[Haskell-cafe] Cannot understand liftM2
Richard Adams
richard.adams at lvvwd.com
Mon Feb 13 19:19:44 CET 2012
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.Adams at lvvwd.com
Tel. (702) 856-3627
-----Original Message-----
From: Ivan Perez [mailto:ivanperezdominguez at gmail.com]
Sent: Friday, February 10, 2012 12:28 PM
To: john at repetae.net
Cc: Richard Adams; haskell-cafe at 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 <john at 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.adams at 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.
>>
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