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[-] spread@programming.dev 27 points 1 year ago

Impressive, very nice. Now let's see LLM's space complexity.

[-] fubo@lemmy.world 22 points 1 year ago

O(all the GPUs, all of them)

[-] VictorPrincipum@vlemmy.net 9 points 1 year ago

Hey now, don’t forget all the memory too

[-] rcmaehl@lemmy.world 8 points 1 year ago
[-] Sotuanduso@lemm.ee 1 points 1 year ago

/UnexpectedGimli

[-] whiskeypickle@lemmy.ml 4 points 1 year ago

Eggshell… and is that… Gothic type?

[-] duncesplayed@lemmy.one 15 points 1 year ago

Any algorithm can be O(n^2) if you only want it to be occasionally right.

[-] Iridium@lemmy.world 13 points 1 year ago
Function isPrime(number):
    return false

Accurate for almost 100% of cases

[-] darcy@sh.itjust.works 2 points 1 year ago

as test count approach infinity

[-] julianh@lemm.ee 5 points 1 year ago

Any algorithm can be O(1) if you cache all the answers beforehand.

[-] MajorHavoc@lemmy.world 2 points 1 year ago

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.

[-] darcy@sh.itjust.works 2 points 1 year ago

isnt O(n³) usually simplified to O(n²) anyway ?

[-] AlmightySnoo@lemmy.world 3 points 1 year ago* (last edited 1 year ago)

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).

[-] darcy@sh.itjust.works 2 points 1 year ago
[-] MajorHavoc@lemmy.world 2 points 1 year ago

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".

this post was submitted on 02 Jul 2023
59 points (92.8% liked)

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