Dear Ivan, A great explanation you have provided! It is very clear. Thank you so much! (You Haskell folks are so willing to help.) Wish there was something I knew that would be useful to you.
Thank you. Sincerely, Richard E. Adams Applications Developer Las Vegas Valley Water District Email: richard.ad...@lvvwd.com Tel. (702) 856-3627 -----Original Message----- From: Ivan Perez [mailto:ivanperezdoming...@gmail.com] Sent: Friday, February 10, 2012 12:28 PM To: j...@repetae.net Cc: Richard Adams; haskell-cafe@haskell.org Subject: 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 <j...@repetae.net> 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 <richard.ad...@lvvwd.com> 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! >> >> -- >> View this message in context: >> http://haskell.1045720.n5.nabble.com/Cannot-understand-liftM2-tp30856 >> 49p5470185.html Sent from the Haskell - Haskell-Cafe mailing list >> archive at Nabble.com. >> >> _______________________________________________ >> Haskell-Cafe mailing list >> Haskell-Cafe@haskell.org >> http://www.haskell.org/mailman/listinfo/haskell-cafe > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe