2 ms·
I feel it’s more likely there is an n^2 algorithm using some radically different approach.
by zeroonetwothree 3y ago
I feel it’s more likely there is an n^2 algorithm using some radically different approach.
- bobbylarrybobby 3y agoI would assume n^(2+epsilon) where epsilon can be made arbitrarily small, but must be positive.