No, n³ cannot be O(n²) as otherwise that would mean that there exists a positive constant K and a positive threshold m such that for any integer n greater than m you would have n³ less than K*n², which would be the same as saying n less than K, which cannot hold for any integer n greater than m. So n³ cannot be an O(n²), which means that something that is an O(n³) is not necessarily an O(n²).
It’s the other way around, if something is an O(n²) then it is necessarily also an O(n³).
Yes. The other answer is technically correct, but yours is pragmatically correct.
If a solution is worse than O(nln(n))* then most of us are going to be looking for a pragmatic and completely alternate way to deal with it, rather than analyzing how to make it mildly less terrible.
So I’m just writing O(n^2) as a quick professional replacement for my original write in answer of “dogshit”.
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Impressive, very nice. Now let’s see LLM’s space complexity.
O(all the GPUs, all of them)Hey now, don’t forget all the memory too
And my cache!
Eggshell… and is that… Gothic type?
Any algorithm can be O(n^2) if you only want it to be occasionally right.
Any algorithm can be O(1) if you cache all the answers beforehand.
Yes.
And depending how occasionally we’re talking, I can code for some very fast solutions when the correctness requirements are low enough.
Alternately, if we want it to only be occasionally fast, I’ve got a very nice looking and very wrong algorithm for that, as well.
Accurate for almost 100% of cases
as test count approach infinity
isnt O(n³) usually simplified to O(n²) anyway ?
No, n³ cannot be O(n²) as otherwise that would mean that there exists a positive constant K and a positive threshold m such that for any integer n greater than m you would have n³ less than K*n², which would be the same as saying n less than K, which cannot hold for any integer n greater than m. So n³ cannot be an O(n²), which means that something that is an O(n³) is not necessarily an O(n²).
It’s the other way around, if something is an O(n²) then it is necessarily also an O(n³).
ok thanks
Yes. The other answer is technically correct, but yours is pragmatically correct.
If a solution is worse than O(nln(n))* then most of us are going to be looking for a pragmatic and completely alternate way to deal with it, rather than analyzing how to make it mildly less terrible.
So I’m just writing O(n^2) as a quick professional replacement for my original write in answer of “dogshit”.