this post was submitted on 30 May 2024
396 points (87.8% liked)
AnarchyChess
5278 readers
383 users here now
Holy hell
Other chess communities:
[email protected]
[email protected]
founded 2 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
More like O(2^n^) and what do you mean by "this"? The number will increase in O(2^n^) until it hits the point where it stops. What exactly is bound? The number of posts in total?
#getyourdefinitionsstraight
Sorry, mixed up n^2^ and 2^n^. But what I meant was that there's eventually going to be a point where the limiting factor is the number of people willing to upvote it, which is asymptotically constant (after crossing the threshold of making it onto the front page.)
Both the number of posts and the width of the posts are limited by a constant in this way, though the latter is a much larger constant. I suppose I was talking about the width of the posts, but it would have been more accurate to say it's bound above by 2^(the number of users on Lemmy.)
In short, I do not think these posts are going to reach a 2048-wide en passant, but I don't think image size is going to be the reason why.
According to that logic, everything is O(1) because at some point you go out of memory or your computer crashes.
"How fast is your sorting algorithm?" – "It can't sort all the atoms in the universe so O(1)."
That's not how O notation works. It is about asymptomatics. It is about "What if it doesn't crash", "what if infinity".
So, again, what exactly is your question if your answer is O(1)?
O notation has a precise definition. A function f : N -> R+ is said to be O(g(x)) (for some g : N -> R) if there exists a constant c so that f(n) <= cg(n) for all sufficiently large n. If f is bounded, then f is O(1).
If n is in O(1), than any O(f(n)) is O(1) but who says the number of lemmy users is bounded? We will grow and we will continue to grow. When economists calculate with infinite growth, so can we