On Wed, 20 Jun 2007, Brent Yorgey wrote:
isSquare :: (Integral a) => a -> Bool isSquare n = (floor . sqrt $ fromIntegral n) ^ 2 == n
Is there any way to write that without the fromIntegral? If you leave out the fromIntegral and the explicit type signature, it type checks, but the type constraints are such that there are no actual types that you can call it on.
This is a good example: You wonder, whether fromIntegral can be avoided. I wonder, whether fromIntegral fulfills the task at all. Actually, it does not. It fails for big integers, because there is no Double that represents 10^1000. That is you have to rescale the number. Even below this number, 'isSquare' will fail due to rounding errors: Prelude> isSquare ((10^100)^2) False That is, 'isSquare' does not do what it promises. Btw. I would at least use 'round' because the Double sqrt might be slightly below the true root. Unfortunately we don't have access to the native sqrt implementation of the GNU multiprecision library GMP so we have to roll our own version: (^!) :: Num a => a -> Int -> a (^!) x n = x^n {- | Compute the floor of the square root of an Integer. -} squareRoot :: Integer -> Integer squareRoot 0 = 0 squareRoot 1 = 1 squareRoot n = let twopows = iterate (^!2) 2 (lowerRoot, lowerN) = last $ takeWhile ((n>=) . snd) $ zip (1:twopows) twopows newtonStep x = div (x + div n x) 2 iters = iterate newtonStep (squareRoot (div n lowerN) * lowerRoot) isRoot r = r^!2 <= n && n < (r+1)^!2 in head $ dropWhile (not . isRoot) iters Btw. I think that 'squareRoot' is the basic problem and I'd like to change the Wiki article accordingly.