Incrementally solving the Applicative instance for Compose

February 04, 2019 - Ben Tefay

Here at Stacktrace, our team have been working our way through the incredible (and incredibly challenging) Haskell data61 course. Last week we started working on the Applicative instance for Compose, and boy, it is much harder to implement than you might expect.

Lets have a look:

newtype Compose f g a = Compose (f (g a))

instance (Applicative f, Applicative g) => Applicative (Compose f g a) where
pure :: a -> Compose f g a
(<*>) :: Compose f g (a -> b) -> Compose f g a -> Compose f g b

There isn’t much going on, right? Just some Compose, f, g, a and b.

Well, I was in agony for what felt like a brain-melting hour before my colleague put me out of my misery and solved the hardest part for me. Just another day in the life of learning Haskell.

In fairness, pure is easy enough.

instance (Applicative f, Applicative g) => Applicative (Compose f g a) where
pure :: a -> Compose f g a

We just need to put an a in a g, and then put that g a in an f. Given g and f are Applicatives, we can call pure to create one of each Applicative and then wrap the result in Compose:

pure a = Compose \$ pure \$ pure a

Or, if we’re feeling a bit point free:

pure = Compose . pure . pure

In contrast <*> (“apply”, or more whimsically, “spaceship”) is hard.

Lets look at that signature again:

instance (Applicative f, Applicative g) => Applicative (Compose f g a) where
(<*>) :: Compose f g (a -> b) -> Compose f g a -> Compose f g b

It doesn’t look that hard, right? The problem is this: how on earth do you reach through f and g, to get hold of a -> b and a, so that you can actually apply them to each other and end up with f (g b).

For the life of me, I couldn’t figure out how to solve this incrementally. That left me trying to solve it all at once, in my head. I failed.

To practice being more piecemeal with Haskell, let’s try to walk through this in a way that my past self would have appreciated.

To start, lets write the easy stuff. Since Compose is a newtype wrapper, we need to unwrap the contents of our two Compose arguments, and then put the result back in Compose at the end:

instance (Applicative f, Applicative g) => Applicative (Compose f g a) where
(<*>) :: Compose f g (a -> b) -> Compose f g a -> Compose f g b
Compose fga2b <*> Compose fga = Compose \$ _fgb

with the types:

fga2b :: f (g (a -> b))
fga :: f (g a)
_fgb :: f (g b)

The good news is that we’re now just dealing with f, g, a and b. The bad news is that it’s not obvious what to do next.

Lets try to break the problem down. We need to figure out what _fgb should be. Given f and g are Applicatives, we have four choices to start us off (ignoring pure, since the last thing we need is more nested Applicatives):

1. _fgb = _ <*> fga2b
2. _fgb = _ <*> fga
3. _fgb = _ <\$> fga2b
4. _fgb = _ <\$> fga

It can’t be <\$> (“fmap”), as we’ll end up inside f twice, once for fga and once for fga2b. So it must be <*>:

_fgb :: f (g b)
_fgb = _ <*> _

Whatever _fgb is, it needs to have a type of f (g b). Given we’re calling <*>, where <*> :: f (a -> b) -> f a -> f b, we probably need a non-function on the right. That means we have to start with option (3): _fgb = ? <*> fga.

Now we’re getting somewhere:

instance (Applicative f, Applicative g) => Applicative (Compose f g a) where
(<*>) :: Compose f g (a -> b) -> Compose f g a -> Compose f g b
Compose fga2b <*> Compose fga = Compose \$ _fga2gb <*> fga

with the types:

fga2b :: f (g (a -> b))
fga :: f (g a)
_fga2gb :: f (g a -> g b)

We’ve already used fga, so that leaves us with fga2b. Can we transform fga2b :: f (g (a -> b)) into f (g a -> g b)?

Lets try to do it in a separate function called lessScary:

Compose fga2b <*> Compose fga = Compose \$ lessScary fga2b <*> fga
where
lessScary :: f (g (a -> b)) -> f (g a -> g b)
lessScary fga2b = _somethingGoesHere

This seems a little more manageable. We need to transform the contents of f from g (a -> b) into g a -> g b. That feels like a job for <\$>!

(Compose fga2b) <*> (Compose fga) = Compose \$ lessScary fga2b <*> fga
where
lessScary :: f (g (a -> b)) -> f (g a -> g b)
lessScary fga2b = _somethingElseGoesHere <\$> fga2b

with the types:

_somethingElseGoesHere :: g (a -> b) -> (g a -> g b)

Yes yes yes! Lets break that _somethingElseGoesHere out into another function:

Compose fga2b <*> Compose fga = Compose \$ lessScary fga2b <*> fga
where
lessScary :: f (g (a -> b)) -> f (g a -> g b)
lessScary fga2b = whatThe <\$> fga2b
whatThe :: g (a -> b) -> g a -> g b
whatThe ga2b = _oneMoreThingGoesHere

Note that I’ve dropped the brackets around (g a -> g b) in g (a -> b) -> (g a -> g b), since that’s actually the same as g (a -> b) -> g a -> g b.

You know what g (a -> b) -> g a -> g b looks like? Yup! It looks like <*> for g:

Compose fga2b <*> Compose fga = Compose \$ lessScary fga2b <*> fga
where
lessScary :: f (g (a -> b)) -> f (g a -> g b)
lessScary fga2b = whatThe <\$> fga2b
whatThe :: g (a -> b) -> g a -> g b
whatThe ga2b = (<*>) ga2b

That’s it! We did it.

We can take things further though, and condense a bunch of this code.

First, lets drop ga2b from (<*>) in whatThe, since it’s the first parameter to both:

Compose fga2b <*> Compose fga = Compose \$ lessScary fga2b <*> fga
where
lessScary :: f (g (a -> b)) -> f (g a -> g b)
lessScary fga2b = whatThe <\$> fga2b
whatThe :: g (a -> b) -> g a -> g b
whatThe = (<*>)

Now, lets inline whatThe:

Compose fga2b <*> Compose fga = Compose \$ lessScary fga2b <*> fga
where
lessScary :: f (g (a -> b)) -> f (g a -> g b)
lessScary fga2b = (<*>) <\$> fga2b

lessScary can be inlined too:

Compose fga2b <*> Compose fga = Compose \$ (<*>) <\$> fga2b <*> fga

And there we have it. The Applicative instance for Compose. In incremental steps.