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Assume I have a list of functions, for example

$\qquad n^{\log \log(n)}, 2^n, n!, n^3, n \ln n, \dots$

How do I sort them asymptotically, i.e. after the relation defined by

$\qquad f \leq_O g \iff f \in O(g)$,

assuming they are indeed pairwise comparable (see also here)? Using the definition of $O$ seems awkward, and it is often hard to prove the existence of suitable constants $c$ and $n_0$.

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Since the OP never came back, I'm removing the localised stuff and make a reference question out of this. – Raphael Feb 17 '13 at 9:40
up vote 28 down vote accepted

If you want rigorous proof, the following lemma is often useful resp. more handy than the definitions.

If $c = \lim_{n\to\infty} \frac{f(n)}{g(n)}$ exists, then

  • $c=0 \qquad \ \,\iff f \in o(g)$,
  • $c \in (0,\infty) \iff f \in \Theta(g)$ and
  • $c=\infty \quad \ \ \ \iff f \in \omega(g)$.

With this, you should be able to order most of the functions coming up in algorithm analysis¹. As an exercise, prove it!

Of course you have to be able to calculate the limits accordingly. Some useful tricks to break complicated functions down to basic ones are:

  • Express both functions as $e^{\dots}$ and compare the exponents; if their ratio tends to $0$ or $\infty$, so does the original quotient.
  • More generally: if you have a convex, continuously differentiable and strictly increasing function $h$ so that you can re-write your quotient as

    $\qquad \displaystyle \frac{f(n)}{g(n)} = \frac{h(f^*(n))}{h(g^*(n))}$,

    with $g^* \in \Omega(1)$ and

    $\qquad \displaystyle \lim_{n \to \infty} \frac{f^*(n)}{g^*(n)} = \infty$,


    $\qquad \displaystyle \lim_{n \to \infty} \frac{f(n)}{g(n)} = \infty$.

    See here for a rigorous proof of this rule (in German).

  • Consider continuations of your functions over the reals. You can now use L'Hôpital's rule; be mindful of its conditions²!

  • Have a look at the discrete equivalent, Stolz–Cesàro.
  • When factorials pop up, use Stirling's formula:

    $\qquad \displaystyle n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$

It is also useful to keep a pool of basic relations you prove once and use often, such as:

  • logarithms grow slower than polynomials, i.e.

    $\qquad\displaystyle (\log n)^\alpha \in o(n^\beta)$ for all $\alpha, \beta > 0$.

  • order of polynomials:

    $\qquad\displaystyle n^\alpha \in o(n^\beta)$ for all $\alpha < \beta$.

  • polynomials grow slower than exponentials:

    $\qquad\displaystyle n^\alpha \in o(c^n)$ for all $\alpha$ and $c > 1$.

It can happen that above lemma is not applicable because the limit does not exist (e.g. when functions oscillate). In this case, consider the following characterisation of Landau classes using limes superior/inferior:

With $c_s := \limsup_{n \to \infty} \frac{f(n)}{g(n)}$ we have

  • $0 \leq c_s < \infty \iff f \in O(g)$ and
  • $c_s = 0 \iff f \in o(g)$.

With $c_i := \liminf_{n \to \infty} \frac{f(n)}{g(n)}$ we have

  • $0 < c_i \leq \infty \iff f \in \Omega(g)$ and
  • $c_i = \infty \iff f \in \omega(g)$.


  • $0 < c_i,c_s < \infty \iff f \in \Theta(g) \iff g \in \Theta(f)$ and
  • $ c_i = c_s = 1 \iff f \sim g$.

Check here and here if you are confused by my notation.

¹ Nota bene: My colleague wrote a Mathematica function that does this successfully for many functions, so the lemma really reduces the task to mechanical computation.

² See also here.

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Is the Mathematica function available somewhere? – Juho Mar 11 '15 at 7:42
@Juho Not publicly, afaik, but it's elementary to write yourself; compute Limit[f[n]/g[n], n -> Infinity] and perform a case distinction. – Raphael Mar 11 '15 at 7:44

Another tip: sometimes applying a monotone function (like log or exp) to the functions makes things clearer.

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This should be done carefully: $2n\in O(n)$, but $2^{2n}\notin O(2^n)$. – Shaull Feb 17 '13 at 11:35
Seconded. The "apply monotone function" thing seems to be some kind of folklore which does not work in general. We have been working on sufficient criteria and have been come up with what I posted in the latest revision of my answer. – Raphael Nov 8 '13 at 17:15

Skiena provides a sorted list of the dominance relations between the most common functions in his book, The Algorithm Design Manual:

$$n!\gg c^n \gg n^3 \gg n^2 \gg n^{1+\epsilon} \gg n \lg n \gg n \gg n^{1/2}$$ $$ \gg \lg^2n \gg \lg n \gg \frac{\lg n}{\lg\lg n} \gg \lg\lg n \gg \alpha(n) \gg 1$$

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That's an oddly specific list. Many of the relations (whatever $\gg$ means exactly) can be summarised to a handful of more general lemmata. – Raphael Nov 8 '13 at 17:16
It's his notation for a dominance relation. – Robert S. Barnes Nov 10 '13 at 10:14

Tip: draw graphs of these functions using something like Wolfram Alpha to get a feeling for how they grow. Note that this is not very precise, but if you try it for sufficiently large numbers, you should see the comparative patterns of growth. This of course is no substitute for a proof.

E.g., try: plot log(log(n)) from 1 to 10000 to see an individual graph or plot log(log(n)) and plot log(n) from 1 to 10000 to see a comparison.

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Should we really recommend vodoo? – Raphael Mar 27 '12 at 17:13
+1 for suggesting to draw graphs of the functions, although the linked graphs are rather confusing unless you know what they mean. – Tsuyoshi Ito Mar 27 '12 at 17:35

I suggest proceeding according to the definitions of various notations. Start with some arbitrary pair of expressions, and determine the order of those, as outlined below. Then, for each additional element, find its position in the sorted list using binary search and comparing as below. So, for example, let's sort $n^{\log\log n}$ and $2^n$, the first two functions of n, to get the list started.

We use the property that $n = 2^{\log n}$ to rewrite the first expression as $n^{\log\log n} = (2^{\log n})^{\log\log n} = 2^{\log n\log\log n}$. We could then proceed to use the definition to show that $n^{\log\log n} = 2^{\log n\log\log n} \in o(2^n)$, since for any constant $c > 0$, there is an $n_0$ such that for $n \geq n_0$, $c(n^{\log\log n}) = c(2^{\log n\log\log n}) < 2^n$.

Next, we try $3^n$. We compare it to $2^n$, the largest element we have placed so far. Since $3^n = (2^{\log 3})^n = 2^{n\log3}$, we similarly show that $2^n \in o(3^n) = o(2^{n \log 3})$.


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