从几何图或排序中学习随机点
Learning random points from geometric graphs or orderings
摘要：
设$n$的随机点$X_v$为$v \in V$，独立均匀地分布在面积$n$的正方形$\left[-\sqrt{n}/2,\sqrt{n}/2\right]^2$上.我们没有看到这些点，但通过以下两种方式之一了解它们。假设我们首先给定相应的随机几何图$G$，其中当欧氏距离$d_E(X_u,X_v)$至多为$r$时，不同的顶点$u$和$v$相邻。如果阈值距离$r$满足$n^{3/14} \ll r \ll n^{1/2}$，则以下以高概率成立。给定图$G$（没有任何几何信息），在多项式时间内，我们可以近似地重建隐藏的嵌入，在这个意义上，“直到对称”，对于每个顶点$v$，我们找到一个点在距离$X_v$的$r$之内;也就是说，我们找到一个“位移”最多关于$r$的嵌入。现在假设，不是给出图$G$，而是对于每个顶点$v$，通过增加与$v$的欧几里德距离来给出其他顶点的排序。然后，以高概率，在多项式时间内，我们可以找到具有小得多的位移误差$O(\sqrt{\log n})$的嵌入。
英文摘要：
Suppose that there is a family of $n$ random points $X_v$ for $v \in V$, independently and uniformly distributed in the square $\left[-\sqrt{n}/2,\sqrt{n}/2\right]^2$ of area $n$. We do not see these points, but learn about them in one of the following two ways. Suppose first that we are given the corresponding random geometric graph $G$, where distinct vertices $u$ and $v$ are adjacent when the Euclidean distance $d_E(X_u,X_v)$ is at most $r$. If the threshold distance $r$ satisfies $n^{3/14} \ll r \ll n^{1/2}$, then the following holds with high probability. Given the graph $G$ (without any geometric information), in polynomial time we can approximately reconstruct the hidden embedding, in the sense that, `up to symmetries', for each vertex $v$ we find a point within distance about $r$ of $X_v$; that is, we find an embedding with `displacement' at most about $r$. Now suppose that, instead of being given the graph $G$, we are given, for each vertex $v$, the ordering of the other vertices by increasing Euclidean distance from $v$. Then, with high probability, in polynomial time we can find an embedding with the much smaller displacement error $O(\sqrt{\log n})$.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
