this post was submitted on 02 Jul 2023
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Programmer Humor

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[–] spread@programming.dev 24 points 1 year ago (2 children)

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

[–] fubo@lemmy.world 20 points 1 year ago (1 children)

O(all the GPUs, all of them)

[–] VictorPrincipum@vlemmy.net 7 points 1 year ago (1 children)

Hey now, don’t forget all the memory too

[–] rcmaehl@lemmy.world 6 points 1 year ago (1 children)
[–] 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 13 points 1 year ago (3 children)

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

[–] Iridium@lemmy.world 12 points 1 year ago (1 children)
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 6 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 (2 children)

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

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

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".