Why is ω(n) smaller than O(n)?
I know what is little omega (for example, n = ω(log n)), but I can't understand why ω(n) is smaller than O(n).
Algorithmic complexity has a mathematic definition.
If f and g are two functions, f = O(g) if you can find two constants c (> 0) and n such as f(x) < c * g(x) for every x > n.
For Ω, it is the opposite: you can find constants such as f(x) > c * g(x).
f = Θ(g) if there are three constants c, d and n such as c * g(x) < f(x) < d * g(x) for every x > n.
Then, O means your function is dominated, Θ your function is equivalent to the other function, Ω your function has a lower limit.
So, when you are using Θ, your approximation is better for you are "wrapping" your function between two edges ; whereas O only set a maximum. Ditto for Ω (minimum).
To sum up:
O(n): in worst situations, your algorithm has a complexity of nΩ(n): in best case, your algorithm has a complexity of nΘ(n): in every situation, your algorithm has a complexity of nTo conclude, your assumption is wrong: it is Θ, not Ω. As you may know, n > log(n) when n has a huge value. Then, it is logic to say n = Θ(log(n)), according to previous definitions.