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Studying for a test and getting this question:

Comparing two algorithms with asymptotic complexities O(n) and O(n + log(n)), 
which one of the following is true?

A) O(n + log(n)) dominates O(n)
B) O(n) dominates O(n + log(n))
C) Neither algorithm dominates the other.

O(n) dominates log(n) correct? So in this do we just take o(n) from both and deduce neither dominate?

Deekor
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[C] is true, because of the summation property of Big-O

Summation 
O(f(n)) + O(g(n)) -> O(max(f(n), g(n))) 
For example: O(n^2) + O(n) = O(n^2)

In Big-O, you only care about the largest-growing function and ignore all the other additives.

Edit: originally I put [A] as an answer, I just didn't put much attention to all the options and misinterpreted the [A] option. Here is more formal proof

O(n) ~ O(n + log(n))      <=>
O(n) ~ O(n) + O(log(n))   <=>
O(n) ~ O(n).
Community
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oleksii
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Yes, that's correct. If runtime is the sum of several runtimes, by order of magnitude, the largest order of magnitude dominates.

La-comadreja
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Assuming that big-O notation is used in the sense of asymptotic tight bound, which really should be denoted with a big-Theta, then I would answer C), because Theta(n) = Theta(n + log(n)). (Because log(n) is dominated by n).

If I am formally (mathematically) correct, then I would say that none of these answers is correct, because O(n) and O(n+log(n)) only give upper bounds, but not lower bounds on the asymptotic behaviour:

Let f(n) in O(n) and g(n) in O(n + log(n)). Then there are the following contra examples:

For A): Let f(n) = n in O(n) and g(n) = 1 in O(n + log(n)). Then g(n) does not dominate f(n).

for B): Let f(n) = 1 in O(n) and g(n) = n in O(n + log(n)). Then f(n) does not dominate g(n).

for C): Let f(n) = 1 in O(n) and g(n) = n in O(n + log(n)). Then g(n) does dominate f(n).

As this would be a very tricky question, I assume that you use the more common sloppy definition, which would give the answer C). (But you might want to check your definitions for big-O).

If my answer confuses you, then you probably didn't use the formal definition and you should probably ignore my answer...