Re: [Haskell-cafe] (a -> [b]) -> [a -> b] ?

Hi, Am Montag, den 04.12.2006, 13:12 -0600 schrieb Nicolas Frisby:

It seems there's an assumption about the range of the parameter function and the range of the entire function. That is, I think we're assuming that the length of the final result is the same as the length of the result of the first function?

Ah, of course, I forgot to mention to say what the function should do :-) If we call in unsequence, the following should be true for f::(a -> [b]) and v::a f v == map (\g -> g v) (unsequence f) Let’s see if I can prove that my suggested solution is correct: f v == map (\g -> g v) (map (\n a -> f a !! n) [1..]) == map ((\g -> g v) . (\n a -> f a !! n)) [1..] == map ((\n -> (f v !! n)) [1..] * here I use that map (\n -> l !!n ) [1..] == l. I hope that is valid == f v Ok, looks good. But still, I don’t like this solution with (!!) for some reason. Thanks, Joachim

On 12/4/06, Joachim Breitner wrote:

...

Hi,

while pondering over the four fours problem, I wondered: Is there a function of type (a -> [b]) -> [a -> b]

It looks a bit like sequence when applied in the ((->) a) Monad: sequence :: [a -> b] -> a -> [b] but I was looking for the other direction.

I came up with: \g -> map (\n a -> g a !! n) [1..] which has the desired type and functionality, but it looks rather inelegant and messy. Any better ideas?

Thanks, Joachim

-- Joachim "nomeata" Breitner mail: mail@joachim-breitner.de | ICQ# 74513189 | GPG-Key: 4743206C JID: joachimbreitner@amessage.de | http://www.joachim-breitner.de/ Debian Developer: nomeata@debian.org _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- Joachim "nomeata" Breitner mail: mail@joachim-breitner.de | ICQ# 74513189 | GPG-Key: 4743206C JID: joachimbreitner@amessage.de | http://www.joachim-breitner.de/ Debian Developer: nomeata@debian.org