How can I re-express the below thress functions of algorithm complexity using the big-O notation?

12

2n^2 + 3 n

5 n lg n + 4 n

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- 06-05-2004aliceUsing big-O notation
How can I re-express the below thress functions of algorithm complexity using the big-O notation?

12

2n^2 + 3 n

5 n lg n + 4 n - 06-06-2004laserlight
I havent studied complexity theory formally yet, but my guesses:

12 => O(1)

2n^2 + 3 n => O(n^2)

5 n lg n + 4 n => O(n log n) - 06-06-2004XSquared
I think that the third one is O(n), but I'm not sure.

- 06-06-2004electRONixthis sounds like homework =d
I'm taking a theory of algorithms class ...

yeah the third one should be O(nlgn) since the nlgn part grows faster than the linear part. You always want to take the big O of what grows the fastest.. for example

n + n^3 would be O(n^3)

the first one is constant time so that is O(1) and the second one is quadratic time, O(nē) .. to be more precise there is also a theta function (for example while O(n) means the function is less than some constant times n, theta(n) means the function is just about the same as n. That's probably more detail then you needed so stick to

O(1)

O(nē) and

O(nlgn) - 06-08-2004Zach L.
You can take a quick limit to see which grows faster. That is, (n * lg n) / n = lg n which is unbounded as n grows. So, O(n lg n) > O(n).

- 06-08-2004Speedy5
In general, take the highest order term in the polynomial, strip its coeficient and thats the Big-O efficiency. Remember, logarithms are exponents (inverse though) so this also applies for them.

- 07-05-2004alice
for the third '5 n lg n + 4 n'

since nlgn<n2<n3

f(n)=5 n lg n + 4 n < 5(n3)+4n

=19n

am I correct? - 07-05-2004laserlight
hmm... I dont understand your reasoning.

f(n)*is*5n lg n + 4n - 07-05-2004Zach L.
n3 should be n^3... That is, n cubed, not three times n