We revisit the sample and computational complexity of completing a rank-1 tensor in $\otimes_{i=1}^{N} \mathbb{R}^{d}$, given a uniformly sampled subset of its entries. We present a characterization of the problem (i.e. nonzero entries) which admits an algorithm amounting to Gauss-Jordan on a pair of random linear systems. For example, when $N = \Theta(1)$, we prove it uses no more than $m = O(d^2 \log d)$ samples and runs in $O(md^2)$ time. Moreover, we show any algorithm requires $\Omega(d\log d)$ samples. By contrast, existing upper bounds on the sample complexity are at least as large as $d^{1.5} \mu^{\Omega(1)} \log^{\Omega(1)} d$, where $\mu$ can be $\Theta(d)$ in the worst case. Prior work obtained these looser guarantees in higher rank versions of our problem, and tend to involve more complicated algorithms.

翻译：我们重新研究了在给定其条目均匀采样子集的条件下，补全 $\otimes_{i=1}^{N} \mathbb{R}^{d}$ 中秩-1张量所需的样本与计算复杂度。我们提出了该问题（即非零条目）的一个特征刻画，该刻画允许一种算法，其本质等同于对一对随机线性系统执行高斯-若尔当消元法。例如，当 $N = \Theta(1)$ 时，我们证明该算法使用的样本数不超过 $m = O(d^2 \log d)$，且运行时间为 $O(md^2)$。此外，我们证明任何算法都需要 $\Omega(d\log d)$ 个样本。相比之下，现有关于样本复杂度的上界至少为 $d^{1.5} \mu^{\Omega(1)} \log^{\Omega(1)} d$，其中 $\mu$ 在最坏情况下可能为 $\Theta(d)$。先前的工作是在我们问题的高秩版本中获得这些较宽松的保证，并且往往涉及更复杂的算法。