[Haskell-cafe] puzzle: prove this floorSqrt correct

-- Here's the discrete version of Newton's method for finding -- the square root. Does it always work? Any literature? floorSqrt :: Integer -> Integer floorSqrt n | n>=1 = detect (iterate (\x->(x+n`quot`x)`quot`2) 1) where detect (x:xs@(y:z:_)) | x==z = min x y | otherwise = detect xs isFloorSqrt n x = x*x <= n && n < (x+1)*(x+1) main = print$ all (\n->isFloorSqrt n (floorSqrt n)) [1..1000000] ________________________________________________________________ Verschicken Sie romantische, coole und witzige Bilder per SMS! Jetzt neu bei WEB.DE FreeMail: http://freemail.web.de/?mc=021193