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

• URL: http://arxiv.org/abs/2608.26552v1
• Date: Thu, 27 Aug 2026 02:43:30 GMT
• Title: Hadamard Flattening and Gaussian Pooling Sketch for Least Squares with Coordinate-wise Guarantee
• Abstract summary: Randomized sketch-and-solve algorithms accelerate overconstrained $ell$ regression by replacing the input with a smaller problem.<n>Our sketch yields the $ell_infty$ guarantee with $m=O(-2dlog d)$ rows, uses one Hadamard pass with a padded internal dimension $N=widetildeO(n+-2d3)$, and is efficient to apply.
• Score: 12.89814955033651
• License: http://creativecommons.org/licenses/by-nc-sa/4.0/
• Abstract: Randomized sketch-and-solve algorithms accelerate overconstrained $\ell_2$ 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 $\ell_\infty$ norm. In particular, we want to find a vector $x'\in \mathbb{R}^d$ such that $\|x'-x^*\|_\infty\leq \fracε{\sqrt d}\cdot \|Ax^\star-b\|_2\cdot \|A^\dagger\|_{\rm op}$. Price, Song and Woodruff initiated the study of this problem and showed that the subsampled randomized Hadamard transform (SRHT) with $O(ε^{-2} d^{1+Θ(\sqrt{\log\log n/\log d})})$ rows achieves this guarantee. A subsequent work of Song, Ye, Yin and Zhang claimed to improve the row count to $O(ε^{-2}d\log^3 n)$. 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-$d$ 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 $\ell_\infty$ guarantee with $m=O(ε^{-2}d\log d)$ rows, uses one Hadamard pass with a padded internal dimension $N=\widetilde{O}(n+ε^{-2}d^3)$, and is efficient to apply: the sketched pair $(SA, Sb)$ can be computed in $O(Nd\log N)=\widetilde{O}(nd+ε^{-2}d^4)$ time.

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