Presentation on theme: "Program Analysis using Random Interpretation Sumit Gulwani UC-Berkeley March 2005."— Presentation transcript:
Program Analysis using Random Interpretation Sumit Gulwani UC-Berkeley March 2005
1 Program Analysis Applications in all aspects of software development, e.g. Program correctness –Software bugs are expensive! Compiler optimizations –Provide people freedom to write code the way they want (leaving performance issues to compilers). Translation validation –Semantic equivalence of programs before and after compilation (difficult to trust o/p of compiler for safety-critical systems).
2 Design choices in Program Analysis Completeness (precision, # of false positives) Computational complexity Ease of implementation Soundness = If analysis says no bugs, it means no bugs. What if we allow probabilistic soundness ? –We get more precise, efficient and even simpler algorithms. –Earlier probabilistic algorithms were used in other areas like networks, but not in program analysis. –We obtain a new class of analyses: random interpretation.
3 Random Interpretation = Random Testing + Abstract Interpretation Random Testing: Test program on random inputs Simple, efficient but unsound (cant prove absence of bugs) Abstract Interpretation: Class of deterministic program analyses Interpret (analyze) an abstraction (approximation) of program Sound but usually complicated, expensive Random Interpretation: Class of randomized program analyses Almost as simple, efficient as random testing Almost as sound as abstract interpretation
4 a := 0; b := i;a := i-2; b := 2; c := b – a; d := i – 2b; assert(c+d = 0); assert(c = a+i) c := 2a + b; d := b – 2i; True False True * * Example 1
5 a := 0; b := i;a := i-2; b := 2; c := b – a; d := i – 2b; assert(c+d = 0); assert(c = a+i) c := 2a + b; d := b – 2i; True False True * * Example 1: Random Testing Need to test blue path to falsify second assertion. Chances of choosing blue path from set of all 4 paths are small. Hence, random testing is unsound.
6 a+b=i a+b=i, c=-d a=i-2, b=2 a+b=i c=2a+b, d=b-2i a+b=i c=b-a, d=i-2b a=0, b=i a := 0; b := i;a := i-2; b := 2; c := b – a; d := i – 2b; assert(c+d = 0); assert(c = a+i) c := 2a + b; d := b – 2i; True False True * * Example 1: Abstract Interpretation Computes invariant at each program point. Operations are usually complicated and expensive.
7 a := 0; b := i;a := i-2; b := 2; c := b – a; d := i – 2b; assert(c+d = 0); assert(c = a+i) c := 2a + b; d := b – 2i; True False True * * Example 1: Random Interpretation Choose random values for input variables. Execute both branches of a conditional. Combine values of variables at join points. Test the assertion.
8 Outline Random Interpretation Linear arithmetic (POPL 2003) –Uninterpreted functions (POPL 2004) –Inter-procedural analysis (POPL 2005) –Other applications
9 Linear relationships in programs with linear assignments Linear relationships (e.g., x=2y+5) are useful for –Program correctness (e.g. buffer overflows) –Compiler optimizations (e.g., constant and copy propagation, CSE, Induction variable elimination etc.) programs with linear assignments does not mean inapplicability to real programs –abstract other program stmts as non-deterministic assignments (standard practice in program analysis)
10 Basic idea in random interpretation Generic algorithm: Choose random values for input variables. Execute both branches of a conditional. Combine the values of variables at join points. Test the assertion.
11 Idea #1: The Affine Join operation w = 7 a = 2 b = 3 a = 4 b = 1 a = 7 (2,4) = -10 b = 7 (3,1) = 15 Affine join 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 (a+b=5) It does not introduce false relationships w.h.p.
12 Idea #1: The Affine Join operation Affine join 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 (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) w = 5 a = 5 (2,4) = -6 b = 5 (3,1) = 11 w = 7 a = 2 b = 3 a = 4 b = 1 a = 7 (2,4) = -10 b = 7 (3,1) = 15
13 Geometric Interpretation of Affine Join a b a + b = 5 b = 2 (a = 2, b = 3) (a = 4, b = 1) : State before the join : State after the join satisfies all the affine relationships that are satisfied by both (e.g. a + b = 5) Given any relationship that is not satisfied by any of (e.g. b=2), also does not satisfy it with high probability
i=3, a=0, b=3 i=3 a := 0; b := i; a := i-2; b := 2; c := b – a; d := i – 2b; assert (c+d = 0); assert (c = a+i) i=3, a=-4, b=7 c=23, d=-23 c := 2a + b; d := b – 2i; i=3, a=1, b=2 i=3, a=-4, b=7 c=-1, d=1 i=3, a=-4, b=7 c=11, d=-11 False w 1 = 5 w 2 = 2 True * * Example 1 Choose a random weight for each join independently. All choices of random weights verify first assertion Almost all choices contradict second assertion
15 Example 2 We need to make use of the conditional x=y on the true branch to prove the assertion. a := x + y b := a b := 2x assert (b = 2x) TrueFalse x = y ?
16 Idea #2: The Adjust Operation Execute multiple runs of the program in parallel. Sample S = Collection of states at a program point Adjust(S, e=0) is the sample obtained by linear combination of states in S such that –The equality conditional is satisfied. –Note that original relationships are preserved. Use Adjust(S, e=0) on true branch of the conditional e=0
17 Geometric Interpretation of Adjust Program states = points Adjust = projection onto the hyperplane Adjust operation loses one point. Algorithm to obtain S = Adjust(S, e=0) S4S4 S2S2 S3S3 S1S1 S3S3 S1S1 S2S2 Hyperplane e = 0
18 Correctness of Random Interpreter R Completeness: If e 1 =e 2, then R ) e 1 =e 2 –assuming non-det conditionals Soundness: If e 1 e 2, then R e 1 = e 2 –error prob. · b, j : number of branches and joins d: size of set from which random values are chosen k: number of points in the sample –If j = b = 10, k = 15, d ¼ 2 32, then error ·
19 Proof Methodology Proving correctness was the most complicated part in this work. We used the following methodology. Design an appropriate deterministic algorithm (need not be efficient) Prove (by induction) that the randomized algorithm simulates each step of the deterministic algorithm with high probability.
21 Problem: Global value numbering a := 5; x := a*b; y := 5*b; z := b*a; a := 5; x := F(a,b); y := F(5,b); z := F(b,a); Abstraction x=y and x=z Reasoning about multiplication is undecidable only x=y Reasoning is decidable but tricky in presence of joins Axiom: If x 1 =y 1 and x 2 =y 2, then F(x 1,x 2 )=F(y 1,y 2 ) Goal: Detect expression equivalence when program operators are abstracted using uninterpreted functions Application: Compiler optimizations, Translation validation
assert(x = y); assert(z = F(y)); * 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 –as affine join operation w x := a; y := a; z := F(a); x := b; y := b; z := F(b); Example True False
23 How to execute uninterpreted functions ? Expression Language 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 r 2 e 2 2 –Preserves all equivalences in straight-line code –But not across join points Lets try linear interpretation
24 Random 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. E.g. e and e have same encodings even though e e Problem: Scalar multiplication is commutative. Solution: Choose r 1 and r 2 to be random matrices and evaluate expressions to vectors F FF abcd e =e =F FF acbd e = 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)
26 Example a := 0; b := i;a := i-2; b := 2; c := b – a; d := i – 2b; assert (c + d = 0); assert (c = a + i) c := 2a + b; d := b – 2i; True False The second assertion is true in the context i=2. Interprocedural Analysis requires computing procedure summaries. True * *
i=2 a=0, b=i a := 0; b := i;a := i-2; b := 2; c := b – a; d := i – 2b; assert (c+d = 0); assert (c = a+i) a=8-4i, b=5i-8 c=21i-40, d=40-21i c := 2a + b; d := b – 2i; a=i-2, b=2 a=8-4i, b=5i-8 c=8-3i, d=3i-8 a=8-4i, b=5i-8 c=9i-16, d=16-9i False w 1 = 5 w 2 = 2 Idea #1: Keep input variables symbolic Do not choose random values for input variables (to later instantiate by any context). Resulting program state at the end is a random procedure summary. a=0, b=2 c=2, d=-2 True * *
29 Experiments Randomized algorithm discovers 10-70% more facts. Randomized algorithm is slower by a factor of 2. Randomized Deterministic
30 Experimental measure of error The % of incorrect relationships decreases with increase in S = size of set from which random values are chosen. N = # of random summaries used S N The experimental results are better than what is predicted by theory
31 Outline Random Interpretation –Linear arithmetic (POPL 2003) –Uninterpreted functions (POPL 2004) –Inter-procedural analysis (POPL 2005) Other applications
32 Other applications of random interpretation Model Checking –Randomized equivalence testing algorithm for FCEDs, which represent conditional linear expressions and are generalization of BDDs. (SAS 04) Theorem Proving –Randomized decision procedure for linear arithmetic and uninterpreted functions. This runs an order of magnitude faster than det. algo. (CADE 03) Ideas for deterministic algorithms –PTIME algorithm for global value numbering, thereby solving a 30 year old open problem. (SAS 04)
Summary Linear ArithmeticAffine Join, Adjust Lessons Learned Randomization buys efficiency, simplicity at cost of prob. soundness. Randomization suggests ideas for deterministic algorithms. Combining randomized and symbolic techniques is powerful. Uninterpreted Fns.Vectors Interproc. AnalysisSymbolic i/p variables Key Idea