Sorry, forgot to cc list
I think this is called taking a good thing too far, but cool too:
f1 u = u + 1
f2 u v = u + v
f3 u v w = u + v + w
-- functions renamed for consistency)
zipWith1 = map
zipWith2 = zipWith
-- and hey presto!
us1 = 3 : zipWith1 f1 us1
us2 = 2 : 3 : zipWith2 f2
Hi all,
I am completely stuck with a problem involving a loop construct from
imperative programming, that I want to translate to Haskell code.
The problem goes as follows:
Based on a value u_n, I can calculate a new value u_{n+1} with a
function f(u). u_n was calculated from u_{n-1} and so on
On Thu, 6 Sep 2007, Axel Gerstenberger wrote:
module Main where
import System.IO
import Text.Printf
main :: IO ()
main = do
let all_results1 = take 2 $ step [1]
--print $ length all_results1 -- BTW: if not commented out,
-- all values of
On 06/09/07, Axel Gerstenberger [EMAIL PROTECTED] wrote:
module Main where
import System.IO
import Text.Printf
main :: IO ()
main = do
let all_results1 = take 2 $ step [1]
--print $ length all_results1 -- BTW: if not commented out,
--
Thanks to all of you. The suggestions work like a charm. Very nice.
I still need to digest the advices, but have already one further
question: How would I compute the new value based on the 2 (or even
more) last values instead of only the last one?
[ 2, 3 , f 3 2, f((f 3 2) 3), f ( f((f 3 2)
Axel Gerstenberger wrote:
Thanks to all of you. The suggestions work like a charm. Very nice.
I still need to digest the advices, but have already one further
question: How would I compute the new value based on the 2 (or even
more) last values instead of only the last one?
[ 2, 3 , f 3 2,
On 06/09/07, Axel Gerstenberger [EMAIL PROTECTED] wrote:
Thanks to all of you. The suggestions work like a charm. Very nice.
I still need to digest the advices, but have already one further
question: How would I compute the new value based on the 2 (or even
more) last values instead of only
On 06/09/07, Axel Gerstenberger [EMAIL PROTECTED] wrote:
however,I don't get it this to work. Is it possible to see the
definition of the iterate function? The online help just shows it's usage...
The Haskell 98 report includes source for the standard prelude. Check 'em out...
When you get to more than two arguments, it will probably be nicer to
do something like this:
fibs = map (\(a,b) - a) $ iterate (\(a,b) - (b, a+b)) (0,1)
or
fibs = unfoldr (\(a,b) - Just (a, (b, a+b))) (0,1) -- this uses
unfoldr to get rid of the map
This is essentially a translation of the
On 06/09/07, Sebastian Sylvan [EMAIL PROTECTED] wrote:
foo = 2 : 3 : zipWith f (drop 1 foo) foo
There's also zipWith3 etc. for functions with more arguments.
I think this is called taking a good thing too far, but cool too:
f1 u = u + 1
f2 u v = u + v
f3 u v w = u + v + w
-- functions
On Thu, 6 Sep 2007, Axel Gerstenberger wrote:
Thanks to all of you. The suggestions work like a charm. Very nice.
I still need to digest the advices, but have already one further question:
How would I compute the new value based on the 2 (or even more) last values
instead of only the last
On 9/6/07, Dougal Stanton [EMAIL PROTECTED] wrote:
On 06/09/07, Sebastian Sylvan [EMAIL PROTECTED] wrote:
[2,3,4,9,16,29,54,99,182,335] -- what's this?
Two times this: http://www.research.att.com/~njas/sequences/A000213
plus this: http://www.research.att.com/~njas/sequences/A001590
plus two
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