ 8 7 ∨ ¯ 9 ∨ 10 ➻ SAT! σ = {1, 7, ¯ 2, ¯ 4, ¯ 5}

• K. Ghorbal (INRIA)

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Unit Propagation

Unit clause is a clause with only one literal, ℓ say Remove all the clauses containing ℓ

• A clause containing ℓ has the form ℓ ∨ w
• (ℓ ↔ 1) ∧ (ℓ ∨ w)

∼SAT 1

• Repeat till no more clauses contain ℓ

Remove all instances of ¬ℓ from all the clauses

• A clause containing ¬ℓ has the form ℓ ∨ w
• ¬ℓσ = 0
• (ℓ ↔ 1) ∧ (¬ℓ ∨ w)

∼SAT w

• Repeat till no more clauses contain ¬ℓ

➻ Augment σ such that ℓσ = 1

• K. Ghorbal (INRIA)

25 SIF M2 25 / 42

Boolean Constraint Propagation (BCP)

• Unit propagation is a typical instance of BCP
• Consumes the most significant runtime of modern solvers

Several heuristics proved efficient

• Counter-based (GRASP) [Marques-Silva, Sakallah, 1996]
• Head/Tail lists (SATO) [Zhang, Stickel, 1996]
• 2-literal watching (Chaff) [Moskewicz et al. 2001]
• K. Ghorbal (INRIA)

26 SIF M2 26 / 42

Counter-Based Algorithm for BCP

• Denote by |C| the total number of literals in C
• Each clause C has two counters:
• C(ℓ = 0) := #ℓ such that ℓσ = 0
• C(ℓ = 1) := #ℓ such that ℓσ = 0
• Each variable s has two lists of clauses:
• Ps: set of clauses where the variable occurs positively
• Ns: set of clauses where the variable occurs negatively

If s is assigned, C(ℓ = 0) and C(ℓ = 1) for all C in Ps ∪ Ns are updated

• If C(ℓ = 0) = |C| then C is a conflicting clause (more later)
• If C(ℓ = 0) = −1 + |C| and C(ℓ = 1) = 0 then it is a unit clause
• K. Ghorbal (INRIA)

27 SIF M2 27 / 42

Counter-Based Algorithm for BCP

• Denote by |C| the total number of literals in C
• Each clause C has two counters:
• C(ℓ = 0) := #ℓ such that ℓσ = 0
• C(ℓ = 1) := #ℓ such that ℓσ = 0
• Each variable s has two lists of clauses:
• Ps: set of clauses where the variable occurs positively
• Ns: set of clauses where the variable occurs negatively

If s is assigned, C(ℓ = 0) and C(ℓ = 1) for all C in Ps ∪ Ns are updated

• If C(ℓ = 0) = |C| then C is a conflicting clause (more later)
• If C(ℓ = 0) = −1 + |C| and C(ℓ = 1) = 0 then it is a unit clause
• K. Ghorbal (INRIA)

27 SIF M2 27 / 42

Counter-Based Algorithm for BCP

• Denote by |C| the total number of literals in C
• Each clause C has two counters:
• C(ℓ = 0) := #ℓ such that ℓσ = 0
• C(ℓ = 1) := #ℓ such that ℓσ = 0
• Each variable s has two lists of clauses:
• Ps: set of clauses where the variable occurs positively
• Ns: set of clauses where the variable occurs negatively

If s is assigned, C(ℓ = 0) and C(ℓ = 1) for all C in Ps ∪ Ns are updated

• If C(ℓ = 0) = |C| then C is a conflicting clause (more later)
• If C(ℓ = 0) = −1 + |C| and C(ℓ = 1) = 0 then it is a unit clause
• K. Ghorbal (INRIA)

27 SIF M2 27 / 42

Counter-Based Algorithm for BCP

• Denote by |C| the total number of literals in C
• Each clause C has two counters:
• C(ℓ = 0) := #ℓ such that ℓσ = 0
• C(ℓ = 1) := #ℓ such that ℓσ = 0
• Each variable s has two lists of clauses:
• Ps: set of clauses where the variable occurs positively
• Ns: set of clauses where the variable occurs negatively

If s is assigned, C(ℓ = 0) and C(ℓ = 1) for all C in Ps ∪ Ns are updated

• If C(ℓ = 0) = |C| then C is a conflicting clause (more later)
• If C(ℓ = 0) = −1 + |C| and C(ℓ = 1) = 0 then it is a unit clause
• K. Ghorbal (INRIA)

27 SIF M2 27 / 42

Counter-Based Algorithm for BCP

• Denote by |C| the total number of literals in C
• Each clause C has two counters:
• C(ℓ = 0) := #ℓ such that ℓσ = 0
• C(ℓ = 1) := #ℓ such that ℓσ = 0
• Each variable s has two lists of clauses:
• Ps: set of clauses where the variable occurs positively
• Ns: set of clauses where the variable occurs negatively

If s is assigned, C(ℓ = 0) and C(ℓ = 1) for all C in Ps ∪ Ns are updated

• If C(ℓ = 0) = |C| then C is a conflicting clause (more later)
• If C(ℓ = 0) = −1 + |C| and C(ℓ = 1) = 0 then it is a unit clause
• K. Ghorbal (INRIA)

27 SIF M2 27 / 42

Counter-Based Algorithm for BCP

• Denote by |C| the total number of literals in C
• Each clause C has two counters:
• C(ℓ = 0) := #ℓ such that ℓσ = 0
• C(ℓ = 1) := #ℓ such that ℓσ = 0
• Each variable s has two lists of clauses:
• Ps: set of clauses where the variable occurs positively
• Ns: set of clauses where the variable occurs negatively

If s is assigned, C(ℓ = 0) and C(ℓ = 1) for all C in Ps ∪ Ns are updated

• If C(ℓ = 0) = |C| then C is a conflicting clause (more later)
• If C(ℓ = 0) = −1 + |C| and C(ℓ = 1) = 0 then it is a unit clause
• K. Ghorbal (INRIA)

27 SIF M2 27 / 42

Counter-Based Algorithm for BCP

• Denote by |C| the total number of literals in C
• Each clause C has two counters:
• C(ℓ = 0) := #ℓ such that ℓσ = 0
• C(ℓ = 1) := #ℓ such that ℓσ = 0
• Each variable s has two lists of clauses:
• Ps: set of clauses where the variable occurs positively
• Ns: set of clauses where the variable occurs negatively

If s is assigned, C(ℓ = 0) and C(ℓ = 1) for all C in Ps ∪ Ns are updated

• If C(ℓ = 0) = |C| then C is a conflicting clause (more later)
• If C(ℓ = 0) = −1 + |C| and C(ℓ = 1) = 0 then it is a unit clause
• K. Ghorbal (INRIA)

27 SIF M2 27 / 42

Counter-Based Algorithm for BCP

• Denote by |C| the total number of literals in C
• Each clause C has two counters:
• C(ℓ = 0) := #ℓ such that ℓσ = 0
• C(ℓ = 1) := #ℓ such that ℓσ = 0
• Each variable s has two lists of clauses:
• Ps: set of clauses where the variable occurs positively
• Ns: set of clauses where the variable occurs negatively

If s is assigned, C(ℓ = 0) and C(ℓ = 1) for all C in Ps ∪ Ns are updated

• If C(ℓ = 0) = |C| then C is a conflicting clause (more later)
• If C(ℓ = 0) = −1 + |C| and C(ℓ = 1) = 0 then it is a unit clause
• K. Ghorbal (INRIA)

27 SIF M2 27 / 42

Watched 2-Literal

Empirical fact: Find and perform literal prorogation is expensive Watched Literals

• For every clause, pick two literals to be watched
• If a literal is assigned, check only those clauses in which the literal is

watched

• When inspecting, if the propagation is not triggered, pick a new literal

to watch

• K. Ghorbal (INRIA)

28 SIF M2 28 / 42

Watched 2-Literal

Empirical fact: Find and perform literal prorogation is expensive Watched Literals

• For every clause, pick two literals to be watched
• If a literal is assigned, check only those clauses in which the literal is

watched

• When inspecting, if the propagation is not triggered, pick a new literal

to watch

• K. Ghorbal (INRIA)

28 SIF M2 28 / 42

Watched 2-Literal

Empirical fact: Find and perform literal prorogation is expensive Watched Literals

• For every clause, pick two literals to be watched
• If a literal is assigned, check only those clauses in which the literal is

watched

• When inspecting, if the propagation is not triggered, pick a new literal

to watch

• K. Ghorbal (INRIA)

28 SIF M2 28 / 42

Watched 2-Literal

Empirical fact: Find and perform literal prorogation is expensive Watched Literals

• For every clause, pick two literals to be watched
• If a literal is assigned, check only those clauses in which the literal is

watched

• When inspecting, if the propagation is not triggered, pick a new literal

to watch

• K. Ghorbal (INRIA)

28 SIF M2 28 / 42

Resolution Rule

If s does not appear in the wff w, then (s ∨ a) ∧ (¬s ∨ b) ∧ w ∼SAT (a ∨ b)

resolvent

∧w

• i

(s ∨ ai) ∧

• j

(¬s ∨ bj) ∧ w ∼SAT  

i

ai ∨

• j

bj   ∧ w  

i

ai ∨

• j

bj   ∧ w ∼  

i

• j

(ai ∨ bj)   ∧ w Resolution Rule

• Simplifies by s. No explicit assignment for s!
• K. Ghorbal (INRIA)

29 SIF M2 29 / 42

Resolution Rule

If s does not appear in the wff w, then (s ∨ a) ∧ (¬s ∨ b) ∧ w ∼SAT (a ∨ b)

resolvent

∧w

• i

(s ∨ ai) ∧

• j

(¬s ∨ bj) ∧ w ∼SAT  

i

ai ∨

• j

bj   ∧ w  

i

ai ∨

• j

bj   ∧ w ∼  

i

• j

(ai ∨ bj)   ∧ w Resolution Rule

• Simplifies by s. No explicit assignment for s!
• K. Ghorbal (INRIA)

29 SIF M2 29 / 42

Resolution Rule

If s does not appear in the wff w, then (s ∨ a) ∧ (¬s ∨ b) ∧ w ∼SAT (a ∨ b)

resolvent

∧w

• i

(s ∨ ai) ∧

• j

(¬s ∨ bj) ∧ w ∼SAT  

i

ai ∨

• j

bj   ∧ w  

i

ai ∨

• j

bj   ∧ w ∼  

i

• j

(ai ∨ bj)   ∧ w Resolution Rule

• Simplifies by s. No explicit assignment for s!
• K. Ghorbal (INRIA)

29 SIF M2 29 / 42

Resolution Rule

If s does not appear in the wff w, then (s ∨ a) ∧ (¬s ∨ b) ∧ w ∼SAT (a ∨ b)

resolvent

∧w

• i

(s ∨ ai) ∧

• j

(¬s ∨ bj) ∧ w ∼SAT  

i

ai ∨

• j

bj   ∧ w  

i

ai ∨

• j

bj   ∧ w ∼  

i

• j

(ai ∨ bj)   ∧ w Resolution Rule

• Simplifies by s. No explicit assignment for s!
• K. Ghorbal (INRIA)

29 SIF M2 29 / 42

Splitting (or Branching) Rule

Davis-Logemann-Loveland 1962

Memory Consumption The resolution rule can cause a quadratic expansion every time it is applied exhausting rapidly the available memory The DLL algorithm replaces the resolution rule with a Splitting Rule

1 Simplify by Unit Propagation and Pure Literals 2 Recursively pick a hard core variable s 3 Test if (w ∧ s) is SAT 4 Otherwise return the result for (w ∧ ¬s)

• K. Ghorbal (INRIA)

30 SIF M2 30 / 42

Splitting (or Branching) Rule

Davis-Logemann-Loveland 1962

Memory Consumption The resolution rule can cause a quadratic expansion every time it is applied exhausting rapidly the available memory The DLL algorithm replaces the resolution rule with a Splitting Rule

1 Simplify by Unit Propagation and Pure Literals 2 Recursively pick a hard core variable s 3 Test if (w ∧ s) is SAT 4 Otherwise return the result for (w ∧ ¬s)

• K. Ghorbal (INRIA)

30 SIF M2 30 / 42

Splitting (or Branching) Rule

Davis-Logemann-Loveland 1962

Memory Consumption The resolution rule can cause a quadratic expansion every time it is applied exhausting rapidly the available memory The DLL algorithm replaces the resolution rule with a Splitting Rule

1 Simplify by Unit Propagation and Pure Literals 2 Recursively pick a hard core variable s 3 Test if (w ∧ s) is SAT 4 Otherwise return the result for (w ∧ ¬s)

• K. Ghorbal (INRIA)

30 SIF M2 30 / 42

Splitting (or Branching) Rule

Davis-Logemann-Loveland 1962

Memory Consumption The resolution rule can cause a quadratic expansion every time it is applied exhausting rapidly the available memory The DLL algorithm replaces the resolution rule with a Splitting Rule

1 Simplify by Unit Propagation and Pure Literals 2 Recursively pick a hard core variable s 3 Test if (w ∧ s) is SAT 4 Otherwise return the result for (w ∧ ¬s)

• K. Ghorbal (INRIA)

30 SIF M2 30 / 42

Splitting (or Branching) Rule

Davis-Logemann-Loveland 1962

Memory Consumption The resolution rule can cause a quadratic expansion every time it is applied exhausting rapidly the available memory The DLL algorithm replaces the resolution rule with a Splitting Rule

1 Simplify by Unit Propagation and Pure Literals 2 Recursively pick a hard core variable s 3 Test if (w ∧ s) is SAT 4 Otherwise return the result for (w ∧ ¬s)

• K. Ghorbal (INRIA)

30 SIF M2 30 / 42

Splitting (or Branching) Rule

Davis-Logemann-Loveland 1962

Memory Consumption The resolution rule can cause a quadratic expansion every time it is applied exhausting rapidly the available memory The DLL algorithm replaces the resolution rule with a Splitting Rule

1 Simplify by Unit Propagation and Pure Literals 2 Recursively pick a hard core variable s 3 Test if (w ∧ s) is SAT 4 Otherwise return the result for (w ∧ ¬s)

• K. Ghorbal (INRIA)

30 SIF M2 30 / 42

DPLL Modern Decision Procedure

Zhang, Malik, 2002

s t a t u s = p r e p r o c e s s ( ) ; i f ( s t a t u s !=UNKNOWN) return s t a t u s ; while ( true ) { decide next branch ( ) ; while ( true ) { s t a t u s = deduce ( ) ; i f ( s t a t u s == CONFLICT) { b l e v e l = a n a l y z e c o n f l i c t ( ) ; i f ( b l e v e l == 0) return UNSATISFIABLE ; else backtrack ( b l e v e l ) ; } else i f ( s t a t u s == SATISFIABLE ) return SATISFIABLE ; else break ; } }

• K. Ghorbal (INRIA)

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Branching Heuristics

Which variable to branch with ? Greedy Algorithms

• Exploit the statistics of the clause database
• Estimate the branching effect on each variable (cost function)
• Ex1: Generate the largest number of implications
• Ex2: Satisfy most clauses

Heuristcs

• Maximum occurences on minimum sized clauses (MOM)
• Literal Count Heuristcs

Dynamic Largest Individual Sum (DLIS) [Marques-Silva, 1999]

• Counts the number of unresolved clauses for each free variable
• Chooses the variable with the largest number
• State-dependent (recalculated each time before branching)
• K. Ghorbal (INRIA)

32 SIF M2 32 / 42

Variable State Independent Decaying Sum

• VSIDS. [Moskewicz et al., 2001]
• Keeps two scores for each variable
• (# of pos occurences, # of neg occurences)
• Increases the score of a variable by a constant if it appears in a

learned conflicting-clause

• Periodically, all the scores are divided by a constant
• Branch with the variable with the highest combined score

➻ Cheap to maintain (State Independent) ➻ Captures the recently active variables

• K. Ghorbal (INRIA)

33 SIF M2 33 / 42

Conflict-Driven Clause Learning (CDCL)

Marques-Silva,Sakallah,1996 and Bayardo,Schrag,1997

Modern SAT solvers essentially implements a backtracking-based search algorithm Two graphs are built iteratively

• Search graph
• Implication graph
• K. Ghorbal (INRIA)

34 SIF M2 34 / 42

Search Graph

DPLL related Graphs

s3 s1 s4 s2

BCP Next Branch BCP

• K. Ghorbal (INRIA)

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Backtracking

Conflicting Clause: a clause with all its literals assigned to 0 Backtrack when a conflict occurs

• No future decisions are possible
• Backtrack to the immediately previous decision made
• Flip the assignement and continue
• If decision level 0 reached, return unsat
• K. Ghorbal (INRIA)

36 SIF M2 36 / 42

Backtracking

Conflicting Clause: a clause with all its literals assigned to 0 Backtrack when a conflict occurs

• No future decisions are possible
• Backtrack to the immediately previous decision made
• Flip the assignement and continue
• If decision level 0 reached, return unsat
• K. Ghorbal (INRIA)

36 SIF M2 36 / 42

Implication Graph

Gomes et al. Handbook of Knowledge Representation, Chapter2. 2008

No incident edges for decision nodes!

a cut corresponding to clause (¬ a ∨ ¬ b) ¬ p ¬ q b a ¬ t ¬ x1 ¬ x2 ¬ x3 y ¬ ¬ ¬ ¬ y Λ reason side conflict side conflict variable

Figure 2.1: A conflict graph

• K. Ghorbal (INRIA)

37 SIF M2 37 / 42

DPLL: Backjump and Learn

Backjump

• Jump to a past decision that caused the conflict
• (not necessarily the latest like in backtracking)
• Not unique in general (heuristics)

Learn

• Add a new clause to avoid reaching the same conflict again
• Not unique in general (heuristics)

Example: ¯ 1 ∨ 2, 3 ∨ 4, ¯ 5 ∨ ¯ 6, ¯ 2 ∨ ¯ 5 ∨ 6 (Satisfiable)

• K. Ghorbal (INRIA)

38 SIF M2 38 / 42

DPLL: Backjump and Learn

Backjump

• Jump to a past decision that caused the conflict
• (not necessarily the latest like in backtracking)
• Not unique in general (heuristics)

Learn

• Add a new clause to avoid reaching the same conflict again
• Not unique in general (heuristics)

Example: ¯ 1 ∨ 2, 3 ∨ 4, ¯ 5 ∨ ¯ 6, ¯ 2 ∨ ¯ 5 ∨ 6 (Satisfiable)

• K. Ghorbal (INRIA)

38 SIF M2 38 / 42

DPLL: Backjump and Learn

Backjump

• Jump to a past decision that caused the conflict
• (not necessarily the latest like in backtracking)
• Not unique in general (heuristics)

Learn

• Add a new clause to avoid reaching the same conflict again
• Not unique in general (heuristics)

Example: ¯ 1 ∨ 2, 3 ∨ 4, ¯ 5 ∨ ¯ 6, ¯ 2 ∨ ¯ 5 ∨ 6 (Satisfiable)

• K. Ghorbal (INRIA)

38 SIF M2 38 / 42

DPLL: Forget and Restart

Mostly used in SMT Solvers

Forget

• When too much clauses are learned
• heuristics: those not frequently used by literal propagations

Restart

• If stuck, restart from the beginning (extreme backjumping)
• Keep the learned clauses
• K. Ghorbal (INRIA)

39 SIF M2 39 / 42

DPLL: Forget and Restart

Mostly used in SMT Solvers

Forget

• When too much clauses are learned
• heuristics: those not frequently used by literal propagations

Restart

• If stuck, restart from the beginning (extreme backjumping)
• Keep the learned clauses
• K. Ghorbal (INRIA)

39 SIF M2 39 / 42

Summary

SAT Problem

• Equisatisfiability
• SAT for proving tautological implications/equivalences
• CNF transformation

CDCL-DPLL Algorithm

• Unit Propagation
• Pure Literal
• Resolution/Splitting/Conflict Learning
• K. Ghorbal (INRIA)

40 SIF M2 40 / 42

Alternative Approaches: Stalmarck’s method

• Designed to detect unsat formulas
• Works on conjunctions of triplet of the form p ↔ q ∧ r
• The transformation works like CNF, but we leave them as equivalences
• Saturation by deduction is performed till reaching a contradiction

(0 ↔ 1)

• Otherwise, Split for each variable to deduce new equivalences
• Continue with higher level of saturations
• K. Ghorbal (INRIA)

41 SIF M2 41 / 42

Saturation Rules (example)

Works on triplet of the form p ↔ q ∧ r

• If r ↔ 1, then p ↔ q
• If p ↔ 1, then q ↔ 1 and r ↔ 1
• If q ↔ 0, then p ↔ 0
• If q ↔ r, then p ↔ q and p ↔ r
• If p ↔ ¬q, then q ↔ 1 and r ↔ 0
• K. Ghorbal (INRIA)

42 SIF M2 42 / 42