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Old 2011-11-07, 13:59   #9
axn's Avatar
Jun 2003

19·271 Posts

Originally Posted by LaurV View Post
I am still waiting for an algorithm with complexity O((log n)^a)), with "a" as big as you want, but fixed... >:P
I am just wondering. What do you think 'n' means in this context? I get the feeling that you consider 'n' to be the number being multiplied rather than the number of bits/digits in the number being multiplied.

Last fiddled with by axn on 2011-11-07 at 14:01 Reason: thing -> think
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