Re: [Haskell-cafe] For Project Euler #1 isn't it more efficient to generate just the numbers you need? <Spoiler>

One doesn't have to touch them to compute their sum. 2011/5/7 KC :

Do you mean O(1) complexity?

Don't you have to touch at least the multiples of 3 & 5 making it O(k) where is the number of multiples of 3 and the number of multiples of 5?

On Fri, May 6, 2011 at 10:10 PM, Lyndon Maydwell wrote:

...

If you're looking for efficiency, I believe you can actually do #1 in constant time:

On Sat, May 7, 2011 at 7:31 AM,   wrote:

...

-- Instead of this -- sumMultiples3or5 s = sum [x | x <- [3..s-1], x `mod` 3 == 0 || x `mod` 5 == 0]

-- Isn't this faster

sumMultiples3or5 s = sum ([x | x <- [3,6..s-1]] `merge` [x | x <- [5,10..s-1]])

merge xs [] = xs merge [] ys = ys merge txs@(x:xs) tys@(y:ys)    | x < y     = x : xs `merge` tys    | x > y     = y : txs `merge` ys    | otherwise = x : xs `merge` ys

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-- -- Regards, KC

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