More on Applicative Effects in Free Monads

In my last post, I explored the problem of Applicative effects in free monads. After some great discussion between Edward Kmett, Dave Menendez, /u/MitchellSalad, and myself, I think I’ve come to a new understanding.

For reference, here’s my “free monad given an applicative.”

data Free f a where
Pure :: a -> Free f a
Free :: f (Free f a) -> Free f a

instance Functor f => Functor (Free f) where
fmap f (Pure a) = Pure (f a)
fmap f (Free a) = Free (fmap (fmap f) a)

instance Applicative f => Applicative (Free f) where
pure = Pure
Pure f <*> a = fmap f a
Free f <*> Pure a = Free \$ fmap (fmap (\$ a)) f
Free f <*> Free a = Free (fmap (<*>) f <*> a)

instance Applicative f => Monad (Free f) where
Pure a >>= k = k a
Free a >>= k = Free (fmap (>>= k) a)

The core of the issue is that under my implementation, (<*>) was purposefully different than ap. Turns out, it’s a law that (<*>) must be morally equivalent to ap. So if the two turn out effectively different, that’s illegal.

Dave Menendez gave a good counter example to the original post using Const.

instance Monoid m => Applicative (Const m) where
pure _ = mempty
Const f <*> Const v = Const (f `mappend` v)

liftFree (Const "a") <*>  liftFree (Const "b") = Free (Const "ab")
liftFree (Const "a") `ap` liftFree (Const "b") = Free (Const "a")

As you can see, (<*>) gives a fundamentally different answer than ap.

Cheating on the Ap Test

The truth is, (<*>) doesn’t have to literally be assigned ap. (<*>) just has to be morally equivalent to ap. In the Const example, morally different results were computed. But in my original example with Haxl, the results were legal. The difference is that with Const, (<*>) is morally different than anything ap could ever do. It basically shouldn’t be legal to interpret Free (Const m).

According to Edward Kmett, it’s all up to interpretation! The solution is to provide a limited view into Free. If the interpreter of the free monad agrees to certain laws, then interpretation can be guaranteed valid. Essentially, we’re shifting the burden of proof for (<*>) = ap to the interpreter.

Let’s recall the Applicative-optimized instance of Applicative for Free.

instance Applicative f => Applicative (Free f) where
pure = Pure
Pure a <*> Pure b = Pure \$ a b
Pure a <*> Free mb = Free \$ fmap a <\$> mb
Free ma <*> Pure b = Free \$ fmap (\$ b) <\$> ma
Free ma <*> Free mb = Free \$ fmap (<*>) ma <*> mb -- <- The important line

In that last line, (<*>) is used on the f Applicative. We’re basically asking f to perform this computation. So we know that as long as f is being responsible with (<*>), we will get a correct applicative computation. The question then, is how to use this to make a correct monadic computation?

For a simple example, let’s turn to retract. This function doesn’t actually need any porting to work with this Free.

retract :: Monad f => Free f a -> f a
retract (Pure a) = return a
retract (Free as) = as >>= retract

If (<*>) = ap in f, which it has to, then all the work is done for us. There’s no way to construct a computation that can misuse (<*>) and ap to break f’s monad laws.

But what about examples where f isn’t a monad? The simplest way to use these fs is with foldFree.

foldFree :: (Functor f, Monad m) => (forall x . f x -> m x) -> Free f a -> m a
foldFree _ (Pure a)  = return a
foldFree f (Free as) = f as >>= foldFree f

In the conventional Free, the first parameter is a natural transformation. Given any f, create a corresponding m, and foldFree will handle the monadic bits. But this interpreter won’t work with the Applicative-optimized Free. Well, it will compile and run just fine. It just won’t produce law-abiding results. Given that f is actually an Applicative, and (<*>) was used in the computation, it’s possible that the natural transformation doesn’t correctly transcribe applicative effects. If this is the case, we don’t know for sure that the ms returned are correct, so we don’t know that the result is equivalent to using ap.

That natural transformation has to be something stronger. It has to preserve the meaning of applicative computations between f and m. So it needs to be an applicative homomorphism. That is, it needs to be a natural transformation that obeys these laws.

hm :: (Applicative f, Applicative m) => f x -> m x
hm (pure a) = pure a
hm (f <*> g) = hm f <*> hm g

In code, the only change we can make is strengthening that Functor constraint.

foldFree :: (Applicative f, Monad m) => (forall x . f x -> m x) -> Free f a -> m a
foldFree _ (Pure a)  = return a
foldFree f (Free as) = f as >>= foldFree f

We just have to take for granted that the user is supplying a natural transformation that obeys those laws. If it does, then we know that the meanings of applicative computations are being preserved. Then, as with retract, the rest is properly guaranteed by the monad laws.

Back to Freer

So what about the Freer monad? Considering, once again, that we have a free applicative, which takes any type of kind * -> *, we can get the freer monad the same way as before.

type Freer f = Free (Ap f)

Now the interpreter’s job is to take Ap, which is effectively a linked list of fs, and use those fs to make an applicative homomorphism. Now, we have a freer monad that turns any * -> * into a monad, and allows the interpreter to optimize applicative effects.

It’s still extremely slow. I plan to look into church encoding it to improve the performance. But I’m not sure if that will work well with (<*>). Anyway, I’m very happy just knowing a solution for this exists. Let me know if I’ve made any other mistakes.