On 04/12/2015 09:56 PM, Paul Rubin wrote:
Marko Rauhamaa <ma...@pacujo.net> writes:And in fact, the sqrt optimization now makes the original version 20% faster: ... bound = int(math.sqrt(n))That could conceivably fail because of floating point roundoff or overflow, e.g. fac(3**1000). A fancier approach to finding the integer square root might be worthwhile though.
If I were trying to get a bound for stopping the divide operation, on a value too large to do exact real representation, I'd try doing just a few iterations of Newton's method. Even if you don't converge it to get an exact value, you can arrange that you have a number that's for sure no less than the square root. And you can get pretty close in just a few times around.
-- DaveA -- https://mail.python.org/mailman/listinfo/python-list
