[Haskell-cafe] Optimization fun
wchogg at gmail.com
Sat Feb 10 16:25:22 EST 2007
On 2/10/07, Lennart Augustsson <lennart at augustsson.net> wrote:
> 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.
Now why is foldl' too strict? I don't think I understand?
* 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.
Yeah, I don't know what the %#*( happened there. I should have proofread.
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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