Big Oh quesiton

[garywoot]

Hi guys,

I have a quick question about the Big Oh for this algorithm:

Algorithm alg
x = 0
for i=0 to n^2
......for j=0 to i
..........x = 3+i
return x


I only recently learned this so I'm not sure if what I'm doing is correct, but I arrived at the answer that the time complexity in Big Oh is O(n^3).

Since outer loop excuses ~n^2 times, inner loop executes ~i times (minus constants), then inner loop total executes 1+2+3+...+n^2 times, therefore sum of k=1 to n of k^2 is n(n+1)(2n+1)/6 = O(n^3)

could someone please tell me if what I'm doing is correct? Thanks a lot!

[Insectoid]

Yeah, you're correct. Complexity of nested loops is just the product of the complexity of each loop. n^2*n = O(N*3)

[garywoot]

answer is actually O(n^4)

nbd only close to failed the assignment on all the other time complexity questions

this is why i never just multiply loops with loops -_-;;

[Dreadnought]

garywoot wrote:

Since outer loop excuses ~n^2 times, inner loop executes ~i times (minus constants), then inner loop total executes 1+2+3+...+n^2 times, therefore sum of k=1 to n of k^2 is n(n+1)(2n+1)/6 = O(n^3)

You had the right idea, I think you just got confused in your logic. You want the sum from k=1 to n^2 of k (which is O(n^4)), not the sum from k=1 to n of k^2 (which is O(n^3)).

[garywoot]

Dreadnought @ Fri Sep 27, 2013 10:50 pm wrote:

garywoot wrote:

Since outer loop excuses ~n^2 times, inner loop executes ~i times (minus constants), then inner loop total executes 1+2+3+...+n^2 times, therefore sum of k=1 to n of k^2 is n(n+1)(2n+1)/6 = O(n^3)

You had the right idea, I think you just got confused in your logic. You want the sum from k=1 to n^2 of k (which is O(n^4)), not the sum from k=1 to n of k^2 (which is O(n^3)).

Ahh makes sense.