1. 1 Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

2. 2 Conflict clauses and Resolution Binary-resolution is a sound inference rule: Example:

3. 3 Consider the following example: Conflict clause: c 5 : ( x 2 Ç : x 4 Ç x 10 ) We show that c 5 is inferred by resolution from c 1,…, c 4 Conflict clauses and resolution

4. 4 Conflict clause: c 5 : ( x 2 Ç : x 4 Ç x 10 ) BCP order: x 4, x 5, x 6, x 7  T1 = Res(c 4,c 3,x 7 ) = ( : x 5 Ç : x 6 )  T2 = Res(T1, c 2, x 6 ) = ( : x 4 Ç : x 5 Ç X 10 )  T3 = Res(T2,c 1,x 5 ) = (x 2 Ç : x 4 Ç x 10 ) Conflict clauses and resolution

5. 5 Applied to our example: Finding the conflict clause: cl is asserting the first UIP

6. 6 Conflict clauses and resolution c2c2 c3c3 c4c4 c5c5 c1c1 This will be part of the (Hyper) Resolution Graph:

7. 7 The resolution graph What is it good for ? Example: for computing an Unsatisfiable core [Picture Borrowed from Zhang, Malik SAT’03]

8. Minimizing the core The proof is not unique.  Different proofs / different cores. Can we find a minimum / minimal / smaller cores/proofs? 8

9. Minimizing the core Core compression [ZM03,…]…  Min-core-biased search [NRS’13] Proof compression:  Exponential-time transformations [GKS’06].  Linear time transformations “Recycle units” [BFHSS’08] 9

10. Core compression (smaller core) A basic approach: run until reaching a fixpoint [chaff] 10 SAT solver  core == last_core ? core last_core = core yes no initially last_core = ;

11. 11 Remove an unmarked clause c 2  SAT(  ) ? mark c yes  := core no Initially  ’s clauses are unmarked Return  All marked Core compression (minimal core)

12. 12 Proof compression Example of a linear-time transformation / “Recycle-units” Observation: When learning (resolving) a new clause in SAT,  The resolving clauses are not satisfied  Hence, the resolution-variable is unassigned l l 1 : l l 2 l 1 l 2

13. 13 Example (cont’d) Suppose that the pivot’s constant value is learned later on. We will use it to simplify the resolution proof.

14. 14 Recycle-units / easy case 1 3 -1 2 5 2 3 51 -2 1 3 5 -1 4-1 -4 3 5

15. 15 Recycle-units / easy case 1 3 -1 2 5 2 3 51 -2 1 3 5 -1 4-1 -4 3 5

16. 16 Recycle-units 1 3 31 -2 1 3 5 -1 4-1 -4 3 5 3 3

17. 17 Recycle-units 1 3 3 -1 4-1 -4 Reduced proof by 4 clauses Reduced core by 2 clauses

18. 18 Recycle-units / beware of cycles 1 3 2 3 51 -2 1 3 5 -1 4-1 -4 By making this connection we created cyclic reasoning

19. 19 Recycle-units / beware of cycles Solution:  mark antecedents of units  apply only to unmarked nodes 1 3 2 3 51 -2 1 3 5 -1 4-1 -4 -1 2 5 3 5

20. 20 Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

21. 21 4. Periodically, all the counters are divided by a constant. 3. The unassigned variable with the highest counter is chosen. 2. When a clause is added, the counters are updated. 1. Each variable in each polarity has a counter initialized to 0. (Implemented in Chaff) Decision heuristics VSIDS (Variable State Independent Decaying Sum)

22. 22 Chaff holds a list of unassigned variables sorted by the counter value. Updates are needed only when adding conflict clauses. Thus - decision is made in constant time. Decision heuristics VSIDS (cont ’ d)

23. 23 VSIDS is a ‘quasi-static’ strategy: - static because it doesn’t depend on current assignment - dynamic because it gradually changes. Variables that appear in recent conflicts have higher priority. This strategy is a conflict-driven decision strategy. “..employing this strategy dramatically (i.e. an order of magnitude) improved performance... “ Decision heuristics VSIDS (cont ’ d)

24. 24 Decision Heuristics - Berkmin Keep conflict clauses in a stack Choose the first unresolved clause in the stack  If there is no such clause, use VSIDS Choose from this clause a variable + value according to some scoring (e.g. VSIDS) This gives absolute priority to conflicts.

25. 25 Berkmin heuristic tail- first conflict clause

26. 26 Deduction() allocates new implied variables and conflicts. How can this be done efficiently ? Observation: More than 90% of the time SAT solvers perform Deduction(). More engineering aspects of SAT solvers

27. 27 Hold 2 counters for each clause  : val1 (  ) - # of negative literals assigned 0 in  + # of positive literals assigned 1 in . val0 (  ) - # of negative literals assigned 1 in  + # of positive literals assigned 0 in . Grasp implements Deduction() with counters

28. 28 Every assignment to a variable x results in updating the counters for all the clauses that contain x. Grasp implements Deduction() with counters  is satisfied iff val1(  ) > 0  is unsatisfied iff val0(  ) = |  |  is unit iff val1(  ) = 0  val0(  ) = |  | - 1  is unresolved iff val1(  ) = 0  val0(  ) < |  | - 1. Backtracking: Same complexity.

29. 29 Observation: during Deduction(), we are only interested in newly implied variables and conflicts. These occur only when the number of literals in  with value ‘ false ’ is greater than |  | - 2 Conclusion: no need to visit a clause unless (val0(  ) > |  | - 2)  How can this be implemented ? Chaff implements Deduction() with 2 observers

30. 30 For each clause ¼, define two ‘ observers ’ : O1(  ), O2(  ). O1(  ) and O2(  ): two  literals which are not ‘ false ’.  is unit if updating one observer leads to O1(  ) = O2(  ). Visit clause  only if O1(  ) or O2(  ) become ‘ false ’. Chaff implements Deduction() with 2 observers

31. 31 V[1]=0 V[5]=0, v[4]= 0 Unit clause Backtrack v[4] = v[5]= X v[1] = 1 Chaff implements Deduction() with 2 observers V[2]=0 O1O2

32. 32 Complexity of Backtracking:  Observers of a unit clause are on the highest decision level present in the clause.  Hence backtracking will un-assign them first. Conclusion: when backtracking, observers stay in place. Chaff implements Deduction() with 2 observers

33. 33 Chaff implements Deduction() with 2 observers Which observing literals?  Strategy: the least frequently updated variables The observers method has a learning curve  Initial observers: chosen arbitrarily.  Then: shifted from variables that were recently updated These variables will most probably be reassigned soon. In our example: the next time v[5] is updated, it will point to a significantly smaller set of clauses.

34. 34 Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

35. 35 GSAT: stochastic SAT solving for i = 1 to max_tries { T := randomly generated truth assignment for j = 1 to max_flips { if T satisfies  return TRUE choose v s.t. flipping v’s value gives largest increase in the # of satisfied clauses (break ties randomly). T := T with v’s assignment flipped. } } Given a CNF formula , choose max_tries and max_flips Many alternative heuristics

36. 36 Numerous progressing heuristics Hill-climbing Tabu-list Simulated-annealing Random-Walk Min-conflicts...

37. 37 Improvement # 1: clause weights Initial weight of each clause: 1 Increase by k the weight of unsatisfied clauses. Choose v according to maximum increase in weight Clause weights is another example of conflict-driven decision strategy.

38. 38 Improvement # 2: Averaging-in Q: Can we reuse information gathered in previous tries in order to speed up the search ? A: Yes! Rather than choosing T randomly each time, repeat ‘good assignments’ and choose randomly the rest.

39. 39 Let X1, X2 and X3 be equally wide bit vectors. Define a function bit_average : X1  X2  X3 as follows: b 1 i b 1 i = b 2 i randomotherwise (where b j i is the i-th bit in Xj, j  {1,2,3}) b 3 i := Improvement # 2: Averaging-in

40. 40 Notation: T i init : The initial assignment T in cycle i. T i best : The assignment with highest # of satisfied clauses in cycle i. T 1 init := random assignment. T 2 init := random assignment.  i > 2, T i init := bit_average(T i-1 best, T i-2 best ) Improvement # 2: Averaging-in