3 ms·
I don't know if he edited his comment, but as it is currently written the log is not in the exponent under standard order of operations. Also, putting a consta
by jessriedel 6y ago
I don't know if he edited his comment, but as it is currently written the log is not in the exponent under standard order of operations.
Also, putting a constant factor like log(2) inside big-O notation is meaningless.
- kadoban 6y agoConstant factors in exponents are _not_ meaningless. One of those cases where the rules-of-thumb bite you.
- jessriedel 6y agoI didn't claim that. The log(2) term is multiplied, and that's implied by my use of "factor", which in this context is always understood to mean multiplicative factor.
- kadoban 6y agoYou're correct. I think this comes down to confusion on the operator precedence. I read it as O(n^(3 log 2)) because otherwise it's kind of very obviously not correct (no way it's n^3, that's already known), but should have stated that, because according to usual notation/precedence it's not that.