haskell - 为什么 Monoidal 和 Applicative 定律告诉我们同样的事情?
问题描述
不久前,我了解到如何Monoidal
成为另一种表示方式。TypeclassopediaApplicative
上有一个有趣的问题:
- (棘手)证明给定您在第一个练习中的实现[
pure
并使用and和相反的方式(<*>)
写下来],通常的定律和上述定律是等价的。unit
(**)
Applicative
Monoidal
以下是这些类别和法律:
-- A note from https://wiki.haskell.org/Typeclassopedia#Alternative_formulation:
-- In this and the following laws, ≅ refers to isomorphism rather than equality.
-- In particular we consider (x,()) ≅ x ≅ ((),x) and ((x,y),z) ≅ (x,(y,z)).
-- Monoidal.
class Functor f => Monoidal f where
unit :: f ()
(**) :: f a -> f b -> f (a,b)
-- unit ** v ≅ v - Left Identity.
-- u ** unit ≅ u - Right Identity.
-- u ** (v ** w) ≅ (u ** v) ** w - Associativity.
-- Applicative.
class Functor f => Applicative f where
pure :: a -> f a
infixl 4 <*>, ...
(<*>) :: f (a -> b) -> f a -> f b
...
-- pure id <*> v = v - Identity.
-- pure f <*> pure x = pure (f x) - Homomorphism.
-- u <*> pure y = pure ($ y) <*> u - Interchange.
-- u <*> (v <*> w) = pure (.) <*> u <*> v <*> w - Composition.
用其他人写下组合子没什么大不了的:
unit = pure ()
f ** g = (,) <$> f <*> g = liftA2 (,) f g
pure x = const x <$> unit
f <*> g = uncurry ($) <$> (f ** g)
以下是我对法律为什么告诉我们同样事情的理解:
u <*> pure y = pure ($ y) <*> u -- Interchange: Applicative law.
我们首先要注意的是($ y) ≅ y
(更正式地说:) (y -> a) -> a ≅ y
。考虑到这一点,互换法简单地告诉我们(a, b) ≅ (b, a)
。
pure id <*> v = v -- Identity: Applicative law.
我认为id
它本身就是一个单元,因为它是唯一的居民类型forall a. a -> a
。因此,这条定律给了我们左恒等式:
unit ** v = v -- Left Identity: Monoidal law.
现在我们可以用它(a, b) ≅ (b, a)
来写下正确的身份:
u ** unit = u -- Right Identity: Monoidal law.
组成法:
u <*> (v <*> w) = pure (.) <*> u <*> v <*> w -- Composition: Applicative law.
我认为这条定律与关联性的含义相同Monoidal
:
u ** (v ** w) ≅ (u ** v) ** w
即,(a, (b, c)) ≅ ((a, b), c)
。Applicative
只是增加了一层应用。
所以,我们已经涵盖了所有的Monoidal
法律。我相信没有必要反其道而行之,因为我们将使用相同的同构。但人们可能会注意到一些奇怪的事情——我们没有使用同态Applicative
定律:
pure f <*> pure x = pure (f x)
我尝试根据自然自由定理来理解同态Monoidal
:
fmap (g *** h) (u ** v) = fmap g u ** fmap h v
但这似乎很奇怪,因为同态不处理副作用,但 Naturality 可以很好地处理它们。
所以,我有3个问题:
- 我的推理对吗?
- 同态在这张图中处于什么位置?
- 我们如何理解自然自由定理
Applicative
?
解决方案
暂时把它放在这里......想讨论这个,但我已经花了很长时间实现它:它是一个 Coq 证明脚本,以绝对防水的方式显示等效性。
Require Import Coq.Program.Basics.
Require Import Coq.Init.Datatypes.
Require Import Coq.Init.Notations.
Notation "f ∘ g" := (compose f g).
Class Functor (F: Type -> Type) : Type :=
{ fmap : forall {x} {y}, (x->y) -> (F x->F y)
; fmap_id : forall x, @fmap x x id = id
; fmap_compose : forall {x} {y} {z} (f: y->z) (g: x->y)
, fmap (f∘g) = fmap f ∘ fmap g
}.
Lemma fmap_twice {F} `{Functor F} {x} {y} {z} (f: y->z) (g: x->y) (xs: F x)
: fmap (f∘g) xs = fmap f (fmap g xs).
Proof.
rewrite fmap_compose. now compute.
Qed.
Definition parallel {a} {b} {c} {d} (f: a->c) (g: b->d)
: (a*b) -> (c*d) := fun xy => match xy with
| (x,y) => (f x, g y)
end.
Notation "f *** g" := (parallel f g) (at level 40, left associativity).
Definition rassoc {a} {b} {c} : ((a*b)*c) -> (a*(b*c))
:= fun xyz => match xyz with | ((x,y),z) => (x,(y,z)) end.
Definition tt_ {a} (x:a) := (tt, x).
Definition _tt {a} (x:a) := (x, tt).
Class Monoidal F `{Functor F} : Type :=
{ funit : F unit
; fzip : forall {a} {b}, F a -> F b -> F (a*b)
; left_identity : forall {a} (v: F a)
, fzip funit v = fmap tt_ v
; right_identity : forall {a} (v: F a)
, fzip v funit = fmap _tt v
; associativity : forall {a} {b} {c} (u: F a) (v: F b) (w: F c)
, fzip u (fzip v w) = fmap rassoc (fzip (fzip u v) w)
; naturality : forall {a} {b} {c} {d}
(g: a->c) (h: b->d) (u: F a) (v: F b)
, fmap (g***h) (fzip u v) = fzip (fmap g u) (fmap h v)
}.
Notation "u ** v" := (fzip u v) (at level 40, left associativity).
Lemma naturalityL {F} `{Monoidal F} {a} {b} {c}
(f: a->c) (u: F a) (v: F b)
: fmap (f***id) (fzip u v) = fzip (fmap f u) v.
Proof.
assert (v = fmap id v) as ->. { now rewrite fmap_id. }
rewrite <- naturality.
assert (v = fmap id v) as <-. { now rewrite fmap_id. }
now trivial.
Qed.
Lemma naturalityR {F} `{Monoidal F} {a} {b} {c}
(f: b->c) (u: F a) (v: F b)
: fmap (id***f) (fzip u v) = fzip u (fmap f v).
Proof.
assert (u = fmap id u) as ->. { now rewrite fmap_id. }
rewrite <- naturality.
assert (u = fmap id u) as <-. { now rewrite fmap_id. }
now trivial.
Qed.
Definition to {a} {b} (y: a) (f: a->b) := f y.
Class Applicative F `{Functor F} : Type :=
{ pure : forall {a}, a -> F a
; app : forall {a} {b}, F (a->b) -> F a -> F b
; identity : forall {a} (v: F a)
, app (pure id) v = v
; homomorphism : forall {a} {b} (f: a->b) (x: a)
, app (pure f) (pure x) = pure (f x)
; interchange : forall {a} {b} (u: F (a->b)) (y: a)
, app u (pure y) = app (pure (to y)) u
; composition : forall {a} {b} {c}
(u: F (b->c)) (v: F (a->b)) (w: F a)
, app u (app v w) = app (app (app (pure compose) u) v) w
; appFtor : forall {a} {b} (g: a->b) (x: F a)
, fmap g x = app (pure g) x
}.
Notation "fs <*> xs" := (app fs xs) (at level 40, left associativity).
Require Import Coq.Program.Tactics.
Require Import Coq.Logic.FunctionalExtensionality.
Definition apl {a} {b} (fx: (a->b)*a)
:= match fx with |(f,x) => f x end.
Program Instance MonoidalIsApplicative {F} `{Monoidal F}
: Applicative F
:= { pure := fun {a} (x: a) => fmap (const x) funit
; app := fun {a} {b} (fs: F (a->b)) (xs: F a)
=> fmap apl (fzip fs xs) }.
Next Obligation. (* identity *)
rewrite <- naturalityL.
rewrite -> left_identity.
repeat (rewrite <- fmap_twice).
rewrite -> fmap_id.
now compute.
Qed.
Next Obligation. (* homomorphism *)
rewrite <- naturality.
rewrite -> left_identity.
repeat (rewrite <- fmap_twice).
now compute.
Qed.
Next Obligation. (* interchange *)
rewrite <- naturalityL.
rewrite <- naturalityR.
repeat (rewrite <- fmap_twice).
rewrite -> right_identity.
rewrite -> left_identity.
repeat (rewrite <- fmap_twice).
now compute.
Qed.
Next Obligation. (* composition *)
rewrite <- naturalityR.
rewrite -> associativity.
repeat (rewrite <- naturalityL).
rewrite -> left_identity.
repeat (rewrite <- naturalityL).
repeat (rewrite <- fmap_twice).
f_equal. (* This part is just about *)
unfold compose. (* convincing Coq that two *)
apply functional_extensionality. (* functions are equal, it *)
intro x. (* has nothing to do with *)
destruct x as ((btc, atb), a0). (* applicative or monoidal *)
now compute. (* functors, specifically. *)
Qed.
Next Obligation. (* appFtor *)
rewrite <- naturalityL.
rewrite -> left_identity.
repeat (rewrite <- fmap_twice).
now compute.
Qed.
Lemma fmapPure {F} `{Applicative F} {a} {b}
(f: a->b) (x: a) : fmap f (pure x: F a) = pure (f x).
Proof.
rewrite -> appFtor.
now apply homomorphism.
Qed.
Lemma fmapBracket {F} `{Applicative F} {a} {b} {c} {d}
(f: c->d) (g: a->b->c) (xs: F a) (ys: F b)
: fmap f (fmap g xs<*>ys) = fmap (fun x y => f (g x y)) xs <*> ys.
Proof.
repeat (rewrite -> appFtor).
rewrite -> composition.
rewrite -> homomorphism.
rewrite -> composition.
repeat (rewrite -> homomorphism).
now compute.
Qed.
Lemma fmap_both {F} `{Applicative F} {a} {b} {c} {d}
(f: a->c->d) (g: b->c) (xs: F a) (ys: F b)
: fmap f xs <*> fmap g ys = fmap (fun x y => f x (g y)) xs <*> ys.
Proof.
repeat (rewrite -> appFtor).
rewrite -> composition.
repeat (rewrite <- appFtor).
rewrite <- fmap_twice.
rewrite -> interchange.
rewrite -> appFtor.
rewrite -> composition.
repeat (rewrite -> homomorphism).
rewrite <- appFtor.
now compute.
Qed.
Definition tup {a} {b} (x:a) (y:b) : (a*b) := (x,y).
Program Instance ApplicativeIsMonoidal {F} `{Applicative F}
: Monoidal F
:= { funit := pure tt
; fzip := fun {a} {b} (u: F a) (v: F b)
=> fmap tup u <*> v }.
Next Obligation. (* left_identity *)
repeat (rewrite -> appFtor).
rewrite -> homomorphism.
now compute.
Qed.
Next Obligation. (* right_identity *)
repeat (rewrite -> appFtor).
rewrite -> interchange.
rewrite -> composition.
repeat (rewrite -> homomorphism).
now compute.
Qed.
Next Obligation. (* associativity *)
repeat (rewrite -> fmapBracket).
rewrite -> composition.
repeat (rewrite <- appFtor).
rewrite <- fmap_twice.
rewrite -> fmap_both.
now compute.
Qed.
Next Obligation. (* naturality *)
rewrite -> fmap_both.
rewrite <- fmap_twice.
rewrite -> fmapBracket.
now compute.
Qed.
使用 Coq 8.9.1 编译。
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