I am trying to wrap my head around these asymptotic notations. Given $f(n)$ and $g(n)$, one can write $f(n) = \Omega(g(n))$ as shorthand for $f(n) \geq c*g(n), n\geq n_0$. But what happens when $n<n_0$. What can we then say about $f(n)$ and $g(n)$? Do we just assume that $f(n)$ is a constant and the relation still works? Or does the question not even make sense to ask?

  • $\begingroup$ What is $n_0$ ? May be, trying to answer that will help you. $\endgroup$
    – babou
    Commented Jun 15, 2015 at 13:13

1 Answer 1


Nothing. If all you know about two functions is their $O$/$\Omega$ relations, you know nothing about how they relate to each other on any finite prefix.

Yes, this makes asymptotics of, say, runtime functions absolutely useless in terms of questions like, "which algorithm should I use in scenario X?" On first sight, that is.

Luckily, we usually do analyse with more detail but throw them away when formulating the theorem with coarse notation like Landau symbols. Many results about "practical" algorithms implicitly say, "and this behaviour can be observed for reasonable $n$".

Sometimes you get better theorems which use Landau notation for "error terms" only; authors like Knuth do not say "algorithm A runs in time $O(n \log n)$", but rather "algorithm A takes $2n\log n - 2n + O(1)$ comparisons". This does not completely do away with your issue, but mitigates it to some extent. If you know that that $O(1)$ stands for $\leq 1377$, you can make informed decisions based on that result.

Some of our reference material discusses facets of this issue.


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