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

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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

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

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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?

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

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6 Overview of the Algorithm 1.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 2.Test them on disequality literals –If the random assignments satisfy e 1 = e 2, then the formula …. Æ e 1 e 2 Æ …. is unsatisfiable

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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) 1.If S ² g=0, then S ² g=0 2.S ² e=0 3.If S ² g=0, then 9 g (S ² g=0 and g=0 Æ e=0 ) g=0) –S satisfies exactly one more linearly independent relationship satisfied by S

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8 Adjust Operation: Geometric Interpretation Algorithm to obtain S = Adjust(S, e=0) S4S4 S2S2 S3S3 S1S1 S3S3 S1S1 S2S2 Hyperplane e = 0. Assignments = points Adjust = projection onto the hyperplane represented by an equality literal S satisfies e=0 and all relationships satisfied by S

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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,e i =0)

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

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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.

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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 2 32, r = 15, then error probability ·

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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(nk 2 )

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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

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15 Retracting Assumptions: Unadjust Operation Remember Unadjust(S,e=0) = S [ { } (S [ { }) ² e=0 iff S ² e=0 S4S4 S2S2 S3S3 S1S1 S3S3 S1S1 S2S2 Hyperplane e = 0 S = Adjust(S, e=0)

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16 Uninterpreted Function Symbols Use Ackerman transformation –Replace uninterpreted term e with new variable V e –For any F(e) and F(e) add if V e = V e then V F(e) = V F(e) Example (x=y) Æ (f(x)=u) Æ (f(y)=w) ! (x=y) Æ (v 1 =u) Æ (v 2 =w) Æ (if x=y then v 1 = v 2 ) Implementation –After adjusting for an equality, check if any of the conditional literals require adjustment.

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17 Experimental Results ExampleArith- dense Arith- sparse Both- dense Both- sparse Uf #equalities2625205035 #adjusts2514294272 #points3020405040 Rand (ms)2.31.33.47.59.6 ICS (ms)386.484.837.073.923.1 ICS/Rand1686511102.5 ICS = Integrated Canonizer and Solver

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18 Comparison with Shostaks Algorithm Symbolic manipulation vs. expression evaluation Shostaks solver » adjust operation Shostaks canonizer » probabilistic canonical form

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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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