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That argument applies to any O(log n) algorithm. The base of the logarithm doesn't matter.
by ScottBurson 12y ago
That argument applies to any O(log n) algorithm. The base of the logarithm doesn't matter.
- TheLoneWolfling 12y agoWell, yes and no. Yes, in that the argument applies. No, in that the size is larger. It's more reasonable to say one algorithm is 80x faster than another of the same complexity than to say one algorithm is 530x faster than another of the same complexity.
- lomnakkus 12y agoConstants matter. To "overcome" O(log_a n) you need a constant of n^a for any given n. That's quite a lot in practice... although it's matematically trivial.