This is actually a perfect case for lazy immutable arrays, if you use a circular
program:
import Data.Array
foo' :: Array Int Int -> Int -> Int
foo' arr 0 = 0
foo' arr 1 = 1
foo' arr 2 = 2
foo' arr n = arr!(n-1) + arr!(n-2) + arr!(n-3)
foo :: Int -> Int
foo n = arr ! n
where
assocs = [(i, foo' arr i) | i <- [0..n]]
arr = array (0,n) assocs
But I haven't checked its performance against your version, so I don't know how
good it is.
/ Emil
Magnus Therning skrev:
On Tue, Dec 16, 2008 at 12:14 PM, Lemmih <lem...@gmail.com> wrote:
You could use a Map or a mutable array. However, this kind of problem
comes up a lot less often than you'd think.
Well, I happen to have a problem just like it right now, hence my
interest :-) In order to better understand the different options I
thought I'd look at solving simpler problems with similar "shape".
Thanks for pointing me in the direction of mutable arrays, I haven't
explored those before.
/M
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