How arbitrary is the “ap” implementation for monads?

Question

I am currently studying the bonds between monad and applicative functors.

I see two implementation for ap:

ap m1 m2 = do { f <- m1 ; x <- m2 ;          


        

Answer 1

Let's start with the obvious fact: such a definition for <*> violates the ap-law in the sense that <*> should ap, where ap is the one defined in the Monad class, i.e. the first one you posted.

Trivialities aside, as far as I can see, the other applicative laws should hold.

More concretely, let's focus on the composition law you mentioned. Your "reversed" ap

(<**>) m1 m2 = do { x <- m2 ; f <- m1 ; return (f x) }

can also be defined as

(<**>) m1 m2 = pure (flip ($)) <*> m2 <*> m1

where <*> is the "regular" ap.

This means that, for instance,

u <**> (v <**> w) =
  { def. <**> }
pure (flip ($)) <*> (v <**> w) <*> u =
  { def. <**> }
pure (flip ($)) <*> (pure (flip ($)) <*> w <*> v) <*> u =
  { composition law }
pure (.) <*> pure (flip ($)) <*> (pure (flip ($)) <*> w) <*> v <*> u =
  { homomorphism law }
pure ((.) (flip ($))) <*> (pure (flip ($)) <*> w) <*> v <*> u =
  { composition law }
pure (.) <*> pure ((.) (flip ($))) <*> pure (flip ($)) <*> w <*> v <*> u =
  { homomorphism law (x2)}
pure ((.) ((.) (flip ($))) (flip ($))) <*> w <*> v <*> u =
  { beta reduction (several) }
pure (\x f g -> g (f x)) <*> w <*> v <*> u

(I hope I got everything OK)

Try doing something similar to the left hand side.

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