复杂度logn和log之间的差异（sqrt（n））

Question

considering

log(sqrt(n)) = (1/2)log(n)

And if for asymptotic analysis we don't consider the constant terms so, is O(log(sqrt(n))) is as good as O(log(n))?

As per my understanding log(sqrt(n)) will grow slowly in comparison to log(n) if we increase the size of n. But I am not able understand the glitch in moving power of (1/2) at front? Is it just this that factor 1/2 only slows down the rate?

consider the case when we have log(n*n) represented as 2log(n) , and log(n)?

Answer 1

它渐近是一样的：

O(log(sqrt(n))) = O(log(n^1/2)) = O(1/2 log(n)) = O(log(n))

Answer 2

你是对的，O（log（sqrt（n）））与你的问题中给出的推理与O（log（n））相同。

Answer 3

Time(A) = log n

Time(B) = log sqrt(n) = log n^(1/2) = 1/2 log n

Asymptotically the same

O(Time(A)) = O(log n)

O(Time(B)) = O(1/2 log n) = O(log n)

O(Time(A)) = O(Time(B))

Insignificantly different

Time(A) = 1   * log n

Time(B) = 1/2 * log n

Time(A) > Time(B)

Time(A) = 2 * Time(B)

Conclusion

log n = 2 log sqrt(n)

Although the difference between log n and log sqrt(n) is in insignificant, log n will always take double the amount of time log sqrt(n) takes

Visual

Answer 4

The big-O notation ignores any constant multiplier.

O(500000.N) is O(N) and is O(0.00001.N) .

For the same reason, O(Log(Sqrt(N))) is O(1/2.Log(N)) is O(Log(N)) , and that in any base.

The big-O notation is not about the speed of your program, it is about the growth of the running time as N increases.