AI 中文总结
该研究针对逆星差构造了随机比特高效的有界独立均匀点，通过有限域上的随机向量值多项式实现，所需随机比特远少于独立网格采样，可将差控制在指定精度内。
AI 中文摘要
我们针对逆星差给出了一种随机比特高效的构造。对任意固定的$u\in(0,1)$，满足$k=O(d(1+\log(1+N/d)))$的$k$-wise独立均匀点$\boldsymbol{X}_1,\ldots,\boldsymbol{X}_N$满足蒙特卡洛界$D_N^*(\boldsymbol{X}_1,\ldots,\boldsymbol{X}_N)=O(\sqrt{d/N})$，且概率至少为$u$。因此，$N=O(d\varepsilon^{-2})$且$k=O(d(1+\log\varepsilon^{-1}))$足以使差不超过$\varepsilon$。证明通过链论证分离出所需的有限阶矩并给出显式常数。有限域上的随机向量值多项式利用$O(d^2(1+\log(1+N/d))\log N)$个随机比特实现了网格上所需的有界独立性，而独立网格采样需用$\Theta(dN\log(dN))$个比特。
英文摘要
We give a random-bit-efficient construction for the inverse star discrepancy. For every fixed $u\in(0,1)$, $k$-wise independent uniform points $\boldsymbol{X}_1,\ldots,\boldsymbol{X}_N$ with $k=O(d(1+\log(1+N/d)))$ satisfy the Monte Carlo bound $D_N^*(\boldsymbol{X}_1,\ldots,\boldsymbol{X}_N) =O(\sqrt{d/N})$ with probability at least $u$. Consequently, $N=O(d\varepsilon^{-2})$ and $k=O(d(1+\log\varepsilon^{-1}))$ suffice to attain discrepancy at most $\varepsilon$. The proof isolates the finitely many moments required by a chaining argument and gives explicit constants. A random vector-valued polynomial over a finite field realizes the required bounded independence on a grid using $O(d^2(1+\log(1+N/d))\log N)$ random bits, rather than the $Θ(dN\log(dN))$ bits used by independent grid sampling.
