[poincaredisk]

I imagine the point of this algorithm, like a lot of algorithm research, is to prove the upper bound of complexity for this problem. Not to be used in practice (despite what this article seem to suggest).

On a similar note, there's a lot of work put into optimal matrix multiplication algorithm. We know the lower bound is N*2, the obvious upper bound is N*3, the best (complexity wise, not practical at all) current algorithm is N*2.37, but we don't know how fast can it really get. Is it possible to write N*2 algorithm? We don't know.

[KolenCh]

Both N^2 and N*2 are fine, but N*2 should really not mean what you want it to mean…

[KolenCh]

Now reading my comment again, I probably understand what happened. I typed N*2, and yet it showed as N*2. Probably this is why.

[nnoremap2]

I mean nobody is stopping me from writing an exponential time algorithm.

[adrianN]

Knowing an upper bound means you know that the best solution for the problem takes at most that much work. It does not mean that you can’t find worse solutions.

[Dylan16807]

Sure? If you're slow on purpose that doesn't affect the upper bound set by the "obvious" method.

[jakeinspace]

I think they’re replying to the claimed upper bound of n^3. I’m not actually sure what that means.

[demurgos]

see schoolbook algorithm for the n^3 bound: https://en.wikipedia.org/wiki/Computational_complexity_of_ma...

It comes from directly applying the definition of matrix multiplication on a square matrix.

[Dylan16807]

I'm not sure how to phrase this better than saying it's the bound set by the obvious method. Are you familiar with the obvious method of multiplying a matrix? It's n operations for each cell, and you have n^2 cells.

Being worse on purpose is always possible, but it doesn't change that bound.

It's the "obvious" upper bound, not the "how did you even get here without noticing the obvious method" upper bound.