Type-Based Reasoning about Efficiency D. Seidel J. Voigtl ander - - PowerPoint PPT Presentation

Type-Based Reasoning about Efficiency

• D. Seidel∗
• J. Voigtl¨

ander

University of Bonn

August 3rd, 2011

∗Supported by the DFG under grant VO 1512/1-1.

Parametric Polymorphism in Haskell A standard function: map g [ ] = [ ] map g (a : as) = (g a) : (map g as)

1

Parametric Polymorphism in Haskell A standard function: map g [ ] = [ ] map g (a : as) = (g a) : (map g as) Some invocations: map succ [1, 2, 3] = [2, 3, 4]

1

Parametric Polymorphism in Haskell A standard function: map g [ ] = [ ] map g (a : as) = (g a) : (map g as) Some invocations: map succ [1, 2, 3] = [2, 3, 4] map not [True, False] = [False, True]

1

Parametric Polymorphism in Haskell A standard function: map g [ ] = [ ] map g (a : as) = (g a) : (map g as) Some invocations: map succ [1, 2, 3] = [2, 3, 4] map not [True, False] = [False, True] map even [1, 2, 3] = [False, True, False]

1

Parametric Polymorphism in Haskell A standard function: map g [ ] = [ ] map g (a : as) = (g a) : (map g as) Some invocations: map succ [1, 2, 3] = [2, 3, 4] map not [True, False] = [False, True] map even [1, 2, 3] = [False, True, False] map not [1, 2, 3]

1

Parametric Polymorphism in Haskell A standard function: map :: (α → β) → [α] → [β] map g [ ] = [ ] map g (a : as) = (g a) : (map g as) Some invocations: map succ [1, 2, 3] = [2, 3, 4] map not [True, False] = [False, True] map even [1, 2, 3] = [False, True, False] map not [1, 2, 3]

1

Parametric Polymorphism in Haskell A standard function: map :: (α → β) → [α] → [β] map g [ ] = [ ] map g (a : as) = (g a) : (map g as) Some invocations: map succ [1, 2, 3] = [2, 3, 4] map not [True, False] = [False, True] map even [1, 2, 3] = [False, True, False] map not [1, 2, 3] rejected at compile-time

1

Parametric Polymorphism in Haskell A standard function: map :: (α → β) → [α] → [β] Some invocations: map succ [1, 2, 3] = [2, 3, 4] map not [True, False] = [False, True] map even [1, 2, 3] = [False, True, False] map not [1, 2, 3] rejected at compile-time

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Another Example reverse :: [α] → [α] reverse [ ] = [ ] reverse (a : as) = (reverse as) + + [a]

2

Another Example reverse :: [α] → [α] reverse [ ] = [ ] reverse (a : as) = (reverse as) + + [a] For every choice of g and l: reverse (map g l) = map g (reverse l) Provable by induction.

2

Another Example reverse :: [α] → [α] reverse [ ] = [ ] reverse (a : as) = (reverse as) + + [a] For every choice of g and l: reverse (map g l) = map g (reverse l) Provable by induction. Or as a “free theorem” [Wadler, FPCA’89].

2

Another Example reverse :: [α] → [α] For every choice of g and l: reverse (map g l) = map g (reverse l) Provable by induction. Or as a “free theorem” [Wadler, FPCA’89].

2

Another Example reverse :: [α] → [α] tail :: [α] → [α] For every choice of g and l: reverse (map g l) = map g (reverse l) tail (map g l) = map g (tail l)

2

Another Example reverse :: [α] → [α] tail :: [α] → [α] f :: [α] → [α] For every choice of g and l: reverse (map g l) = map g (reverse l) tail (map g l) = map g (tail l) f (map g l) = map g (f l)

2

Automatic Generation of Free Theorems At http://www-ps.iai.uni-bonn.de/ft:

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Automatic Generation of Free Theorems

3

Applications

◮ Short Cut Fusion [Gill et al., FPCA’93]

4

Applications

◮ Short Cut Fusion [Gill et al., FPCA’93] ◮ The Dual of Short Cut Fusion

[Takano & Meijer, FPCA’95], [Svenningsson, ICFP’02]

◮ . . .

4

Applications

◮ Short Cut Fusion [Gill et al., FPCA’93] ◮ The Dual of Short Cut Fusion

[Takano & Meijer, FPCA’95], [Svenningsson, ICFP’02]

◮ . . . ◮ Knuth’s 0-1-principle and the like

[Day et al., Haskell’99], [V., POPL’08]

4

Applications

◮ Short Cut Fusion [Gill et al., FPCA’93] ◮ The Dual of Short Cut Fusion

[Takano & Meijer, FPCA’95], [Svenningsson, ICFP’02]

◮ . . . ◮ Knuth’s 0-1-principle and the like

[Day et al., Haskell’99], [V., POPL’08]

◮ Bidirectionalization [V., POPL’09]

4

Applications

◮ Short Cut Fusion [Gill et al., FPCA’93] ◮ The Dual of Short Cut Fusion

[Takano & Meijer, FPCA’95], [Svenningsson, ICFP’02]

◮ . . . ◮ Knuth’s 0-1-principle and the like

[Day et al., Haskell’99], [V., POPL’08]

◮ Bidirectionalization [V., POPL’09] ◮ . . . ◮ Testing polymorphic properties

[Bernardy et al., ESOP’10]

4

What About Efficiency? f :: α → Nat f :: α → α → α f :: α → (α, α) f (g x) = f x f (g x) (g y) = g (f x y) f (g x) = let y = f x in (g (fst y), g (snd y))

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What About Efficiency? f :: α → Nat f :: α → α → α f :: α → (α, α) f (g x) = f x f (g x) (g y) = g (f x y) f (g x) = let y = f x in (g (fst y), g (snd y)) call-by- value

> > <

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What About Efficiency? f :: α → Nat f :: α → α → α f :: α → (α, α) f (g x) = f x f (g x) (g y) = g (f x y) f (g x) = let y = f x in (g (fst y), g (snd y)) call-by- value

> > <

call-by- name

= = < (?)

5

What About Efficiency? f :: α → Nat f :: α → α → α f :: α → (α, α) f (g x) = f x f (g x) (g y) = g (f x y) f (g x) = let y = f x in (g (fst y), g (snd y)) call-by- value

> > <

call-by- name

= = < (?)

call-by- need

= = <

5

What About Efficiency? f :: α → Nat f :: α → α → α f :: α → (α, α) f (g x) = f x f (g x) (g y) = g (f x y) f (g x) = let y = f x in (g (fst y), g (snd y)) call-by- value

> > <

call-by- name

= = < (?)

call-by- need

= = <

5

What About Efficiency? Now, how about, say f :: α → α → (α, α) ?

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What About Efficiency? Now, how about, say f :: α → α → (α, α) ? Or f :: [α] → [α] ?

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What About Efficiency? Now, how about, say f :: α → α → (α, α) ? Or f :: [α] → [α] ? And how to actually establish such statements?

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But isn’t it Somehow Trivial? Recall f :: α → Nat, with standard free theorem: f (g x) = f x for all choices of types τ1, τ2, function g :: τ1 → τ2, and x :: τ1.

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But isn’t it Somehow Trivial? Recall f :: α → Nat, with standard free theorem: f (g x) = f x for all choices of types τ1, τ2, function g :: τ1 → τ2, and x :: τ1. Clearly, this means that f is a constant function, i.e., for all x and y: f y = f x

7

But isn’t it Somehow Trivial? Recall f :: α → Nat, with standard free theorem: f (g x) = f x for all choices of types τ1, τ2, function g :: τ1 → τ2, and x :: τ1. Clearly, this means that f is a constant function, i.e., for all x and y: f y = f x So, “obviously”, f (g x) ⊒ f x where ⊒ means “the same result, and equally fast

• r slower”.

7

But isn’t it Somehow Trivial? Recall f :: α → Nat, with standard free theorem: f (g x) = f x for all choices of types τ1, τ2, function g :: τ1 → τ2, and x :: τ1. Clearly, this means that f is a constant function, i.e., for all x and y: f y = f x So, “obviously”, f (g x) ⊒ f x where ⊒ means “the same result, and equally fast

• r slower”. What’s wrong with this reasoning?

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But isn’t it Somehow Trivial? Consider: f x = if x == 0 g x = 0 then 0 else f (x − 1)

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But isn’t it Somehow Trivial? Consider: f x = if x == 0 g x = 0 then 0 else f (x − 1) Certainly, for all x and y: f y = f x

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But isn’t it Somehow Trivial? Consider: f x = if x == 0 g x = 0 then 0 else f (x − 1) Certainly, for all x and y: f y = f x But certainly not, for x > 0: f (g x) ⊒ f x

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But isn’t it Somehow Trivial? Consider: f :: Nat → Nat g :: α → Nat f x = if x == 0 g x = 0 then 0 else f (x − 1) Certainly, for all x and y: f y = f x But certainly not, for x > 0: f (g x) ⊒ f x

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But isn’t it Somehow Trivial? Consider: f :: Nat → Nat g :: α → Nat f x = if x == 0 g x = 0 then 0 else f (x − 1) Certainly, for all x and y: f y = f x But certainly not, for x > 0: f (g x) ⊒ f x Exploiting polymorphism is really essential, and not just for the extensional statements!

8

Free Theorems, Formally — In a Nutshell

Syntax for a typed λ-calculus: τ ::= α | Nat | τ → τ | . . . t ::= x | n | t + t | λx :: τ.t | t t | . . .

9

Free Theorems, Formally — In a Nutshell

Syntax for a typed λ-calculus: τ ::= α | Nat | τ → τ | . . . t ::= x | n | t + t | λx :: τ.t | t t | . . . Semantics: αθ = θ(α) Natθ = N τ1 → τ2θ = τ2θ

τ1θ

θ ∈ SetTVar

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Free Theorems, Formally — In a Nutshell

Syntax for a typed λ-calculus: τ ::= α | Nat | τ → τ | . . . t ::= x | n | t + t | λx :: τ.t | t t | . . . Semantics: αθ = θ(α) xσ = σ(x) Natθ = N nσ = n τ1 → τ2θ = τ2θ

τ1θ

λx :: τ.tσ = λv.tσ[x→v] θ ∈ SetTVar t1 t2σ = t1σ t2σ

9

Free Theorems, Formally — In a Nutshell

Syntax for a typed λ-calculus: τ ::= α | Nat | τ → τ | . . . t ::= x | n | t + t | λx :: τ.t | t t | . . . Semantics: αθ = θ(α) xσ = σ(x) Natθ = N nσ = n τ1 → τ2θ = τ2θ

τ1θ

λx :: τ.tσ = λv.tσ[x→v] θ ∈ SetTVar t1 t2σ = t1σ t2σ Logical relation: ∆α,ρ = ρ(α) ∆Nat,ρ = idN ∆τ1→τ2,ρ = {(f, g) | ∀(x, y) ∈ ∆τ1,ρ. (f x, g y) ∈ ∆τ2,ρ} with ρ ∈ RelTVar

9

Free Theorems, Formally — In a Nutshell

Syntax for a typed λ-calculus: τ ::= α | Nat | τ → τ | . . . t ::= x | n | t + t | λx :: τ.t | t t | . . . Semantics: αθ = θ(α) xσ = σ(x) Natθ = N nσ = n τ1 → τ2θ = τ2θ

τ1θ

λx :: τ.tσ = λv.tσ[x→v] θ ∈ SetTVar t1 t2σ = t1σ t2σ Logical relation: ∆α,ρ = ρ(α) ∆Nat,ρ = idN ∆τ1→τ2,ρ = {(f, g) | ∀(x, y) ∈ ∆τ1,ρ. (f x, g y) ∈ ∆τ2,ρ} with ρ ∈ RelTVar Theorem: for closed term t of type τ, (t∅, t∅) ∈ ∆τ,ρ.

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An Example Derivation, for f :: α → Nat ∀ρ, t closed with t :: τ. (t∅, t∅) ∈ ∆τ,ρ

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An Example Derivation, for f :: α → Nat ∀ρ, t closed with t :: τ. (t∅, t∅) ∈ ∆τ,ρ ⇒ (t = f and τ = α → Nat) ∀ρ. (f∅, f∅) ∈ ∆α→Nat,ρ

10

An Example Derivation, for f :: α → Nat ∀ρ, t closed with t :: τ. (t∅, t∅) ∈ ∆τ,ρ ⇒ (t = f and τ = α → Nat) ∀ρ. (f∅, f∅) ∈ ∆α→Nat,ρ ⇔ (∆τ1→τ2,ρ = {(f, g) | ∀(x, y) ∈ ∆τ1,ρ. (f x, g y) ∈ ∆τ2,ρ}) ∀ρ, (x, y) ∈ ∆α,ρ. (f∅ x, f∅ y) ∈ ∆Nat,ρ

10

An Example Derivation, for f :: α → Nat ∀ρ, t closed with t :: τ. (t∅, t∅) ∈ ∆τ,ρ ⇒ (t = f and τ = α → Nat) ∀ρ. (f∅, f∅) ∈ ∆α→Nat,ρ ⇔ (∆τ1→τ2,ρ = {(f, g) | ∀(x, y) ∈ ∆τ1,ρ. (f x, g y) ∈ ∆τ2,ρ}) ∀ρ, (x, y) ∈ ∆α,ρ. (f∅ x, f∅ y) ∈ ∆Nat,ρ ⇒ (∆α,ρ = ρ(α), ρ(α) := g ∈ τ2∅

τ1∅, ∆Nat,ρ = idN)

∀g ∈ τ2∅

τ1∅, x ∈ τ1∅. f∅ x = f∅ (g x)

10

An Example Derivation, for f :: α → Nat ∀ρ, t closed with t :: τ. (t∅, t∅) ∈ ∆τ,ρ ⇒ (t = f and τ = α → Nat) ∀ρ. (f∅, f∅) ∈ ∆α→Nat,ρ ⇔ (∆τ1→τ2,ρ = {(f, g) | ∀(x, y) ∈ ∆τ1,ρ. (f x, g y) ∈ ∆τ2,ρ}) ∀ρ, (x, y) ∈ ∆α,ρ. (f∅ x, f∅ y) ∈ ∆Nat,ρ ⇒ (∆α,ρ = ρ(α), ρ(α) := g ∈ τ2∅

τ1∅, ∆Nat,ρ = idN)

∀g ∈ τ2∅

τ1∅, x ∈ τ1∅. f∅ x = f∅ (g x)

⇒ (term semantics) ∀g :: τ1 → τ2, x :: τ1. f x∅ = f (g x)∅

10

Bringing Costs into the Picture New semantics: x¢

σ

= (σ(x), 0) n¢

σ

= (n, 0) λx :: τ.t¢

σ = (λv.1 ⊲t¢ σ[x→v], 0)

t1 t2¢

σ

= t1¢

σ ¢ t2¢ σ

where: c ⊲(v, c′) = (v, c + c′) and f ¢ x = (c + c′) ⊲(g v) if f = (g, c), x = (v, c′)

11

Bringing Costs into the Picture New semantics: x¢

σ

= (σ(x), 0) n¢

σ

= (n, 0) λx :: τ.t¢

σ = (λv.1 ⊲t¢ σ[x→v], 0)

t1 t2¢

σ

= t1¢

σ ¢ t2¢ σ

where: c ⊲(v, c′) = (v, c + c′) and f ¢ x = (c + c′) ⊲(g v) if f = (g, c), x = (v, c′) Note: t :: α → Nat implies t¢

∅ ∈ C(C(N)θ(α)),

for every θ ∈ SetTVar, where C(S) = {(v, c) | v ∈ S, c ∈ Z}.

11

How about the Logical Relation? Tempting would be: ∆¢

α,ρ

= ρ(α) ∆¢

Nat,ρ = idN×Z

∆¢

τ1→τ2,ρ = {(f, g) | ∀(x, y) ∈ ∆¢ τ1,ρ.

(f ¢ x, g ¢ y) ∈ ∆¢

τ2,ρ}

with ρ(α1), ρ(α2), . . . ⊆ C(Si) × C(Ti)

12

How about the Logical Relation? Tempting would be: ∆¢

α,ρ

= ρ(α) ∆¢

Nat,ρ = idN×Z

∆¢

τ1→τ2,ρ = {(f, g) | ∀(x, y) ∈ ∆¢ τ1,ρ.

(f ¢ x, g ¢ y) ∈ ∆¢

τ2,ρ}

with ρ(α1), ρ(α2), . . . ⊆ C(Si) × C(Ti) But NO, does not work!

12

How about the Logical Relation? Tempting would be: ∆¢

α,ρ

= ρ(α) ∆¢

Nat,ρ = idN×Z

∆¢

τ1→τ2,ρ = {(f, g) | ∀(x, y) ∈ ∆¢ τ1,ρ.

(f ¢ x, g ¢ y) ∈ ∆¢

τ2,ρ}

with ρ(α1), ρ(α2), . . . ⊆ C(Si) × C(Ti) But NO, does not work! It would allow us to conclude from: ∀ρ, f :: α → Nat. (f¢

∅, f¢ ∅) ∈ ∆¢ α→Nat,ρ

that: ∀g :: τ1 → τ2, x :: τ1. f x¢

∅ = f (g x)¢ ∅

12

How about the Logical Relation? Much more disciplined: ∆¢

α,ρ

= C(ρ(α)) ∆¢

Nat,ρ = idN×Z

∆¢

τ1→τ2,ρ = {(f, g) | cost(f) = cost(g)

∧ ∀(x, y) ∈ ∆¢

τ1,ρ. (f ¢ x, g ¢ y) ∈ ∆¢ τ2,ρ}

with ρ(α1), ρ(α2), . . . ⊆ Si ×Ti, where C(R) = {((u, c), (v, c) | (u, v) ∈ R, c ∈ Z}

13

How about the Logical Relation? Much more disciplined: ∆¢

α,ρ

= C(ρ(α)) ∆¢

Nat,ρ = idN×Z

∆¢

τ1→τ2,ρ = {(f, g) | cost(f) = cost(g)

∧ ∀(x, y) ∈ ∆¢

τ1,ρ. (f ¢ x, g ¢ y) ∈ ∆¢ τ2,ρ}

with ρ(α1), ρ(α2), . . . ⊆ Si ×Ti, where C(R) = {((u, c), (v, c) | (u, v) ∈ R, c ∈ Z} Now indeed . . . Theorem: for closed term t of type τ, (t¢

∅, t¢ ∅) ∈ ∆¢ τ,ρ

13

Now, What about our Example f :: α → Nat ? ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→Nat,ρ

14

Now, What about our Example f :: α → Nat ? ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→Nat,ρ

⇒ (∆¢

τ1→τ2,ρ= {(f, g) | cost(f) = cost(g)

∧∀(x, y) ∈ ∆¢

τ1,ρ. (f ¢ x, g ¢ y) ∈ ∆¢ τ2,ρ})

∀ρ, (x, y) ∈ ∆¢

α,ρ. (f¢ ∅ ¢ x, f¢ ∅ ¢ y) ∈ ∆¢ Nat,ρ

14

Now, What about our Example f :: α → Nat ? ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→Nat,ρ

⇒ (∆¢

τ1→τ2,ρ= {(f, g) | cost(f) = cost(g)

∧∀(x, y) ∈ ∆¢

τ1,ρ. (f ¢ x, g ¢ y) ∈ ∆¢ τ2,ρ})

∀ρ, (x, y) ∈ ∆¢

α,ρ. (f¢ ∅ ¢ x, f¢ ∅ ¢ y) ∈ ∆¢ Nat,ρ

⇒ (∆¢

α,ρ = C(ρ(α)), ρ(α) := R ∈ Rel, ∆¢ Nat,ρ = idN×Z)

∀R ∈ Rel, (x, y) ∈ C(R). f¢

∅ ¢ x = f¢ ∅ ¢ y

14

Now, What about our Example f :: α → Nat ? ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→Nat,ρ

⇒ (∆¢

τ1→τ2,ρ= {(f, g) | cost(f) = cost(g)

∧∀(x, y) ∈ ∆¢

τ1,ρ. (f ¢ x, g ¢ y) ∈ ∆¢ τ2,ρ})

∀ρ, (x, y) ∈ ∆¢

α,ρ. (f¢ ∅ ¢ x, f¢ ∅ ¢ y) ∈ ∆¢ Nat,ρ

⇒ (∆¢

α,ρ = C(ρ(α)), ρ(α) := R ∈ Rel, ∆¢ Nat,ρ = idN×Z)

∀R ∈ Rel, (x, y) ∈ C(R). f¢

∅ ¢ x = f¢ ∅ ¢ y

How to choose R to bring g :: τ1 → τ2 into play?

14

Now, What about our Example f :: α → Nat ? ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→Nat,ρ

⇒ (∆¢

τ1→τ2,ρ= {(f, g) | cost(f) = cost(g)

∧∀(x, y) ∈ ∆¢

τ1,ρ. (f ¢ x, g ¢ y) ∈ ∆¢ τ2,ρ})

∀ρ, (x, y) ∈ ∆¢

α,ρ. (f¢ ∅ ¢ x, f¢ ∅ ¢ y) ∈ ∆¢ Nat,ρ

⇒ (∆¢

α,ρ = C(ρ(α)), ρ(α) := R ∈ Rel, ∆¢ Nat,ρ = idN×Z)

∀R ∈ Rel, (x, y) ∈ C(R). f¢

∅ ¢ x = f¢ ∅ ¢ y

How to choose R to bring g :: τ1 → τ2 into play? Problem: {(x, y) | g¢

∅ ¢ x = y} not C(R) for any R

14

Now, What about our Example f :: α → Nat ? Problem: {(x, y) | g¢

∅ ¢ x = y} not C(R) for any R

Solution: but {(c ⊲ appCost(g, x) ⊲ x, c ⊲ y) | g¢

∅ ¢ x = y, c ∈ Z} = C(Rg)

for Rg = {(val(x), val(g¢

∅ ¢ x)) | x ∈ τ1¢ ∅}

where appCost(g, x) = cost(g¢

∅ ¢ x) − cost(x)

15

Now, What about our Example f :: α → Nat ? Problem: {(x, y) | g¢

∅ ¢ x = y} not C(R) for any R

Solution: but {(c ⊲ appCost(g, x) ⊲ x, c ⊲ y) | g¢

∅ ¢ x = y, c ∈ Z} = C(Rg)

for Rg = {(val(x), val(g¢

∅ ¢ x)) | x ∈ τ1¢ ∅}

where appCost(g, x) = cost(g¢

∅ ¢ x) − cost(x)

Now: ∀R ∈ Rel, (x, y) ∈ C(R). f¢

∅ ¢ x = f¢ ∅ ¢ y

⇒ ∀x ∈ τ1¢

∅. f¢ ∅ ¢ (appCost(g, x) ⊲ x)

= f¢

∅ ¢ (g¢ ∅ ¢ x)

15

Now, What about our Example f :: α → Nat ? Now: ∀R ∈ Rel, (x, y) ∈ C(R). f¢

∅ ¢ x = f¢ ∅ ¢ y

⇒ ∀x ∈ τ1¢

∅. f¢ ∅ ¢ (appCost(g, x) ⊲ x)

= f¢

∅ ¢ (g¢ ∅ ¢ x)

15

Now, What about our Example f :: α → Nat ? Now: ∀R ∈ Rel, (x, y) ∈ C(R). f¢

∅ ¢ x = f¢ ∅ ¢ y

⇒ ∀x ∈ τ1¢

∅. f¢ ∅ ¢ (appCost(g, x) ⊲ x)

= f¢

∅ ¢ (g¢ ∅ ¢ x)

⇒ ∀x :: τ1. appCost(g, x¢

∅) ⊲(f¢ ∅ ¢ x¢ ∅)

= f¢

∅ ¢ (g¢ ∅ ¢ x¢ ∅)

15

Now, What about our Example f :: α → Nat ? Now: ∀R ∈ Rel, (x, y) ∈ C(R). f¢

∅ ¢ x = f¢ ∅ ¢ y

⇒ ∀x ∈ τ1¢

∅. f¢ ∅ ¢ (appCost(g, x) ⊲ x)

= f¢

∅ ¢ (g¢ ∅ ¢ x)

⇒ ∀x :: τ1. appCost(g, x¢

∅) ⊲(f¢ ∅ ¢ x¢ ∅)

= f¢

∅ ¢ (g¢ ∅ ¢ x¢ ∅)

⇒ ∀x :: τ1. appCost(g, x¢

∅) ⊲f x¢ ∅ = f (g x)¢ ∅

15

Now, What about our Example f :: α → Nat ? Now: ∀R ∈ Rel, (x, y) ∈ C(R). f¢

∅ ¢ x = f¢ ∅ ¢ y

⇒ ∀x ∈ τ1¢

∅. f¢ ∅ ¢ (appCost(g, x) ⊲ x)

= f¢

∅ ¢ (g¢ ∅ ¢ x)

⇒ ∀x :: τ1. appCost(g, x¢

∅) ⊲(f¢ ∅ ¢ x¢ ∅)

= f¢

∅ ¢ (g¢ ∅ ¢ x¢ ∅)

⇒ ∀x :: τ1. appCost(g, x¢

∅) ⊲f x¢ ∅ = f (g x)¢ ∅

Hence: f x ⊏ f (g x)

15

Let’s Try another Example, f :: α → α ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→α,ρ

16

Let’s Try another Example, f :: α → α ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→α,ρ

⇒ ∀ρ, (x, y) ∈ ∆¢

α,ρ. (f¢ ∅ ¢ x, f¢ ∅ ¢ y) ∈ ∆¢ α,ρ

16

Let’s Try another Example, f :: α → α ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→α,ρ

⇒ ∀ρ, (x, y) ∈ ∆¢

α,ρ. (f¢ ∅ ¢ x, f¢ ∅ ¢ y) ∈ ∆¢ α,ρ

⇒ ∀g :: τ1 → τ2, (x, y) ∈ C(Rg). (f¢

∅ ¢ x, f¢ ∅ ¢ y) ∈ C(Rg)

16

Let’s Try another Example, f :: α → α ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→α,ρ

⇒ ∀ρ, (x, y) ∈ ∆¢

α,ρ. (f¢ ∅ ¢ x, f¢ ∅ ¢ y) ∈ ∆¢ α,ρ

⇒ ∀g :: τ1 → τ2, (x, y) ∈ C(Rg). (f¢

∅ ¢ x, f¢ ∅ ¢ y) ∈ C(Rg)

⇒ ∀g :: τ1 → τ2, x ∈ τ1¢

∅.

(f¢

∅ ¢ (appCost(g, x) ⊲ x),

f¢

∅ ¢ (g¢ ∅ ¢ x)) ∈ C(Rg)

16

Let’s Try another Example, f :: α → α ∀ρ. (f¢

∅, f¢ ∅) ∈ ∆¢ α→α,ρ

⇒ ∀ρ, (x, y) ∈ ∆¢

α,ρ. (f¢ ∅ ¢ x, f¢ ∅ ¢ y) ∈ ∆¢ α,ρ

⇒ ∀g :: τ1 → τ2, (x, y) ∈ C(Rg). (f¢

∅ ¢ x, f¢ ∅ ¢ y) ∈ C(Rg)

⇒ ∀g :: τ1 → τ2, x ∈ τ1¢

∅.

(f¢

∅ ¢ (appCost(g, x) ⊲ x),

f¢

∅ ¢ (g¢ ∅ ¢ x)) ∈ C(Rg)

Does this imply g¢

∅ ¢ (f¢ ∅ ¢ x) = f¢ ∅ ¢ (g¢ ∅ ¢ x) ?

16

Let’s Try another Example, f :: α → α ∀g :: τ1 → τ2, x ∈ τ1¢

∅.

(appCost(g, x) ⊲(f¢

∅ ¢ x),

f¢

∅ ¢ (g¢ ∅ ¢ x)) ∈ C(Rg)

Does this imply g¢

∅ ¢ (f¢ ∅ ¢ x) = f¢ ∅ ¢ (g¢ ∅ ¢ x) ?

Let’s see: C(Rg) = {(c ⊲ appCost(g, x) ⊲ x, c ⊲ y) | g¢

∅ ¢ x = y, c ∈ Z}

16

Let’s Try another Example, f :: α → α ∀g :: τ1 → τ2, x ∈ τ1¢

∅.

(appCost(g, x) ⊲(f¢

∅ ¢ x),

f¢

∅ ¢ (g¢ ∅ ¢ x)) ∈ C(Rg)

Does this imply g¢

∅ ¢ (f¢ ∅ ¢ x) = f¢ ∅ ¢ (g¢ ∅ ¢ x) ?

Let’s see: C(Rg) = {(c ⊲ appCost(g, x′) ⊲ x′, c ⊲ y) | g¢

∅ ¢ x′ = y, c ∈ Z}

16

Let’s Try another Example, f :: α → α ∀g :: τ1 → τ2, x ∈ τ1¢

∅.

(appCost(g, x) ⊲(f¢

∅ ¢ x),

f¢

∅ ¢ (g¢ ∅ ¢ x)) ∈ C(Rg)

Let’s see: C(Rg) = {(c ⊲ appCost(g, x′) ⊲ x′, c ⊲ y) | g¢

∅ ¢ x′ = y, c ∈ Z}

Actually, the above only imply: ∀x.∃x′. appCost(g, x) ⊲(g¢

∅ ¢ (f¢ ∅ ¢ x))

= appCost(g, x′) ⊲(f¢

∅ ¢ (g¢ ∅ ¢ x))

16

Let’s Try another Example, f :: α → α Define: Rg

x = {(val(x¢ ∅), val(g¢ ∅ ¢ x¢ ∅))}

17

Let’s Try another Example, f :: α → α Define: Rg

x = {(val(x¢ ∅), val(g¢ ∅ ¢ x¢ ∅))}

Then: C(Rg

x ) = {(c ⊲ appCost(g, x¢ ∅) ⊲x¢ ∅,

c ⊲(g¢

∅ ¢ x¢ ∅)) | c ∈ Z}

17

Let’s Try another Example, f :: α → α Define: Rg

x = {(val(x¢ ∅), val(g¢ ∅ ¢ x¢ ∅))}

Then: C(Rg

x ) = {(c ⊲ appCost(g, x¢ ∅) ⊲x¢ ∅,

c ⊲(g¢

∅ ¢ x¢ ∅)) | c ∈ Z}

Thus: · · · ⇒ ∀g :: τ1 → τ2, x :: τ1, (x, y) ∈ C(Rg

x ).

(f¢

∅ ¢ x, f¢ ∅ ¢ y) ∈ C(Rg x )

⇒ ∀g :: τ1 → τ2, x :: τ1. (f¢

∅ ¢ (appCost(g, x¢ ∅) ⊲x¢ ∅),

f¢

∅ ¢ (g¢ ∅ ¢ x¢ ∅)) ∈ C(Rg x )

17

Let’s Try another Example, f :: α → α Then: C(Rg

x ) = {(c ⊲ appCost(g, x¢ ∅) ⊲x¢ ∅,

c ⊲(g¢

∅ ¢ x¢ ∅)) | c ∈ Z}

Thus: · · · ⇒ ∀g :: τ1 → τ2, x :: τ1, (x, y) ∈ C(Rg

x ).

(f¢

∅ ¢ x, f¢ ∅ ¢ y) ∈ C(Rg x )

⇒ ∀g :: τ1 → τ2, x :: τ1. (f¢

∅ ¢ (appCost(g, x¢ ∅) ⊲x¢ ∅),

f¢

∅ ¢ (g¢ ∅ ¢ x¢ ∅)) ∈ C(Rg x )

⇒ ∀g :: τ1 → τ2, x :: τ1. g¢

∅ ¢ (f¢ ∅ ¢ x¢ ∅) = f¢ ∅ ¢ (g¢ ∅ ¢ x¢ ∅)

17

Other Examples For f :: α → α → α, g (f x y) ⊏ f (g x) (g y)

18

Other Examples For f :: α → α → α, g (f x y) ⊏ f (g x) (g y) For f :: α → (α, α), mapPair (g, g) (f x) ⊐ f (g x)

18

Other Examples For f :: α → α → α, g (f x y) ⊏ f (g x) (g y) For f :: α → (α, α), mapPair (g, g) (f x) ⊐ f (g x) For f :: [α] → Nat, f l ⊏ f (map g l)

18

Other Examples For f :: α → α → α, g (f x y) ⊏ f (g x) (g y) For f :: α → (α, α), mapPair (g, g) (f x) ⊐ f (g x) For f :: [α] → Nat, f l ⊏ f (map g l) For f :: [α] → [α], get conditional statements about relative efficiency of map g (f l) and f (map g l).

18

Conclusion We did:

◮ develop of “cost-aware” logical relation and its

parametricity theorem,

19

Conclusion We did:

◮ develop of “cost-aware” logical relation and its

parametricity theorem,

◮ realize that more effort has to go into the actual

deriving of concrete free theorems,

19

Conclusion We did:

◮ develop of “cost-aware” logical relation and its

parametricity theorem,

◮ realize that more effort has to go into the actual

deriving of concrete free theorems,

◮ establish a number of building blocks/heuristics

for that.

19

Conclusion We did:

◮ develop of “cost-aware” logical relation and its

parametricity theorem,

◮ realize that more effort has to go into the actual

deriving of concrete free theorems,

◮ establish a number of building blocks/heuristics

for that. Further plans:

◮ mechanize derivation of cost-aware statements,

19

Conclusion We did:

◮ develop of “cost-aware” logical relation and its

parametricity theorem,

◮ realize that more effort has to go into the actual

deriving of concrete free theorems,

◮ establish a number of building blocks/heuristics

for that. Further plans:

◮ mechanize derivation of cost-aware statements, ◮ apply to automatic program transformations,

19

Conclusion We did:

◮ develop of “cost-aware” logical relation and its

parametricity theorem,

◮ realize that more effort has to go into the actual

deriving of concrete free theorems,

◮ establish a number of building blocks/heuristics

for that. Further plans:

◮ mechanize derivation of cost-aware statements, ◮ apply to automatic program transformations, ◮ extend to full recursion, other language features,

19

Conclusion We did:

◮ develop of “cost-aware” logical relation and its

parametricity theorem,

◮ realize that more effort has to go into the actual

deriving of concrete free theorems,

◮ establish a number of building blocks/heuristics

for that. Further plans:

◮ mechanize derivation of cost-aware statements, ◮ apply to automatic program transformations, ◮ extend to full recursion, other language features, ◮ investigate call-by-name/call-by-need,

19

Conclusion We did:

◮ develop of “cost-aware” logical relation and its

parametricity theorem,

◮ realize that more effort has to go into the actual

deriving of concrete free theorems,

◮ establish a number of building blocks/heuristics

for that. Further plans:

◮ mechanize derivation of cost-aware statements, ◮ apply to automatic program transformations, ◮ extend to full recursion, other language features, ◮ investigate call-by-name/call-by-need, ◮ use more realistic cost measures?

19

References I

J.-P. Bernardy, P. Jansson, and K. Claessen. Testing polymorphic properties. In European Symposium on Programming, Proceedings, volume 6012 of LNCS, pages 125–144. Springer-Verlag, 2010.

• B. Bjerner and S. Holmstr¨
• m.

A compositional approach to time analysis of first order lazy functional programs. In Functional Programming Languages and Computer Architecture, Proceedings, pages 157–165. ACM Press, 1989. N.A. Day, J. Launchbury, and J. Lewis. Logical abstractions in Haskell. In Haskell Workshop, Proceedings. Technical Report UU-CS-1999-28, Utrecht University, 1999.

20

References II

• A. Gill, J. Launchbury, and S.L. Peyton Jones.

A short cut to deforestation. In Functional Programming Languages and Computer Architecture, Proceedings, pages 223–232. ACM Press, 1993.

• Y. Liu and G. G´
• mez.

Automatic accurate cost-bound analysis for high-level languages. IEEE Transactions on Computers, 50(12):1295–1309, 2001. J.C. Reynolds. Types, abstraction and parametric polymorphism. In Information Processing, Proceedings, pages 513–523. Elsevier, 1983.

21

References III

• M. Rosendahl.

Automatic complexity analysis. In Functional Programming Languages and Computer Architecture, Proceedings, pages 144–156. ACM Press, 1989.

• D. Sands.

A na¨ ıve time analysis and its theory of cost equivalence. Journal of Logic and Computation, 5(4):495–541, 1995.

• J. Svenningsson.

Shortcut fusion for accumulating parameters & zip-like functions. In International Conference on Functional Programming, Proceedings, pages 124–132. ACM Press, 2002.

22

References IV

• A. Takano and E. Meijer.

Shortcut deforestation in calculational form. In Functional Programming Languages and Computer Architecture, Proceedings, pages 306–313. ACM Press, 1995.

• J. Voigtl¨

ander. Much ado about two: A pearl on parallel prefix computation. In Principles of Programming Languages, Proceedings, pages 29–35. ACM Press, 2008.

• J. Voigtl¨

ander. Bidirectionalization for free! In Principles of Programming Languages, Proceedings, pages 165–176. ACM Press, 2009.

23

References V

• P. Wadler.

Strictness analysis aids time analysis. In Principles of Programming Languages, Proceedings, pages 119–132. ACM Press, 1988.

• P. Wadler.

Theorems for free! In Functional Programming Languages and Computer Architecture, Proceedings, pages 347–359. ACM Press, 1989.

24

A “Real” Example: Fusion [Gill et al., FPCA’93] Extensional free theorem: For every f :: (τ → α → α) → α → α, foldr k z (f (:) [ ]) = f k z The whole point of fusion: We expect, foldr k z (f (:) [ ]) ⊒ f k z A counterexample (in call-by-value setting): f :: (Nat → α → α) → α → α f k z = case [k 5 z] of {[ ] → z; x : xs → z}

25