Numbers as multiplicative functions (weird but entertaining)

Question

In the comments of the question Tacit function composition in Haskell, people mentioned making a Num instance for a -> r, so I thought I\'d play

Answer 1

In an expression like 2 3 4 with your instances, both 2 and 3 are functions. So 2 is actually (2 *) and has a type Num a => a -> a. 3 is the same. 2 3 is then (2 *) (3 *) which is the same as 2 * (3 *). By your instance, this is liftM2 (*) 2 (3 *) which is then liftM2 (*) (2 *) (3 *). Now this expression works without any of your instances.

So what does this mean? Well, liftM2 for functions is a sort of double composition. In particular, liftM2 f g h is the same as \ x -> f (g x) (h x). So liftM2 (*) (2 *) (3 *) is then \ x -> (*) ((2 *) x) ((3 *) x). Simplifying a bit, we get: \ x -> (2 * x) * (3 * x). So now we know that 2 3 4 is actually (2 * 4) * (3 * 4).

Now then, why does liftM2 for functions work this way? Let's look at the monad instance for (->) r (keep in mind that (->) r is (r ->) but we can't write type-level operator sections):

instance Monad ((->) r) where  
    return x = \_ -> x  
    h >>= f = \w -> f (h w) w  

So return is const. >>= is a little weird. I think it's easier to see this in terms of join. For functions, join works like this:

join f = \ x -> f x x

That is, it takes a function of two arguments and turns it into a function of one argument by using that argument twice. Simple enough. This definition also makes sense. For functions, join has to turn a function of two arguments into a function of one; the only reasonable way to do this is to use that one argument twice.

>>= is fmap followed by join. For functions, fmap is just (.). So now >>= is equal to:

h >>= f = join (f . h)

which is just:

h >>= f = \ x -> (f . h) x x

now we just get rid of . to get:

h >>= f = \ x -> f (h x) x

So now that we know how >>= works, we can look at liftM2. liftM2 is defined as follows:

liftM2 f a b = a >>= \ a' -> b >>= \ b' -> return (f a' b')

We can simply this bit by bit. First, return (f a' b') turns into \ _ -> f a' b'. Combined with the \ b' ->, we get: \ b' _ -> f a' b'. Then b >>= \ b' _ -> f a' b' turns into:

 \ x -> (\ b' _ -> f a' b') (b x) x

since the second x is ignored, we get: \ x -> (\ b' -> f a' b') (b x) which is then reduced to \ x -> f a' (b x). So this leaves us with:

a >>= \ a' -> \ x -> f a' (b x)

Again, we substitute >>=:

\ y -> (\ a' x -> f a' (b x)) (a y) y

this reduces to:

 \ y -> f (a y) (b y)

which is exactly what we used as liftM2 earlier!

Hopefully now the behavior of 2 3 4 makes sense completely.