Lets say we consider the Erdős-Renyi undirected random graph $G(n,p)$ with $V(G) = \{1,2,\cdots,n\}$ and $\displaystyle{P((u,v)\in E(G)) = p} \quad \forall u,v \in V $.

Is there anything we can say about the probability of the $G$ containing a $HAM$ Cycle?

This seems like a helpful quantity to figure out for certain computations

Generating random graphs and seeing what fraction contains a $HAM$ cycle can of-course be done.

What is the rate of increase of $P(G\text{ contains }HAMCYC)$ as $p$ increases?


1 Answer 1


The classical version of this question is for Hamiltonian cycles, but there is probably little difference. I will only consider the version with cycles.

In order for a graph to contain a Hamiltonian cycle, the minimal degree should be at least 2. This is essentially the only obstruction for Hamiltonicity. To state this we need to define the following process:

  • For every pair $\{x,y\} \in [n]$, let $\theta_{xy} \sim U([0,1])$ (independently).
  • Define $G_p = \{ \{x,y\} : \theta_{xy} \leq p \}$.

By construction, $G_p \sim G(n,p)$. As $p$ goes from 0 to 1, more and more edges are "exposed". Bollobás proved the following result:

Let $p_2$ be the minimum $p$ such that the minimal degree of $G_p$ is at least 2.

Let $p_H$ be the minimum $p$ such that $G_p$ is Hamiltonian.

With high probability, $p_2 = p_H$.

The expected degree of a vertex is $p(n-1) \approx pn$. This suggests looking at $p = c/n$. A simple calculation shows that for appropriate values of $c$, the degree of a vertex has distribution roughly Poisson with expectation $c$. In particular, the probability that a vertex has degree less than 2 is roughly $q = e^{-c}(1 + c)$. A further calculation shows that these events for different vertices are roughly independent, and so the distribution of the number of vertices of degree less than 2 is roughly Poisson with expectation $nq$. In particular, the probability that the minimal degree is at least 2 is roughly $e^{-nq}$, for appropriate values of $c$. If $c = \log n + \log \log n + r$ then $q = \frac{\log n + \log \log n + r + 1}{e^rn\log n} \approx \frac{e^{-r}}{n}$, and so $e^{-nq} \approx e^{-e^{-r}}$. This suggests the following result:

The probability that $G(n,p)$ is Hamiltonian for $p = \frac{\log n + \log\log n + r}{n}$ tends to $e^{-e^{-r}}$ for constant $r$.

If $r \to -\infty$ the probability tends to zero, and if $r \to \infty$ it tends to one.

This result indeed holds.

You can see a final project of Brunet for some pointers. For more information, consult any decent textbook on random graphs; Hamiltonicity is a classical topic which would be covered in many of them.

  • $\begingroup$ This is brilliant! Thank you so much. $\endgroup$ Commented Aug 20, 2019 at 8:45

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