Program Analysis and Verification Noam Rinetzky Lecture 11: Abstract Interpretation III Slides credit: Roman Manevich, Mooly Sagiv, Eran Yahav
Abstract Interpretation [Cousot’77] Mathematical framework for approximating semantics (aka abstraction) – Allows designing sound static analysis algorithms Usually compute by iterating to a fixed-point – Computes (loop) invariants Can be interpreted as axiomatic verification assertions Generalizes Hoare Logic & WP / SP calculus
Abstract Interpretation [Cousot’77] Mathematical foundation of static analysis – Abstract domains Abstract states Join ( ) – Transformer functions Abstract steps – Chaotic iteration Structured Programs Abstract computation
Abstract (conservative) interpretation set of states operational semantics (concrete semantics) statement S abstract representation abstract representation abstract semantics statement S abstract representation abstraction
Abstract (conservative) interpretation set of states operational semantics (concrete semantics) statement S set of states abstract representation abstract semantics statement S abstract representation concretization
Abstract Interpretation [Cousot’77] Mathematical foundation of static analysis – Abstract domains Abstract states Join ( ) – Transformer functions Abstract steps – Chaotic iteration Abstract computation Structured Programs Lattices (D, , , , , ) Lattices (D, , , , , ) Monotonic functions Fixpoints
A taxonomy of semantic domain types Complete Lattice (D, , , , , ) Lattice (D, , , , , ) Join semilattice (D, , , ) Meet semilattice (D, , , ) Complete partial order (CPO) (D, , ) Partial order (poset) (D, ) Preorder (D, )
Partial order We say that a binary order relation over a set D is a preorder if the following conditions hold for every d, d’, d’’ D – Reflexive: d d – Transitive: d d’ and d’ d’’ implies d d’’ – Anti-symmetric: d d’ and d’ d implies d = d’
Chains d d’ means d d’ and d d’ An ascending chain is a sequence x 1 x 2 … x k … A descending chain is a sequence x 1 x 2 … x k … The height of a poset (D, ) is the length of the maximal ascending chain in D
Join: Least upper bound (LUB) (D, ) is a poset b ∊ D is an upper bound of A ⊆ D if ∀ a A: a b b ∊ D is the least upper bound of A ⊆ D if – b is an upper bound of A – If b’ is an upper bound of A then b b’ – Join: X = LUB of X x y = {x, y} May not exist
Meet: Greatest lower bound (GLB) (D, ) is a poset b ∊ D is an lower bound of A ⊆ D if ∀ a A: b a b ∊ D is the greatest lower bound of A ⊆ D if – b is an lower bound of A – If b’ is an lower bound of A then b’ b – Meet: X = GLB of X x y = {x, y} May not exist
Complete partial order (CPO) A poset (D, ) is a complete partial if every ascending chain x 1 x 2 … x k … has a LUB
Lattices (D, , , , , ) is a lattice if – (D, ) is a partial order – ∀ X FIN D. X is defined – A top element – ∀ X FIN D. X is defined – A bottom element A lattice (D, , , , , ) is a complete lattice if – X and Y are defined for arbitrary sets
Example: Powerset lattices (2 X, , , , , X) is the powerset lattice of X – A complete lattice
Towards a recipe for static analysis
Collecting semantics For a set of program states State, we define the collecting lattice (2 State, , , , , State) The collecting semantics accumulates the (possibly infinite) sets of states generated during the execution – Not computable in general
Abstract (conservative) interpretation set of states operational semantics (concrete semantics) statement S set of states abstract representation abstract semantics statement S abstract representation concretization
Abstract Domain: Sign lattice x=0 x0x0 x<0x>0 x0x0
Abstract (conservative) interpretation {x ↦ 1, x ↦ 2, …}{x ↦ 0, x ↦ 1, …} operational semantics (concrete semantics) x=x-1 {x ↦ 0, x ↦ 1, …} 0 < x abstract semantics x = x -1 0 ≤ x concretization
But … what if we have x & y? – Define lattice (semantics) for each variable – Compose lattices Goal: compositional definition What if we have more than 1 statement? – Define semantics for entire program via CFG – Different “abstract states” at every CFG node
One lattice per variable true false x=0 x0x0 x<0x>0 x0x0 true false y=0 y0y0 y<0y>0 y0y0
Domain Constructors
Cartesian product of complete lattices For two complete lattices L 1 = (D 1, 1, 1, 1, 1, 1 ) L 2 = (D 2, 2, 2, 2, 2, 2 ) Define the poset L cart = (D 1 D 2, cart, cart, cart, cart, cart ) as follows: – (x 1, x 2 ) cart (y 1, y 2 ) iff x 1 1 y 1 x 2 2 y 2 – cart = ? cart = ? cart = ? cart = ? Lemma: L is a complete lattice Define the Cartesian constructor L cart = Cart(L 1, L 2 )
Disjunctive completion For a complete lattice L = (D, , , , , ) Define the powerset lattice L = (2 D, , , , , ) = ? = ? = ? = ? = ? Lemma: L is a complete lattice L contains all subsets of D, which can be thought of as disjunctions of the corresponding predicates Define the disjunctive completion constructor L = Disj(L)
Relational product of lattices L 1 = (D 1, 1, 1, 1, 1, 1 ) L 2 = (D 2, 2, 2, 2, 2, 2 ) L rel = (2 D 1 D 2, rel, rel, rel, rel, rel ) as follows: – L rel = Disj(Cart(L 1, L 2 )) Lemma: L is a complete lattice
Finite maps For a complete lattice L = (D, , , , , ) and finite set V Define the poset L V L = (V D, V L, V L, V L, V L, V L ) as follows: – f 1 V L f 2 iff for all v V f 1 (v) f 2 (v) – V L = ? V L = ? V L = ? V L = ? Lemma: L is a complete lattice Define the map constructor L V L = Map(V, L)
Concrete Domain: (for the Collecting semantics) How should we represent the set of states at a given control-flow node by a lattice? How should we represent the sets of states at all control-flow nodes by a lattice?
The collecting lattice Lattice for a given control-flow node v: L v =(2 State, , , , , State) Lattice for entire control-flow graph with nodes V: L CFG = Map(V, L v ) We will use this lattice as a baseline for static analysis and define abstractions of its elements
Equational definition of the semantics Define variables of type set of states for each control-flow node Define constraints between them if x > 0 x := x entry exit R[entry] R[2] R[3] R[exit]
Equational definition of the semantics R[2] = R[entry] x:=x-1 R[3] R[3] = R[2] {s | s(x) > 0} R[exit] = R[2] {s | s(x) 0} A system of recursive equations How can we approximate it using what we have learned so far? if x > 0 x := x - 1 entry exit R[entry] R[2] R[3] R[exit]
An abstract semantics R[2] = R[entry] x:=x-1 # R[3] R[3] = R[2] {s | s(x) > 0} # R[exit] = R[2] {s | s(x) 0} # A system of recursive equations if x > 0 x := x - 1 entry exit R[entry] R[2] R[3] R[exit] Abstract transformer for x:=x-1 Abstract representation of {s | s(x) < 0}
Abstract interpretation via concretization set of states collecting semantics statement S set of states abstract representation of sets of states abstract semantics statement S abstract representation of sets of states concretization
Abstract interpretation via abstraction set of states collecting semantics statement S abstract representation of sets of states abstract representation of sets of states abstract semantics statement S abstract representation of sets of states abstraction
Required knowledge Collecting semantics Abstract semantics Connection between collecting semantics and abstract semantics Algorithm to compute abstract semantics
The collecting lattice (sets of states) Lattice for a given control-flow node v: L v =(2 State, , , , , State) Lattice for entire control-flow graph with nodes V: L CFG = Map(V, L v ) We will use this lattice as a baseline for static analysis and define abstractions of its elements
Equation systems in general Let L be a complete lattice (D, , , , , ) Let R be a vector of variables R[0, …, n] D … D Let F be a vector of functions of the type F[i] : R[0, …, n] R[0, …, n] A system of equations R[0] = f[0](R[0], …, R[n]) … R[n] = f[n](R[0], …, R[n]) In vector notation R = F(R) Questions: 1.Does a solution always exist? 2.If so, is it unique? 3.If so, is it computable?
Equation systems in general Let L be a complete lattice (D, , , , , ) Let R be a vector of variables R[0, …, n] D … D Let F be a vector of functions of the type F[i] : R[0, …, n] R[0, …, n] A system of equations R[0] = f[0](R[0], …, R[n]) … R[n] = f[n](R[0], …, R[n]) In vector notation R = F(R) Questions: 1.Does a solution always exist? 2.If so, is it unique? 3.If so, is it computable?
Monotone functions Let L 1 =(D 1, ) and L 2 =(D 2, ) be two posets A function f : D 1 D 2 is monotone if for every pair x, y D 1 x y implies f(x) f(y) A special case: L 1 =L 2 =(D, ) f : D D
Important cases of monotonicity f(X, Y) = X Y For a set X and any function g: F(X) = { g(x) | x X } – Notice that the collecting semantics function is defined in terms of Join (set union) Semantic function for atomic statements lifted to sets of states
Fixed-points L = (D, , , , , ) f : D D monotone Fix(f) = { d | f(d) = d } Red(f) = { d | f(d) d } Ext(f) = { d | d f(d) } Theorem [Tarski 1955] – lfp(f) = Fix(f) = Red(f) Fix(f) – gfp(f) = Fix(f) = Ext(f) Fix(f) Red(f) Ext(f) Fix(f) lfp gfp fn()fn() fn()fn() 1.Does a solution always exist? Yes 2.If so, is it unique? No, but it has least/greatest solutions 3.If so, is it computable? Under some conditions…
Fixed point example for program R[0] = { x Z} R[1] = R[0] R[4] R[2] = R[1] {s | s(x) > 0} R[3] = R[1] {s | s(x) 0} R[4] = x:=x-1 R[2] if x>0 x := x entry exit xZxZ xZxZ { x >0}{ x <0} if x>0 x := x entry exit xZxZ xZxZ { x >0}{ x <0} F(d) : Fixed-point = d
Continuity and ACC condition Let L = (D, , , ) be a complete partial order – Every ascending chain has an upper bound A function f is continuous if for every increasing chain Y D*, f( Y) = { f(y) | y Y } L satisfies the ascending chain condition (ACC) if every ascending chain eventually stabilizes: d 0 d 1 … d n = d n+1 = …
Fixed-point theorem [Kleene] Let L = (D, , , ) be a complete partial order and a continuous function f: D D then lfp(f) = n N f n ( ) Lemma: Monotone functions on posets satisfying ACC are continuous
Resulting algorithm Kleene’s fixed point theorem gives a constructive method for computing the lfp lfp fn()fn() f()f() f2()f2() … d := while f(d) d do d := d f(d) return d Algorithm lfp(f) = n N f n ( ) Mathematical definition
Chaotic iteration Input: – A cpo L = (D, , , ) satisfying ACC – L n = L L … L – A monotone function f : D n D n – A system of equations { X[i] = f i (X) | 1 i n } Output: lfp(f) A worklist-based algorithm for i:=1 to n do X[i] := WL = {1,…,n} while WL do j := pop WL // choose index non-deterministically N := F[i](X) if N X[i] then X[i] := N add all the indexes that directly depend on i to WL (X[j] depends on X[i] if F[j] contains X[i]) return X
Chaotic iteration for static analysis Specialize chaotic iteration for programs Create a CFG for program Choose a cpo of properties for the static analysis to infer: L = (D, , , ) Define variables R[0,…,n] for input/output of each CFG node such that R[i] D For each node v let v out be the variable at the output of that node: v out = F[v]( u | (u,v) is a CFG edge) – Make sure each F[v] is monotone Variable dependence determined by outgoing edges in CFG
Constant propagation example x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit x := 4; while (y 5) do z := x; x := 4
Constant propagation lattice For each variable x define L as For a set of program variables Var=x 1,…,x n L n = L L … L no information not-a-constant
Write down variables x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit x := 4; while (y 5) do z := x; x := 4
Write down equations x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R2R2 R2R2 R2R2 R3R3 R4R4 R6R6 R1R1 R5R5 R0R0 x := 4; while (y 5) do z := x; x := 4
Collecting semantics equations x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R2R2 R2R2 R2R2 R3R3 R4R4 R6R6 R 0 = State R 1 = x:=4 R 0 R 2 = R 1 R 5 R 3 = assume y 5 R 2 R 4 = z:=x R 3 R 5 = x:=4 R 4 R 6 = assume y=5 R 2 R1R1 R5R5 R0R0
Constant propagation equations x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R2R2 R2R2 R2R2 R3R3 R4R4 R6R6 R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R1R1 R5R5 R0R0 abstract transformer
Abstract operations for CP R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 Lattice elements have the form: (v x, v y, v z ) x:=4 # (v x,v y,v z ) = (4, v y, v z ) z:=x # (v x,v y,v z ) = (v x, v y, v x ) assume y 5 # (v x,v y,v z ) = (v x, v y, v x ) assume y=5 # (v x,v y,v z ) = if v y = k 5 then ( , , ) else (v x, 5, v z ) R 1 R 5 = (a 1, b 1, c 1 ) (a 5, b 5, c 5 ) = (a 1 a 5, b 1 b 5, c 1 c 5 ) CP lattice for a single variable
Chaotic iteration for CP: initialization x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R 2 =( , , ) R2R2 R2R2 R 3 =( , , ) R 4 =( , , ) R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 1 =( , , ) R 5 =( , , ) R 0 =( , , ) WL = {R 0, R 1, R 2, R 3, R 4, R 5, R 6 }
Chaotic iteration for CP x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R 2 =( , , ) R2R2 R2R2 R 3 =( , , ) R 4 =( , , ) R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 1 =( , , ) R 5 =( , , ) R 0 =( , , ) WL = {R 1, R 2, R 3, R 4, R 5, R 6 }
Chaotic iteration for CP x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R 2 =( , , ) R2R2 R2R2 R 3 =( , , ) R 4 =( , , ) R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 1 =(4, , ) R 5 =( , , ) R 0 =( , , ) WL = {R 2, R 3, R 4, R 5, R 6 }
Chaotic iteration for CP x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R 2 =( , , ) R2R2 R2R2 R 3 =( , , ) R 4 =( , , ) R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 1 =(4, , ) R 5 =( , , ) R 0 =( , , ) 3 4 WL = {R 2, R 3, R 4, R 5, R 6 }
Chaotic iteration for CP x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R 2 =(4, , ) R2R2 R2R2 R 3 =( , , ) R 4 =( , , ) R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 1 =(4, , ) R 5 =( , , ) R 0 =( , , ) 3 4 WL = {R 2, R 3, R 4, R 5, R 6 }
Chaotic iteration for CP x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R 2 =(4, , ) R2R2 R2R2 R 3 =( , , ) R 4 =( , , ) R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 1 =(4, , ) R 5 =( , , ) R 0 =( , , ) WL = {R 3, R 4, R 5, R 6 }
Chaotic iteration for CP x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R2R2 R2R2 R 3 =(4, , ) R 4 =( , , ) R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 5 =( , , ) R 1 =(4, , ) R 0 =( , , ) R 2 =(4, , ) WL = {R 4, R 5, R 6 }
Chaotic iteration for CP x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R2R2 R2R2 R 3 =(4, , ) R 4 =(4, , 4) R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 5 =( , , ) R 1 =(4, , ) R 0 =( , , ) R 2 =(4, , ) WL = {R 5, R 6 }
Chaotic iteration for CP x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R2R2 R2R2 R 6 =( , , ) R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 R 5 =(4, , 4) R 4 =(4, , 4) R 3 =(4, , ) R 1 =(4, , ) R 0 =( , , ) R 2 =(4, , ) WL = {R 2, R 6 } added R 2 back to worklist since it depends on R 5
Chaotic iteration for CP R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R2R2 R2R2 R 6 =( , , ) R 5 =(4, , 4) R 4 =(4, , 4) R 3 =(4, , ) R 1 =(4, , ) R 0 =( , , ) R 2 =(4, , ) WL = {R 6 }
Chaotic iteration for CP R 0 = R 1 = x:=4 # R 0 R 2 = R 1 R 5 R 3 = assume y 5 # R 2 R 4 = z:=x # R 3 R 5 = x:=4 # R 4 R 6 = assume y=5 # R 2 x := 4 if (*) assume y 5 assume y=5 z := x x := 4 entry exit R2R2 R2R2 R 6 =(4, 5, ) R 5 =(4, , 4) R 4 =(4, , 4) R 3 =(4, , ) R 1 =(4, , ) R 0 =( , , ) R 2 =(4, , ) Fixed-point WL = {} In practice maintain a worklist of nodes
Chaotic iteration for static analysis Specialize chaotic iteration for programs Create a CFG for program Choose a cpo of properties for the static analysis to infer: L = (D, , , ) Define variables R[0,…,n] for input/output of each CFG node such that R[i] D For each node v let v out be the variable at the output of that node: v out = F[v]( u | (u,v) is a CFG edge) – Make sure each F[v] is monotone Variable dependence determined by outgoing edges in CFG
Complexity of chaotic iteration Parameters: – n the number of CFG nodes – k is the maximum in-degree of edges – Height h of lattice L – c is the maximum cost of Applying F v Checking fixed-point condition for lattice L Complexity: O(n h c k) Incremental (worklist) – Implement worklist by priority queue and order nodes by reversed topological order
Required knowledge Collecting semantics Abstract semantics (over lattices) Algorithm to compute abstract semantics (chaotic iteration) Connection between collecting semantics and abstract semantics Abstract transformers
Recap We defined a reference semantics – the collecting semantics We defined an abstract semantics for a given lattice and abstract transformers We defined an algorithm to compute abstract least fixed-point when transformers are monotone and lattice obeys ACC Questions: 1.What is the connection between the two least fixed- points? 2.Transformer monotonicity is required for termination – what should we require for correctness?
Recap We defined a reference semantics – the collecting semantics We defined an abstract semantics for a given lattice and abstract transformers We defined an algorithm to compute abstract least fixed-point when transformers are monotone and lattice obeys ACC Questions: 1.What is the connection between the two least fixed- points? 2.Transformer monotonicity is required for termination – what should we require for correctness?
Galois Connection Given two complete lattices C = (D C, C, C, C, C, C )– concrete domain A = (D A, A, A, A, A, A )– abstract domain A Galois Connection (GC) is quadruple (C, , , A) that relates C and A via the monotone functions – The abstraction function : D C D A – The concretization function : D A D C for every concrete element c D C and abstract element a D A ( (a)) a and c ( (c)) Alternatively (c) a iff c (a) – Homework
Galois Connection: c ( (c)) 1 c 2 (c)(c) 3 ( (c)) The most precise (least) element in A representing c CA
Galois Connection: ( (a)) a 1 3 ( (a)) 2 (a)(a) a CA What a represents in C (its meaning)
Example: lattice of equalities Concrete lattice: C = (2 State, , , , , State) Abstract lattice: EQ = { x=y | x, y Var} A = (2 EQ, , , , EQ, ) – Treat elements of A as both formulas and sets of constraints Useful for copy propagation – a compiler optimization (X) = ? (Y) = ?
Example: lattice of equalities Concrete lattice: C = (2 State, , , , , State) Abstract lattice: EQ = { x=y | x, y Var} A = (2 EQ, , , , EQ, ) – Treat elements of A as both formulas and sets of constraints Useful for copy propagation – a compiler optimization (X) = { (s) | s X} = A { (s) | s X} – (s) = ({s}) = { x=y | s x = s y} that is s x=y (Y) = { s | s Y } = models( Y)
Galois Connection: c ( (c)) 1 [x 5, y 5, z 5] 2 x=x, y=y, z=z, x=y, y=x, x=z, z=x, y=z, z=y 3 … [x 6, y 6, z 6] [x 5, y 5, z 5] [x 4, y 4, z 4] … 4 x=x, y=y, z=z The most precise (least) element in A representing [x 5, y 5, z 5] CA
Most precise abstract representation 1 c 5 CA 8 9 (c)(c) (c) = {c’ | c (c’)}
Most precise abstract representation 1 c 5 CA 8 9 (c)= x=x, y=y, z=z, x=y, y=x, x=z, z=x, y=z, z=y (c) = {c’ | c (c’)} [x 5, y 5, z 5] x=y, y=z x=y, z=y x=y
Galois Connection: ( (a)) a 1 3 x=x, y=y, z=z, x=y, y=x, x=z, z=x, y=z, z=y 2 … [x 6, y 6, z 6] [x 5, y 5, z 5] [x 4, y 4, z 4] … x=y, y=z What a represents in C (its meaning) is called a semantic reduction CA
Galois Insertion a: ( (a))=a 1 x=x, y=y, z=z, x=y, y=x, x=z, z=x, y=z, z=y 2 … [x 6, y 6, z 6] [x 5, y 5, z 5] [x 4, y 4, z 4] … CA How can we obtain a Galois Insertion from a Galois Connection? All elements are reduced
Properties of a Galois Connection The abstraction and concretization functions uniquely determine each other: (a) = {c | (c) a} (c) = {a | c (a)}
Abstracting (disjunctive) sets It is usually convenient to first define the abstraction of single elements (s) = ({s}) Then lift the abstraction to sets of elements (X) = A { (s) | s X}
The case of symbolic domains An important class of abstract domains are symbolic domains – domains of formulas C = (2 State, , , , , State) A = (D A, A, A, A, A, A ) If D A is a set of formulas then the abstraction of a state is defined as (s) = ({s}) = A { | s } the least formula from D A that s satisfies The abstraction of a set of states is (X) = A { (s) | s X} The concretization is ( ) = { s | s } = models( )
Inducing along the connections Assume the complete lattices C = (D C, C, C, C, C, C ) A = (D A, A, A, A, A, A ) M = (D M, M, M, M, M, M ) and Galois connections GC C,A =(C, C,A, A,C, A) and GC A,M =(A, A,M, M,A, M) Lemma: both connections induce the GC C,M = (C, C,M, M,C, M) defined by C,M = C,A A,M and M,C = M,A A,C
Inducing along the connections 1 C,AC,A A,CA,C c 2 C,A(c)C,A(c) 5 CA 3 M A,MA,M 4 M,AM,A c’c’ a’ = A,M ( C,A (c))
Sound abstract transformer Given two lattices C = (D C, C, C, C, C, C ) A = (D A, A, A, A, A, A ) and GC C,A =(C, , , A) with A concrete transformer f : D C D C an abstract transformer f # : D A D A We say that f # is a sound transformer (w.r.t. f) if c: f(c)=c’ f # ( (c)) (c’) For every a and a’ such that (f( (a))) A f # (a)
Transformer soundness condition 1 12 CA f 3 4 f#f# 5 c: f(c)=c’ (f # (c)) (c’)
Transformer soundness condition 2 CA 12 f#f# 3 5 f 4 a: f # (a)=a’ f( (a)) (a’)
Best (induced) transformer CA 2 3 f f # (a)= (f( (a))) 1 f#f# 4 Problem: incomputable directly
Best abstract transformer [CC’77] Best in terms of precision – Most precise abstract transformer – May be too expensive to compute Constructively defined as f # = f – Induced by the GC Not directly computable because first step is concretization We often compromise for a “good enough” transformer – Useful tool: partial concretization
Negative property of best transformers Let f # = f Best transformer does not compose (f(f( (a)))) f # (f # (a))
(f(f( (a)))) f # (f # (a)) CA 2 3 f 1 f#f# f 7 f#f# 8 9 f
Soundness theorem 1 1.Given two complete lattices C = (D C, C, C, C, C, C ) A = (D A, A, A, A, A, A ) and GC C,A =(C, , , A) with 2.Monotone concrete transformer f : D C D C 3.Monotone abstract transformer f # : D A D A 4. a D A : f( (a)) (f # (a)) Then lfp(f) (lfp(f # )) (lfp(f)) lfp(f # )
Soundness theorem 1 CA f fn fn … lpf(f) f2 f2 f3f3 f #n … lpf(f # ) f#2 f#2 f#3f#3 f# f# a D A : f( (a)) (f # (a)) a D A : f n ( (a)) (f #n (a)) a D A : lfp(f n )( (a)) (lfp(f #n )(a)) lfp(f) lfp(f # )
Soundness theorem 2 1.Given two complete lattices C = (D C, C, C, C, C, C ) A = (D A, A, A, A, A, A ) and GC C,A =(C, , , A) with 2.Monotone concrete transformer f : D C D C 3.Monotone abstract transformer f # : D A D A 4. c D C : (f(c)) f # ( (c)) Then (lfp(f)) lfp(f # ) lfp(f) (lfp(f # ))
Soundness theorem 2 CA f fn fn … lpf(f) f2 f2 f3f3 f #n … lpf(f # ) f#2 f#2 f#3f#3 f# f# c D C : (f(c)) f # ( (c)) c D C : (f n (c)) f #n ( (c)) c D C : (lfp(f)(c)) lfp(f # )( (c)) lfp(f) lfp(f # )
A recipe for a sound static analysis Define an “appropriate” operational semantics Define “collecting” structural operational semantics Establish a Galois connection between collecting states and abstract states Local correctness: show that the abstract interpretation of every atomic statement is sound w.r.t. the collecting semantics Global correctness: conclude that the analysis is sound
Completeness Local property: – forward complete: c: (f # (c)) = (f(c)) – backward complete: a: f( (a)) = (f # (a)) A property of domain and the (best) transformer Global property: – (lfp(f)) = lfp(f # ) – lfp(f) = (lfp(f # )) Very ideal but usually not possible unless we change the program model (apply strong abstraction) and/or aim for very simple properties
Forward complete transformer 12 CA f 3 4 f#f# c: (f # (c)) = (f(c))
Backward complete transformer CA 12 f#f# 3 5 f a: f( (a)) = (f # (a))
Global (backward) completeness CA f fn fn … lpf(f) f2 f2 f3f3 f #n … lpf(f # ) f#2 f#2 f#3f#3 f# f# a: f( (a)) = (f # (a)) a: f n ( (a)) = (f #n (a)) a D A : lfp(f n )( (a)) = (lfp(f #n )(a)) lfp(f) = lfp(f # )
Global (forward) completeness CA f fn fn … lpf(f) f2 f2 f3f3 f #n … lpf(f # ) f#2 f#2 f#3f#3 f# f# c D C : (f(c)) = f # ( (c)) c D C : (f n (c)) = f #n ( (c)) c D C : (lfp(f)(c)) = lfp(f # )( (c)) lfp(f) = lfp(f # )
Three example analyses Abstract states are conjunctions of constraints Variable Equalities – VE-factoids = { x=y | x, y Var} false VE = (2 VE-factoids, , , , false, ) Constant Propagation – CP-factoids = { x=c | x Var, c Z} false CP = (2 CP-factoids, , , , false, ) Available Expressions – AE-factoids = { x=y+z | x Var, y,z Var Z} false A = (2 AE-factoids, , , , false, )
Lattice combinators reminder Cartesian Product – L 1 = (D 1, 1, 1, 1, 1, 1 ) L 2 = (D 2, 2, 2, 2, 2, 2 ) – Cart(L 1, L 2 ) = (D 1 D 2, cart, cart, cart, cart, cart ) Disjunctive completion – L = (D, , , , , ) – Disj(L) = (2 D, , , , , ) Relational Product – Rel(L 1, L 2 ) = Disj(Cart(L 1, L 2 ))
Cartesian product of complete lattices For two complete lattices L 1 = (D 1, 1, 1, 1, 1, 1 ) L 2 = (D 2, 2, 2, 2, 2, 2 ) Define the poset L cart = (D 1 D 2, cart, cart, cart, cart, cart ) as follows: – (x 1, x 2 ) cart (y 1, y 2 ) iff x 1 1 y 1 and x 2 2 y 2 – cart = ? cart = ? cart = ? cart = ? Lemma: L is a complete lattice Define the Cartesian constructor L cart = Cart(L 1, L 2 )
Cartesian product of GCs GC C,A =(C, C,A, A,C, A) GC C,B =(C, C,B, B,C, B) Cartesian Product GC C,A B = (C, C,A B, A B,C, A B) – C,A B (X)= ? – A B,C (Y) = ?
Cartesian product of GCs GC C,A =(C, C,A, A,C, A) GC C,B =(C, C,B, B,C, B) Cartesian Product GC C,A B = (C, C,A B, A B,C, A B) – C,A B (X) = ( C,A (X), C,B (X)) – A B,C (Y) = A,C (X) B,C (X) What about transformers?
Cartesian product transformers GC C,A =(C, C,A, A,C, A)F A [st] : A A GC C,B =(C, C,B, B,C, B)F B [st] : B B Cartesian Product GC C,A B = (C, C,A B, A B,C, A B) – C,A B (X) = ( C,A (X), C,B (X)) – A B,C (Y) = A,C (X) B,C (X) How should we define F A B [st] : A B A B
Cartesian product transformers GC C,A =(C, C,A, A,C, A)F A [st] : A A GC C,B =(C, C,B, B,C, B)F B [st] : B B Cartesian Product GC C,A B = (C, C,A B, A B,C, A B) – C,A B (X) = ( C,A (X), C,B (X)) – A B,C (Y) = A,C (X) B,C (X) How should we define F A B [st] : A B A B Idea: F A B [st](a, b) = (F A [st] a, F B [st] b) Are component-wise transformers precise?
Cartesian product analysis example Abstract interpreter 1: Constant Propagation Abstract interpreter 2: Variable Equalities Let’s compare – Running them separately and combining results – Running the analysis with their Cartesian product a := 9; b := 9; c := a; CP analysisVE analysis {a=9} {a=9, b=9} {a=9, b=9, c=9} {} {} {c=a}
Cartesian product analysis example Abstract interpreter 1: Constant Propagation Abstract interpreter 2: Variable Equalities Let’s compare – Running them separately and combining results – Running the analysis with their Cartesian product CP analysis + VE analysis a := 9; b := 9; c := a; {a=9} {a=9, b=9} {a=9, b=9, c=9, c=a}
Cartesian product analysis example Abstract interpreter 1: Constant Propagation Abstract interpreter 2: Variable Equalities Let’s compare – Running them separately and combining results – Running the analysis with their Cartesian product CP VE analysis Missing {a=b, b=c} a := 9; b := 9; c := a; {a=9} {a=9, b=9} {a=9, b=9, c=9, c=a}
Transformers for Cartesian product Naïve (component-wise) transformers do not utilize information from both components – Same as running analyses separately and then combining results Can we treat transformers from each analysis as black box and obtain best transformer for their combination?
Can we combine transformer modularly? No generic method for any abstract interpretations
Reducing values for CP VE X = set of CP constraints of the form x=c (e.g., a=9 ) Y = set of VE constraints of the form x=y Reduce CP VE (X, Y) = (X’, Y’) such that (X’, Y’) (X’, Y’) Ideas?
Reducing values for CP VE X = set of CP constraints of the form x=c (e.g., a=9 ) Y = set of VE constraints of the form x=y Reduce CP VE (X, Y) = (X’, Y’) such that (X’, Y’) (X’, Y’) ReduceRight: – if a=b X and a=c Y then add b=c to Y ReduceLeft: – If a=c and b=c Y then add a=b to X Keep applying ReduceLeft and ReduceRight and reductions on each domain separately until reaching a fixed-point
Transformers for Cartesian product Do we get the best transformer by applying component-wise transformer followed by reduction? – Unfortunately, no (what’s the intuition?) – Can we do better? – Logical Product [Gulwani and Tiwari, PLDI 2006]
Product vs. reduced product CP VE lattice {a=9}{c=a}{c=9}{c=a} {a=9, c=9}{c=a} {[a 9, c 9]} collecting lattice {}
Reduced product For two complete lattices L 1 = (D 1, 1, 1, 1, 1, 1 ) L 2 = (D 2, 2, 2, 2, 2, 2 ) Define the reduced poset D 1 D 2 = {(d 1,d 2 ) D 1 D 2 | (d 1,d 2 ) = (d 1,d 2 ) } L 1 L 2 = (D 1 D 2, cart, cart, cart, cart, cart )
Transformers for Cartesian product Do we get the best transformer by applying component-wise transformer followed by reduction? – Unfortunately, no (what’s the intuition?) – Can we do better? – Logical Product [Gulwani and Tiwari, PLDI 2006]
Logical product-- Assume A=(D,…) is an abstract domain that supports two operations: for x D – inferEqualities(x) = { a=b | (x) a=b } returns a set of equalities between variables that are satisfied in all states given by x – refineFromEqualities(x, {a=b}) = y such that (x)= (y) y x
Example
Information loss example if (…) b := 5 else b := -5 if (b>0) b := b-5 else b := b+5 assert b==0 {} {b=5} {b=-5} {b= } can’t prove
Disjunctive completion of a lattice For a complete lattice L = (D, , , , , ) Define the powerset lattice L = (2 D, , , , , ) = ? = ? = ? = ? = ? Lemma: L is a complete lattice L contains all subsets of D, which can be thought of as disjunctions of the corresponding predicates Define the disjunctive completion constructor L = Disj(L)
Disjunctive completion for GCs GC C,A =(C, C,A, A,C, A) GC C,B =(C, C,B, B,C, B) Disjunctive completion GC C,P(A) = (C, P(A), P(A), P(A)) – C,P(A) (X) = ? – P(A),C (Y) = ?
Disjunctive completion for GCs GC C,A =(C, C,A, A,C, A) GC C,B =(C, C,B, B,C, B) Disjunctive completion GC C,P(A) = (C, P(A), P(A), P(A)) – C,P(A) (X) = { C,A ({x}) | x X} – P(A),C (Y) = { P(A) (y) | y Y} What about transformers?
Information loss example if (…) b := 5 else b := -5 if (b>0) b := b-5 else b := b+5 assert b==0 {} {b=5} {b=-5} {b= 5 b=-5 } {b= 0 } proved
The base lattice CP {x=0} true {x=-1}{x=-2}{x=1}{x=2} …… false
The disjunctive completion of CP {x=0} true {x=-1}{x=-2}{x=1}{x=2} …… false {x=-2 x=-1}{x=-2 x=0}{x=-2 x=1}{x=1 x=2} ……… {x=0 x=1 x=2}{x=-1 x=1 x=-2} ……… … What is the height of this lattice?
Taming disjunctive completion Disjunctive completion is very precise – Maintains correlations between states of different analyses – Helps handle conditions precisely – But very expensive – number of abstract states grows exponentially – May lead to non-termination Base analysis (usually product) is less precise – Analysis terminates if the analyses of each component terminates How can we combine them to get more precision yet ensure termination and state explosion?
Taming disjunctive completion Use different abstractions for different program locations – At loop heads use coarse abstraction (base) – At other points use disjunctive completion Termination is guaranteed (by base domain) Precision increased inside loop body
With Disj(CP) while (…) { if (…) b := 5 else b := -5 if (b>0) b := b-5 else b := b+5 assert b==0 } Doesn’t terminate
With tamed Disj(CP) while (…) { if (…) b := 5 else b := -5 if (b>0) b := b-5 else b := b+5 assert b==0 } terminates CP Disj(CP) What MultiCartDomain implements
Reducing disjunctive elements A disjunctive set X may contain within it an ascending chain Y=a b c… We only need max(Y) – remove all elements below
Relational product of lattices L 1 = (D 1, 1, 1, 1, 1, 1 ) L 2 = (D 2, 2, 2, 2, 2, 2 ) L rel = (2 D 1 D 2, rel, rel, rel, rel, rel ) as follows: – L rel = ?
Relational product of lattices L 1 = (D 1, 1, 1, 1, 1, 1 ) L 2 = (D 2, 2, 2, 2, 2, 2 ) L rel = (2 D 1 D 2, rel, rel, rel, rel, rel ) as follows: – L rel = Disj(Cart(L 1, L 2 )) Lemma: L is a complete lattice What does it buy us? – How is it relative to Cart(Disj(L 1 ), Disj(L 2 ))? What about transformers?
Relational product of GCs GC C,A =(C, C,A, A,C, A) GC C,B =(C, C,B, B,C, B) Relational Product GC C,P(A B) = (C, C,P(A B), P(A B),C, P(A B)) – C,P(A B) (X) = ? – P(A B),C (Y) = ?
Relational product of GCs GC C,A =(C, C,A, A,C, A) GC C,B =(C, C,B, B,C, B) Relational Product GC C,P(A B) = (C, C,P(A B), P(A B),C, P(A B)) – C,P(A B) (X) = {( C,A ({x}), C,B ({x})) | x X} – P(A B),C (Y) = { A,C (y A ) B,C (y B ) | (y A,y B ) Y}
Cartesian product example Correlations preserved
Function space GC C,A =(C, C,A, A,C, A) GC C,B =(C, C,B, B,C, B) Denote the set of monotone functions from A to B by A B Define for elements of A B as follows (a 1, b 1 ) (a 2, b 2 ) = if a 1 =a 2 then {(a 1, b 1 B b 1 )} else {(a 1, b 1 ), (a 2, b 2 )} Reduced cardinal power GC C,A B = (C, C,A B, A B,C, A B) – C,A B (X) = {( C,A ({x}), C,B ({x})) | x X} – A B,C (Y) = { A,C (y A ) B,C (y B ) | (y A,y B ) Y} Useful when A is small and B is much larger – E.g., typestate verification
Widening/Narrowing
How can we prove this automatically? RelProd(CP, VE)
Intervals domain One of the simplest numerical domains Maintain for each variable x an interval [L,H] – L is either an integer of - – H is either an integer of + A (non-relational) numeric domain
Intervals lattice for variable x [0,0][-1,-1][-2,-2][1,1][2,2]... [- ,+ ] [0,1][1,2][2,3][-1,0][-2,-1] [-10,10] [1, + ][ - ,0 ]... [2, + ][0, + ][ - ,-1 ]... [-20,10]
Intervals lattice for variable x D int [x] = { (L,H) | L - ,Z and H Z,+ and L H} =[- ,+ ] = ? – [1,2] [3,4] ? – [1,4] [1,3] ? – [1,3] [1,4] ? – [1,3] [- ,+ ] ? What is the lattice height?
Intervals lattice for variable x D int [x] = { (L,H) | L - ,Z and H Z,+ and L H} =[- ,+ ] = ? – [1,2] [3,4] no – [1,4] [1,3] no – [1,3] [1,4] yes – [1,3] [- ,+ ]yes What is the lattice height? Infinite
Joining/meeting intervals [a,b] [c,d] = ? – [1,1] [2,2] = ? – [1,1] [2, + ] = ? [a,b] [c,d] = ? – [1,2] [3,4] = ? – [1,4] [3,4] = ? – [1,1] [1,+ ] = ? Check that indeed x y if and only if x y=y
Joining/meeting intervals [a,b] [c,d] = [min(a,c), max(b,d)] – [1,1] [2,2] = [1,2] – [1,1] [2,+ ] = [1,+ ] [a,b] [c,d] = [max(a,c), min(b,d)] if a proper interval and otherwise – [1,2] [3,4] = – [1,4] [3,4] = [3,4] – [1,1] [1,+ ] = [1,1] Check that indeed x y if and only if x y=y
Interval domain for programs D int [x] = { (L,H) | L - ,Z and H Z,+ and L H} For a program with variables Var={x 1,…,x k } D int [Var] = ?
Interval domain for programs D int [x] = { (L,H) | L - ,Z and H Z,+ and L H} For a program with variables Var={x 1,…,x k } D int [Var] = D int [x 1 ] … D int [x k ] How can we represent it in terms of formulas?
Interval domain for programs D int [x] = { (L,H) | L - ,Z and H Z,+ and L H} For a program with variables Var={x 1,…,x k } D int [Var] = D int [x 1 ] … D int [x k ] How can we represent it in terms of formulas? – Two types of factoids x c and x c – Example: S = {x 9, y 5, y 10} – Helper operations c + + = + remove(S, x) = S without any x-constraints lb(S, x) =
Assignment transformers x := c # S = ? x := y # S = ? x := y+c # S = ? x := y+z # S = ? x := y*c # S = ? x := y*z # S = ?
Assignment transformers x := c # S = remove(S,x) {x c, x c} x := y # S = remove(S,x) {x lb(S,y), x ub(S,y)} x := y+c # S = remove(S,x) {x lb(S,y)+c, x ub(S,y)+c} x := y+z # S = remove(S,x) {x lb(S,y)+lb(S,z), x ub(S,y)+ub(S,z)} x := y*c # S = remove(S,x) if c>0 {x lb(S,y)*c, x ub(S,y)*c} else {x ub(S,y)*-c, x lb(S,y)*-c} x := y*z # S = remove(S,x) ?
assume transformers assume x=c # S = ? assume x<c # S = ? assume x=y # S = ? assume x c # S = ?
assume transformers assume x=c # S = S {x c, x c} assume x<c # S = S {x c-1} assume x=y # S = S {x lb(S,y), x ub(S,y)} assume x c # S = ?
assume transformers assume x=c # S = S {x c, x c} assume x<c # S = S {x c-1} assume x=y # S = S {x lb(S,y), x ub(S,y)} assume x c # S = (S {x c-1}) (S {x c+1})
Effect of function f on lattice elements L = (D, , , , , ) f : D D monotone Fix(f) = { d | f(d) = d } Red(f) = { d | f(d) d } Ext(f) = { d | d f(d) } Theorem [Tarski 1955] – lfp(f) = Fix(f) = Red(f) Fix(f) – gfp(f) = Fix(f) = Ext(f) Fix(f) Red(f) Ext(f) Fix(f) lfp gfp fn()fn() fn()fn()
Effect of function f on lattice elements L = (D, , , , , ) f : D D monotone Fix(f) = { d | f(d) = d } Red(f) = { d | f(d) d } Ext(f) = { d | d f(d) } Theorem [Tarski 1955] – lfp(f) = Fix(f) = Red(f) Fix(f) – gfp(f) = Fix(f) = Ext(f) Fix(f) Red(f) Ext(f) Fix(f) lfp gfp fn()fn() fn()fn()
Continuity and ACC condition Let L = (D, , , ) be a complete partial order – Every ascending chain has an upper bound A function f is continuous if for every increasing chain Y D*, f( Y) = { f(y) | y Y } L satisfies the ascending chain condition (ACC) if every ascending chain eventually stabilizes: d 0 d 1 … d n = d n+1 = …
Fixed-point theorem [Kleene] Let L = (D, , , ) be a complete partial order and a continuous function f: D D then lfp(f) = n N f n ( )
Resulting algorithm Kleene’s fixed point theorem gives a constructive method for computing the lfp lfp fn()fn() f()f() f2()f2() … d := while f(d) d do d := d f(d) return d Algorithm lfp(f) = n N f n ( ) Mathematical definition
Chaotic iteration Input: – A cpo L = (D, , , ) satisfying ACC – L n = L L … L – A monotone function f : D n D n – A system of equations { X[i] | f(X) | 1 i n } Output: lfp(f) A worklist-based algorithm for i:=1 to n do X[i] := WL = {1,…,n} while WL do j := pop WL // choose index non-deterministically N := F[i](X) if N X[i] then X[i] := N add all the indexes that directly depend on i to WL (X[j] depends on X[i] if F[j] contains X[i]) return X
Concrete semantics equations R[0] = { x Z} R[1] = x:=7 R[2] = R[1] R[4] R[3] = R[2] {s | s(x) < 1000} R[4] = x:=x+1 R[3] R[5] = R[2] {s | s(x) 1000} R[6] = R[5] {s | s(x) 1001} R[0] R[2] R[3] R[4] R[1] R[5] R[6]
Abstract semantics equations R[0] = ({ x Z}) R[1] = x:=7 # R[2] = R[1] R[4] R[3] = R[2] ({s | s(x) < 1000}) R[4] = x:=x+1 # R[3] R[5] = R[2] ({s | s(x) 1000}) R[6] = R[5] ({s | s(x) 1001}) R[5] ({s | s(x) 999}) R[0] R[2] R[3] R[4] R[1] R[5] R[6]
Abstract semantics equations R[0] = R[1] = [7,7] R[2] = R[1] R[4] R[3] = R[2] [- ,999] R[4] = R[3] + [1,1] R[5] = R[2] [1000,+ ] R[6] = R[5] [999,+ ] R[5] [1001,+ ] R[0] R[2] R[3] R[4] R[1] R[5] R[6]
Too many iterations to converge
How many iterations for this one?
Widening Introduce a new binary operator to ensure termination – A kind of extrapolation Enables static analysis to use infinite height lattices – Dynamically adapts to given program Tricky to design Precision less predictable then with finite- height domains (widening non-monotone)
Formal definition For all elements d 1 d 2 d 1 d 2 For all ascending chains d 0 d 1 d 2 … the following sequence is finite – y 0 = d 0 – y i+1 = y i d i+1 For a monotone function f : D D define – x 0 = – x i+1 = x i f(x i ) Theorem: – There exits k such that x k+1 = x k – x k Red(f) = { d | d D and f(d) d }
Analysis with finite-height lattice A f #n = lpf(f # ) … f#2 f#2 f#3f#3 f# f# Red(f) Fix(f)
Analysis with widening A f#2 f#3f#2 f#3 f#2 f#2 f#3f#3 f# f# Red(f) Fix(f) lpf(f # )
Widening for Intervals Analysis [c, d] = [c, d] [a, b] [c, d] = [ if a c then a else - , if b d then b else
Semantic equations with widening R[0] = R[1] = [7,7] R[2] = R[1] R[4] R[2.1] = R[2.1] R[2] R[3] = R[2.1] [- ,999] R[4] = R[3] + [1,1] R[5] = R[2] [1001,+ ] R[6] = R[5] [999,+ ] R[5] [1001,+ ] R[0] R[2] R[3] R[4] R[1] R[5] R[6]
Choosing analysis with widening Enable widening
Non monotonicity of widening [0,1] [0,2] = ? [0,2] [0,2] = ?
Non monotonicity of widening [0,1] [0,2] = [0, ] [0,2] [0,2] = [0,2]
Analysis results with widening Did we prove it?
Analysis with narrowing A f#2 f#3f#2 f#3 f#2 f#2 f#3f#3 f# f# Red(f) Fix(f) lpf(f # )
Formal definition of narrowing Improves the result of widening y x y (x y) x For all decreasing chains x 0 x 1 … the following sequence is finite – y 0 = x 0 – y i+1 = y i x i+1 For a monotone function f: D D and x k Red(f) = { d | d D and f(d) d } define – y 0 = x – y i+1 = y i f(y i ) Theorem: – There exits k such that y k+1 =y k – y k Red(f) = { d | d D and f(d) d }
Narrowing for Interval Analysis [a, b] = [a, b] [a, b] [c, d] = [ if a = - then c else a, if b = then d else b ]
Semantic equations with narrowing R[0] = R[1] = [7,7] R[2] = R[1] R[4] R[2.1] = R[2.1] R[2] R[3] = R[2.1] [- ,999] R[4] = R[3]+[1,1] R[5] = R[2] # [1000,+ ] R[6] = R[5] [999,+ ] R[5] [1001,+ ] R[0] R[2] R[3] R[4] R[1] R[5] R[6]
Analysis with widening/narrowing Two phases – Phase 1: analyze with widening until converging – Phase 2: use values to analyze with narrowing Phase 2: R[0] = R[1] = [7,7] R[2] = R[1] R[4] R[2.1] = R[2.1] R[2] R[3] = R[2.1] [- ,999] R[4] = R[3]+[1,1] R[5] = R[2] # [1000,+ ] R[6] = R[5] [999,+ ] R[5] [1001,+ ] Phase 1: R[0] = R[1] = [7,7] R[2] = R[1] R[4] R[2.1] = R[2.1] R[2] R[3] = R[2.1] [- ,999] R[4] = R[3] + [1,1] R[5] = R[2] [1001,+ ] R[6] = R[5] [999,+ ] R[5] [1001,+ ]
Analysis with widening/narrowing
Analysis results widening/narrowing Precise invariant