Dan, Thanks - you're probably right - just my intuition said to me that rather than calculating that the 13th root of 4021503534212915433093809093996098953996019232 is 3221.2904208350265.... there must be a quicker way of finding out its between 3221 and 3222.... ....but perhaps not. Tim
________________________________ From: Dan Goodman [mailto:dg.gm...@thesamovar.net] Sent: Sat 31-Jan-09 3:11 PM To: python-list@python.org Subject: Re: nth root Takes less than 1 sec here to do (10**100)**(1./13) a million times, and only about half as long to do (1e100)**(1./13), or about 14 times as long as to do .2**2. Doesn't look like one could hope for it to be that much quicker as you need 9 sig figs of accuracy to get the integer part of (10**100)**(1./13) (floats have about 7 and doubles about 16). Dan Tim wrote: > In PythonWin I'm running a program to find the 13th root (say) of > millions of hundred-digit numbers. I'm using > n = 13 > root = base**(1.0/n) > which correctly computes the root to a large number of decimal > places, but therefore takes a long time. All I need is the integer > component. Is there a quicker way? > > > > ------------------------------------------------------------------------ > > -- > http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
