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引用次数: 1

摘要

基于内积相似连接中候选对的过滤问题,我们研究了以下内积估计问题:给定参数$d\in {\bf N}$, $\alpha>\beta\geq 0$和单位向量$x,y\in {\bf R}^{d}$,考虑区分$\langle x, y\rangle\leq\beta$和$\langle x, y\rangle\geq \alpha$的情况,其中$\langle x, y\rangle = \sum_{i=1}^d x_i y_i$是向量$x$和$y$的内积。目标是根据在尽可能短的位串中独立编码的每个向量上的信息来区分这些情况。与使用随机降维压缩向量的许多工作相反,我们寻求确定性地解决问题,没有错误的概率。一般来说,内积估计可以通过以$\varepsilon = \alpha - \beta$为界的加性误差估计$\langle x, y\rangle$来解决。我们证明了每个向量的$d \log_2 \left(\tfrac{\sqrt{1-\beta}}{\varepsilon}\right) \pm \Theta(d)$位信息是必要和充分的。我们的上界是建设性的,当$\beta$接近$1$时,它将已知的$d \log_2(1/\varepsilon) + O(d)$上界提高了2倍。即使在一个更强的模型中,其中一个向量是确切已知的,并且允许任意估计函数,下界也成立。
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The space complexity of inner product filters
Motivated by the problem of filtering candidate pairs in inner product similarity joins we study the following inner product estimation problem: Given parameters $d\in {\bf N}$, $\alpha>\beta\geq 0$ and unit vectors $x,y\in {\bf R}^{d}$ consider the task of distinguishing between the cases $\langle x, y\rangle\leq\beta$ and $\langle x, y\rangle\geq \alpha$ where $\langle x, y\rangle = \sum_{i=1}^d x_i y_i$ is the inner product of vectors $x$ and $y$. The goal is to distinguish these cases based on information on each vector encoded independently in a bit string of the shortest length possible. In contrast to much work on compressing vectors using randomized dimensionality reduction, we seek to solve the problem deterministically, with no probability of error. Inner product estimation can be solved in general via estimating $\langle x, y\rangle$ with an additive error bounded by $\varepsilon = \alpha - \beta$. We show that $d \log_2 \left(\tfrac{\sqrt{1-\beta}}{\varepsilon}\right) \pm \Theta(d)$ bits of information about each vector is necessary and sufficient. Our upper bound is constructive and improves a known upper bound of $d \log_2(1/\varepsilon) + O(d)$ by up to a factor of 2 when $\beta$ is close to $1$. The lower bound holds even in a stronger model where one of the vectors is known exactly, and an arbitrary estimation function is allowed.
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