Achim Schneider <[EMAIL PROTECTED]> wrote: > "Rafael Almeida" <[EMAIL PROTECTED]> wrote: > > > perfectSquares :: [Integer] > > perfectSquares = zipWith (*) [1..] [1..] > > > isPerfectSquare :: Integer -> Bool > > isPerfectSquare x = (head $ dropWhile (<x) perfectSquares) == x > > what about > > module Main where > > isPerfectSquare :: Integer -> Bool > isPerfectSquare n = sqrrt == fromIntegral (truncate sqrrt) > where sqrrt = sqrt $ fromIntegral n > > ? It's a hell alot faster, but I have no idea if some numerical > property of square roots could make it give different results than > your version, in rare cases. > Well, even if so, I bet calculating the square root by successive approximation using integers would still be faster, it's O(ld(n)) instead of O(n).
-- (c) this sig last receiving data processing entity. Inspect headers for past copyright information. All rights reserved. Unauthorised copying, hiring, renting, public performance and/or broadcasting of this signature prohibited. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
