How do I evaluate alrorithms using theta, big O etc. ?

[XodoX]

Homework Statement

Loglog n and 2^log*2

Homework Equations

O(g(n)) (big O)

Ω(g(n))

ω(g(n))

The Attempt at a Solution

Theta is worse case and Big O gives you the upper bound of the function f(n) etc. Question is, how do I evaluate those 2 algorithms using them? Do I just plug in numbers or what? I don't quite get it. I want to compare the 2 algorithms by using those parameters, so I need to get the values and compare them.

 

[lippyka]

How do I go about it? O(log log n) and O(2^log*2) The first algorithm is O(log log n). This means that the time complexity of the algorithm is in the order of log log n, which is a polynomial function. The second algorithm is O(2^log*2), which means that the time complexity of the algorithm is in the order of 2^log*2, which is an exponential function. Ω(log log n) and Ω(2^log*2) The first algorithm is Ω(log log n). This means that the time complexity of the algorithm is in the order of log log n, which is a polynomial function. The second algorithm is Ω(2^log*2), which means that the time complexity of the algorithm is in the order of 2^log*2, which is an exponential function. ω(log log n) and ω(2^log*2) The first algorithm is ω(log log n). This means that the time complexity of the algorithm is in the order of log log n, which is a polynomial function. The second algorithm is ω(2^log*2), which means that the time complexity of the algorithm is in the order of 2^log*2, which is an exponential function. In terms of comparison, the first algorithm will always be slower than the second algorithm as the exponential function grows faster than the polynomial function.