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haskell - 为什么 Monoidal 和 Applicative 定律告诉我们同样的事情?

转载 作者:行者123 更新时间:2023-12-02 02:10:44 33 4
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我了解了Monoidal作为表示 Applicative 的另一种方式不久前。 Typeclassopedia上有一个有趣的问题:

  1. (Tricky) Prove that given your implementations from the first exercise [pure and (<*>) written down using unit and (**) and the other way around], the usual Applicative laws and the Monoidal laws stated above are equivalent.

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,互换法只是告诉我们 (a, b) ≅ (b, a) .
pure id <*> v = v -- Identity: Applicative law.
我认为 id成为一个单元本身,因为它是 forall a. a -> a 类型的唯一居民.因此,这条定律给了我们左恒等式:
unit ** v = v -- Left Identity: Monoidal law.
现在我们可以使用 (a, b) ≅ (b, a)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 Monoidal 的关联性相同。 :
u ** (v ** w) ≅ (u ** v) ** w
也就是说, (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 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。但是人们可能注意到了一些奇怪的事情——我们没有使用同态 Applicative法律:
pure f <*> pure x = pure (f x)
我尝试根据 Monoidal 的自然自由定理来理解同态。 :
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 ?
  • 最佳答案

    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。

    关于haskell - 为什么 Monoidal 和 Applicative 定律告诉我们同样的事情?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62949200/

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