Timeline for Algorithm complexity: $\mathcal O(n\cdot m)$ vs. $\mathcal O(max(n,m)^2)$

9 events

when toggle format	what		by	license	comment
Nov 8, 2018 at 15:22	answer	added	kelalaka		timeline score: 1
Nov 8, 2018 at 15:20	comment	added	SEJPM		Actually $\mathcal O(n\cdot m)$ denotes "complexity is at most linear in $n$ and $m$" and thus I'd say $\mathcal O(n\cdot m)$ is much easier to understand and actually a better bound than $\mathcal O(\max(n,m)^2)$, because the latter tells you "if we fix $n$ and increase $m$ linearly, work wil increase quadratically" whereas the former will say that the work will grow linearly.
Nov 8, 2018 at 12:17	history	edited	Alpha Bravo	CC BY-SA 4.0	deleted 4 characters in body
Nov 8, 2018 at 12:12	history	edited	Alpha Bravo	CC BY-SA 4.0	added 89 characters in body
Nov 8, 2018 at 12:01	history	edited	Alpha Bravo	CC BY-SA 4.0	added 663 characters in body
Nov 8, 2018 at 10:36	history	edited	Alpha Bravo	CC BY-SA 4.0	added 14 characters in body; edited title
Nov 8, 2018 at 10:24	comment	added	kelalaka		The `some efficient computation` that is the only part belongs to secuirty parameter $k$.
Nov 8, 2018 at 10:23	history	edited	Alpha Bravo	CC BY-SA 4.0	added 36 characters in body
Nov 8, 2018 at 9:58	history	asked	Alpha Bravo	CC BY-SA 4.0