Non-Linear Reasoning for Invariant Synthesis
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1 Non-Linear Reasoning for Invariant Synthesis Zachary Kincaid 1 John Cyphert 2 Jason Brec 2 Thomas Reps 2,3 1 Princeton University 2 University of Wisconsin-Madison 3 GrammaTech, Inc January 12, 15:50
2 The problem: generating non-linear numerical loop invariants Resource-bound analysis Side channel analysis Secure information flow...
3 while(i < n): x = x + i i = i + 1 i () = i ( 1) + 1 x () = x ( 1) + i ( 1) Loop analyzer Recurrence solver ( i ) = i +. 0 x ( 1) = x + + i 2 = i (0) + x () = x (0) ( 1) + + i (0) 2 i ()
4 while(i < n): x = x + i i = i + 1 branching nested loops non-determinism Loop analyzer i () = i ( 1) + 1 x () = x ( 1) + i ( 1) Recurrence solver ( i = i +. 0 x ( 1) = x + + i 2 ) = i (0) + x () = x (0) ( 1) + + i (0) 2 i ()
5 while(i < n): x = x + i i = i + 1 branching nested loops non-determinism Loop analyzer i () = i ( 1) + 1 x () = x ( 1) + i ( 1) Recurrence solver ( = i (0) + i ) = i + x () = x (0) ( 1) i (0) x ( 1) = x + + i 2 algebraic numbers 2 i ()
6 binary-search(a,target): lo = 1, hi = size(a), tics = 0 while (lo <= hi): tics++; mid = lo + (hi-lo)/2 if A[mid] == target: return mid else if A[mid] < target: lo = mid+1 else : hi = mid-1 log(a) times
7 binary-search(a,target): lo = 1, hi = size(a), tics = 0 while (lo <= hi): tics++; mid = lo + (hi-lo)/2 if A[mid] == target: return mid else if A[mid] < target: lo = mid+1 else : hi = mid-1 tics = tics + 1 mid = lo + (hi lo)/2 ((A[mid] < target lo = mid + 1 hi = hi) (A[mid] > target lo = lo hi = mid 1))
8 binary-search(a,target): tics (+1) = tics () + 1 lo = 1, hi = (hi size(a), lo ) (+1) tics (hi= 0 lo) () /2 1 while (lo <= hi): tics++; tics = tics + 1 mid = lo + (hi-lo)/2 mid = lo + (hi lo)/2 if A[mid] == target: ((A[mid] < target return mid lo = mid + 1 else if A[mid] < target: hi = hi) lo = mid+1 (A[mid] > target lo else : = lo hi = mid 1)) hi = mid-1
9 binary-search(a,target): tics () = tics ( (0) + lo = 1, hi = size(a), tics 1 ) (hi lo+2) (hi lo ) () = 0 (0) 2 while (lo <= hi): 2 tics++; tics = tics + 1 mid = lo + (hi-lo)/2 mid = lo + (hi lo)/2 if A[mid] == target: ((A[mid] < target return mid lo = mid + 1 else if A[mid] < target: hi = hi) lo = mid+1 (A[mid] > target lo else : = lo hi = mid 1)) hi = mid-1
10 binary-search(a,target): lo = 1, hi = size(a), tics = 0 while (lo <= hi): tics++; mid = lo + (hi-lo)/2 if A[mid] == target: return mid else if A[mid] < target: lo = mid+1 else : hi = mid-1. 0 tics = tics ( + 1 ) (hi lo+2) 2 (hi lo ) 2
11 for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++
12 for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++ j < i j = j + 1 tics = tics + 1 i = i n = n
13 tics (+1) = tics () + 1 j (+1) = j () + 1 i (+1) = i () n (+1) = n () for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++ j < i j = j + 1 tics = tics + 1 i = i n = n
14 tics () = tics (0) + j () = j (0) + i () = i (0) n () = n (0) for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++ j < i j = j + 1 tics = tics + 1 i = i n = n
15 for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++ i = i n = n j i. 0 tics = tics + j = j +
16 for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++ i < n i = i + 1 n = n j = i. 0 tics = tics + j =
17 tics (+1) = tics () +i () i (+1) = i () + 1 n (+1) = n () for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++ i < n i = i + 1 n = n j = i. 0 tics = tics + j =
18 tics () = tics (0) + ( + 1)/2 + i (0) i () = i (0) + n () = n (0) for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++ i < n i = i + 1 n = n j = i. 0 tics = tics + j =
19 for (i = 0; i < n; i++): for (j = 0; j < i; j++): tics++ i = n n = n j = i. 0 tics = tics + i = i + ( + 1) 2 + i
20 Warm up: the linear case Suppose loop body formula F(x, x ) is linear. Goal: find a linear system y = Ay + b + linear transformation T s.t F(x, x ) = (Tx ) = A(Tx) + b
21 Warm up: the linear case Binary search: project onto tics, (hi lo) Suppose loop body formula F(x, x ) is linear. Goal: find a linear system y = Ay + b + linear transformation T s.t F(x, x ) = (Tx ) = A(Tx) + b
22 Warm up: the linear case Suppose loop body formula F(x, x ) is linear. Goal: find a linear system y = Ay + b + linear transformation T s.t Algorithm: F(x, x ) = (Tx ) = A(Tx) + b 1 Compute the affine hull of F by sampling linearly independent models of F using an SMT solver. Result is system of (all) equations Ax = Bx + c entailed by F(x, x )
23 Warm up: the linear case Suppose loop body formula F(x, x ) is linear. Goal: find a linear system y = Ay + b + linear transformation T s.t Algorithm: F(x, x ) = (Tx ) = A(Tx) + b 1 Compute the affine hull of F by sampling linearly independent models of F using an SMT solver. Result is system of (all) equations Ax = Bx + c entailed by F(x, x ) 2 Fixpoint computation: We have: Ax = Bx + c Linear transformation T We need: y = By + c
24 Warm up: the linear case Suppose loop body formula F(x, x ) is linear. Goal: find a linear system y = Ay + b + linear transformation T s.t Algorithm: F(x, x ) = (Tx ) = A(Tx) + b 1 Compute the affine hull of F by sampling linearly independent models of F using an SMT solver. Result is system of (all) equations Ax = Bx + c entailed by F(x, x ) 2 Fixpoint computation: We have: Ax = Bx + c T 0 x = T 0 Bx + T 0 c We need: y = By + c
25 Warm up: the linear case Suppose loop body formula F(x, x ) is linear. Goal: find a linear system y = Ay + b + linear transformation T s.t Algorithm: F(x, x ) = (Tx ) = A(Tx) + b 1 Compute the affine hull of F by sampling linearly independent models of F using an SMT solver. Result is system of (all) equations Ax = Bx + c entailed by F(x, x ) 2 Fixpoint computation: We have: Ax = Bx + c T 0 x = T 0 Bx + T 0 c computes best abstraction T 1 x = T 1 Bx + T 1 c We need: y. = By + c
26 Reasoning about non-linear arithmetic
27 for (i = 0; i < n; i++): if (*): continue for (j = 0; j < n; j++): for ( = 0; < n; ++): tics++ i = i + 1 i < n n = n tics = tics j = j = y 0. tics = tics + y n j = y = n = n
28 for (i = 0; i < n; i++): i = i + 1 i < n if (*): continue n for (j = 0; j < n; j++): = n for ( = 0; < n; ++): tics = tics j = j tics++ = y 0. tics = tics + y n j = y = n = n tics tics + n 2
29 The wedge abstract domain The wedge domain is an abstract domain for reasoning about non-linear integer/rational arithmetic The properties expressible by wedges correspond to the conjunctive fragment of non-linear arithmetic (x y, x/y, x y, log x (y), x mod y,...)
30 The wedge abstract domain The wedge domain is an abstract domain for reasoning about non-linear integer/rational arithmetic The properties expressible by wedges correspond to the conjunctive fragment of non-linear arithmetic (x y, x/y, x y, log x (y), x mod y,...) treat non-linear terms as independent dimensions Wedge ignore, treat non-polynomial terms as independent dimensions Polyhedron Algebraic variety
31 The wedge abstract domain The wedge domain is an abstract domain for reasoning about non-linear integer/rational arithmetic The properties expressible by wedges correspond to the conjunctive fragment of non-linear arithmetic (x y, x/y, x y, log x (y), x mod y,...) treat non-linear terms as independent dimensions Wedge ignore, treat non-polynomial terms as independent dimensions Polyhedron linear equations Algebraic variety
32 The wedge abstract domain The wedge domain is an abstract domain for reasoning about non-linear integer/rational arithmetic The properties expressible by wedges correspond to the conjunctive fragment of non-linear arithmetic (x y, x/y, x y, log x (y), x mod y,...) treat non-linear terms as independent dimensions Wedge ignore, treat non-polynomial terms as independent dimensions Polyhedron linear equations Algebraic variety Inference rules Congruence closure
33 Symbolic abstraction i = i + 1 i < n n = n tics = tics j = j = y 0. tics = tics + y n j = y = n = n
34 Symbolic abstraction i i = i + 1 = i + 1 i < n i < n n = n tics = tics n = n tics = tics + s y n j j = s = j y = n = n
35 Symbolic abstraction i i = i + 1 = i + 1 i < n i < n n = n n = n tics = tics tics = tics + n n j = j j = n = n 0 n n 0 n n
36 Symbolic abstraction i = i + 1 i < n n = n tics tics tics + n n j = j 0 n n
37 Extracting recurrences Given: non-linear transition formula F(x, x ) 1 Compute wedge w that over-approximates F 2 Extract recurrences from w
38 Extracting recurrences Given: non-linear transition formula F(x, x ) 1 Compute wedge w that over-approximates F 2 Extract recurrences from w Class of extractable recurrences: (Tx ) = A(Tx) + t Additive term t involves polynomials & exponentials.
39 How can we solve recurrence equations?
40 Operational Calculus Recurrence Solver [Berg 1967] Operational calculus is an algebra of infinite sequences. Idea: 1 Translate recurrence into equation in operational calculus x (+1) = x () + 1 qx (q 1)x 0 = x Solve the equation x = x q 1 3 Translate solution bac x () = x (0) +
41 Operational Calculus Field of operators: Operator is a sequence with finitely many negative positions a = (a 2, a 1 a 0, a 1, a 2,...) b = ( b 0, b 1, b 2,...) Addition is pointwise: (a + b) i a i + b i Multiplication is convolution difference: n n 1 (ab) n = a i b n i + a i b n i 1 i= i=
42 Operational Calculus Field of operators: Operator is a sequence with finitely many negative positions a = (a 2, a 1 a 0, a 1, a 2,...) b = ( b 0, b 1, b 2,...) Addition is pointwise: (a + b) i a i + b i Multiplication is convolution difference: n n 1 (ab) n = a i b n i + a i b n i 1 i= i= Left shift operator q = (1 1, 1, 1,...) qa = (a 2 a 1 a 0 a 1, a 2, a 3,...)
43 Recurrence operational calculus Recurrences are equations in operational calculus x (+1) = x + t qx (q 1)x 0 = x + T (t) Thin of x as an sequence ( x 0, x 1, x 2,...)
44 Recurrence operational calculus Recurrences are equations in operational calculus x (+1) = x + t qx (q 1)x 0 = x + T (t) Thin of x as an sequence ( x 0, x 1, x 2,...) Use left-shift operator to write recurrence as an equation qx = (x 0 x 1, x 2, x 3,...) (q 1)x 0 = (x 0 0, 0, 0,...) qx (q 1)x 0 = ( x 1, x 2, x 3,...)
45 Recurrence operational calculus Recurrences are equations in operational calculus x (+1) = x + t qx (q 1)x 0 = x + T (t) Thin of x as an sequence ( x 0, x 1, x 2,...) Use left-shift operator to write recurrence as an equation qx = (x 0 x 1, x 2, x 3,...) (q 1)x 0 = (x 0 0, 0, 0,...) qx (q 1)x 0 = ( x 1, x 2, x 3,...) Can translate any expression in the grammar s, t Expr() ::= c Q c s + t st T (c) = c T (ct) = ct (t). T (s + t) = T (s) + T (t) T () = 1 q 1.
46 Operational Calculus classical algebra T (c) = c T (ct) = ct (t) T (s + t) = T (s) + T (t) T () = 1 q 1. T 1 T 1 ( T 1 (c) = c T 1 (ct) = ct 1 (t) (s + t) = T 1 1 q 1 ) =. (s) + T 1 (t)
47 Operational Calculus classical algebra T (c) = c T (ct) = ct (t) T (s + t) = T (s) + T (t) T () = 1 q 1. T 1 T 1 ( T 1 (c) = c T 1 (ct) = ct 1 (t) (s + t) = T 1 1 q 1 ) = T 1 (t) =?. (s) + T 1 (t) Operational Calculus classical algebra translation is not complete!
48 Operational Calculus classical algebra T (c) = c T (ct) = ct (t) T (s + t) = T (s) + T (t) T () = 1 q 1. T 1 T 1 ( T 1 (c) = c T 1 (ct) = ct 1 (t) (s + t) = T 1 1 q 1 ) =. T 1 (t) = f t () (s) + T 1 (t) Operational Calculus Implicitly classical algebra interpreted translation function is not complete!
49 Operational Calculus classical algebra T (c) = c T (ct) = ct (t) T (s + t) = T (s) + T (t) T () = 1 q 1 T (f t ()) = t. T 1 T 1 ( T 1 (c) = c T 1 (ct) = ct 1 (t) (s + t) = T 1 1 q 1 ) =. T 1 (t) = f t () (s) + T 1 (t) Operational Calculus Implicitly classical algebra interpreted translation function is not complete!
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51 Contributions: Wedge abstract domain Algorithm for extracting recurrences from loop bodies with control flow & non-determinism Recurrence solver that avoids algebraic numbers Result: non-linear invariant generation for arbitrary loops
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