That’s just not how complexity analysis works, if there’s a bound anywhere, it is O(1). The behaviour below that bound may be of more practical importance, but it’s not how O-notation works. The parent comment doesn’t make a great deal of sense for other reasons, but your objections don’t hold either.
By this logic, all programs that we run are O(1), since any data structure is in practice bounded in size by your available address space (+ disk space if we want to be pedantic). So, you can replace any loop over the elements of a DS with a loop over all memory in the address space, and you will get an O(1) algorithm.
In the end, it's just not a useful way of modelling the problem.
Yes, so ordinarily we analyse an idealised algorithm running on idealised machine. However if an upper bound is clearly introduced to this idealised setting, as in the root comment, then it can’t be ignored.
Re:The true cost of linked lists
by waynesonfire on Monday June 06, 2022 @02:01PM
(#31643630)
So an idealised machine doesn't have an arbitrary INT_MAX limit. The algorithms are O(n).
But playing tricks with the INT_MAX limit ignores(or, more accurately, spits in the face of) the fact that we want our computers to pretend to be idealized machines.
Floats aren’t real numbers but we treat them like the mathematical object that they model for most purposes.
Of course, being aware of the limits of our models are important, but abusing the tragic finitness of our models to “well actually” someone is generally not helpful.
“The heat death of the universe will occur in finite time, therefore all algorithms you can ever use are O(1)” isn’t a very helpful metric.
It takes the same (huge) amount of time for all inputs - there's no early exit it always scans to INT_MAX. Hence O(1)
This is incorrect. Putting a finite bound on n does not reduce n to 1. n is always meant to be a finite arbitrarily large value. n=INT_MAX qualifies as arbitrarily large, especially on a 64-bit system.
Re:The true cost of linked lists
by Sohcahtoa82 on Monday June 06, 2022 @05:30PM
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Big-O notation says nothing about the overall absolute size. All that matters is how the amount of work scales with the input value.
If an algorithm will always take 1 billion years to complete regardless of the input, then it's still O(1).
O(x) = O(1) if x is some finite constant.
Consider n=INT_MAX*2 or INT_MAX^2 though.
The behaviour does flatline as n approaches infinity
Re:The true cost of linked lists
by PhineasRex on Monday June 06, 2022 @12:30PM
(#31642563)
There is no bound on n, it always searches to INT_MAX regardless of n.
Functionally, though, I would describe this algorithm as O(N). It's just a straight up linear scan of the problem space. It's unbounded, with a worst case of O(Infinity) and no upper limit on what searched can be, but it's functionally O(N) regardless.
O(1) needs to basically be something like a hash key/pointer to memory address to get the result you're looking for. It will not increase linearly as the problem space increases, whereas your first example increases exactly linearly.