1. Carnegie Mellon University Boolean Satisfiability with Transitivity Constraints Boolean Satisfiability with Transitivity Constraints http://www.cs.cmu.edu/~bryant Randal E. Bryant Miroslav N. Velev

2. – 2 – Outline Application Domain Verify correctness of a pipelined processor Based on Burch-Dill correspondence checking Burch & Dill, CAV ‘94 Verification Task Decide validity of formula in logic of equality with uninterpreted functions Translate into equational logic Propositional logic with equations of form v i = v j Bryant, German & Velev, CAV ’99 Goel, Sahid, Zhou, Aziz, & Singhal, CAV ‘98 New Contribution Efficient handling of transitivity constraints

3. – 3 – Decision Problem Logic of Equality with Uninterpreted Functions (EUF) Truth Values Dashed Lines Model control signals Domain Values Solid lines Model data wordsTask Determine whether formula is universally valid True for all interpretations of variables and function symbols

4. – 4 – Eliminating Function Applications Verification Task Prove: x = f(f(x))  x = f(f(f(x)))  x = f(x) Instance of: x = y  x = f(y)  x = f(x) Ackermann’s Method Replace: f(x)  f 1 f( f(x))  f 2 f(f( f(x)))  f 3 Gives: x = f 2  x = f 3  x = f 1 Functional Consistency Constraints x = f 1  f 1 = f 2 f 1 = f 2  f 2 = f 3 x = f 2  f 1 = f 3

5. – 5 – Eliminating Funct. Apps. (cont.) Equational Formula Complement of substituted formula + consistency constraints ClausesOrigin x = f 2  x = f 3  x  f 1  [x = f 2  x = f 3  x = f 1 ]  ( x  f 1  f 1 = f 2 ) x = f 1  f 1 = f 2  ( f 1  f 2  f 2 = f 3 ) f 1 = f 2  f 2 = f 3  ( x  f 2  f 1 = f 3 ) x = f 2  f 1 = f 3 Verification Task Prove that equational formula is not satisfiable x = f 2  x = f 3  x  f 1  ( x  f 1  f 1 = f 2 )  ( f 1  f 2  f 2 = f 3 )  ( x  f 2  f 1 = f 3 )

6. – 6 – Solving Equational Formulas Historically E.g., Nelson & Oppen ‘80 Create special purpose search engine Davis-Putnam search Data structure to maintain equivalence classesQuestion Can we translate problem into pure propositional logic? Would enable use of BDDs or SAT checkers

7. – 7 – Replacing Equations by Variables Relational Variables Goel, Sahid, Zhou, Aziz, & Singhal, CAV ‘98 Replace v i = v j by propositional variable e i,j Propositional Formula F sat Relabeling: x  v 1 f 1  v 2 f 2  v 3 f 3  v 4 ClausesOrigin e 13  e 14   e 12 x = f 2  x = f 3  x  f 1  (  e 12  e 23 )  ( x  f 1  f 1 = f 2 )  (  e 23  e 34 )  ( f 1  f 2  f 2 = f 3 )  (  e 13  e 24 )  ( x  f 2  f 1 = f 3 ) e 13  e 14   e 12  (  e 12  e 23 )  (  e 23  e 34 )  (  e 13  e 24 )

8. – 8 – Need for Transitivity Constraints Propositional Formula F sat e 13  e 14   e 12  (  e 12  e 23 )  (  e 23  e 34 )  (  e 13  e 24 )Solution e 13 = true e 14 = true e 12 = false e 23 = true e 34 = true e 24 = true Transitivity Violation in Solution e 13 = true e 23 = true e 12 = false Corresponds to x = f 2 and f 2 = f 1 but x  f 1

9. – 9 – Handling Transitivity Constraints: Goel, et al., CAV ’98 Complexity Finding solution to F sat that satisfies transitivity constraints is NP-Hard Even when F sat represented as OBDD Their method Enumerate implicants of F sat from OBDD representation Discard any implicant that contains transitivity violation Eventually find solution or run out of implicants Our Experiments Works well for small benchmarks Far too many implicants for larger benchmarks

10. – 10 – Handling Transitivity Constraints: Our Method Idea Generate propositional formula F trans expressing transitivity constraints Satisfy formula F sat  F trans Using OBDDs or SAT checker Sources of Efficiency Equational structure very sparse Far fewer than n(n-1)/2 relational variables Only need to enforce limited set of transitivity constraints With OBDDs, can reduce set of relational variables Only those in true support of F sat

11. – 11 – Benchmark Circuits Single Issue Pipeline: 1xDLX-C Analogous to DLX model in Hennessy & Patterson Verified in ‘94 by Burch & Dill Dual Issue Pipeline #1: 2xDLX-CA Second pipeline can only handle R-R and R-I instructions Burch (DAC ‘96) required 28 manual case splits, 3 commutative diagrams, and 1800s. Dual Issue Pipeline #2: 2xDLX-CC Second pipeline can also handle all instructions

12. – 12 – Verifying Original Benchmarks None Require Transitivity Constraints F sat is unsatisfiable in every case Circuits don’t make use of transitivity in forwarding or stall decisionsPerformance CircuitOBDD Secs.FGRASP Secs. 1xDLX-C0.23 2xDLX-CA11.176 2xDLX-CC29.5,035

13. – 13 – Transitivity Benchmarks Modified, but Correct Circuits Modify forwarding logic ESrc1 = MDest  ESrc1 = MDest  ( ESrc1 = ESrc2  ESrc2 = MDest ) Equivalent under transitivity Circuit names 1xDLX-Ct, 2xDLX-CAt, 2xDLX-CCt Buggy Circuits 100 buggy versions of 2xDLX-CC Each contains single modification of control logic Must ensure that counterexample satisfies transitivity constraints

14. – 14 – 1xDLX-C Equation Structure Vertices For each v i 13 different register identifiersEdges For each equation Control stalling and forwarding logic 27 relational variables Out of 78 possible

15. – 15 – 2DLX-CCt Equation Structure Equations Between 25 different register identifiers 143 relational variables Out of 300 possible

16. – 16 – Graph Interpretation of Transitivity Transitivity Violation Cycle in graph Exactly one edge has e i,j = false  === = == =

17. – 17 – Exploiting Chords Chord Edge connecting two non- adjacent vertices in cycle Property Sufficient to enforce transitivity constraints for all chord-free cycles If transitivity holds for all chord-free cycles, then holds for arbitrary cycles   

18. – 18 – Enumerating Chord-Free Cycles Strategy Enumerate chord-free cycles in graph Each cycle of length k yields k transitivity constraints 1 2k Problem Potentially exponential number of chord-free cycles 2 k +k chord-free cycles

19. – 19 – Adding Chords Strategy Add edges to graph to reduce number of chord-free cycles 1 2k 2 k +k chord-free cycles 2k+1 chord-free cycles Trade-Off Reduces formula size Increases number of relational variables

20. – 20 – Chordal Graph Definition Every cycle of length > 3 has a chordGoal Add minimum number of edges to make graph chordal Relation to Sparse Gaussian Elimination Choose pivot ordering that minimizes fill-in NP-hard Simple heuristics effective

21. – 21 – Adding Chordal Edges to 1xDLX-C Original 27 relational variables 286 cycles 858 clauses Augmented 33 relational variables 40 cycles 120 clauses

22. – 22 – Adding Chordal Edges to 2xDLX-CCt Original 143 relational variables 2,136 cycles 8,364 clauses Augmented 193 relational variables 858 cycles 2,574 clauses

23. – 23 – SAT Checker on Good Circuits Strategy Run on clauses encoding F sat and F trans FGRASP Performance (Secs.) Circuit F sat F sat  F trans 1xDLX-C34 1xDLX-Ct---9 2xDLX-CA1761,275 2xDLX-CAt---896 2xDLX-CC5,0359,932 2xDLX-CCt---15,003Observation Much more challenging with transitivity constraints imposed

24. – 24 – SAT Checker on Buggy Circuits Performance Penalty with Transitivity Constraints Geometric average slowdown = 2.3X

25. – 25 – Using OBDDs Possible Strategy Build OBDDs for F sat and F trans Compute F sat  F trans Find satisfying solution

26. – 26 – Limitation of OBDDs OBDD for F trans can be of exponential size Regardless of variable ordering Formal result Relational variables forming k X k mesh OBDD representation has  (2 k/4 ) nodes Experimental Results Unable to build OBDD of F trans for large benchmarks 6 X 6 mesh

27. – 27 – Better Use of OBDDs Strategy Build OBDD for F sat Determine relational variables in true support Easy with OBDD Generate F trans for these variables Compute conjunction and find satisfying solutionPerformance When F sat unsatisfiable, no further steps required For other benchmarks, yields tractable F trans

28. – 28 – 2DLX-CCt Reduced Constraints Relational variables 46 original 6 chordal OBDD Representation 7,168 nodes

29. – 29 – Reduced Constraints: Average-Case Buggy Circuit Relational Variables 17 original 3 chordal OBDD Representation 70 nodes

30. – 30 – Reduced Constraints: Worst-Case Buggy Circuit Relational variables 52 original 16 chordal OBDD Representation 93,937 nodes

31. – 31 – OBDDs on Good Circuits CUDD Performance (Secs.) Circuit Time 1xDLX-C0.2 1xDLX-Ct2 2xDLX-CA11 2xDLX-CAt109 2xDLX-CC29 2xDLX-CCt441Observation Significantly more effort with transitivity constraints Better performance than FGRASP

32. – 32 – OBDDs on Buggy Circuits Performance Penalty with Transitivity Constraints Geometric average slowdown = 1.01X

33. – 33 – Conclusion Equational Formulas can be Solved by Propositional Methods Exploit sparse structure of equations Reduces number of variables Reduces formula size With OBDDs, can identify essential relational variables In true support of F sat Can use either SAT checker or OBDDs OBDDs do best for unsatisfiable formulas

34. – 34 – Extension Formulas with Ordering Constraints Constraints of form v i  v j Symbolic Solution Introduce variables a i,j and a j,i for each constraint v i rel v j a i,j true when v i  v j Solution defines partial orderingApplication Scheduling problems