Re: [Haskell-cafe] (a - [b]) - [a - b] ?
On 12/4/06, Joachim Breitner [EMAIL PROTECTED] wrote: \g - map (\n a - g a !! n) [1..] I think that's about as good as it gets. -- Taral [EMAIL PROTECTED] You can't prove anything. -- Gödel's Incompetence Theorem ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] (a - [b]) - [a - b] ?
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? If I'm correct in presuming that constraint, then I think this indicates that a more elegant solution might involve using the lightweight-dependently-typed vectors approach. Though I can't promise it will actually be nicer! Nick On 12/4/06, Joachim Breitner [EMAIL PROTECTED] 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: [EMAIL PROTECTED] | ICQ# 74513189 | GPG-Key: 4743206C JID: [EMAIL PROTECTED] | http://www.joachim-breitner.de/ Debian Developer: [EMAIL PROTECTED] ___ 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
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 [EMAIL PROTECTED] 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: [EMAIL PROTECTED] | ICQ# 74513189 | GPG-Key: 4743206C JID: [EMAIL PROTECTED] | http://www.joachim-breitner.de/ Debian Developer: [EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- Joachim nomeata Breitner mail: [EMAIL PROTECTED] | ICQ# 74513189 | GPG-Key: 4743206C JID: [EMAIL PROTECTED] | http://www.joachim-breitner.de/ Debian Developer: [EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] (a - [b]) - [a - b] ?
On Mon, Dec 04, 2006 at 07:00:23PM +, Joachim Breitner wrote: 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? I like sequence a2bs = (head . a2bs) : sequence (tail . a2bs) This also makes it explicit that by golly your function had better return an infinite list, or we're in trouble. -- David Roundy Department of Physics Oregon State University ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] (a - [b]) - [a - b] ?
Joachim Breitner: here I use that map (\n - l !!n ) [1..] == l. I hope that is valid map (\n - l !! n) [1..] is more like (tail l). Did you mean to use [0..]? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] (a - [b]) - [a - b] ?
Hi, Am Dienstag, den 05.12.2006, 05:59 +1000 schrieb Matthew Brecknell: Joachim Breitner: here I use that map (\n - l !!n ) [1..] == l. I hope that is valid map (\n - l !! n) [1..] is more like (tail l). Did you mean to use [0..]? Probably. I hardly use (!!), so I did not remember if it starts with 0 or 1. Thanks for the hint, Joachim -- Joachim Breitner e-Mail: [EMAIL PROTECTED] Homepage: http://www.joachim-breitner.de ICQ#: 74513189 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
