On Thu, Jan 28, 2010 at 11:18:12AM +0000, Zbigniew Stanasiuk wrote:
Hallo all,
I have two functions f and g of one arguments:
f :: (Num a) => [a] -> [a] g :: (Num a) => [a] -> [a]
I can do composition (f .f .f.) [a] or (g .g .g.) [a] for a few repeated functions. And what about if I want to nest f, say, many many times?
Use this: iterate :: (t -> t) -> t -> [t] For example Prelude> take 10 $ iterate (*2) 1 [1,2,4,8,16,32,64,128,256,512] That means that if you want to nest your function 'x' times, then you just take the x-th element of the list. Prelude> let x = 10 in iterate (*2) 1 !! x 1024
The matter is more complicated (from my prespective :-) ) if I would like to get e.g.: the composition (g .f .g. g.f .f) [a] if t == 0 then f else g and having the list [1,0,1,1,0,0] as a pattern to construct above composition.
How to make above, generally for arbitrary length ???
You could roll on your own functions, for example composition :: [t -> t] -> t -> t composition = foldr (.) id composition' :: [t -> t] -> [Int] -> t -> t composition' fs is = composition [fs !! i | i <- is] Note that you would them as composition [g, f, g, g, f, f] or composition' [f, g] [1, 0, 1, 1, 0, 0] You could also write 'composition' as import Data.Monoid composition :: [t -> t] -> t -> t composition = appEndo . mconcat . map Endo 'map Endo' just puts everything in the Endo monoid, while 'appEndo' takes the result from the monoid. In other words, 'composition' is the same as a concatenation in the Endo monoid. It may be defined as: newtype Endo a = Endo {appEndo :: a -> a} instance Monoid (Endo a) where mempty = Endo id mappend (Endo f) (Endo g) = Endo (f.g) Do you see the pattern? :) HTH, -- Felipe.