3 ms·
I suspect the limit will be O(n^2xlog(n)). It might also be easier to work out the complexity by considering the size of the problem to be the number of element
by phkahler 3y ago
I suspect the limit will be O(n^2xlog(n)). It might also be easier to work out the complexity by considering the size of the problem to be the number of elements in a matrix instead, so O(nlog(n)) which would then be equivalent to O(n^2xlog(n)) using the n we use today. I also find it interesting that the current exponents seem to be closing in on 2+1/e.