Re: Big-Theta( log(base 3) n )


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Posted by Pete Schoenhoff on July 29, 1999 at 22:38:10:

In Reply to: Big-Theta( log(base 3) n ) posted by Michael Ashwell on July 29, 1999 at 19:51:22:


___BOOM!___

Man, I didn't think of that! You're right!
log(base b)n = [log(base a)n]/[log(base a)b]
so log(base 3)n = [log(base 2)n]/[log(base 2)3]
= approx. [log(base 2)n]/1.6
so log(base 3)n = approx. (.63) log(base 2)n
--which reduces it down to a constant!-

Wow - you're pretty good. That one flew right by me!




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