Hello, On 14 May 2008, at 02:06, Ronald Guida wrote:

I have a few questions about commutative monads and applicative functors.

...

From what I have read about applicative functors, they are weaker than monads because with a monad, I can use the results of a computation to select between alternative future computations and their side effects, whereas with an applicative functor, I can only select between the results of computations, while the structure of those computations and their side effects are fixed in advance.

But then there are commutative monads. I'm not exactly sure what a commutative monad is, but my understanding is that in a commutative monad the order of side effects does not matter.

This leads me to wonder, are commutative monads still stronger than applicative functors, or are they equivalent?

I would say that they are stronger because they still support: concat :: Monad m => m (m a) -> m a or (>>=) :: Monad m => m a -> (a -> m b) -> m b which are not supported, in general, by applicative functors. In fact, I would probably risk to say that they are even stronger than monads (as there are less commutative monads than regular monads).

And by the way, what exactly is a commutative monad?

Here is a possible characterization: The monad m is commutative if, for all mx and my: do {x <- mx; y <- my; return (x,y)} = do {y <- my; x <- mx; return (x,y)} As you mentioned above, the basic idea is that the order of the side effects does not matter. This law is not true in general for monads. I am not sure if you know about the paper entitled "The essence of the Iterator Pattern" (by Jeremy Gibbons and myself): http://www.comlab.ox.ac.uk/people/Bruno.Oliveira/iterator-jfp.pdf But, you may be interested to read it as it discusses related ideas. In particular you may want to take a look at Section 5.4. Cheers, Bruno

On Tue, May 13, 2008 at 9:06 PM, Ronald Guida wrote:

I have a few questions about commutative monads and applicative functors.

...

From what I have read about applicative functors, they are weaker than monads because with a monad, I can use the results of a computation to select between alternative future computations and their side effects, whereas with an applicative functor, I can only select between the results of computations, while the structure of those computations and their side effects are fixed in advance.

But then there are commutative monads. I'm not exactly sure what a commutative monad is, but my understanding is that in a commutative monad the order of side effects does not matter.

This leads me to wonder, are commutative monads still stronger than applicative functors, or are they equivalent?

And by the way, what exactly is a commutative monad?

A monad is commutative if the expression "a >>= \x -> b >>= \y -> f x y" is equivalent to "b >>= \y -> a >>= \x -> f x y". The identity, state reader, and maybe monads are commutative, for example. To put it another way, a monad is commutative if liftM2 f a b = liftM2 (flip f) b a Since liftA2 generalizes liftM2, we can also say that an applicative functor is commutative if liftA2 f a b = liftA2 (flip f) b a Or, put another way, if a <*> b = flip ($) <$> b <*> a If w is a commutative monoid (that is, if mappend w1 w2 == mappend w2 w1), then Const w is a commutative applicative functor. To summarize: some applicative functors are commutative, some applicative functors are monads, and the ones that are both are commutative monads. -- Dave Menendez http://www.eyrie.org/~zednenem/