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Hadamard Flattening and Gaussian Pooling Sketch for Least Squares with Coordinate-wise Guarantee

Published 27 Aug 2026 in cs.DS, cs.LG, and stat.ML | (2608.26552v1)

Abstract: Randomized sketch-and-solve algorithms accelerate overconstrained ℓ2\ell_2 regression by replacing the input with a smaller problem. Standard subspace embeddings guarantee that the cost of the regression is nearly preserved, but coordinate-wise accuracy of the solution is more delicate: we want the solution vector itself to be close to the optimal solution in ℓ∞\ell_\infty norm. In particular, we want to find a vector $x&#39;\in \mathbb{R}<sup>d$ such that $|x&#39;-x<sup>*|_\infty\leq</sup> \fracε{\sqrt d}\cdot |Ax<sup>\star-b|_2\cdot</sup> |A<sup>\dagger|_{\rm</sup> op}$. Price, Song and Woodruff initiated the study of this problem and showed that the subsampled randomized Hadamard transform (SRHT) with O(ε<sup>−2</sup>d<sup>1+Θ(log⁡log⁡</sup>n/log⁡d))O(ε<sup>{-2}</sup> d<sup>{1+Θ(\sqrt{\log\log</sup> n/\log d})}) rows achieves this guarantee. A subsequent work of Song, Ye, Yin and Zhang claimed to improve the row count to O(ε<sup>−2dlog⁡<sup>3</sup></sup>n)O(ε<sup>{-2}d\log<sup>3</sup></sup> n). Unfortunately, their proof relies on an independence assumption that does not hold in general, and we exhibit an explicit instance on which it fails. To achieve a truly nearly-linear-in-dd row count, we introduce a new fast, dense randomized transform, which combines a randomized Hadamard flattening, a random permutation, and balanced, disjoint Gaussian pooling. Conditioned on the Hadamard-and-permutation stage, the sketched problem becomes an exact Gaussian regression in which the noise is independent of the entire sketched design; this conditional independence is exactly what the earlier argument was missing. Our sketch yields the ℓ∞\ell_\infty guarantee with m=O(ε<sup>−2dlog⁡</sup>d)m=O(ε<sup>{-2}d\log</sup> d) rows, uses one Hadamard pass with a padded internal dimension N=O~(n+ε<sup>−2d<sup>3)N=\widetilde{O}(n+ε<sup>{-2}d<sup>3), and is efficient to apply: the sketched pair (SA,Sb)(SA, Sb) can be computed in O(Ndlog⁡N)=O~(nd+ε<sup>−2d<sup>4)O(Nd\log N)=\widetilde{O}(nd+ε<sup>{-2}d<sup>4) time.

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