3 ms·
I think they’re replying to the claimed upper bound of n^3. I’m not actually sure what that means.
by jakeinspace 2y ago
I think they’re replying to the claimed upper bound of n^3. I’m not actually sure what that means.
- Dylan16807 2y agoI'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.
- demurgos 2y agosee schoolbook algorithm for the n^3 bound: https://en.wikipedia.org/wiki/Computational_complexity_of_matrix_multiplication https://en.wikipedia.org/wiki/Computational_complexity_of_ma... It comes from directly applying the definition of matrix multiplication on a square matrix.