Suppose we are given an $n$-dimensional order-3 symmetric tensor $T \in (\mathbb{R}^n)^{\otimes 3}$ that is the sum of $r$ random rank-1 terms. The problem of recovering the rank-1 components is possible in principle when $r \lesssim n^2$ but polynomial-time algorithms are only known in the regime $r \ll n^{3/2}$. Similar "statistical-computational gaps" occur in many high-dimensional inference tasks, and in recent years there has been a flurry of work on explaining the apparent computational hardness in these problems by proving lower bounds against restricted (yet powerful) models of computation such as statistical queries (SQ), sum-of-squares (SoS), and low-degree polynomials (LDP). However, no such prior work exists for tensor decomposition, largely because its hardness does not appear to be explained by a "planted versus null" testing problem. We consider a model for random order-3 tensor decomposition where one component is slightly larger in norm than the rest (to break symmetry), and the components are drawn uniformly from the hypercube. We resolve the computational complexity in the LDP model: $O(\log n)$-degree polynomial functions of the tensor entries can accurately estimate the largest component when $r \ll n^{3/2}$ but fail to do so when $r \gg n^{3/2}$. This provides rigorous evidence suggesting that the best known algorithms for tensor decomposition cannot be improved, at least by known approaches. A natural extension of the result holds for tensors of any fixed order $k \ge 3$, in which case the LDP threshold is $r \sim n^{k/2}$.
翻译:假设我们给出一个$n$维三阶对称张量$T \in (\mathbb{R}^n)^{\otimes 3}$,它是$r$个随机秩-1项的和。在$r \lesssim n^2$的情况下原则上可以恢复秩-1项,但是只有在$r \ll n^{3/2}$的范围内才能使用多项式时间算法。类似的“统计 - 计算间隙”在许多高维推理任务中都存在,在最近几年,有大量研究工作致力于通过针对受限的(但强大的)计算模型如统计查询(SQ)、平方和(SoS)和低阶多项式(LDP)来证明这些问题的表面计算困难性。然而,这种张量分解的难度似乎不能被“已知与未知”测试问题解释。我们考虑一种随机三阶张量分解模型,其中一个分量的范数略大于其余分量(以打破对称性),并且分量是从超立方体中均匀抽取的。我们解决了LDP模型中的计算复杂性问题:当$r \ll n^{3/2}$时,张量条目的$O(\log n)$-次多项式函数可以准确地估计最大分量,但是当$r \gg n^{3/2}$时,它们不能这样做。这提供了严格的证据,表明张量分解的最佳已知算法至少不能通过已知的方法进行改进。结果的自然扩展适用于任意固定阶数$k \ge 3$的张量,其中LDP阈值为$r \sim n^{k/2}$。