What is average number of cycles in an undirected ordered graph of size $n$? I've tried finding out sum of number of cycles in all sorts of a graph of size n but I couldn't find that out.

  • 1
    $\begingroup$ What specifically have you tried resp where have you looked? $\endgroup$
    – Raphael
    Apr 1, 2014 at 11:45

1 Answer 1


Hint: Suppose that there are $Z_m$ cycles of length $m$, and that $p_m$ is the probability that a specific cycle of length $m$ is in the graph. Then the expected number of cycles is $$ \sum_{m=3}^n Z_m p_m $$ due to linearity of expectation.

Here is the entire calculation. In the $G(n,p)$ model, in which each edge is present with probability $p$ independently, clearly $p_m = p^m$. The number of ordered $m$-tuples of points is $n!/(n-m)!$. Each cycle of length $m$ is described by $2m$ such tuples: choose a starting point and an orientation. Therefore the total number of cycles of length $m$ is $n!/2m(n-m)!$, and the expected number of cycles is $$ E = \sum_{m=3}^n \frac{n!}{2m(n-m)!} p^m. $$

In order to estimate this number, let's find the largest term. The ratio of two successive terms is roughly $$ \frac{p^{m+1}}{(n-m-1)!} \frac{(n-m)!}{p^m} = (n-m)p. $$ When $p$ is constant, we see that the most important term corresponds to $m = n$, namely $n! p^n/(2n)$, and we deduce that for large enough $n$, the expected number of cycles satisfies $$ 2(n-1)! p^n \leq E \leq 2n! p^n. $$


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