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I have been thinking over this problem for a few days now and am hung up on calculating the number of times the second nested for-loop will run. I believe that I have the correct formula for determining the running time for the other two for-loops, but this third one has me hung up. I have the first loop running n-1 times. The equation to determine the number of times loop #2 runs is; The summation of 1 to n-1. If anyone could help me understand how to find the number of times loop #3 runs it would be greatly appreciated.

    for ( int i=1; i<=n-1; i++ ) {
        for ( int j=i+1; j<=n; j++ ) {
            for ( int k=1; k<=j; k++ ) {
            }
        }
    }
John
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    Why don't you just try it? Add `x++` to the inside of the loop and try with various `n`. Also, this question is tagged [tag:big-o], why? Computing exact counts has nothing to do with Big-O. – svick Sep 10 '11 at 23:17
  • That is what I have in my little program. I left out a few lines to make code more readable. Adding x++ will show me how many times the loop will run, but I was looking for a mathematical expression using n that will give me the answer. – John Sep 10 '11 at 23:20

1 Answers1

2

The third loop runs C times:

C = Sum( Sum ( Sum ( 1 , k = 1 .. j ) , j = i+1 .. n ) , i = 1 .. n-1 )
  = Sum( Sum (           j            , j = i+1 .. n ) , i = 1 .. n-1 )
  = 2 + 3 + 4 + ... + n
      + 3 + 4 + ... + n
    ...
                    + n
  = 2*1 + 3*2 + 4*3 + 5*4 + ... + n*(n-1)
  = (1*1 + 1) + (2*2 + 2) + (3*3 + 3) + ... + ((n-1)*(n-1) + n-1)
  = (1^2 + 2^2 + ... (n-1)^2) + (1 + 2 + 3 + ... + (n-1))
  = (n-1)*n*(2*n-1)/6 + (n-1)*n/2 
  = (n-1)*n*(2*n+2)/6
  = O(n^3)

Here I used the formulas: 

1^2 + 2^2 + ... + m^2 = m*(m+1)*(2*m+1)/6

and

1 + 2 + ... + m = m*(m+1)/2
Petar Ivanov
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