Solvers for the Problem of Boolean Satisfiability SAT

Solvers for the Problem of Boolean Satisfiability (SAT) Will Klieber 15 -414 Aug 31, 2011

Why study SAT solvers? n Many problems reduce to SAT. n n n n Formal verification CAD, VLSI Optimization AI, planning, automated deduction Modern SAT solvers are often fast. Other solvers (QBF, SMT, etc. ) borrow techniques from SAT solvers and related solvers are still active areas of research. 2

Negation-Normal Form (NNF) n A formula is in negation-normal form iff: n n all negations are directly in front of variables, and the only logical connectives are: “ ”, “ ”. A literalis a variable or its negation. Convert to NNF by pushing negations inward: (De Morgan’s Laws) 3

Disjunctive Normal Form (DNF) n Recall: A literal is a variable or its negation. n A formula is in DNF iff: n n n it is a disjunction of conjunctions of literals. Every formula in DNF is also in NNF. A simple (but inefficient) way convert to DNF: n n Make a truth table for the formula φ. Each row where φ is true corresponds to a conjunct. 4

Conjunctive Normal Form (CNF) n A formula is in CNF iff: n n n Modern SAT solvers use CNF. Any formula can be converted to CNF. n n it is a conjunction of disjunctions of literals. Equivalent CNF can be exponentially larger. Equi-satisfiable CNF (Tseitin encoding): n Only linearly larger than original formula. 5

Tseitin transformation to CNF n Introduce new variables to represent subformulas. n E. g, to convert (A (B C)): n n Replace (B C) with a new variable g 1. Add clauses to equate g 1 with (B C). n n Gives value of g 1 for all 4 possible assignments to {B, C}. 6

Tseitin transformation to CNF Convert (A (B C)) to CNF by introducing new variable g 1 for (B C). 7

SAT Solvers -- Representation n A CNF formula is represented by a set of clauses. n n A clause is represented by a set of literals n n Empty set represents a true formula. Empty set represents a false clause. A variable is represented by a positive integer. The logical negation of a variable is represented by the arithmetic negation of its number. E. g. , ((x 1 x 2) ( x 1 x 2)) is represented by {{1, 2}, {-1, -2}} 8

Naïve Approach n n n SAT problem: Given a boolean formula φ, does there exist an assignment that satisfies φ? Naïve approach: Search all assignments! n n variables 2 n possible assignments n Explosion! ^ SAT is NP-complete: n n Worst case is likely O(2 n), unless P=NP. But for many cases that arise in practice, we can do much better. ^ 9

Unit Propagation n Davis-Putnam-Logemann-Loveland (DPLL) Unit Clause: Clause with exactly one literal. Algorithm: n n n If a clause has exactly one literal, then assign it true. Repeat until there are no more unit clauses. Example: n n ((x 1 x 2) ( x 1 x 2) (x 1)) (( T x 2) ( F x 2) (T)) (( T ) ( x 2 )) T 10

Helper function from copy import copy, deepcopy def Assign. Lit (Clause. List, lit): Clause. List = deepcopy(Clause. List) for clause in copy(Clause. List): if lit in clause: Clause. List. remove(clause) if -lit in clause: clause. remove(-lit) return Clause. List >>> Assign. Lit([[1, 2, -3], [-1, -2, 4], [3, 4]], 1) [[-2, 4], [3, 4]] >>> Assign. Lit([[1, 2, -3], [-1, -2, 4], [3, 4]], -1) [[2, -3], [3, 4]] Assumption: No clause contains both a variable and its negation. 11

Naïve Solver def Assign. Lit (Clause. List, lit): Clause. List = deepcopy(Clause. List) for clause in copy(Clause. List): if lit in clause: Clause. List. remove(clause) if -lit in clause: clause. remove(-lit) return Clause. List def Is. Satisfiable (Clause. List): # Test if no unsatisfied clauses remain if len(Clause. List) == 0: return True # Test for presense of empty clause if [ ] in Clause. List: return False # Split on an arbitrarily decided literal Dec. Lit = Clause. List[0][0] return (Is. Satisfiable(Assign. Lit(Clause. List, Dec. Lit)) or Is. Satisfiable(Assign. Lit(Clause. List, -Dec. Lit))) 12

DPLL Solver def Is. Satisfiable (Clause. List): # Unit propagation repeat until fixed point: for each unit clause UC in Clause. List: unc han ged Forced. Lit = UC[0] Clause. List = Assign. Lit(Clause. List, Forced. Lit) # Test if no unsatisfied clauses remain if len(Clause. List) == 0: return True # Test for presense of empty clause if [ ] in Clause. List: return False # Split on an arbitrarily decided literal Dec. Lit = (choose a variable occuring in Clause. List) return (Is. Satisfiable(Assign. Lit(Clause. List, Dec. Lit)) or Is. Satisfiable(Assign. Lit(Clause. List, -Dec. Lit))) 13

GRASP: an efficient SAT solver Original Slides by Pankaj Chauhan Modified by Will Klieber Please interrupt me if anything is not clear!

Terminology n CNF formula n n n x 1, …, xn: n variables 1, …, m: m clauses Assignment A n n n n Set of (variable, value) pairs. Notation: {(x 1, 1), (x 2, 0)}, {x 1: 1, |A| < n partial assignment |A| = n complete assignment |A= 0 falsifying assignment |A= 1 satisfying assignment |A= X unresolved asgnment φ = 1 2 3 1 = (x 2 x 3) 2 = ( x 1 x 4) 3 = ( x 2 x 4) A = {x 1=0, x 2=1, x 3=0, x 4=1} x 2: 0}, {x 1=1, x 2=0}, {x 1, x 2} {x 1=0, x 2=1, x 4=1} {x 1=0, x 2=1, x 3=0, x 4=1} {x 1=1, x 4=1} {x 1=0, x 2=0, x 4=1} 15

Terminology n An assignment partitions the clause database into three classes: n n n Satisfied, falsified, unresolved Free literal: an unassigned literal Unit clause: has exactly one free literal 16

Basic Backtracking Search Organize the search in the form of a decision tree. n n n Each node is a decision variable. Outgoing edges: assignment to the decision variable. Depth of node in decision tree is decision level (x). “ x=v @ d ” means variable x is assigned value v at decision level d. x 1 = 1@1 x 2 = 1@2 x 1 = 0@1 x 2 = 0@2 17

Basic Backtracking Search Make new decision assignments. 2. Infer implied assignments by a deduction process (unit propagation). 1. n n 3. May lead to falsifying clauses, conflict! The assignment is called “conflicting assignment”. Conflicting assignments leads to backtrack. 18

Backtracking Search in Action Example 1 x 1 = 0@1 x 2 = 0@2 1 = (x 2 x 3) 2 = ( x 1 x 4) 3 = ( x 2 x 4) x 2 x 3 = 1@2 {(x 1, 0), (x 2, 0), (x 3, 1)} No backtrack in this example! 19

Backtracking Search in Action Example 2 x 1 1 = (x 2 x 3) 2 = ( x 1 x 4) 3 = ( x 2 x 4) x 1 = 1@1 x 4 = 0@1 x 2 = 0@1 x 3 = 1@1 {(x 1, 1), (x 2, 0), (x 3, 1) , (x 4, 0)} No backtrack in this example! 20

Backtracking Search in Action Example 3 1 = (x 2 x 3) x 1 = 1@1 x 4 = 0@1 x 2 = 0@1 x 3 = 1@1 conflict x 1 = 0@1 x 2 2 = ( x 1 x 4) 3 = ( x 2 x 4) 4 = ( x 1 x 2 x 3) x 2 = 0@2 x 3 = 1@2 {(x 1, 0), (x 2, 0), (x 3, 1)} 21

GRASP n n GRASP is Generalized sea. Rch Algorithm for the Satisfiability Problem (Silva, Sakallah, ’ 96). Features: n n n Implication graphs for Unit Propagation and conflict analysis. Learning of new clauses. Non-chronological backtracking! 22

Learning n n n GRASP can learn new clauses that are logically implied by the original formula. Goal is to allow Unit Prop to deduce more forced literals, pruning the search space. Example: n n n φ contains clauses (x y z) and (x y z). Resolvingon z yields a new clause (x y). If y is false, then x must be true for φ to be true. n n n But not discoverable by simple Unit Prop w/o resolvent clause. Clause (x y) allows Unit Prop to force x=1 when y=0. New clauses learned from conflicting assignments. 23

Resolution 24

Top-level of GRASP-like solver 1. Cur. Asgn = {}; 2. while(true) { 3. while(value of φ under Cur. Asgn is unknown) { 4. Decide. Lit(); // Add decision literal to Cur. Asgn. 5. Propagate(); // Add forced literals to Cur. Asgn. 6. } 7. if (Cur. Asgn satisifies φ) {return true; } 8. Analyze conflict and learn a new clause; 9. if (the learned clause is empty) {return false; } 10. Backtrack(); 11. Propagate(); // Learned clause will force a literal 12. } 25

GRASP Decision Heuristics n n n Procedure Decide. Lit() Choose the variable that satisfies the most clauses Other possibilities exist 26

GRASP Deduction n n Unit Propagation is a type of Boolean Constraint Propagation (BCP). Grasp does Unit Prop using implication graphs: E. g. , for the clause = (x y), if y=1, then x=1 is forced; the antecedent of x is {y=1}. n If a variable x is forced by a clause during BCP, then assignment of 0 to all other literals in the clause is called the antecedent assignment A(x). n E. g. , for = (x y z), n A(x) = {y: 0, z: 1}, A(y) = {x: 0, z: 1}, A(z) = {x: 0, y: 0} Variables directly responsible forcing the value of x. n Antecedent assignment of a decision variable is empty. 27

Implication Graphs n n Depicts the antecedents of assigned variables. A node is an assignment to a variable. n (decision or implied) Predecessors of x correspond to antecedent A(x). n No predecessors for decision assignments! For special conflict vertex , antecedent A( ) is assignment to vars in the falsified clause. x 2=1@6 x 10=0@3 x 5=1@6 x 4=1@6 x 1=1@6 conflict x 6=1@6 x 3=1@6 x 9=0@1 x 11=0@3 28

Example Implication Graph Current truth assignment: {x 9=0@1, x 12=1@2, x 13=1@2, x 10=0@3, x 11=0@3} Current decision assignment: {x 1=1@6} x 10=0@3 1 = ( x 1 x 2) x 2=1@6 2 = ( x 1 x 3 x 9) 1 3 = ( x 2 x 3 x 4) 4 = ( x 4 x 5 x 10) 5 = ( x 4 x 6 x 11) 6 = ( x 5 x 6) 7 = (x 1 x 7 x 12) 8 = (x 1 x 8) x 1=1@6 2 2 x 9=0@1 3 3 x 3=1@6 4 4 x 4=1@6 5 5 x 5=1@6 6 6 conflict x 6=1@6 x 11=0@3 9 = ( x 7 x 8 x 13) 29

GRASP Conflict Analysis n n n After a conflict arises, analyze the implication graph. Add new clause that would prevent the occurrence of the same conflict in the future. Learning Determine decision level to backtrack to; this might not be the immediate one. Non-chronological backtrack 30

Learning Algorithm 1. Let CA be the assignment of False to all literals in the 2. 3. 4. 5. falsified clause. (“CA” is short for “conflict assignment”. ) 1. Example: CA= {x 5=1@6, x 6 = 1@6} A literal l 2 CA is a unique implication point (UIP) iff every other literal in CA has an earlier decision level than l. loop: 1. Remove the most recently assigned literal from CA and replace it by its antecedent. 2. if (CA is empty or has a UIP): break; Let {L 1, . . . , Ln} = CA; learn clause ( L 1 . . . Ln). Backtrack to the earliest decision level at which the learned clause will force the UIP to be false. 1. Why is this guaranteed to be possible? 31

Example Implication Graph Current truth assignment: {x 9=0@1, x 12=1@2, x 13=1@2, x 10=0@3, x 11=0@3} Current decision assignment: {x 1=1@6} x 10=0@3 1 = ( x 1 x 2) x 2=1@6 2 = ( x 1 x 3 x 9) 1 3 = ( x 2 x 3 x 4) 4 = ( x 4 x 5 x 10) 5 = ( x 4 x 6 x 11) 6 = ( x 5 x 6) 7 = (x 1 x 7 x 12) 8 = (x 1 x 8) x 1=1@6 2 2 x 9=0@1 3 3 4 4 x 4=1@6 5 5 x 3=1@6 x 5=1@6 6 6 conflict x 6=1@6 x 11=0@3 9 = ( x 7 x 8 x 13) 32

Example 1 = ( x 1 x 8 x 9 ) 2 = ( x 1 x 8 x 9) 3 = ( x 1 x 8 x 9 ) 4 = ( x 1 x 8 x 9) 5 = (x 1 x 3) 6 = (x 1 x 3) 33

Is that all? n n Huge overhead for boolean constraint propagation (BCP) Better decision heuristics Better learning, problem specific Better engineering! Chaff 34