Question

for (int i = 1; i < N; i *= 2) { ... }

Things like that are the signatures of logarithmic complexity.

But how get log(N)?

Could you give mathematical evidence?

Answer 1

Useful reference on algorithmic complexity: http://en.wikipedia.org/wiki/Big_O_notation

On the nth iteration,

i = 2^n

We know that it iterates until i >= N

Therefore,

i < N 

Now,

2^n = i < N

N > 2^n

log2 N > log2 (2^n)

log2 N > n

We know it iterates n times, being less than log2 N.

Thus # iterations < log2 N, or # iterations is O(log N)

QED. Logarithmic complexity.