Juggling existentials without unsafeCoerce

Question

Lately I have been playing with this type, which I understand to be an encoding of the free distributive functor (for tangential background on that, see this answer):

data Ev g a where
    Ev :: ((g x -> x) -> a) -> Ev g a

deriving instance Functor (Ev g)

The existential constructor ensures I can only consume an Ev g by supplying a polymorphic extractor forall x. g x -> x, and that the lift and lower functions of the free construction can be given compatible types:

runEv :: Ev g a -> (forall x. g x -> x) -> a
runEv (Ev s) f = s f

evert :: g a -> Ev g a
evert u = Ev $ \f -> f u

revert :: Distributive g => Ev g a -> g a
revert (Ev s) = s <$> distribute id

However, there is a difficulty upon trying to give Ev g a Distributive instance. Given that Ev g is ultimately just a function with a weird argument type, one might hope that just threading distribute for functions (which amounts to (??) :: Functor f => f (a -> b) -> a -> f b from lens, and doesn't inspect the argument type in any way) through the Ev wrapper:

instance Distributive (Ev g) where
    distribute = Ev . distribute . fmap (\(Ev s) -> s)

That, however, does not work:

Flap3.hs:95:53: error:
    • Couldn't match type ‘x’ with ‘x0’
      ‘x’ is a rigid type variable bound by
        a pattern with constructor:
          Ev :: forall (g :: * -> *) x a. ((g x -> x) -> a) -> Ev g a,
        in a lambda abstraction
        at Flap3.hs:95:44-47
      Expected type: (g x0 -> x0) -> a
        Actual type: (g x -> x) -> a
    • In the expression: s
      In the first argument of ‘fmap’, namely ‘(\ (Ev s) -> s)’
      In the second argument of ‘(.)’, namely ‘fmap (\ (Ev s) -> s)’
    • Relevant bindings include
        s :: (g x -> x) -> a (bound at Flap3.hs:95:47)
   |
95 |     distribute = Ev . distribute . fmap (\(Ev s) -> s) 
   |                                                     ^
Failed, no modules loaded.

GHC objects to rewrapping the existential, even though we do nothing untoward with it between unwrapping and rewrapping. The only way out I found was resorting to unsafeCoerce:

instance Distributive (Ev g) where
    distribute = Ev . distribute . fmap (\(Ev s) -> unsafeCoerce s)

Or, spelling it in a perhaps more cautious manner:

instance Distributive (Ev g) where
    distribute = eevee . distribute . fmap getEv
        where
        getEv :: Ev g a -> (g Any -> Any) -> a
        getEv (Ev s) = unsafeCoerce s
        eevee :: ((g Any -> Any) -> f a) -> Ev g (f a)
        eevee s = Ev (unsafeCoerce s)

Is it possible to get around this problem without unsafeCoerce? or there truly is no other way?

Additional remarks:

• I believe Ev is the most correct type I can give to the construction, though I would be happy to be proved wrong. All my attempts to shift the quantifiers elsewhere lead either to needing unsafeCoerce somewhere else or to evert and revert having types that don't allow them to be composed.

• This situation looks, at first glance, similar to the one described in this blog post by Sandy Maguire, which ends up sticking with unsafeCoerce as well.

---

The following take at giving Ev g a Representable instance might put the problem into sharper relief. As dfeuer notes, this isn't really supposed to be possible; unsurprisingly, I had to use unsafeCoerce again:

-- Cf. dfeuer's answer.
newtype Goop g = Goop { unGoop :: forall y. g y -> y }

instance Representable (Ev g) where
    type Rep (Ev g) = Goop g
    tabulate f = Ev $ \e -> f (Goop (goopify e))
        where
        goopify :: (g Any -> Any) -> g x -> x
        goopify = unsafeCoerce
    index (Ev s) = \(Goop e) -> s e

While goopify sure looks alarming, I think there is a case for it being safe here. The existential encoding means any e that gets passed to the wrapped function will necessarily be an extractor polymorphic on the element type, that gets specialised to Any merely because I asked for that to happen. That being so, forall x. g x -> x is a sensible type for e. This dance of specialising to Any only to promptly undo it with unsafeCoerce is needed because GHC forces me to get rid of the existential by making a choice. This is what happens if I leave out the unsafeCoerce in this case:

Flap4.hs:64:37: error:
    • Couldn't match type ‘y’ with ‘x0’
      ‘y’ is a rigid type variable bound by
        a type expected by the context:
          forall y. g y -> y
        at Flap4.hs:64:32-37
      Expected type: g y -> y
        Actual type: g x0 -> x0
    • In the first argument of ‘Goop’, namely ‘e’
      In the first argument of ‘f’, namely ‘(Goop e)’
      In the expression: f (Goop e)
    • Relevant bindings include
        e :: g x0 -> x0 (bound at Flap4.hs:64:24)
   |
64 |     tabulate f = Ev $ \e -> f (Goop e) 
   |                                     ^
Failed, no modules loaded.

---

Prolegomena needed to run the code here:

{-# LANGUAGE DeriveFunctor #-}
{-# LANGUAGE StandaloneDeriving #-}
{-# LANGUAGE GADTs #-}
{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE TypeFamilies #-}

import Data.Distributive
import Data.Functor.Rep
import Unsafe.Coerce
import GHC.Exts (Any)

-- A tangible distributive, for the sake of testing.
data Duo a = Duo { fstDuo :: a, sndDuo :: a }
    deriving (Show, Eq, Ord, Functor)

instance Distributive Duo where
    distribute u = Duo (fstDuo <$> u) (sndDuo <$> u)

Answer 1

The suggestions by danidiaz and dfeuer have led me to a tidier encoding, though unsafeCoerce is still necessary:

{-# LANGUAGE DeriveFunctor #-}
{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE TypeFamilies #-}

import Unsafe.Coerce
import GHC.Exts (Any)
import Data.Distributive
import Data.Functor.Rep

-- Px here stands for "polymorphic extractor".
newtype Px g = Px { runPx :: forall x. g x -> x }

newtype Ev g a = Ev { getEv :: Px g -> a }
    deriving Functor

runEv :: Ev g a -> (forall x. g x -> x) -> a
runEv s e = s `getEv` Px e

evert :: g a -> Ev g a
evert u = Ev $ \e -> e `runPx` u

revert :: Distributive g => Ev g a -> g a
revert (Ev s) = (\e -> s (mkPx e)) <$> distribute id
    where
    mkPx :: (g Any -> Any) -> Px g
    mkPx e = Px (unsafeCoerce e) 

instance Distributive (Ev g) where
    distribute = Ev . distribute . fmap getEv

instance Representable (Ev g) where
    type Rep (Ev g) = Px g
    tabulate = Ev 
    index = getEv

The x variable in my original formulation of Ev was, at heart, being universally quantified; I had merely disguised it as an existential behind a function arrow. While that encoding makes it possible to write revert without unsafeCoerce, it shifts the burden to the instance implementations. Directly using universal quantification is ultimately better in this case, as it keeps the magic concentrated in one place.

The unsafeCoerce trick here is essentially the same demonstrated with tabulate in the question: the x in distribute id :: Distributive g => g (g x -> x) is specialised to Any, and then the specialisation is immediately undone, under the fmap, with unsafeCoerce. I believe the trick is safe, as I have sufficient control over what is being fed to unsafeCoerce.

As for getting rid of unsafeCoerce, I truly can't see a way. Part of the problem is that it seems I would need some form of impredicative types, as the unsafeCoerce trick ultimately amounts to turning forall x. g (g x -> x) into g (forall x. g x -> x). For the sake of comparison, I can write a vaguely analogous, if much simpler, scenario using the subset of the impredicative types functionality that would fall under the scope of the mooted ExplicitImpredicativeTypes extension (see GHC ticket #14859 and links therein for discussion):

{-# LANGUAGE ImpredicativeTypes #-}
{-# LANGUAGE TypeApplications #-}
{-# LANGUAGE RankNTypes #-}

newtype Foo = Foo ([forall x. Num x => x] -> Int)

testFoo :: Applicative f => Foo -> f Int
testFoo (Foo f) = fmap @_ @[forall x. Num x => x] f 
    $ pure @_ @[forall x. Num x => x] [1,2,3]

GHCi> :set -XImpredicativeTypes 
GHCi> :set -XTypeApplications 
GHCi> testFoo @Maybe (Foo length)
Just 3

distribute id, however, is thornier than [1,2,3]. In id :: g x -> g x, the type variable I'd like to keep quantified appears in two places, with one of them being the second type argument to distribute (the (->) (g x) functor). To my untrained eye at least, that looks utterly intractable.

Answer 2

Every Distributive functor can be made Representable, though we can't prove that in Haskell (I imagine it's not constructive). But one approach to addressing your problem is to just switch classes.

newtype Evv f a = Evv
  {unEvv :: forall g. Representable g
         => (forall x. f x -> g x) -> g a}

instance Functor (Evv g) where
  fmap f (Evv q) = Evv $ \g -> fmap f (q g)

evert :: g a -> Evv g a
evert ga = Evv $ \f -> f ga

revert :: Representable g => Evv g a -> g a
revert (Evv f) = f id

newtype Goop f = Goop
  {unGoop :: forall x. f x -> x}

instance Distributive (Evv g) where
  collect = collectRep

instance Representable (Evv g) where
  type Rep (Evv g) = Goop g
  tabulate f = Evv $ \g -> fmap (\rg -> f (Goop $ \fx -> index (g fx) rg)) $ tabulate id
  index (Evv g) (Goop z) = runIdentity $ g (Identity . z)

I haven't yet tried this with Distributive directly (as HTNW suggests), but I wouldn't be surprised if it were simply impossible for some reason.

Warning: I have not proven that this is actually the free Representable!