Calculate the complexity of a nested for loop

问题

I want to calculate the complexity of this nested for loop:

s = 0;
for(i=1; i<=n; i*=2)
   for(j=1; j<=i; j*=2)
      s++;

What strategy do I use to find the Big O complexity of this piece of code?

回答1:

The outer loop marches through 1, 2, 4, 8, ... n, which takes O(lg n) steps because you can only double one O(lg n) times until you hit n.

The inner loop does the same. It only goes up to i, but in the final iteration of the outer loop, i reaches its maximum value which is again n, so that's also O(lg n).

Putting this together gives an upper bound of O((lg n)²), which is commonly abbreviated O(lg² n).

回答2:

Strategy for getting the answer yourself

Plug in different values of n into the equation, and make a chart of how many times the innermost part of the loop runs:

s = 0;
for(i=1; i<=n; i*=2)
  for(j=1; j<=i; j*=2)
    s++;

Something like this:

n     num_times_inner_loop_part_runs
1     1
2     3
3     3
4     6
5     6
6     6
7     6
8     10
9     10
...
15    10
16    15
...
31    15
32    21

You can get these data points with a program like this:

int n = 9;  //change this n
int counter = 0;
for(i=1; i<=n; i*=2){
  for(j=1; j<=i; j*=2){
    s++;
    counter++;
  }
}
cout << "counter is: " <<  counter << endl;

Plot the num_times_inner_loop_part runs on an X/Y coordinate plane and you'll see a curve.

Name the curve that fits closest. In this case, it is X = (log(Y)^2)

If you plot your data and X = (log(Y)^2), you'll find they should overlap each other.

Therefore, the complexity of this function is O((log(n))^2) which is an improvement over O(n)