On 14/04/2016 09:13, blindanag...@nowhere.net wrote: > On 14/04/2016 08:59, blindanag...@nowhere.net wrote: >> On 14/04/2016 07:52, ast wrote: > >> This means that the result will be correct provided it has 53 or less >> bits - just short of 16 decimal digits (i.e for square numbers with less >> than 32 digits). >> >> With an integer square root function (isqrt), this program: >> >> for exp in count(20): >> v = 2 ** exp - 1 >> if isqrt(v) != sqrt(v): >> print(exp, isqrt(v), sqrt(v)) >> break >> >> terminates with: >> >> 54 18014398509481983 1.8014398509481982e+16 >> >> showing a first error for a 54 bit square root >> > I should also have said that the square root of integer squares with > between 15 and 30 decimal digits will only be correct if the square > numbers themselves are exactly representable in 53 bits. So we can > expect failures for squares with 16 or more digits.
However, if a number with 31 or less digits is known to be the square of an integer, the IEEE754 sqrt function will (I believe) give the correct result. -- https://mail.python.org/mailman/listinfo/python-list
