Documentation

Monads provide substitution (fmap) and renormalization (join):

m >>= f = join (fmap f m)

A free Monad is one that does no work during the normalization step beyond simply grafting the two monadic values together.

[] is not a free Monad (in this sense) because join [[a]] smashes the lists flat.

On the other hand, consider:

 data Tree a = Bin (Tree a) (Tree a) | Tip a

 instance Monad Tree where
   return = Tip
   Tip a >>= f = f a
   Bin l r >>= f = Bin (l >>= f) (r >>= f)

This Monad is the free Monad of Pair:

 data Pair a = Pair a a

And we could make an instance of MonadFree for it directly:

 instance MonadFree Pair Tree where
    wrap (Pair l r) = Bin l r

Or we could choose to program with Free Pair instead of Tree and thereby avoid having to define our own Monad instance.

Moreover, Control.Monad.Free.Church provides a MonadFree instance that can improve the asymptotic complexity of code that constructs free monads by effectively reassociating the use of (>>=). You may also want to take a look at the kan-extensions package (http://hackage.haskell.org/package/kan-extensions).

See Free for a more formal definition of the free Monad for a Functor.

The Free Monad for a Functor f.

Formally

A Monad n is a free Monad for f if every monad homomorphism from n to another monad m is equivalent to a natural transformation from f to m.

Why Free?

Every "free" functor is left adjoint to some "forgetful" functor.

If we define a forgetful functor U from the category of monads to the category of functors that just forgets the Monad, leaving only the Functor. i.e.

U (M,return,join) = M

then Free is the left adjoint to U.

Being Free being left adjoint to U means that there is an isomorphism between

Free f -> m in the category of monads and f -> U m in the category of functors.

Morphisms in the category of monads are Monad homomorphisms (natural transformations that respect return and join).

Morphisms in the category of functors are Functor homomorphisms (natural transformations).

Given this isomorphism, every monad homomorphism from Free f to m is equivalent to a natural transformation from f to m

Showing that this isomorphism holds is left as an exercise.

In practice, you can just view a Free f a as many layers of f wrapped around values of type a, where (>>=) performs substitution and grafts new layers of f in for each of the free variables.

This can be very useful for modeling domain specific languages, trees, or other constructs.

This instance of MonadFree is fairly naive about the encoding. For more efficient free monad implementation see Control.Monad.Free.Church, in particular note the improve combinator. You may also want to take a look at the kan-extensions package (http://hackage.haskell.org/package/kan-extensions).

A number of common monads arise as free monads,

• Given data Empty a, Free Empty is isomorphic to the Identity monad.
• Free Maybe can be used to model a partiality monad where each layer represents running the computation for a while longer.

retract is the left inverse of lift and liftF

 retract . lift = id
 retract . liftF = id

A version of lift that can be used with just a Functor for f.

Tear down a Free Monad using iteration.

Like iter for monadic values.

Lift a natural transformation from f to g into a natural transformation from FreeT f to FreeT g.

Convert a Free monad from Control.Monad.Free to a FreeT monad from Control.Monad.Trans.Free.

This is Prism' (Free f a) a in disguise

>>> preview _Pure (Pure 3)
Just 3

>>> review _Pure 3 :: Free Maybe Int
Pure 3

This is Prism' (Free f a) (f (Free f a)) in disguise

>>> preview _Free (review _Free (Just (Pure 3)))
Just (Just (Pure 3))

>>> review _Free (Just (Pure 3))
Free (Just (Pure 3))