SearcharxivSearch

arXiv · 1909.01821

Almost Optimal Tensor Sketch

Abstract

We construct a matrix $M\in R^{m\otimes d^c}$ with just $m=O(c\,\lambda\,\varepsilon^{-2}\text{poly}\log1/\varepsilon\delta)$ rows, which preserves the norm $\|Mx\|_2=(1\pm\varepsilon)\|x\|_2$ of all $x$ in any given $\lambda$ dimensional subspace of $ R^d$ with probability at least $1-\delta$. This matrix can be applied to tensors $x^{(1)}\otimes\dots\otimes x^{(c)}\in R^{d^c}$ in $O(c\, m \min\{d,m\})$ time -- hence the name "Tensor Sketch". (Here $x\otimes y = \text{asvec}(xy^T) = [x_1y_1, x_1y_2,\dots,x_1y_m,x_2y_1,\dots,x_ny_m]\in R^{nm}$.) This improves upon earlier Tensor Sketch constructions by Pagh and Pham~[TOCT 2013, SIGKDD 2013] and Avron et al.~[NIPS 2014] which require $m=\Omega(3^c\lambda^2\delta^{-1})$ rows for the same guarantees. The factors of $\lambda$, $\varepsilon^{-2}$ and $\log1/\delta$ can all be shown to be necessary making our sketch optimal up to log factors. With another construction we get $\lambda$ times more rows $m=\tilde O(c\,\lambda^2\,\varepsilon^{-2}(\log1/\delta)^3)$, but the matrix can be applied to any vector $x^{(1)}\otimes\dots\otimes x^{(c)}\in R^{d^c}$ in just $\tilde O(c\, (d+m))$ time. This matches the application time of Tensor Sketch while still improving the exponential dependencies in $c$ and $\log1/\delta$. Technically, we show two main lemmas: (1) For many Johnson Lindenstrauss (JL) constructions, if $Q,Q'\in R^{m\times d}$ are independent JL matrices, the element-wise product $Qx \circ Q'y$ equals $M(x\otimes y)$ for some $M\in R^{m\times d^2}$ which is itself a JL matrix. (2) If $M^{(i)}\in R^{m\times md}$ are independent JL matrices, then $M^{(1)}(x \otimes (M^{(2)}y \otimes \dots)) = M(x\otimes y\otimes \dots)$ for some $M\in R^{m\times d^c}$ which is itself a JL matrix. Combining these two results give an efficient sketch for tensors of any size.

Explore related subjects

Keep this discovery

BibTeXRIS

Thomas D. Ahle, Jakob B. T. Knudsen. 2019-09-03. Almost Optimal Tensor Sketch. https://arxiv.org/abs/1909.01821

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

Quasi-Monte Carlo Beyond Hardy-Krause II: $(1 + \varepsilon)n$ Samples Suffice

Numerical integration studies how well one can estimate the integral of a function $f$ over $[0,1)^d$ using $n$ sample points. The two classical methods, Monte Carlo (MC) and quasi-Monte Carlo (QMC), have complementary strengths and weaknesses, and a fundamental question is to design an approach that combines the benefits of both. Recently, building on the transference principle in discrepancy theory, Bansal and Jiang~\cite{BJ25a} gave a randomized QMC method that bridges MC and QMC guarantees using only i.i.d.\ samples. Their method also goes beyond the classical Koksma--Hlawka inequality: it achieves integration error $\widetilde{O}_d(\sigma_{\mathsf{SO}}(f)/n)$, where the smoothed-out variation $\sigma_{\mathsf{SO}}(f)$ can be substantially smaller than the Hardy--Krause variation that governs the classical bound. However, their algorithm requires $n^2$ i.i.d.\ samples as input, and this quadratic blowup is inherent to any method based on the transference principle. In this work, we bypass the quadratic blowup: for any constant $\varepsilon > 0$, we show that $(1+\varepsilon)n$ i.i.d.\ samples suffice to both obtain the beyond-Hardy--Krause guarantee of~\cite{BJ25a}, resolving an open problem posed there, and to produce low-discrepancy point sequences. Our algorithms are variants of the online Haar-thinning method of Dwivedi, Feldheim, Gurel-Gurevich, and Ramdas~\cite{DFG+19}.

cs.DS

Single-Exponential Algorithms and a Polynomial Kernel for Strong Connectivity Augmentation

Strong Connectivity Augmentation (SCA) asks whether a directed acyclic graph can be made strongly connected by adding at most $k$ prescribed links whose total weight is within a given budget. Klinkby, Misra, and Saurabh (SODA 2021) gave an $O^*(2^{O(k\log k)})$-time algorithm and asked whether the problem admits a single-exponential parameterized algorithm and a polynomial kernel. We answer both questions affirmatively: SCA can be solved in $O^*(9^k)$ time and admits a polynomial kernel with $O(k^4)$ vertices and $O(k^{16})$ bits. For unweighted SCA, we obtain $O^*(4^k)$ time and a kernel with $O(k^3)$ vertices. Our algorithms are based on a particularly simple reduction to Strongly Connected Spanning Subgraph with two edge costs.

cs.DS