O(1)
< O(log(n))
< O(sqrt(n))
< O(n)
< O(n log(n))
< O(n²)
< O(2ⁿ)
< O(n!)
@0foldcv I don't remember ever seeing this combination of < and O. Is it supposed to mean something different than = O(...)?
@axiom @0foldcv So according to your formulation, if \(S\) is the set of all functions that grow proportionally to \(n\) but with constant of proportionality greater than 1000, and \(T\) is the larger set of linearly-growing functions with constant of proportionality greater than 100, then \(S<T<O(n)\)? Why is this a helpful variation on the existing notation?
@11011110 @0foldcv no, I mean what I said