Computational complexity of a piece of code

Question

I have got a program, and trying to compute its complexity. I want to be sure i am not mistaken

for(int i=4; i<=n; i=i*4)
{
    cout<<"counter for first loop: "<<++count1<<endl;
    for(int j=i;j>=0;j=j-4)
    {
        cout<<"counter for second loop: "<<++count2<<endl;
        for(int k=0;k<=n;k++)
        {
            cout<<"counter for third loop: "<<++count3<<endl;
        }
    }
}

Here, the complexity of third loop is O(n), then together with the second loop, the complexity becomes O(n.log₄i), and the complexity of whole program is O(n.(log₄i)²). Am i right in my answer? Thanks

score 2 · Accepted Answer · answered Feb 26 '13 at 11:16

The complexity of the inner most loop is O(n). The complexity of the middle one is O(i/4), which in turn is O(i). The complexity of the outer most loop is O(log₄n). There for the total complexity of the code is O(n.i.log₄n) which is equal to O (n.(log₄n)²).

Mohamed Ennahdi El Idrissi · Answer 2 · 2014-04-13T21:29:48.183

0

You can proceed formally like the following:

enter image description here

Executing this fragment:

sum = 0;
for( i = 4 ;  i <= n;  i = i * 4 ) {
    for( j = i ; j >= 0 ; j = j - 4 ) {
        for( k = 0 ; k <= n ; k ++ ) {
            sum ++;
        }
    }
}

We obtain:

enter image description here