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Lifting Abstract Interpreters to Quantified Logical Domains

Lifting Abstract Interpreters to Quantified Logical Domains. Sumit Gulwani, MSR Bill McCloskey, UCB Ashish Tiwari, SRI. Motivating Example. a[0] = 0; for (i=1; i<n; i++) a[i] = 0;. Postcondition : i  n  a[0] = 0.  k (0 ≤ k < i  a[k] = 0). How Are Quantifiers Useful?.

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Lifting Abstract Interpreters to Quantified Logical Domains

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  1. Lifting Abstract Interpreters to Quantified Logical Domains Sumit Gulwani, MSR Bill McCloskey, UCB Ashish Tiwari, SRI

  2. Motivating Example a[0] = 0; for (i=1; i<n; i++) a[i] = 0; Postcondition: i  n  a[0] = 0  k (0 ≤ k < i  a[k] = 0)

  3. How Are Quantifiers Useful? • Reasoning about arrays • k (0 ≤ k < STRLEN(s)  s[k]  '!') • j, k (0 ≤ j < k < n  a[j] ≤ a[k]) • Reasoning about pointer-based data structures • u (R(hd, u)  R(u, tl)  udata = 0) means list is initialized from hd to tl Security properties Sorting u v R(u, v)

  4. What Do Quantifiers Look Like? k ( 0 ≤ k < n  a[k] = 0 ) • Goal: Create a universally quantified domain parameterized by base domains • Take advantage of existing domains, transfer functions Typically see only universal quantifiers Comes from some domain, e.g. linear arithmetic Belongs to another domain, e.g. equality of uninterpreted functions Quantifier-Free Domain Quantified Domain

  5. Universally Quantified Domain Domain Element Definition A  V1.(B1  C1)  ...  Vn.(Bn  Cn) Partial Order Definition A  V.(B  C) vA'  V.(B'  C') if 1. Av A' A CvC' 2. V.(B  C) V. (B'  C') A B' vB

  6. Transfer Function Example true A[0] := 0; i := 1 i = 1  A[0] = 0 ? i = 2  A[0] = 0  A[1] = 0 ? i = 1  A[0] = 0 ? i < n T F ? i = 1  A[0] = 0 ? A[i] := 0; i := i+1 6

  7. Transfer Function Example true A[0] := 0; i := 1 Join Algorithm i = 1  A[0] = 0 i = 1  A[0] = 0 i = 2  A[0] = 0  A[1] = 0 i = 1  A[0] = 0 i = 1  A[0] = 0 i = 1  A[0] = 0 i = 2  A[0] = 0  A[1] = 0 i < n i < n i = 1  A[0] = 0 T T F F 1  i  2  A[0] = 0 ? i = 1  A[0] = 0 A[i] := 0; i := i+1 7

  8. Transfer Function Example true A[0] := 0; i := 1 Join Algorithm i = 1  A[0] = 0 i = 1  A[0] = 0 i = 2  A[0] = 0  A[1] = 0 i = 1  A[0] = 0 i = 2  A[0] = 0  A[1] = 0 i = 1  A[0] = 0 i = 1  A[0] = 0 i = 1  k(k = 0  A[k] = 0) i = 2  k(0  k  1  A[k] = 0) i < n i < n i = 1  A[0] = 0 T T F F ? i = 1  A[0] = 0 1  i  2  k(0  k < i  A[k] = 0) A[i] := 0; i := i+1 8

  9. Transfer Function Example true A[0] := 0; i := 1 2  i  n  k(0  k < i  A[k] = 0) i = 1  k(k = 0  A[k] = 0) 1  i  k(0  k < i  A[k] = 0) i < n T F 1  i < n  k(0  k < i  A[k] = 0) i  n  k(0  k < i  A[k] = 0) A[i] := 0; i := i+1 9

  10. Outline • Join Algorithm • Quantifier introduction • Joining quantifiers • Experiments • Conclusion

  11. Quantifier Introduction • Quantified facts are drawn from standard facts in A • User gives set of templates to guide quantification • Experiments show that few templates are needed b[0] = 0 b[0] ≤ b[1] k (k = 0  b[k] = 0) j, k (j = 0  k = 1  b[j] ≤ b[k]) Env fact Template Quantified fact (result) A[*] = c b[0] = 0 k(k = 0  b[k] = 0) A[*] ≤A[*] j, k (j = 0  k = 1  b[j] ≤ b[k]) b[0] ≤ b[1]

  12. Outline • Join Algorithm • Quantifier introduction • Joining quantifiers • Experiments • Conclusion

  13. Transfer Function Example true A[0] := 0; i := 1 Join Algorithm i = 1  A[0] = 0 i = 1  A[0] = 0 i = 2  A[0] = 0  A[1] = 0 i = 1  A[0] = 0 i = 2  A[0] = 0  A[1] = 0 i = 1  A[0] = 0 i = 1  A[0] = 0 i = 1  k(k = 0  A[k] = 0) i = 2  k(0  k  1  A[k] = 0) i < n i < n i = 1  A[0] = 0 T T F F ? i = 1  A[0] = 0 1  i  2  k(0  k < i  A[k] = 0) A[i] := 0; i := i+1 13

  14. Joining Quantifiers • Goal: (AL  V.(BL  CL))t (AR  V. (BR  CR)) • Result must be above both inputs in v, so: • AL  V.(BL  CL)vA  V.(B  C) • AR  V. (BR  CR) vA  V.(B  C) • Based on v definition: 1. ALv A and ARv A so A = ALtAR 2. AL CLvC AR CRvC V.(BL  CL) V.(BR  CR) V. (B  C) AL BvBL AR BvBR

  15. Joining Quantifiers AL CLvC AR CRvC • C = (AL CL) t (AR CR) • Rewriting for B: • Best solution for B = (AL BL)  (AR BR) • If it's not in domain, pick best under-approximation V.(BL  CL) V.(BR  CR) V. (B  C) AL BvBL AR BvBR B vALBLand B v AR BR or, B v ALBLand B v AR BR

  16. Under-Approximation Example • Compute (i = 1 k = 0) (i = 2 0  k  1) in LA • 1st step: guess an over-approximation of the answer • 2nd step: Check if (0  k < i) is correct; refine if not (i = 1 k = 0) t(i = 2 0  k  1) = (1  i  2  0  k < i) Many details skipped. See paper! ? (0  k < i)  (i = 1 k = 0) (i = 2 0  k  1) YES

  17. Outline • Join Algorithm • Quantifier introduction • Joining quantifiers • Experiments • Conclusion

  18. Experiments Invariant: a[k] = b[k] for all k Invariant: All data fields of list are zero

  19. Quantified Domain Construction Works! • Base domain D • partial order • transfer functions Under-approximation operators for D (optional) Under- approximation • Quantified domain Q • 3x slowdown relative to D • transfer functions relatively complete

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