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A Randomized Satisfiability Procedure for Arithmetic and Uninterpreted Function Symbols

A Randomized Satisfiability Procedure for Arithmetic and Uninterpreted Function Symbols. Sumit Gulwani George Necula EECS Department University of California, Berkeley. Introduction. Problem Check satisfiability of conjunction of literals Example: x = 2y+3 Æ F(x-3)  F(2y)

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A Randomized Satisfiability Procedure for Arithmetic and Uninterpreted Function Symbols

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  1. A Randomized Satisfiability Procedure for Arithmetic and Uninterpreted Function Symbols Sumit Gulwani George Necula EECS Department University of California, Berkeley

  2. Introduction • Problem • Check satisfiability of conjunction of literals • Example: x = 2y+3 Æ F(x-3)  F(2y) • Application: program verification • Existing algorithms • Linear arithmetic: Gaussian elimination, Simplex • Uninterpreted function terms: congruence closure • Combination: Nelson-Oppen, Shostak • Our proposal • A randomized algorithm • We hope to gain: simplicity and efficiency

  3. Outline • Linear arithmetic • Retracting assumptions • Extension to uninterpreted function symbols • Experimentation

  4. Algebraic Interpretation of Satisfiability 1 : (z = x+y) Æ (x = y) Æ (z  0) 2 : (z = x+y) Æ (x = y) Æ (z  2x) • 1 is satisfiable. For e.g. x=1, y=1, z=2 • 2 is not satisfiable since (z=x+y) Æ (x=y) ) (z=2x) • Can we "test" the satisfiability of these formulae with low error probability?

  5. Geometric Interpretation of Satisfiability 1 : (z = x+y) Æ (x = y) Æ (z  0) 2 : (z = x+y) Æ (x = y) Æ (z  2x) L L P R1: z = 0 R2: z = 2x Line L: solution space for (z = x+y) Æ (x = y) IDEA: If we choose points randomly on L, we can easily tell that L ) R1 and L ) R2

  6. Overview of the Algorithm • Generate random assignments that satisfy all equality literals • We do this incrementally • Start with a set of completely random assignments • Adjust them to satisfy each equality literal one by one • Test them on disequality literals • If the random assignments satisfy e1 = e2, then the formula ….Æ e1 e2Æ …. is unsatisfiable

  7. Adjust Operation: Algebraic Interpretation Notation • Sample S = collection of assignments • S ² g = 0 means all assignments in S satisfy g=0 Properties of S’ = Adjust(S, e=0) • If S ² g=0, then S’ ² g=0 • S’ ² e=0 • If S’ ² g’=0, then 9g (S ² g=0 and g=0 Æ e=0 ) g’=0) • S’ satisfies exactly one more linearly independent relationship satisfied by S

  8. Adjust Operation: Geometric Interpretation Algorithm to obtain S’ = Adjust(S, e=0) S4 S1  . Hyperplane e =0 S’1 S’2 S’3 S2 S3 • Assignments = points • Adjust = projection onto the hyperplane represented by an equality literal • S’ satisfies e=0 and all relationships satisfied by S

  9. The Satisfiability Procedure • IsSatisfiable() = • let  be • S Ã R, where R is a random sample • for i = 1 to k: • S Ã Adjust(S,ei=0)

  10. The Satisfiability Procedure • IsSatisfiable() = • let  be • S Ã R, where R is a random sample • for i = 1 to k: • if S ² ei+c=0 for some c  0, then return Unsatisfiable • else if S ² ei=0 then S Ã Adjust(S,ei=0) • for j = 1 to m: • if S ² ej’ = 0, then return Unsatisfiable • return Satisfiable

  11. Completeness Theorem • “If IsSatisfiable() returns true, then  is satisfiable” • Proof: • The final sample satisfies all the equality literals and the disequality literals in the formula.

  12. Soundness Theorem • “If  is satisfiable, then IsSatisfiable() returns true with high-probability” • Error probability · • m: #disequalities • |F|: size of set from which random values are chosen • r: #assignments in the initial sample R • k: #equality literals • If m = k = 10, |F| ¼ 232, r = 15, then error probability ·

  13. Complexity r = #assignments in the initial sample R n = #variables k = #equality literals • Each adjust operation has cost O(nr) • Number of adjust operations = O(k) • Total cost = O(nkr) = O(nk2)

  14. Retracting Assumptions: Motivation • if z=x+y then if x=y then assert (z=2x) else assert (x=z-y) • ) decide satisfiability of (z=x+y) Æ (x=y) Æ (z2x) and (z=x+y) Æ (xy) Æ (xz-y) • One easy way to retract is to remember old samples • Space overhead

  15. S4 S1  Hyperplane e =0 S’1 S’2 S’3 S2 S3 Retracting Assumptions: Unadjust Operation • Remember  • Unadjust(S’,e=0) = S’ [ { } • (S’ [ { }) ² e=0 iff S ² e=0 S’ = Adjust(S, e=0)

  16. Uninterpreted Function Symbols • Use Ackerman transformation • Replace uninterpreted term e with new variable Ve • For any F(e) and F(e’) add if Ve = Ve’ then VF(e) = VF(e’) • Example (x=y) Æ (f(x)=u) Æ (f(y)=w) ! (x=y) Æ (v1=u) Æ (v2=w) Æ (if x=y then v1 = v2) • Implementation • After adjusting for an equality, check if any of the conditional literals require adjustment.

  17. Experimental Results ICS = Integrated Canonizer and Solver

  18. Comparison with Shostak’s Algorithm • Symbolic manipulation vs. expression evaluation • Shostak’s solver » adjust operation • Shostak’s canonizer » probabilistic canonical form

  19. Conclusion and Future Work • Randomization can help achieve simplicity and efficiency at the expense of making soundness probabilistic • Generate proofs • Can we extend these ideas to other theories – inequalities, arrays? • Integrate symbolic techniques with randomized ones

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