[Haskell-cafe] Monad instance for partially applied type constructor?

[Ryan Ingram]

On Wed, Sep 29, 2010 at 11:08 AM, DavidA <polyomino at f2s.com> wrote:
> Hi,
>
> I have the following code:
>
> {-# LANGUAGE TypeSynonymInstances #-}
>
> data Vect k b = V [(k,b)]
> -- vector space over field k with basis b
> -- for example, V [(5, E 1), (7, E 2)] would represent the vector 5 e1 + 7 e2
>
> data Monomial v = M [(v,Int)]
> -- monomials over variables v
> -- for example, M [(A,3), (B,5)] would represent the monomial a^3 b^5
>
> type Poly k v = Vect k (Monomial v)
> -- multivariate polynomials over field k and variables v
>
> instance Monad (Poly k) where
>    return v = V [(1, M [(v,1)])]
>    p >>= f = ... -- variable substitution
>
> So my thinking is:
> 1. The Monad type class is for one parameter type constructors (eg [], IO)
> 2. Poly is a two-parameter type constructor
> 3. So Poly k is a one-parameter type constructor
> 4. What I'm trying to express, that polynomials over field k are a monad,
> is true.
>
> However, GHC 6.12.2 complains:
>
>    Type synonym `Poly' should have 2 arguments, but has been given 1
>    In the instance declaration for `Monad (Poly k)'
>
> What is going wrong?

Haskell doesn't have true type functions; what you are really saying is

instance Monad (\v -> Vect k (Monomial v))

TypeSynonymInstances just lets you write stuff like this

type Foo = [Int]
instance C Foo where ...

instead of

type Foo = [Int]
instance C [Int] where ...

But it doesn't let you partially apply the type synonym.

On the other hand, if you did this:

newtype Compose f g a = O { unO :: f (g a) }
type Poly k = Compose (Vect k) Monomial

instance Monad (Poly k) where ...

would work, but now you have to wrap/unwrap Compose in the instance definition.

  -- ryan