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by drbaskin
5671 days ago
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I'm sure that you have an implicit statement about the constants when you imply that O(log_{k}(|T|)) is an improvement over O(log_{2}(|T|)), but log_{2}(|T|) = {log_{k}(|T|)} / {log_{k} (2)}. Correct me if I'm wrong, but in terms of big-O notation, I think this makes O(log_{2}(|T|)) the same as O(log_{k}(|T|)). |
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