Question

The code fragment I am to analyse is below:

int sum = 0;
for (int i = 0; i < n; i++)
   for (int j = 0; j < i * i; j++)
      for (int k = 0; k < j; k++)
         sum++;

I know that the first loop is O(n) but that’s about as far as I’ve gotten. I think that the second loop may be O(n^2) but the more I think about it the less sense it makes. Any guidance would be much appreciated.

Answer 1

The first loop executes n times. Each time, the value i grows. For each such i, the second loop executes i*i times. That means the second loop executes 1*1 + 2*2 + 3*3 + ... + n*n times.

This is a summation of squares, and the formula for this is well-known. Hence we have the second loop executing (n(1 + n)(1 + 2 n))/6 times.

Thus, we know that in total there will be (n(1 + n)(1 + 2 n))/6 values of j, and that for each of these the third loop will execute 1 + 2 + ... + j = j(j+1)/2 times. Actually calculating how many times the third loop executes in total would be very difficult. Luckily, all you really need is a least upper bound for the order of the function.

You know that for each of the (n(1 + n)(1 + 2 n))/6 values of j, the third loop will execute less than n(n+1)/2 times. Therefore you can say that the operation sum++ will execute less than [(n(1 + n)(1 + 2 n))/6] * [n(n+1)/2] times. After some quick mental math, that amounts to a polynomial of maximal degree 5, therefore your program is O(n^5).