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garywoot

Thu Sep 26, 2013 1:41 am


Big Oh quesiton

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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!


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Insectoid

Thu Sep 26, 2013 6:03 am


RE:Big Oh quesiton

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


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garywoot

Fri Sep 27, 2013 7:39 pm


RE:Big Oh quesiton

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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 -_-;;

http://puu.sh/4BTVI.png


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Dreadnought

Fri Sep 27, 2013 10:50 pm


Re: Big Oh quesiton

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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)).


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garywoot

Sat Sep 28, 2013 12:47 am


Re: Big Oh quesiton

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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.