Writing in pointfree style f x = g x x

Question

I am learning Haskell. I\'m sorry for asking a very basic question but I cant seem to find the answer. I have a function f defined by :

f x = g x x
         


        

Answer 1

I got here by pure chance, and I want to offer my solution, as nobody has mentioned lifting yet in this thread, not explicitly at least.

This is a solution:

f = liftM2 g id id

How to look at it?

• g has type a -> a -> b, i.e. it takes two values of some type (the same type for both, otherwise the definition the OP gave of f wouldn't make sense), and gives back another value of some type (not necessarily of the same type as the arguments);

• lift2M g is the lifted version of g and it has type (Monad m) => m a -> m a -> m b: it takes two monadic values, which are each a value in a so-far-unspecified context, and gives back a monadic value;

• when passing two functions to liftM2 g, the die is cast on what the context of the Monad is: it is that the values are not in there yet, but will eventually be, when the function will receive the arguments it needs; in other words, functions are monads that store their own future values; therefore, lift2M g takes in input two functions (or, the future values of two functions), and gives back the another function (or, its future value); knowing this, its type is the same as above if you change m to (->) r, or r ->: (r -> a) -> (r -> a) -> (r -> b)

• the two functions that we pass are both id, which promises it'll give back the same value it receives;

• liftM2 g id id is therefore a function of type r -> b that passes its argument to those two ids, which let it unchanged and forward it to g.

In a similar fashion, one can exploit that functions are applicative functors, and use this solution:

f = g <$> id <*> id