On 12/29/06, Neil Mitchell <[EMAIL PROTECTED]> wrote:
Hi
> f1 :: [Int] -> [[Int]]
> f1 [] = []
> f1 (a:as) = [a] : f1 as
f1 is simply a map
> f3 la lb = let a = head la
> b = head lb
> in if sum a <= sum b then
> a : f3 (tail la) lb
> else
> b : f3 la (tail lb)
Why not use pattern matching to split up la and lb, rather than head/tail?
I would have thought the whole function could be written as a nice
foldr merge, where merge :: [Int] -> [Int] -> [Int]. Thats only a
guess at the top of my head though, not worked out properly.
Is this homework? If so its useful to state when you post the question :)
Hi Neil,
I am not sure how to express f1 with map? how do I say
(lambda (ls)
(map (lambda (x) (list x))
ls))
in Haskell? map ([]) ?
Here is my new f3:
f3 :: [[Int]] -> [[Int]] -> [[Int]]
f3 [] lb = lb
f3 la [] = la
f3 la@(a:as) lb@(b:bs) = if sum a <= sum b then
a : f3 as lb
else
b : f3 la bs
(btw, yes this is homework assigned by prof. Quan in quantum computing class
:))
Thanks,
Quan
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