< 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(...)?

@11011110 @0foldcv (your formulation is true, and stronger, but not equivalent in meaning)

@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 I was using existing notation. Strict subset is the natural notion of \(<\) for sets. This isn't a variation.

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