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Chaff: Engineering An Efficient SAT Solver

Chaff: Engineering An Efficient SAT Solver. Presented by Monissa Mohan. Two Strategies that make Chaff work better:. A highly optimized BCP algorithm Two watched literals Fast Backtracking Efficient Decision Heuristic Focused on recently added clauses Highly optimized for speed.

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Chaff: Engineering An Efficient SAT Solver

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  1. Chaff: Engineering An Efficient SAT Solver Presented by Monissa Mohan

  2. Two Strategies that make Chaff work better: • A highly optimized BCP algorithm • Two watched literals • Fast Backtracking • Efficient Decision Heuristic • Focused on recently added clauses • Highly optimized for speed

  3. Some Definitions • Boolean Constraint Propagation (BCP): Identify all the variable assignments required by the current variable state to satisfy f • For f to be sat, every clause must be sat

  4. Implication • Unit Clause rule: • All literals are False(F) with one unassigned literal • These clauses are called UNIT CLAUSES • This literal has to take on the value of True(T) to make fsat f= v1+v2+v3 Now v1 and v2 has been assigned to false(F) • f=F+F+v3 Now f is called a unit clause v3 takes the value of True(T) to make f sat.

  5. Implication for N-literal clause Implication occurs after N-1 assignments of False to its literals • Goal of BCP: • Visit the clause only when the number of zero literals goes from N-2 to N-1 • How is this done? • Theoretically, we could ignore the first N-2 assignments to this clause • The first N-2 assignments won’t have any effect on the BCP (…….+x+y+…….) N literals

  6. Watched Literals • Each clause has two watched literals • Ignore any assignments to the other literals in the clause. • BCP Maintains the following rule • By the end of BCP, one of the watched literal is true or both are undefined. • Guaranteed to find all implications (…….+x+y+…….) N literals

  7. Advantages of Watched Literals • While backtracking, no need to modify watched literals • Variable may be watched only in a small subset of clauses. This reduces the total number of memory accesses which is critical for SAT solvers

  8. Example(1/13) Four clauses are present: • v2 + v3 + v1 + v4 • v1 + v2 + v3’ • v1 + v2’ • v1’+ v4

  9. Example(2/13) Watched Literals: • v2 + v3 + v1 + v4 • v1 + v2+ v3’ • v1 + v2’ • v1’+ v4

  10. Example (3/13) v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:(v1=F) • Assume we decide to set v1 the value F

  11. Example (4/13) v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:(v1=F) • Ignore clauses with a watched literal whose value is T

  12. Example (5/13) v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:(v1=F) • Ignore clauses where neither watched literal value changes

  13. Example (6/13) v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:(v1=F) • Examine clauses with a watched literal whose value is F

  14. Example (7/13) v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:(v1=F) Stack:(v1=F) • In the second clause, replace the watched literal v1 with v3’

  15. Example (8/13) v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:(v1=F) Pending: (v2=F) Stack:(v1=F) • The third clause is a unit and implies v2=F • We record the new implication, and add it to a • queue of assignments to process.

  16. Example (9/13) v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:(v1=F, v2=F) Pending: (v3=F) Stack:(v1=F, v2=F) • Next, we process v2. • We only examine the first 2 clauses

  17. Example (10/13) v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 v2 + v3 + v1 + v4 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:(v1=F, v2=F) Pending: (v3=F) Stack:(v1=F, v2=F) • In the first clause, we replace v2 with v4 • The second clause is a unit and implies v3=F • We record the new implication, and add it to the queue

  18. Example (11/13) v2 +v3+ v1 + v4 v1 +v2+ v3’ v1 + v2’ v1’+ v4 v2 +v3+ v1 + v4 v1 +v2+ v3’ v1 + v2’ v1’+ v4 Stack:(v1=F, v2=F, v3=F) Pending: () Stack:(v1=F, v2=F, v3=F) • Next, we process v3’. We only examine the first clause.

  19. Example (12/13) v2 +v3+ v1 + v4 v1 +v2+ v3’ v1 + v2’ v1’+ v4 v2 +v3+ v1 + v4 v1 +v2+ v3’ v1 + v2’ v1’+ v4 Stack:(v1=F, v2=F, v3=F) Pending: (v4=T) Stack:(v1=F, v2=F, v3=F) • The first clause is a unit and implies v4=T. • We record the new implication, and add it to the queue.

  20. Example (13/13) v2 +v3+ v1 + v4 v1 +v2+ v3’ v1 + v2’ v1’+ v4 Stack:(v1=F, v2=F, v3=F, v4=T) • There are no pending assignments, and no conflict • Therefore, BCP terminates and so does the SAT solver

  21. Identify Conflicts v2+v3+ v1 v1 +v2+ v3’ v1 + v2’ v1’+ v4 Stack:(v1=F, v2=F, v3=F) • What if the first clause does not have v4? • When processing v3’, we examine the first clause. • This time, there is no alternative literal to watch. • BCP returns a conflict

  22. Backtrack v2 + v3 + v1 v1 + v2 + v3’ v1 + v2’ v1’+ v4 Stack:() • We do not need to move any watched literal

  23. Chaff Decision Heuristics • Variable State Independent Decaying Sum(VSIDS) • Rank variables based on literal count in the initial clause database. • Only increment counts as new clauses are added. • Periodically, divide all counts by a constant.

  24. VSIDS Example (1/2) New clause added x1 + x4 x1 + x3’ + x8’ x1 + x8 + x12 x2 + x11 x7’ + x3’ + x9 x7’ + x8 + x9’ x7 + x8 + x10’ x7 + x10 + x12’ Scores: 4: x8,x7 3: x1 2: x3,x10,x12 1: x2,x4,x9,x11 Initial data base x1 + x4 x1 + x3’ + x8’ x1 + x8 + x12 x2 + x11 x7’ + x3’ + x9 x7’ + x8 + x9’ x7 + x8 + x10’ Scores: 4: x8 3: x1,x7 2: x3 1: x2,x4,x9,x10,x11,x12 watch what happens to x8, x7 and x1

  25. VSIDS Example (2/2) Counters divided by 2 x1 + x4 x1 + x3’ + x8’ x1 + x8 + x12 x2 + x11 x7’ + x3’ + x9 x7’ + x8 + x9’ x7 + x8 + x10’ x7 + x10 + x12’ Scores: 2: x8,x7 1: x3,x10,x12,x1 0: x2,x4,x9,x11 New clause added x1 + x4 x1 + x3’ + x8’ x1 + x8 + x12 x2 + x11 x7’ + x3’ + x9 x7’ + x8 + x9’ x7 + x8 + x10’ x7 + x10 + x12’ x12’ + x10 Scores: 2: x8,x7,x12,x10 1: x3,x1 0: x2,x4,x9,x11 watch what happens to x8, x10

  26. Features of Chaff(1/2) • Clause Deletion • Avoids memory explosion • Determines when a clause can be deleted • When more than N literals in the clause becomes unassigned for the first time, the clause will be marked as deleted

  27. Features of Chaff(2/2) • Restarts • Halt in the procedure and a restart of the analysis using some previous information • Clearing the state and decisions of all variables and proceeding as normal • The clauses learned prior to the restart are still there after the restart and can help pruning the search space

  28. Experimental Results • One to two orders of magnitude faster than the best public domain solvers • Works equally good on difficult problems prevalent in EDA domain • Speedup simply comes because of efficient engineering in basic search algorithm

  29. References • Chaff: Engineering an Efficient SAT Solver Matthew W. Moskewicz , Conor F. Madigan, Ying Zhao, Lintao Zhang, Sharad Malik • Boolean Satisfiability in Electronic Design Automation João P. Marques-Silva Karem A. Sakallah

  30. Thank you!Questions?

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