Re: Discussion: should we make liftA2 an Applicative method?

I'm also all for adding liftA2 to the class and have noticed this inefficiency/asymmetry when working on the class hierarchies for other languages On Sun, Jan 15, 2017 at 8:11 AM, Kris Nuttycombe wrote:

I'm in favor of this change. From my perspecive, liftA2 is actually the fundamental Applicative operation, an <*> is merely a convenient isomorphism. When I'm teaching, showing the symmetry between the following always seems to help students:

fmap :: (a -> b) -> f a -> f b liftA2 :: (a -> b -> c) -> f a -> f b -> f c flip (>>=) :: (a -> f b) -> f a -> f b

<*> is obviously exceptionally useful in practice. But liftA2 seems like the more essential shape of that operation.

Kris

On Sat, Jan 14, 2017 at 2:49 PM, David Feuer wrote:

...

Right now, we define

liftA2 :: Applicative f => (a -> b -> c) -> f a -> f b -> f c liftA2 f x y = f <$> x <*> y

For some functors, like IO, this definition is just dandy. But for others, it's not so hot. For ZipList, for example, we get

liftA2 f (ZipList xs) (ZipList ys) = ZipList $ zipWith id (map f xs) ys

In this particular case, rewrite rules will likely save the day, but for many similar types they won't. If we defined a custom liftA2, it would be the obviously-efficient

liftA2 f (ZipList xs) (ZipList ys) = ZipList $ zipWith f xs ys

The fmap problem shows up a lot in Traversable instances. Consider a binary leaf tree:

data Tree a = Bin (Tree a) (Tree a) | Leaf a

The obvious way to write the Traversable instance today is

instance Traversable Tree where traverse _f Tip = pure Tip traverse f (Bin p q) = Bin <$> traverse f p <*> traverse f q

In this definition, every single internal node has an fmap! We could end up allocating a lot more intermediate structure than we need. It's possible to work around this by reassociating. But it's complicated (see Control.Lens.Traversal.confusing[1]), it's expensive, and it can break things in the presence of infinite structures with lazy applicatives (see Dan Doel's blog post on free monoids[2] for a discussion of a somewhat related issue). With liftA2 as a method, we don't need to reassociate!

traverse f (Bin p q) = liftA2 Bin (traverse f p) (traverse f q)

The complication with Traversable instances boils down to an efficiency asymmetry in <*> association. According to the "composition" law,

(.) <$> u <*> v <*> w = u <*> (v <*> w)

But the version on the left has an extra fmap, which may not be cheap. With liftA2 in the class, we get a more balanced law:

If for all x and y, p (q x y) = f x . g y, then liftA2 p (liftA2 q u v) = liftA2 f u . liftA2 g v

[1] https://hackage.haskell.org/package/lens-4.15.1/docs/Control-Lens-Traversal....

[2] http://comonad.com/reader/2015/free-monoids-in-haskell/

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