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Global Value Numbering using Random Interpretation Sumit Gulwani George C. Necula CS Department University of California, Berkeley

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2 Global Value Numbering Problem –To detect equivalences of expressions in a program –To obtain a complete algorithm under the assumptions: Conditionals are non-deterministic Operators are uninterpreted –F(e 1,e 2 ) = F(e 1,e 2 ), F=F, e 1 =e 1, e 2 =e 2 Existing algorithms –Precise but expensive –Efficient but imprecise Use randomization to obtain a precise, efficient but probabistically sound algorithm –Complements our POPL 03 algorithm, which handles only arithmetic

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3 Outline Two key ideas in the algorithm –The affine join operation –K-linear interpretations Correctness of the algorithm Termination of the algorithm

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4 assert(x = y); assert(z = F(y)); Example * x = (a,b) y = (a,b) z = (F(a),F(b)) F(y) = F( (a,b)) Typical algorithms treat as uninterpreted –Hence cannot verify the second assertion The randomized algorithm interprets –Similar to the randomized algorithm for linear arithmetic x := a; y := a; z := F(a); x := b; y := b; z := F(b);

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5 Review: Randomized Algorithm for Linear Arithmetic a := 0; b := 1;a := 1; b := 0; c := b – a; d := 1 – 2b; assert (c + d = 0); assert (c = a + 1) c := 2a + b; d := b – 2; T T F F Between random testing and abstract interpretation Choose random values for input variables Execute both branches Combine the values of a variable at join points using a random affine combination

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6 Review: The Affine Join Operation Affine combination of v 1 and v 2 w.r.t. weight w w (v 1,v 2 ) ´ w v 1 + (1-w) v 2 Affine join preserves common linear relationships (e.g. a+b=5) It does not introduce false relationships w.h.p. Unfortunately, non-linear relationships are not preserved (e.g. a (1+b) = 8) a := 2; b := 3; a := 4; b := 1; a = 7 (2,4) = -10 b = 7 (3,1) = 15 (w = 7)

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7 a := 0; b := 1;a := 1; b := 0; c := b – a; d := 1 – 2b; assert (c + d = 0); assert (c = a + 1) a = -4, b = 5 c = -39, d = 39 c := 2a + b; d := b – 2; a = 1, b = 0a = 0, b = 1 a = -4, b = 5 c = -3, d = 3 a = -4, b = 5 c = 9, d = -9 T T F F w 1 = 5 w 2 = -3 Review: Example Choose a random weight for each join independently. All choices of random weights verify the first assertion Almost all choices contradict the second assertion

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8 Uninterpreted Functions e := y | F(e 1,e 2 ) Choose a random interpretation for F Non-linear interpretation –E.g. F(e 1,e 2 ) = r 1 e 1 2 + r 2 e 2 2 –Preserves all equivalences in straight-line code –But not across join points Lets try linear interpretation

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9 (Naïve) Linear Interpretation Encode F(e 1,e 2 ) = r 1 e 1 + r 2 e 2 Preserves all equivalences across a join point Introduces false equivalences in straight-line code F FF abcd e =e =F FF acbd e = E.g. e and e have same encodings even though e e Problem: too few random coefficients! Encodings e = r 1 (r 1 a+r 2 b) + r 2 (r 1 c+r 2 d) = r 1 2 (a)+r 1 r 2 (b)+r 2 r 1 (c)+r 2 2 (d) e = r 1 2 (a)+r 1 r 2 (c)+r 2 r 1 (b)+r 2 2 (d)

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10 k-linear Interpretations Encode F(e 1,e 2 ) = R 1 e 1 + R 2 e 2 –Every expression evaluates to a vector of length k –R 1 and R 2 are random k £ k matrices –2k 2 random variables, k = o(n) Works since matrix multiplication is not commutative –e = R 1 2 (a) + R 1 R 2 (b) + R 2 R 1 (c) + R 2 2 (d) –e = R 1 2 (a) + R 1 R 2 (c) + R 2 R 1 (b) + R 2 2 (d) F(e 1,e 2 ) 1 F(e 1,e 2 ) k e11e11 …e1ke1k e21e21 …e2ke2k …

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11 The Random Interpreter R y := e V1V1 V * True False V V1V1 V2V2 V1V1 V2V2 V V 1 = V[y Ã V(e)] V 1 = V V 2 = V V: Variables ! Vectors V(e): defined inductively as V(F(e 1,e 2 )) = R 1 V(e 1 ) + R 2 V(e 2 ) V j (e): the j th component of vector V(e) V j (y) = w (V 1 (y),V 2 (y)) for all y,j jj

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12 Outline Two key ideas in the algorithm –The affine join operation –K-linear interpretations Correctness of the algorithm Termination of the algorithm

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13 Completeness and soundness of R We compare the random interpreter R with a suitable abstract interpreter A R mimics A with high probability –R is as complete as A –R is (probabilistically) as sound as A

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14 The Abstract Interpreter A * TrueFalse S S1S1 S2S2 S1S1 S2S2 S S 1 = S S 2 = S S = { e 1 =e 2 | S 1 ) e 1 =e 2, S 2 ) e 1 =e 2 } S 1 = S[y/y] [ { y = e[y/y] } S: set of symbolic equivalences y := e S1S1 S

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15 Completeness Theorem If S ) e 1 = e 2, then V(e 1 ) = V(e 2 ) Proof: –Uninterpreted operators are modeled as linear functions –The affine join operation preserves linear relationships

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16 Soundness Theorem If S ) e 1 = e 2, then with high probability V(e 1 ) V(e 2 ) Error probability · –n: number of function applications –d: size of set from which random values are chosen –t : number of repetitions If n = 100, d ¼ 2 32, t = 5, then error probability ·

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17 Outline Two key ideas in the algorithm –The affine join operation –K-linear interpretations Correctness of the algorithm Termination of the algorithm

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18 Loops and Fixed Point Computation The lattice of sets of equivalences has finite height n. Thus, the abstract interpreter A converges to a fixed point. Thus, the random interpreter R also converges (probabilistically) We can detect convergence by comparing the set of symbolic relationships implied by vectors in two successive iterations

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19 Related Work Efficient but imprecise algorithms –Congruence partitioning [Rosen, Wegman, Zadeck, POPL 88] –Rewrite rules [Ruthing, Knoop, Steffen, SAS 99] - Balanced algorithms [Gargi PLDI 2002] Precise but inefficient algorithms –Abstract interpretation on uninterpreted functions [Kildall 73] Affine join operation –Random interpretation for linear arithmetic [Gulwani, Necula POPL 03]

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20 Conclusion and Future Work Key ideas in the paper (e 1,e 2 ) = w e 1 + (1-w) e 2 –Linearity, Preserves equivalences across a join point F(e 1,e 2 ) = R 1 e 1 + R 2 e 2 –Vectors ) Introduce no false equivalence Random interpretation vs. deterministic algorithms –Linear arithmetic O(n 2 ) vs. O(n 4 ) [POPL 2003] –Uninterpreted functions O(n 3 ) vs. O(n 5 log n) [this talk] Future work –Inter-procedural analysis using random interpretation –Random interpretation for other theories –Combining two random interpreters

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DIVIDING INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.

DIVIDING INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.

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