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?

I think this would be better suited in [computerScience.se](http://cs.stackexchange.com/), maybe? — Natan Streppel, Feb 15 '15 at 00:51
The inner loop is `O(n*f(x))` where `f(x)` is the complexity of executing `x`. The outer loop is `O(n)` which makes it: `O(n^2 * f(x))` — Nir Alfasi, Feb 15 '15 at 00:51
O(n*f(x))? Could you explain, how did you come to this conclusion? Since I have two for-loops. And what about the i? — Arthur, Feb 15 '15 at 01:11
You have two inner for-loops, each of which runs in `O(n*f(x))`: which is `n` iterations and the "work" that each iteration requires is a function of `x`: `f(x)`. Since both inner for-loops have the same complexity, we sum them to `O(n*f(x))`, and since the outer-loop runs `n` times we get total of: ` `O(n^2*f(x))` — Nir Alfasi, Feb 15 '15 at 04:27

ROMANIA_engineer · Accepted Answer · 2015-02-15T01:51:38.097

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)).

P.S.:

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).

Time-complexity of code containing nested loops

1 Answers1