Assume that $G\in G(n,p)$; if $p=\frac{\ln n +\ln \ln n +c(n)}{n}$, the following fact is well known:

\begin{eqnarray} Pr [G\mbox{ has a Hamiltonian cycle}]= \begin{cases} 1 & (c(n)\rightarrow \infty) \\ 0 & (c(n)\rightarrow - \infty) \\ e^{-e^{-c}} & (c(n)\rightarrow c) \end{cases} \end{eqnarray}

That is, a probability $p=\ln n (1+o(1)) / n$ is enough to get an Hamiltonian cycle.

For a given parameter $k\in\{1,\ldots,n\}$, I'm wondering what is the critical probability $p$ for which it's likely that the graph will have a $k$-path.

It seems that $p=\ln k(1+o(1)) / k$ is too high, as we only need one simple path of length $k$ in an $n$-nodes graph.


1 Answer 1


When $k$ is a constant, the threshold probability for a path of length $k$ is $p \sim \dfrac{1}{n^{1+1/k}}$. The same is true for any tree with $k$ edges. The general result is that the threshold probability for the appearance of a subgraph $H$ is $max_{J \subseteq H} n^{-|V(J)|/E(J)|}$, which happens to be the same as $n^{-|V(H)|/E|(H)|}$ for balanced graphs $H$ (such as trees, complete graphs, cycles).

The situation is different when $k$ grows with $n$. I don't know what the threshold is, but looking at the Hamiltonian path case, I'd guess that an extra $\log n$ factor suffices.


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