问题
I understand how to calculate a function's complexity for the most part. The same goes for determining the order of growth for a mathematical function. [I probably don't understand it as much as I think I do, which is why I'm probably asking this.] For instance:
an^3 + bn^2 + cn + d could be written as O(n^3) in big-o notation, since for large enough nthe values of the term bn^2 + cn + d are insignificant in comparison to an^3 (the constant coefficients a, b, c and d are left out as well, as their contribution to the value become insignificant, too).
What I don't understand is, how does this work when the leading term is involved in some sort of division? For instance:
a/n^3 + bn^2 or n^3/a + bn^2
Let n=100, a=1000 and b=10 for the former formula, then we have
n^3/a = 100^3/1000 = 1000 and bn^2 = 10*100^2 = 100,000
or even more dramatic for the latter - in this case the leading term is not only growing slowly as above, but it's also shrinking, isn't it?:
a/n^3 = 1000/100^3 = 0.001 and bn^2 = 100,000 as above.
In both cases the second term contributes much more, so isn't it n^2that actually determines the order of growth?
It gets even more complicated (for me, at least) when the leading term is followed by a subtraction (a/n^3 - bn^2) or when the second term is also a division (n^3/a + n^2/b) or when both are divisions but in mixed order (a/n^3 + n^2/b), etc.
The list seems endless, so my general question is, how to understand and handle formulas that involve division (and subtraction) in order to determine the order of growth for a given function?
回答1:
A division is just a multiplication by the multiplicative inverse, so n^3/a == n^3 * a^-1, and you can handle it the same way as any other coefficient.
With regards to substraction a*n^3 - b*n^2 <= a*n^3, so it is also in O(n^3). Also, a*n^3 - b*n^2 >= a/2 * n^3 for large enough values of n, and it is also in Omega(n^3). A more detailed explanation about substraction can be found in: Algorithm complexity when faced with substraction in value
big O notation is generally used for increasing (don't have to be monotonically) functions, and a decreasing function such as a/n is not a good fit for it, though O(1/n) seems to be still perfectly defined, AFAIK, and it is a subset of O(1) (unless you take into account only discrete functions). However, this has very little value for algorithm's analysis, as a complexity of an algorithm cannot really shrink..
回答2:
There's a very simple rule for the type of questions you posted.
Suppose you're trying to find the order of growth of f(n), and you find some simple function g(n) such that
lim {n -> inf} f(n) / g(n) = k
where k is a positive finite constant. Then
f(n) = Theta(g(n))
(It's easy to see this from the calculus definitions.)
Now let's see how this applies to your examples:
lim {n -> inf} (a/n^3 + bn^2) / n^2 = b
so it's Theta(n^2).
lim {n -> inf} (a n^3 - bn^2) / n^3 = a
so it's Theta(n^2).
(of course, assuming a and b are positive.)
来源：https://stackoverflow.com/questions/29983533/complexity-determining-the-order-of-growth