均匀点集间距 II

设最小两点距离为 $M$。

对 $n$ 个点，$M>t$ 等价于所有相邻间距都 $>t$，可得

$$P(M>t)=(1-(n-1)t)^n,\quad 0\le t\le \frac{1}{n-1}.$$

于是

$$\mathbb{E}[M]=\int_0^{1/(n-1)}(1-(n-1)t)^n\,dt=\frac{1}{(n-1)(n+1)}=\frac{1}{n^2-1}.$$

代入 $n=101$ 得

$$\boxed{\mathbb{E}[M]=\frac{1}{10200}}.$$

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英文解析

Set the minimum distance between two points to $M$.

For $n$ points, $M>t$ is equivalent to $>t$ for all adjacent spacing.

$$P(M>t)=(1-(n-1)t)^n,\quad 0\le t\le \frac{1}{n-1}.$$

as a result

$$\mathbb{E}[M]=\int_0^{1/(n-1)}(1-(n-1)t)^n\,dt=\frac{1}{(n-1)(n+1)}=\frac{1}{n^2-1}.$$

Substitute $n=101$ to get

$$\boxed{\mathbb{E}[M]=\frac{1}{10200}}.$$