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Let $diam(G)$ denote the diameter of a graph $G$; i.e.

$$\mathrm{diam}(G) := \max_{ { i,j} \in \binom{[n]}{2}} d(i, j)$$

, where $d(i,j)$ denotes the length of the shortest path between the vertices $i$ and $j$. Show that for any $h(n) = \omega(1)$ and (at the same time) $h(n) = \mathcal{o}(\log(n))$ holds

$$\mathbb{P}\bigg[ \mathrm{diam} \biggl(G \biggl(n,\sqrt{\frac{2\log(n)+h}{n}} \biggr)\biggr) > 2 \biggr] = \mathcal{o}(1).$$

So by the definiton of small-Oh we need to show that

$$\mathbb{P}\bigg[\mathrm{diam} \biggl(G \biggl(n,\sqrt{\frac{2\log(n)+h}{n}} \biggr)\biggr) > 2 \biggr] \underset{n \rightarrow \infty}{\rightarrow} 0.$$

However, I do not see how to evaluate the probability for the size of $diam$. Could you please give me a hint?

Edit:

According to this question the probability, say $f(n)$, for $G(n,p)$ to be disconnected can be recursively described as:

$$f(n) = 1-\sum\limits_{i=1}^{n-1}f(i){n-1 \choose i-1}(1-p)^{i(n-i)}$$

Furthermore, I think the probability of two vertices $i$ and $j$ to have exactly one common neighbour should be

$$\frac{2\log(n)+h}{n} + \sum_{k=0}^{n-3} \biggl( 1- \sqrt{\frac{2\log(n)+h}{n}} \biggr).$$

3nondatur
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    You're really estimating the probability that every pair of vertices is adjacent or has a common neighbor. You want this probability to be $1-o(1)$. Can you figure out what the probability is for a fixed pair, as a function of $p$? Now try to show that with high probability every pair is good. That's probably what the mysterious $h$ in your expression is for. – Jamie Radcliffe Jan 05 '23 at 00:10
  • Thanks for your comment, I made an edit to my question. – 3nondatur Jan 05 '23 at 18:03
  • Did you manage to do it, or should I post a proof? (Remember to tag me if you're replying to my comment) – Bruno Andrades Dec 17 '24 at 01:59

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