Time-complexity of code containing nested loops

Question

i := n;
WHILE i > 1
    FOR j := i to n DO
        X;
    END
    FOR j := 3*i to 3*n DO
        X;
    END
    DEC(i); (* i decrement *)
END

For this pseudocode, I have to calculate a function f: N -> N, depending on n. I'm doing something like this for the first time, so I don't even know if my approach is correct. So, in the first line I have 1 constant time. The while loop runs n-1 times + n times comparison and n-1 constant time for decrement. The first for loop runs n-i+1 times. The second for loop runs 3n-3i+1 times. So (I think) that would be the formula: f(n)=1+((n-1)+n+(n-1))*((n-i+1)+(3n-3i+1)) and that would be f(n)=12n^2 -12ni -2n +8i -3

But now I have n and i variables? How do I get rid of the i ?

Answer 1

Let's suppose that "to n" means "the last value is n-1" (it is not important in terms of time complexity - see the final explanation).

Let's take f(X) = the time complexity to run X .

The total amount of time is:

1 +
0 + 0 +                   // i=n
f(X) + 3*f(X) +           // i=n-1
2*f(X) + 6*f(X) +         // i=n-2
3*f(X) + 9*f(X) +         // i=n-3
...
(n-2)*f(X) + 3*(n-2)*f(X) // i=2
------------------------------
1 + 4*f(X) + 8*f(X) + 12*f(X) + ... + 4*(n-2)*f(X)

And this sum is equal to:

1 + 4*f(X)*(1+2+3+...+n-2) = 1 + 4*f(X)*(n-2)*(n-1)/2 

But the constants don't influence the trend for that function => the result is:

O(2*f(X)*(n-2)(n-1)) = O(f(X)*(n^2))

which is O(n^2) if X is just a simple operation ( O(1) ).

PS :

If you are not sure that the constants doesn't influence the final result, just try to calculate:

          1 + 4*f(X)*(n-2)*(n-1)/2 
   lim  ---------------------------
n->infinite      f(X)*n^2

and you'll obtain a finite number (it means that the numerator and the denominator are similar).