By that, every path can be analyzed :-) A given program may admit several paths :-( For any given input, another path may be chosen :-((

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1 [x = M[e];] A = (A {e})\expr x [M[e 1 ] = e 2 ;] A = A {e 1,e 2 } By that, every path can be analyzed :-) A given program may admit several paths :-( For any given input, another path may be chosen :-(( == We require the set: A[v] = {[π] π : start v} 42

2 Concretely: We consider all pathsπ which reach v. For every path π, we determine the set of expressions which are available along π. Initially at program start, nothing is available :-) We compute the intersection == safe information How do we exploit this information??? 43

3 Concretely: We consider all pathsπ which reach v. For every path π, we determine the set of expressions which are available along π. Initially at program start, nothing is available :-) We compute the intersection == safe information How do we exploit this information??? 44

4 Transformation 1.1: We provide novel registers T e as storage for thee: u v x = e; u v T e = e; x = T e ; 45

5 Transformation 1.1: We provide novel registers T e as storage for thee: u u x = e; T e = e; v x = T e ; v Neg(e) u Pos(e) u T e = e; v v Neg(T e ) Pos(T e ) v v 46

6 ... analogously for R = M[e]; and M[e 1 ] = e 2 ;. Transformation 1.2: If e is available at program point u, then e need not be re-evaluated: u e A[u] u T e = e; ; We replace the assignment withnop :-) 47

7 Example: x = y +3; x = y +3; x = 7; z = y +3; x = 7; z = y +3; 48

8 Example: T = y +3; x = y +3; x = 7; z = y +3; x = T; x = 7; T = y +3; z = T; 49

9 Example: x = y +3; x = 7; z = y +3; {y +3} {y +3} {y +3} {y +3} {y +3} T = y +3; x = T; x = 7; T = y +3; z = T; 50

10 Example: x = y +3; x = 7; z = y +3; {y +3} {y +3} {y +3} {y +3} {y +3} T = y +3; x = T; x = 7; ; z = T; 51

11 Correctness: (Idea) Transformation 1.1 preserves the semantics and A[u] for all program pointsu :-) Assume π : start u is the path taken by a computation. If e A[u], then alsoe [π]. Therefore, π can be decomposed into: with the following properties: π 1 k π 2 start u 1 u 2 u 52

12 The expression e is evaluated at the edge k; The expression e is not removed from the set of available expressions at any edge inπ 2, i.e., no variable ofereceives a new value :-) == The register T e contains the value of e whenever u is reached :-)) 53

13 The expression e is evaluated at the edge k; The expression e is not removed from the set of available expressions at any edge inπ 2, i.e., no variable ofereceives a new value :-) == The register T e contains the value of e whenever u is reached :-)) 54

14 Warning: Transformation 1.1 is only meaningful for assignments x = e; where: e Vars; the evaluation of e is non-trivial :-} Which leaves open whether... 55

15 Warning: Transformation 1.1 is only meaningful for assignments x = e; where: x Vars(e); e Vars; the evaluation of e is non-trivial :- } Which leaves us with the following question... 56

16 Question: How do we computea[u] for every program point u?? Idea: We collect all restrictions to the values ofa[u] into a system of constraints: A[start] A[v] [k] (A[u]) k = (u, _,v) edge 57

17 Question: How can we compute A[u] for every program point u?? Idea: We collect all restrictions to the values ofa[u] into a system of constraints: A[start] A[v] [k] (A[u]) k = (u, _,v) edge 58

18 Wanted: a maximally large solution (??) an algorithm which computes this :-) Example: Neg(x > 1) y = 1; Pos(x > 1) 2 y = x y; 3 x = x 1; 4 59

19 Wanted: a maximally large solution (??) an algorithm which computes this :-) Example: 0 y = 1; A[0] Neg(x > 1) 5 1 Pos(x > 1) 2 y = x y; 3 x = x 1; 4 60

20 Wanted: a maximally large solution (??) an algorithm which computes this :-) Example: 0 y = 1; A[0] Neg(x > 1) 1 Pos(x > 1) A[1] (A[0] {1})\Expr y A[1] A[4] 5 2 y = x y; 3 x = x 1; 4 61

21 Wanted: a maximally large solution (??) an algorithm which computes this :-) Example: 0 y = 1; A[0] Neg(x > 1) 1 Pos(x > 1) A[1] (A[0] {1})\Expr y A[1] A[4] 5 2 y = x y; A[2] A[1] {x > 1} 3 x = x 1; 4 62

22 Wanted: a maximally large solution (??) an algorithm which computes this :-) Example: 0 y = 1; A[0] Neg(x > 1) 1 Pos(x > 1) A[1] (A[0] {1})\Expr y A[1] A[4] 5 2 y = x y; A[2] A[1] {x > 1} A[3] (A[2] {x y})\expr y 3 x = x 1; 4 63

23 Wanted: a maximally large solution (??) an algorithm which computes this :-) Example: 0 y = 1; A[0] Neg(x > 1) 1 Pos(x > 1) A[1] (A[0] {1})\Expr y A[1] A[4] 5 2 y = x y; A[2] A[1] {x > 1} A[3] (A[2] {x y})\expr y 3 A[4] (A[3] {x 1})\Expr x x = x 1; 4 64

24 Wanted: a maximally large solution (??) an algorithm which computes this :-) Example: 0 y = 1; A[0] Neg(x > 1) 1 Pos(x > 1) A[1] (A[0] {1})\Expr y A[1] A[4] 5 2 y = x y; A[2] A[1] {x > 1} A[3] (A[2] {x y})\expr y 3 A[4] (A[3] {x 1})\Expr x 4 x = x 1; A[5] A[1] {x > 1} 65

25 Wanted: a maximally large solution (??) an algorithm which computes this :-) Example: 0 Solution: y = 1; Neg(x > 1) 1 Pos(x > 1) A[0] = A[1] = {1} y = x y; x = x 1; A[2] = {1,x > 1} A[3] = {1,x > 1} A[4] = {1} A[5] = {1,x > 1} 66

26 Observation: The possible values for A[u] form a complete lattice: D = 2 Expr with B 1 B 2 iff B 1 B 2 The functions [k] : D D are monotonic, i.e., [k] (B 1 ) [k] (B 2 ) gdw. B 1 B 2 67

27 Observation: The possible values for A[u] form a complete lattice: D = 2 Expr with B 1 B 2 iff B 1 B 2 The functions [k] : D D are monotonic, i.e., [k] (B 1 ) [k] (B 2 ) iff B 1 B 2 68

28 Background 2: Complete Lattices A set D together with a relation all a,b,c D, D D is a partial order if for a a a b b a = a = b a b b c = a c reflexivity anti symmetry transitivity Examples: 1. D = 2 {a,b,c} with the relation : a,b,c a,b a,c b,c a b c 69

29 2. Z with the relation = : Z with the relation : Z = Z { } with the ordering:

30 d D is called upper bound for X D if x d for all x X d is called least upper bound (lub) if 1. d is an upper bound and 2. d y for every upper bound y of X. Caveat: has no upper bound! has the upper bounds 71

31 d D is called upper bound for X D if x d for all x X d is called least upper bound (lub) if 1. d is an upper bound and 2. d y for every upper bound y of X. Caveat: has no upper bound! has the upper bounds 72

32 d D is called upper bound for X D if x d for all x X d is called least upper bound (lub) if 1. d is an upper bound and 2. d y for every upper bound y of X. Caveat: {0,2,4,...} Z has no upper bound! {0,2,4} Z has the upper bounds 4,5,6,... 73

33 A complete lattice (cl) D is a partial ordering where every subset X D has a least upper bound X D. Note: Every complete lattice has a least element = D; a greatest element = D D. 74

34 Examples: 1. D = 2 {a,b,c} is a cl :-) 2. D = Z with = is not. 3. D = Z with is neither. 4. D = Z is also not :-( 5. With an extra element, we obtain the flat lattice Z = Z {, } :

35 We have: Theorem: If D is a complete lattice, then every subset X D has a greatest lower bound X. Proof: Construct U = {u D x X : u x}. // the set of all lower bounds of X :-) Set: g := U Claim: g = X 76

36 We have: Theorem: If D is a complete lattice, then every subset X D has a greatest lower bound X. Proof: Construct U = {u D x X : u x}. // the set of all lower bounds of X :-) Set: g := U Claim: g = X 77

37 We have: Theorem: If D is a complete lattice, then every subset X D has a greatest lower bound X. Proof: Construct U = {u D x X : u x}. // the set of all lower bounds of X :-) Set: g := U Claim: g = X 78

38 (1) g is a lower bound ofx : Assume x X. Then: u x for all u U == x is an upper bound of U == g x :-) (2) g is the greatest lower bound of X : Assume u is a lower bound ofx. Then: u U == u g :-)) 79

39 (1) g is a lower bound ofx : Assume x X. Then: u x for all u U == x is an upper bound of U == g x :-) (2) g is the greatest lower bound of X : Assume u is a lower bound ofx. Then: u U == u g :-)) 80

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43 We are looking for solutions for systems of constraints of the form: x i f i (x 1,...,x n ) ( ) 84

44 We are looking for solutions for systems of constraints of the form: x i f i (x 1,...,x n ) ( ) where: x i unknown here: A[u] D values here: 2 Expr D D ordering relation here: f i :D n D constraint here:... Constraint for A[v] : A[v] {[k] (A[u]) k = (u, _,v) edge} Because: x d 1... x d k iff x {d 1,...,d k } :-) 85

45 We are looking for solutions for systems of constraints of the form: x i f i (x 1,...,x n ) ( ) where: x i unknown here: A[u] D values here: 2 Expr D D ordering relation here: f i :D n D constraint here:... Constraint for A[v] (v start): A[v] {[k] (A[u]) k = (u, _,v) edge} Because: x d 1... x d k iff x {d 1,...,d k } :-) 86

46 We are looking for solutions for systems of constraints of the form: x i f i (x 1,...,x n ) ( ) where: x i unknown here: A[u] D values here: 2 Expr D D ordering relation here: f i :D n D constraint here:... Constraint for A[v] (v start): A[v] {[k] (A[u]) k = (u, _,v) edge} Because: x d 1... x d k iff x {d 1,...,d k } :-) 87

47 A mapping f : D 1 D 2 is called monotonic, if f(a) f(b) for all a b. 88

48 A mapping f : D 1 D 2 is called monotonic, if f(a) f(b) for all a b. Examples: (1) D 1 = D 2 = 2 U for a set U and f x = (x a) b. Obviously, every such f is monotonic :-) 89

49 A mapping f : D 1 D 2 is called monotonic, is f(a) f(b) for all a b. Examples: (1) D 1 = D 2 = 2 U for a set U and f x = (x a) b. Obviously, every such f is monotonic :-) (2) D 1 = D 2 = Z (with the ordering ). Then: incx = x+1 is monotonic. decx = x 1 is monotonic. textbullet 90

50 A mapping f : D 1 D 2 is called monotonic, is f(a) f(b) for all a b. Examples: (1) D 1 = D 2 = 2 U for a set U and f x = (x a) b. Obviously, every such f is monotonic :-) (2) D 1 = D 2 = Z (with the ordering ). Then: incx = x+1 is monotonic. decx = x 1 is monotonic. invx = x is not monotonic :-) 91

51 Theorem: If f 1 : D 1 D 2 and f 2 : D 2 D 3 are monotonic, then also f 2 f 1 : D 1 D 3 :-) 92