Re: [Haskell-cafe] Cannot understand liftM2

To understand how liftM2 achieves the cartesian product, I think one way is to find liftM2's implementation and (>>=) implementation as part of []'s instantiation of the Monad class. You can find the first in Control.Monad, and the second in the standard prelude. Lists are monads, and as John (almost) said, liftM2 f x y is equivalent to liftM2 f m1 m2 = do x1 <- m1 x2 <- m2 return (f x1 x2) Which is syntactic sugar (fancy Haskell) for liftM2 f m1 m2 = m1 >>= (\x1 -> m2 >>= (\x2 -> return (f x1 x2))) In the prelude, you can find instance Monad [] where m >>= k = foldr ((++) . k) [] m Fhe right-hand side of (>>=) here is roughly equivalent to concat (map k m). The last step, which I leave as an exercise to the reader (I always wanted to say that), is use the right hand side of the definition of (>>=) for lists in the right hand side of liftM2 when applied to (,) and two lists. You can see the type of the function (,) (yes, comma is a function!) by executing, in ghci: :type (,) Cheers, Ivan. On 9 February 2012 19:23, John Meacham wrote:

A good first step would be understanding how the other entry works:

cartProd :: [a] -> [b] -> [(a,b)] cartProd xs ys = do        x <- xs        y <- ys        return (x,y)

It is about halfway between the two choices.

   John

On Thu, Feb 9, 2012 at 9:37 AM, readams wrote:

...

Nice explanation.  However, at http://stackoverflow.com/questions/4119730/cartesian-product it was pointed out that this

cartProd :: [a] -> [b] -> [(a, b)] cartProd = liftM2 (,)

is equivalent to the cartesian product produced using a list comprehension:

cartProd xs ys = [(x,y) | x <- xs, y <- ys]

I do not see how your method of explanation can be used to explain this equivalence?  Nevertheless, can you help me to understand how liftM2 (,) achieves the cartesian product?  For example,

Prelude Control.Monad.Reader> liftM2 (,) [1,2] [3,4,5] [(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)]

Thank you!

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