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GRASP SAT solver

GRASP SAT solver. J. Marques-Silva and K. Sakallah. Presented by Constantinos Bartzis Slides borrowed from Pankaj Chauhan. What is SAT?. Given a propositional formula in CNF, find an assignment to boolean variables that makes the formula true E.g.  1 = (x 2  x 3 )

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GRASP SAT solver

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  1. GRASP SAT solver J. Marques-Silva and K. Sakallah Presented by Constantinos Bartzis Slides borrowed from Pankaj Chauhan

  2. What is SAT? • Given a propositional formula in CNF, find an assignment to boolean variables that makes the formula true • E.g. 1 = (x2  x3) 2 = (x1  x4) 3 = (x2  x4) A = {x1=0, x2=1, x3=0, x4=1} SATisfying assignment!

  3. Why is SAT important? • Fundamental problem from theoretical point of view • Numerous applications • CAD, VLSI • Optimization • Model Checking and other kinds of formal verification • AI, planning, automated deduction

  4. Terminology • CNF formula  • x1,…, xn: n variables • 1,…, m: m clauses • Assignment A • Set of (x,v(x)) pairs • |A| < n  partial assignment {(x1,0), (x2,1), (x4,1)} • |A| = n  complete assignment {(x1,0), (x2,1), (x3,0), (x4,1)} • |A= 0  unsatisfyingassignment {(x1,1), (x4,1)} • |A= 1  satisfying assignment {(x1,0), (x2,1), (x4,1)} • |A= X  unresolved {(x1,0), (x2,0), (x4,1)} 1 = (x2  x3) 2 = (x1  x4) 3 = (x2  x4) A = {x1=0, x2=1, x3=0, x4=1}

  5. Terminology • An assignment partitions the clause databaseinto three classes • Satisfied • Unsatisfied • Unresolved • Free literals: unassigned literals of a clause • Unitclause: unresolved with only one free literal

  6. Basic Backtracking Search • Organize the search in the form of a decision tree • Each node is an assignment, called decision assignment • Depth of the node in the decision tree decision level (x) • x=v@d  xis assigned tovat decision leveld

  7. Basic Backtracking Search • Iteration: • Make new decision assignments to explore new regions of search space • Infer implied assignments by a deduction process. • May lead to unsatisfied clauses, conflict. The assignment is called conflicting assignment. • If there is a conflict backtrack

  8. x1 x2 DPLL in action x1 = 0@1  1 = (x1  x2  x3  x4) 2 = (x1  x3  x4) 3 = (x1  x2  x3) 4 = (x3  x4) 5 = (x4  x1  x3) 6 = (x2  x4) 7 = (x2  x4 )   U / /  U / x2 = 0@2  / / /   x3 = 1@2   x4 = 1@2 conflict

  9. x1 x2 DPLL in action x1 = 0@1  1 = (x1  x2  x3  x4) 2 = (x1  x3  x4) 3 = (x1  x2  x3) 4 = (x3  x4) 5 = (x4  x1  x3) 6 = (x2  x4) 7 = (x2  x4 )   / x2 = 0@2 / x2 = 1@2 / U   x3 = 1@2  x4 = 1@2  / /  x4 = 1@2 conflict conflict

  10. x1 x4 x2 x2 DPLL in action x1 = 0@1  1 = (x1  x2  x3  x4) 2 = (x1  x3  x4) 3 = (x1  x2  x3) 4 = (x3  x4) 5 = (x4  x1  x3) 6 = (x2  x4) 7 = (x2  x4 ) / / x1 = 1@1 /    x2 = 0@2 x2 = 0@2  x2 = 1@2   x3 = 1@2  x4 = 1@2   x4 = 1@2 x4 = 1@2 conflict conflict {(x1,1), (x2,0), (x4,1)}

  11. DPLL Algorithm Deduction Decision Backtrack

  12. GRASP • GRASP stands for Generalized seaRch Algorithm for the Satisfiability Problem (Silva, Sakallah, ’96) • Features: • Implication graphs for BCP and conflict analysis • Learning of new clauses • Non-chronological backtracking

  13. GRASP search template

  14. GRASP Decision Heuristics • Procedure decide() • Which variable to split on • What value to assign • Default heuristic in GRASP: Choose the variable and assignment that directly satisfies the largest number of clauses • Other possibilities exist

  15. GRASP Deduction • Boolean Constraint Propagation using implication graphs E.g. for the clause  = (x y), ify=1, then we must havex=1 • For a variable xoccuring in a clause, assignment 0 to all other literals is calledantecedent assignmentA(x) • E.g. for  = (x y  z), A(x) = {(y,0), (z,1)}, A(y) = {(x,0),(z,1)}, A(z) = {(x,0), (y,0)} • Variables directly responsible for forcing the value ofx • Antecedent assignment of a decision variable is empty

  16. Implication Graphs • Nodes are variable assignments x=v(x) (decision or implied) • Predecessors of x are antecedent assignments A(x) • No predecessors for decision assignments! • Special conflict vertices  have A() = assignments to variables in the unsatisfied clause • Decision level for an implied assignment is (x) = max{(y)|(y,v(y))A(x)}

  17. 4 x2=1@6 x5=1@6 1 4 3 6 x4=1@6  conflict 6 3 2 5 2 5 x6=1@6 x3=1@6 Example Implication Graph Current truth assignment: {x9=0@1 ,x10=0@3, x11=0@3, x12=1@2, x13=1@2} Current decision assignment: {x1=1@6} 1 = (x1  x2) 2 = (x1  x3  x9) 3 = (x2  x3  x4) 4 = (x4  x5  x10) 5 = (x4  x6  x11) 6 = (x5  x6) 7 = (x1  x7  x12) 8 = (x1 x8) 9 = (x7  x8   x13) x10=0@3 x1=1@6 x9=0@1 x11=0@3

  18. GRASP Deduction Process

  19. GRASP Conflict Analysis • After a conflict arises, analyze the implication graph at current decision level • Add new clauses that would prevent the occurrence of the same conflict in the future  Learning • Determine decision level to backtrack to, might not be the immediate one  Non-chronological backtracking

  20. Learning • Determine the assignment that caused conflict  • Backward traversal of the IG, find the roots of the IG in the transitive fanin of  • This assignment is necessary condition for  • Negation of this assignment is called conflict induced clause C() • Adding C() to the clause database will prevent the occurrenceof  again

  21. 4 x2=1@6 x5=1@6 1 4 3 6 x4=1@6  conflict 6 3 2 5 2 5 x6=1@6 x3=1@6 Learning x10=0@3 x1=1@6 x9=0@1 x11=0@3 C() = (x1  x9  x10  x11)

  22. (x,v(x)) if A(x) =  causesof(x) = (x) [ causesof(y)]o/w (y,v(y))  (x) Learning • For any node of an IG x, partition A(x) into (x) = {(y,v(y)) A(x)|(y)<(x)} (x) = {(y,v(y)) A(x)|(y)=(x)} • Conflicting assignment AC() = causesof(), where

  23. Learning • Learning of new clauses increases clause database size • Increase may be exponential • Heuristically delete clauses based on a user provided parameter • If size of learned clause > parameter, don’t include it

  24. Backtracking Failure driven assertions (FDA): • If C() involves current decision variable, a different assignment for the current variable is immediately tried. • In our IG, after erasing the assignment at level 6, C() becomes a unit clause x1 • This immediately implies x1=0

  25. 4 x2=1@6 x5=1@6 1 4 3 6 x4=1@6  conflict 6 3 2 5 2 5 x6=1@6 x3=1@6 Example Implication Graph Current truth assignment: {x9=0@1 ,x10=0@3, x11=0@3, x12=1@2, x13=1@2} Current decision assignment: {x1=1@6} 1 = (x1  x2) 2 = (x1  x3  x9) 3 = (x2  x3  x4) 4 = (x4  x5  x10) 5 = (x4  x6  x11) 6 = (x5   x6) 7 = (x1  x7  x12) 8 = (x1 x8) 9 = (x7  x8   x13) C()=(x1  x9  x10  x11) x10=0@3 x1=1@6 x9=0@1 x11=0@3

  26. Example Implication Graph Current truth assignment: {x9=0@1 ,x10=0@3, x11=0@3, x12=1@2, x13=1@2} Current decision assignment: {x1=0@6} 1 = (x1  x2) 2 = (x1  x3  x9) 3 = (x2  x3  x4) 4 = (x4  x5  x10) 5 = (x4  x6  x11) 6 = (x5  x6) 7 = (x1  x7  x12) 8 = (x1 x8) 9 = (x7  x8   x13) C() = (x1  x9  x10  x11) x9=0@1 C() x10=0@3 x1=0@6 C() C() x11=0@3

  27. Decision level 3 4 5 x1 6  ' Non-chronological backtracking x8=1@6 x9=0@1 C() 8 9 x10=0@3 ’ x13=1@2 x1=0@6 C() 9 9 C() 7 x11=0@3 x7=1@6 7 x12=1@2 AC(’) = {x9 =0@1, x10 = 0@3, x11 = 0@3, x12=1@2, x13=1@2} C() = (x9  x10  x11   x12   x13)

  28. Backtracking • Backtrack level is given by • = d-1chronological backtrack • < d-1non-chronological backtrack  = max{(x)|(x,v(x))AC(’)}

  29. Procedure Diagnose()

  30. What’s next? • Reduce overhead for constraint propagation • Better decision heuristics • Better learning, problem specific • Better engineering Chaff

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