AI 中文总结
本研究肯定回答了ADRS提出的问题，证明可在$2^{n/2+o(n)}$时间内生成任意参数下的单个离散高斯样本，并基于高斯质量比较得到更高效的精确CVP和SVP算法。
AI 中文摘要
Aggarwal、Dadush、Regev和Stephens-Davidowitz（简称ADRS；发表于STOC 2015）能在$2^{n+o(n)}$时间内针对任意参数生成$2^{n/2}$个离散高斯样本，在平滑参数以上时则可在$2^{n/2+o(n)}$时间内完成。他们提出疑问：后者的时间界是否足以生成任意参数下的单个样本。我们对此给出肯定答案：对于由有理基指定的任意秩为$n$的格$L\subseteq\R^n$，以及任意有理$s^2>0$，我们能以期望$2^{n/2+o(n)}$时间和每次执行$2^{n/2+o(n)}$空间，生成一个与$D_{L,s}$统计距离为$\exp(-\Omega(n^3))$的样本。该算法从随机超格中采样，这类超格在所需尺度下以常数概率满足平滑性，并输出落在$L$中的第一个点；高斯质量比较表明，单次ADRS调用生成的$2^{n/2}$个样本中包含$L$中点的概率为逆多项式级。在该高斯质量比较中，$2^{n/2}$这一因子是紧的。对于任意固定的有理$\alpha<1.4697$，同样的比较方法可得到针对满足$\text{dist}(y,L)\le\alpha\lambda_1(L)$的目标的亚$2^n$精确CVP（最近向量问题）算法，无需唯一性假设，还能得到时间复杂度为$2^{0.7315n+o(n)}$的精确SVP（最短向量问题）算法。
英文摘要
Aggarwal, Dadush, Regev, and Stephens-Davidowitz (ADRS; STOC 2015) sample $2^{n/2}$ discrete Gaussians at an arbitrary parameter in $2^{n+o(n)}$ time, and above smoothing in $2^{n/2+o(n)}$ time. They ask whether the latter bound suffices for one sample at an arbitrary parameter. We answer this question affirmatively: for every rank-$n$ lattice $L\subseteq\R^n$ specified by a rational basis and every rational $s^2>0$, we produce one sample from $D_{L,s}$ within statistical distance $\exp(-Ω(n^3))$ in expected $2^{n/2+o(n)}$ time and $2^{n/2+o(n)}$ space on every execution. The algorithm samples from random superlattices that are smooth at the required scale with constant probability and outputs the first point in $L$; a Gaussian-mass comparison shows that the $2^{n/2}$ samples produced by one ADRS call contain a point of $L$ with inverse-polynomial probability. The factor $2^{n/2}$ is tight in this Gaussian-mass comparison. For every fixed rational $α<1.4697$, the same comparison gives a sub-$2^n$ algorithm for exact CVP on targets satisfying $\dist(y,L)\leαλ_1(L)$, without a uniqueness assumption, and an exact-SVP algorithm in $2^{0.7315n+o(n)}$ time.
