There are many things that makes your code slow. * The default for Haskell is to compute with Integer, not Int. So that makes from Integral and floor very slow. * foldl' is a bad choice, because it is too strict, you want to abort the loop as soon as possible. * your take is really wrong. The number of primes you need to take cannot be computed like that. You want to take primes while the sqrt of x is larger than the prime. Also, your code is not idiomatic Haskell. Here's a different version: primes :: [Int] primes = 2:filter isPrime [3,5..] where isPrime x = all (\ y -> x `mod` y /= 0) $ takeWhile (\ p -
p*p <= x) primes
On Feb 10, 2007, at 21:02 , Creighton Hogg wrote:
primes = 2:(foldr (\x y -> if isPrime x then x:y else y) [] [3..]) where isPrime x = foldl' (\z y -> z && (if x `mod` y == 0 then False else True)) True (take (floor $ sqrt $ fromIntegral x) primes)