On Sunday, 6 December 2015 at 03:24:24 UTC, Timon Gehr wrote:
Some upper bounds are incomparable, so there would not be a total order. But that is not a problem.
It is a problem in all cases as you usually dont have an optimal bound. And with your approach that will most certainly be guaranteed to happen. Comparing bounds does not mean you are comparing running time.
O(1) implies O(f(x)), O(N) implies O(N^2). You can also get tighter bounds for specific input models.
