We show hardness of improperly learning halfspaces in the agnostic model based on worst-case lattice problems, e.g., approximating shortest vectors within polynomial factors. In particular, we show that under this assumption there is no efficient algorithm that outputs any binary hypothesis, not necessarily a halfspace, achieving misclassfication error better than $\frac 1 2 - \epsilon$ even if the optimal misclassification error is as small is as small as $\delta$. Here, $\epsilon$ can be smaller than the inverse of any polynomial in the dimension and $\delta$ as small as $\mathrm{exp}\left(-\Omega\left(\log^{1-c}(d)\right)\right)$, where $0 < c < 1$ is an arbitrary constant and $d$ is the dimension. Previous hardness results [Daniely16] of this problem were based on average-case complexity assumptions, specifically, variants of Feige's random 3SAT hypothesis. Our work gives the first hardness for this problem based on a worst-case complexity assumption. It is inspired by a sequence of recent works showing hardness of learning well-separated Gaussian mixtures based on worst-case lattice problems.
翻译:在基于最坏情况悬浮问题的不可知模型中,我们不适当地学习半空空间,例如,在多元因素中,近似于最短矢量。特别是,我们显示,在这个假设下,没有任何有效的算法可以输出任何二进假设,不一定是半空,实现错误分类错误的好于$\frac 1 2 -\ epsilon$,即使最佳错误分类错误小于$delta$。这里,美元可以小于在维度上任何多盘式的反差,而美元则小于美元。我们显示,在这种假设中,没有有效的算法能够输出任何二进制假设,不一定是一个半空格,即使最佳错误分类错误的错误比$\frac 1 2 -\ epsilonlon$要小,即使最差的错误小于$deltataltal 16] 。这里的问题以平均复杂程度假设为基础,特别是Fegges的变差值值值=left left ⁇ left lift lift lift (-cleas mission of last reclegress mission) a robleas missional pas missional pas missional pas missional