How to write the Functor instance for this type?

Question

I'm struggling with an exercise from Haskell Book (Chapter 16. Functor). Given the following, it expects me to define fmap:

{-# LANGUAGE FlexibleInstances #-}

newtype Flip f a b = 
   Flip (f b a) 
   deriving (Eq, Show)

newtype K a b = 
  K a

instance Functor (Flip K a) where
   fmap = undefined

In a previous exercise I've already done the following:

 data K a b =
   K a

 instance Functor (K a) where
   fmap f (K x) = K x

But for this Flip a b, I can't even understand how to get started, e.g. how to continue fmap f Flip ....

I thought maybe before doing that I should also write a functor for the newtype K a b, similar to what I did data K a b:

instance Functor (K a) where
  fmap f (K x) = K x

But I can't understand how to proceed to the Functor instance for Flip f a b.

Any ideas, hints?

Answer 1

Let's have a look at your solution:

instance Functor (Flip K a) where
  fmap f (Flip (K a)) = Flip (K (f b))

What does Flip (K a) actually mean?

K is the constant functor, which you implemented correctly. It disregards the function f and always returns K a. But what actually happens behind the scenes is that the value K a b is transformed into K a c. Even though the function f was disregarded, the second type of K changed according to the type of f.

Now if we Flip K, we simply turn around the arguments:

Flip (K a b) == K b a

Instead of having a constant value and ignoring the function, we have a changing value and a constant "ignored" type. Looking at the implementation with explicit type signatures we get:

instance Functor (Flip K a) where
  fmap :: (b -> c) -> Flip K a b -> Flip K a c -- or:
  fmap :: (b -> c) -> K b a -> K c a
  fmap f (Flip (K a)) = Flip (K (f b))

As you already concluded, the only possible implementation is the one above.

You can actually see the use cases of a Constant functor in this question.