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)
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