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The DPLL algorithm Combinatorial Problem Solving (CPS) Albert - - PowerPoint PPT Presentation

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The DPLL algorithm Combinatorial Problem Solving (CPS) Albert Oliveras Enric Rodr guez-Carbonell May 22, 2020 Overview of the session Designing an efficient SAT solver DPLL: A Bit of History Abstract DPLL: Rules

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SLIDE 1

The DPLL algorithm

Combinatorial Problem Solving (CPS)

Albert Oliveras Enric Rodr´ ıguez-Carbonell

May 22, 2020

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SLIDE 2

Overview of the session

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Designing an efficient SAT solver

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DPLL: A Bit of History

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Abstract DPLL:

◆

Rules

◆

Examples

◆

Theoretical Results

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SLIDE 3

Designing an efficient SAT solver

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INPUT: formula F in CNF OUTPUT:

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If F is SAT: YES + model

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If F is UNSAT: NO + refutation (proof of unsatisfiability) Two possible methods:

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resolution-based:

  • not direct to obtain model

+ straightforward to give refutation

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DPLL-based: + straightforward to obtain model

  • not direct to give refutation

Due to their efficiency, DPLL-based solvers are the method of choice

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SLIDE 4

Overview of the session

3 / 11

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Designing an efficient SAT solver

■

DPLL: A Bit of History

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Abstract DPLL:

◆

Rules

◆

Examples

◆

Theoretical Results

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SLIDE 5

DPLL - A Bit of History

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Original DPLL was incomplete method for FOL satisfiability

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First paper (Davis and Putnam) in 1960: memory problems

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Second paper (Davis, Logemann and Loveland) in 1962: Depth-first-search with backtracking

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Late 90’s and early 00’s improvements make DPLL efficient:

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Break-through systems: GRASP, SATO, Chaff, MiniSAT

Stalmarck:1k var 1996 SATO:1k var 1996 GRASP:1k var 1996 DLL:10 var 1962 Chaff:10k var 2001 BDD:100 var 1986 DP:10 var 1960 MiniSAT:100k var 2003

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SLIDE 6

Overview of the session

4 / 11

■

Designing an efficient SAT solver

■

DPLL: A Bit of History

■

Abstract DPLL:

◆

Rules

◆

Examples

◆

Theoretical Results

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SLIDE 7

Our Abstraction of DPLL

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Given F in CNF, DPLL tries to build assignment M s.t. M | = F

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Assignments M are represented as sequences of literals (those to be true): EXAMPLE: sequence pqr is M(p) = 1, M(q) = 0, M(r) = 1 (overlining bar ¯ may be used to represent negation, like ¬)

◆

Order in M matters

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No literal appears twice in M

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No contradictory literals in M

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Sequences may have decision literals, denoted ld.

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We will introduce a transition system modelling DPLL

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States in the transition system are pairs M | | F, where M is a (partial) assignment and F is a CNF

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The algorithm starts with an empty assignment

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The rules in the transition system indicate which steps M | | F = ⇒ M′ | | F ′ are allowed.

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SLIDE 8

Abstract DPLL - Rules

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Extending the model:

Decide M | | F = ⇒ M ld | | F if

  • l or ¯

l occurs in F l is undefined in M UnitProp M | | F, C ∨ l = ⇒ M l | | F, C ∨ l if

  • M |

= ¬ C l is undefined in M

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SLIDE 9

Abstract DPLL - Rules (2)

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Repairing the model:

Fail M | | F, C = ⇒ fail if

  • M |

= ¬C M contains no decision literals Backtrack M ld N | | F, C = ⇒ M ¯ l | | F, C if

  • M ld N |

= ¬C N contains no decision lits

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SLIDE 10

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒

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SLIDE 11

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide)

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SLIDE 12

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒

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SLIDE 13

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp)

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SLIDE 14

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒

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SLIDE 15

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide)

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SLIDE 16

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒

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SLIDE 17

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp)

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SLIDE 18

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒

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SLIDE 19

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide)

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SLIDE 20

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒

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SLIDE 21

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp)

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SLIDE 22

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 5d 6 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒

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SLIDE 23

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 5d 6 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Backtrack)

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SLIDE 24

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 5d 6 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Backtrack) 1d 2 3d 4 5 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒

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SLIDE 25

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 5d 6 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Backtrack) 1d 2 3d 4 5 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide)

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SLIDE 26

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 5d 6 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Backtrack) 1d 2 3d 4 5 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5 6d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2

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SLIDE 27

Abstract DPLL - Example 1

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∅ | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (UnitProp) 1d 2 3d 4 5d 6 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Backtrack) 1d 2 3d 4 5 | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 = ⇒ (Decide) 1d 2 3d 4 5 6d | | 1 ∨ 2, 3 ∨ 4, 5 ∨ 6, 6 ∨ 5 ∨ 2 Final state found!

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SLIDE 28

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒

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SLIDE 29

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp)

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SLIDE 30

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒

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SLIDE 31

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide)

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SLIDE 32

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒

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SLIDE 33

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp)

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SLIDE 34

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2d 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒

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SLIDE 35

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2d 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Backtrack)

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SLIDE 36

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2d 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Backtrack) 1 2 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒

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SLIDE 37

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2d 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Backtrack) 1 2 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp)

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SLIDE 38

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2d 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Backtrack) 1 2 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒

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SLIDE 39

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2d 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Backtrack) 1 2 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Fail)

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SLIDE 40

Abstract DPLL - Example 2

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∅ | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Decide) 1 2d | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2d 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Backtrack) 1 2 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (UnitProp) 1 2 3 | | 1 ∨ 2 ∨ 3, 1, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3, 2 ∨ 3 = ⇒ (Fail) fail

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SLIDE 41

Abstract DPLL

10 / 11

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There are no infinite sequences of the form ∅ | | F = ⇒ . . .

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If ∅ | | F = ⇒∗ M | | F with state M | | F final, then

◆

F is satisfiable

◆

M is a model of F

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If ∅ | | F = ⇒∗ fail then F is unsatisfiable Hence the transition system gives a decision procedure for SAT

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SLIDE 42

Bibliography - Some further reading

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Martin Davis, Hilary Putnam. A Computing Procedure for Quantification Theory. J. ACM 7(3): 201-215 (1960)

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Martin Davis, George Logemann, Donald W. Loveland. A machine program for theorem-proving.

  • Commun. ACM 5(7): 394-397 (1962)

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Robert Nieuwenhuis, Albert Oliveras, Cesare Tinelli. Solving SAT and SAT Modulo Theories: From an abstract Davis–Putnam–Logemann–Loveland procedure to DPLL(T).

  • J. ACM 53(6): 937-977 (2006)