1. ENGG3190 Logic Synthesis “Boolean Satisfiability” Winter 2014 S. Areibi School of Engineering University of Guelph

2. Outline SAT ProblemSAT Problem TerminologyTerminology How do we solve it?How do we solve it? ApplicationsApplications Logic Circuit RepresentationLogic Circuit Representation EfficiencyEfficiency 2

3.  What is SAT?  How to make a Boolean Function=1  Why is it important  Application (Network Repair)  Conjunction Normal Form Boolean Satisfiability

4. SAT in a Nutshell Given a Boolean formula (propositional logic formula), find a variable assignment such that the formula evaluates to 1, or prove that no such assignment exists. For n variables, there are 2 n possible truth assignments to be checked. First established NP-Complete problem. S. A. Cook, The complexity of theorem proving procedures, Proceedings, Third Annual ACM Symp. on the Theory of Computing,1971, 151-158 F = (a + b)(a’ + b’ + c) a bb cccc 0 1 0 0 00001 1 1 111

6. Using BDDs to Solve SAT R. Bryant. “Graph-based algorithms for Boolean function manipulation”. IEEE Trans. on Computers, C-35, 8:677-691, 1986. Store the function in a Directed Acyclic Graph (DAG) representation. Compacted form of the function decision tree. Reduction rules guarantee canonicity under fixed variable order. Provides for efficient Boolean function manipulation.  Overkill for SAT.  Overkill for SAT.

7. Why Bother? Core computational engine for major applications  EDA Testing and Verification Logic synthesis FPGA routing Path delay analysis And more…  AI Knowledge base deduction Automatic theorem proving

8. For what values of d0,d1,d2,d3 can Z = 1??

9. Conjunctive Normal Form F = (a + b)(a’ + b’ + c) Simple representation (more efficient data structures) Logic circuit representation Circuits have structural and direction information Circuit – CNF conversion is straightforward!!! Problem Representation a b d e c (a + b + d’) (a’ + d) (b’ + d) d  (a + b) (c’ + d’ + e) (d + e’) (c + e’) e  (c  d) literal clause

10. Clauses Positive Literal Negative Literal

11.  If a clause evaluates to a “0” we call it conflicting  It is conflicting because it conflicts with the goal of Boolean Satisfiability.  If a clause evaluates to a “1” we call it satisfied.  In the third clause since we only know the value of “a” it is unresolved! a = 0, b = 1  Clause = 0 Conflicting!! a = 0, b = 1  Clause = 1 Satisfied!! a = 0, b = 1 Unresolved!! a = 0, b = 1  Clause = 1 Satisfied!!

12.  Deduction is at the heart of Boolean Satisfiability.  If we can deduce that other values must be assigned to some value for SAT  Example?

13.  Can BDDs help?  Yes but many Boolean functions are large and using BDDs will be complex!

14.  Recursion to solve Satisfiability.  Boolean Constraint Propagation (BCP).  The Algorithm.  Limitation.  Code for SAT problem. Boolean Satisfiability

15.  However, only C is the deciding factor (exactly one unassigned literal found in the third clause).  What do we know?  c has to be a “0” so that c’=1 deduced by implication  We deduced by implication that c=0 11 00 This clause has become unit C must be ‘0’

16. Satisfied Literal Unsatisfied Literal Unassigned Literal (a +b’+  c)(b +  c’)(a’ +  c’) a = T, b = T, c is unassigned Implication Implication A variable is forced to be assigned to be True or False based on previous assignments. Unit clause rule Unit clause rule (rule for elimination of one literal clauses) An unsatisfied clause is a unit clause if it has exactly one unassigned literal. The unassigned literal is implied because of the unit clause. Boolean Constraint Propagation Boolean Constraint Propagation (BCP) Iteratively apply the unit clause rule until there is no unit clause available. a.k.a. Unit Propagation Workhorse of DLL based algorithms. Implications and Boolean Constraint Propagation

17.  Phi has nine clauses (Omega1 – Omega9)  Not easy to say  Not easy to say if Phi is Satisfiable!! Imagine thousands of clauses partial assignment  If we perform the partial assignment then what happens?

18.  Nothing much happens?  I can’t satisfy any of these clauses yet.  So what should we do?

19.  Assigning x 1 =1 implies that x2 has to equal to 1 and x3 also.!!  Which clauses imply this decision? Don’t worry about these..

20.  W7 and W8 are already satisfied! (don’t worry about them)  What next?

21.  In gigantic circuits the BCP goes on and on..

22.  If X5=1 and X6=1 then w4 and w5 are unit (Satisfiable).  So we are making great progress …  We continue to uncover clauses that make unit to reach a solution.  So far so good!  What next?  Unit

23.  Unfortunately we have a conflict …  This is one reason of using recursion.  What should we do now?  Undo some previous decision taken that might have led to this situation!!

24. what is the name  So what is the name of the algorithm for the recursive framework?

25.  This the heart and soul of every SAT engine!! (50 years old)  It is famous and has an entry in Wikipedia.

26.  Lots of work on SAT.  What do you think research concentrated on to make the algorithm more efficient?

27. DLL Algorithm Davis, Logemann and Loveland M. Davis, G. Logemann and D. Loveland, “A Machine Program for Theorem-Proving", Communications of ACM, Vol. 5, No. 7, pp. 394-397, 1962 Also known as DPLL for historical reasons Basic framework for many modern SAT solvers

28. Basic DLL Procedure - DFS (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c)

29. Basic DLL Procedure - DFS (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) a

30. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c)  Decision

31. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0  Decision

32. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0  Decision

33. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 d=1 c=0 (a + c + d) a=0 d=0 (a + c + d’) Conflict! Implication Graph

34. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 d=1 c=0 (a + c + d) a=0 d=0 (a + c + d’) Conflict! Implication Graph

35. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0  Backtrack

36. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 d=1 c=1 (a + c’ + d) a=0 d=0 (a + c’ + d’) Conflict! 1  Forced Decision

37. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1  Backtrack

38. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 1  Forced Decision

39. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 d=1 c=0 (a + c’ + d) a=0 d=0 (a + c’ + d’) Conflict! 1 c 0 1  Decision

40. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1  Backtrack

41. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 d=1 c=1 (a + c’ + d) a=0 d=0 (a + c’ + d’) Conflict! 1 c 0 1 1  Forced Decision

42. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1 1  Backtrack

43. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1 1 1  Forced Decision

44. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1 1 1 b 0  Decision

45. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1 1 1 b 0 c=1 b=0 (a’ + b + c) a=1 c=0 (a’ + b + c’) Conflict!

46. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1 1 1 b 0  Backtrack

47. Basic DLL Procedure - DFS a 0 (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1 1 1 b 0 1 a=1 b=1 c=1 (a’ + b’ + c)  Forced Decision

48. Basic DLL Procedure - DFS a (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1 1 1 b 0 1 a=1 b=1 c=1 (a’ + b’ + c)(b’ + c’ + d) d=1 0

49. Basic DLL Procedure - DFS a (a + c + d) (a + c + d’) (a + c’ + d) (a + c’ + d’) (a’ + b + c) (b’ + c’ + d) (a’ + b + c’) (a’ + b’ + c) b 0 c 0 1 c 0 1 1 1 b 0 1 a=1 b=1 c=1 (a’ + b’ + c)(b’ + c’ + d) d=1  SAT 0

50.  Since problems are large (Vars, Literals, Clauses) we cannot use BDDs!!  What are available SAT code available?

51.  Many Good SAT Solvers.  Examples:  MiniSat, from NiKlas Een, Sweden  CHAFF, Sharad Malik (Princeton)  GRASP (Greedy Randomized Adaptive Search) from Karem Sakallah, University of Michigan  … Many others … SAT Solvers

52.  Lots of information on the web and lots of papers and research  What next?  Applications in VLSI Design and Logic Synthesis …

53.  Using SAT for Logic.  Real Netlists logic of gates  CNF  Applications:  Similarity of logic functions  Gate Consistency Function. Boolean Satisfiability

54.  BDDs are not equivalent to SAT!  BDDs are a canonical representation of a Boolean Function.  We can do SAT with a BDD but it is not the most efficient way!

55. Exactly the same inputs! Is F == G?  For every pair of output  X = F 1 XOR G 1, Y = F 2 XOR G 2  Z = X OR Y, Z == 1 IFF F NOT EQUAL TO G!

56.  For NAND gate, (Phi) d = d XNOR (ab)’  (Phi) d = (a + d) (b + d)(a’ + b’ + d’)  How? PHI

57.  Y = AB + A’B’  Y = A’B + AB’

58.  You enter a pattern for inputs and output and then you say either  “YES” that is what the circuit does, or  “NO” that is not what the circuit does.  (Phi) d = d XNOR (ab)’ == 1  Consistent  (Phi) d = d XNOR (ab)’ == 0  inconsistent

59.  (Phi) g = g XNOR (d + e) = (d’ + g)(e’ + g)(d + e + g’) (d’ + g)(e’ + g)(d + e + g’)

60.  Five gate consistency functions listed!  No Boolean simplification necessary for the whole function …  How can we remember the CNF for each gate?? rules  Are there rules for all kinds of basic gates?

61.  Just need to remember them!!

64.  SAT has largely displaced BDDs for “just solve it” apps  Reason is scalability: can do large problems faster, more reliably  Still, SAT, like BDDs, not guaranteed to find a solution in reasonable time or space.  However, BDDs are still used in many important applications apart from SAT.  Many important applications of SAT for Logic Synthesis and Verification.  40 years old, but still “the” big idea: DPLL  Many recent engineering advances make it extremely fast!! Summary