为给定的运行时 function f(n)=O(n^2)+nlog(n) 找到可能的大 theta？

Question

Suppose that f( n ) = O( n ² ) + n log n . Which of the following are possible?

1. f( n ) = Θ(log n )
2. f( n ) = Θ( n )
3. f( n ) = Θ( n ² )
4. f( n ) = Θ( n ³ )

I am a little confused about the runtime function, because of the included O( n ² ). I believe the answers are 2 and 3, because each of them can be multiplied by a number to reach the O( n ² ). Specifically, Θ( n ² ) can be multiplied by 1 to reach the upper bound O( n ² ), and Θ( n ) can be multiplied by n to reach the upper bound O( n ² ).

Am I correct?

Answer 1

I guess the only correct answer is (3). O(n^2) is any function that grows as fast as n^2 or slower. n log n = O(n^2) , so O(n^2) + n log n is any function that is asymptotically "between" n log n and n^2 . Among all Thetas in your question only the third one fits into these bounds.

Answer 2

f(n) = O(n^2) + nlogn means that there's ag(n) in O(n^2) such that f(n) = g(n) + nlogn.

g(n) in O(n^2) means that |g(n)| < cn^2 for some positive constant c, and all large enough n. The absolute value || bars in the definition allow for the possibility that g(n) is negative.

This means 1, 2, 3 can be the right answer. f(n) can't be negative because it describes a running time, but there's no reason why the O(n^2) term can't be negative.

1. g(n) = logn - nlogn is O(n^2), and g(n) + nlogn = logn.
2. g(n) = n - nlogn is O(n^2), and g(n) + nlogn = n
3. g(n) = n^2 - nlogn is O(n^2), and g(n) + nlogn = n^2
4. is not possible.

(Note the question is phrased in terms of big-theta, but it's possible for f(n) to match the bounds exactly in 1, 2, and 3).

3 is the only solution if you assume the O(n^2) term to be non-negative.

Answer 3

Make sure you go back to the definition of big theta when answering these types of questions. For a function f to be in Θ(g(n)) of a function g a few things have to be true:

1. There exists a constant k ₁ that is greater than 0
2. There exists a constant k ₂ that is greater than 0
3. There exists some start value, n ₀
4. For all n > n ₀ , k ₁ * g(n) <= f(n) <= k ₂ * g(n)

(4) basically means that as f(n) grows, k ₁ * g(n) grows slower and k ₂ * g(n) grows faster for the same n.

Luckily the relationship between these functions is really easy to see when they are plotted beside each other:)

Below we can see all the functions plotted alongside one another:

• blue is f(n)
• green is log n
• purple is n
• black is n^2
• red is n^3

based on this plot, we can immediately discard log n , n , and n^3 because there are no two constants we could bound these functions by that would grow in such a way that it bound f(n) as n grows.

n^2 however looks promising. We just need to find two constants that allow n^2 to bound the growth of f(n) .

Below we can see two such constants:

• blue is f(n)
• green is 1 * n^2
• purple is 4 * n^2

By finding these constants we can say definitively that n^2 + n (log n) is Θ(n^2)