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
Yeah, you’re right, I’m not being rigorous here. I’m just co-opting big O notation somewhat inaccurately to express that this isn’t going to get as big as it seems because the number of upvotes isn’t going to increase all that much.
I guess I have a different intuition. For me, O notation applied normally shows that it will not get far because it rises exponentially or if you count the steps in relation to the number of users, it’s O(log n) which doesn’t go high. No need to apply it inaccurately
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
Yeah, you’re right, I’m not being rigorous here. I’m just co-opting big O notation somewhat inaccurately to express that this isn’t going to get as big as it seems because the number of upvotes isn’t going to increase all that much.
I guess I have a different intuition. For me, O notation applied normally shows that it will not get far because it rises exponentially or if you count the steps in relation to the number of users, it’s O(log n) which doesn’t go high. No need to apply it inaccurately