Applicative Archery

July 6, 2015

It is widely agreed that the laws of the Applicative class are not pretty to look at.

pure id <*> v = v                            -- identity
pure f <*> pure x = pure (f x)               -- homomorphism
u <*> pure y = pure ($ y) <*> u              -- interchange
pure (.) <*> u <*> v <*> w = u <*> (v <*> w) -- composition

Monad laws, in comparison, not only look less odd to begin with but can also be stated in a much more elegant way in terms of Kleisli composition (<=<). Shouldn’t there be an analogous nice presentation for Applicative as well? That became a static question in my mind while I was studying applicative functors many moons ago. After finding surprisingly little commentary on this issue, I decided to try figuring it out by myself. 1

Let’s cast our eye over Applicative:

class Functor t => Applicative t where
    pure  :: a -> t a
    (<*>) :: t (a -> b) -> t a -> t b

If our inspiration for reformulating Applicative is Kleisli composition, the only sensible plan is to look for a category in which the t (a -> b) functions-in-a-context from the type of (<*>) are the arrows, just like a -> t b functions are arrows in a Kleisli category. Here is one way to state that plan in Haskell terms:

class Applicative t => Starry t where
    idA  :: t (a -> a)
    (.*) :: t (b -> c) -> t (a -> b) -> t (a -> c)

    infixl 4 .*
-- The Applicative constraint is wishful thinking:
-- When you wish upon a star...

The laws of Starry are the category laws for the t (a -> b) arrows:

idA .* v = v                -- left identity
u .* idA = u                -- right identity
u .* v .* w = u .* (v .* w) -- associativity

The question, then, is whether it is possible to reconstruct Applicative and its laws from Starry. The answer is a resounding yes! The proof is in this manuscript, which I have not transcribed here as it is a little too long for a leisurely post like this one 2. The argument is set in motion by establishing that pure is an arrow mapping of a functor from Hask to a Starry category, and that both (<*>) and (.*) are arrow mappings of functors in the opposite direction. That leads to several naturality properties of those functors, from which the Applicative laws can be obtained. Along the way, we also get definitions for the Starry methods in terms of the Applicative ones…

    idA = pure id
    u .* v = fmap (.) u <*> v

… and vice-versa:

pure x = fmap (const x) idA
u <*> v = fmap ($ ()) (u .* fmap const v)

Also interesting is how the property relating fmap and (<*>)

fmap f u = pure f <*> u

… now tells us that a Functor results from composing the pure functor with the (<*>) functor. That becomes more transparent if we write it point-free:

fmap = (<*>) . pure

In order to ensure Starry is equivalent to Applicative we still need to prove the converse, that is, obtain the Starry laws from the Applicative laws plus the definitions of idA and (.*) just above. That is not difficult; all it takes is substituting the definitions in the Starry laws and:

  • For left identity, noticing that (id .) = id.

  • For right identity, applying the interchange law and noticing that ($ id) . (.) is id in a better disguise.

  • For associativity, using the laws to move all (.) to the left of the (<*>) and then verifying that the resulting messes of dots in both sides are equivalent.

As a tiny example, here is the Starry instance of Maybe

instance Starry Maybe where
    idA              = Just id
    Just g .* Just f = Just (g . f)
    _      .* _      = Nothing

… and the verification of the laws for it:

-- Left identity:
idA .* u = u
Just id .* u = u
-- u = Nothing
Just id .* Nothing = Nothing
Nothing = Nothing
-- u = Just f
Just id .* Just f = Just f
Just (id . f) = Just f
Just f = Just f

-- Right identity:
u .* idA = u
u .* Just id = u
-- u = Nothing
Nothing .* Just id = Nothing
Nothing = Nothing
-- u = Just g
Just g .* Just id = Just g
Just (g .* id) = Just g
Just g = Just g

-- Associativity:
u .* v .* w = u .* (v .* w)
-- If any of u, v and w are Nothing, both sides will be Nothing.
Just h .* Just g .* Just f = Just h .* (Just g .* Just f)
Just (h . g) .* Just f = Just h .* (Just (g . f))
Just (h . g . f) = Just (h . (g . f))
Just (h . g . f) = Just (h . g . f)

It works just as intended:

GHCi> Just (2*) .* Just (subtract 3) .* Just (*4) <*> Just 5
Just 34
GHCi> Just (2*) .* Nothing .* Just (*4) <*> Just 5

I do not think there will be many opportunities to use the Starry methods in practice. We are comfortable enough with applicative style, through which we see most t (a -> b) arrows as intermediates generated on demand, rather than truly meaningful values. Furthermore, the Starry laws are not really easier to prove (though they are certainly easier to remember!). Still, it was an interesting exercise to do, and it eases my mind to know that there is a neat presentation of the Applicative laws that I can relate to.

This post is Literate Haskell, in case you wish to play with Starry in GHCi (here is the raw .lhs file ).

instance Starry Maybe where
instance Starry [] where
instance Starry ((->) a) where
instance Starry IO where

As for proper implementations in libraries, the closest I found was Data.Semigroupoid.Static, which lives in Edward Kmett’s semigroupoids package. “Static arrows” is the actual technical term for the t (a -> b) arrows. The module provides…

newtype Static f a b = Static { runStatic :: f (a -> b) }

… which uses the definitions shown here for idA and (.*) as id and (.) of its Category instance.

  1. There is a reasonably well-known alternative formulation of Applicative: the Monoidal class as featured in this post by Edward Z. Yang. It is quite handy to work with when it comes to checking whether an instance follows the laws.↩︎

  2. Please excuse some oddities in the manuscript, such as off-kilter terminology and weird conventions (e.g. consistently naming arguments in applicative style as w <*> v <*> u rather than u <*> v <*> w in applicative style). The most baffling choice was using id rather than () as the throwaway argument to const. I guess I did that because ($ ()) looks bad in handwriting.↩︎