cs.DSAug 27, 2026

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

Authors: Zhao Song, Lichen Zhang

Organizations: Independent Researcher. · Massachusetts Institute of Technology.

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′∈Rdx'\in \mathbb{R}^d such that ∥x′−x∗∥∞≤εd⋅∥Ax⋆−b∥2⋅∥A†∥op\|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−2d1+Θ(log⁡log⁡n/log⁡d))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−2dlog⁡3n)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-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−2dlog⁡d)m=O(ε^{-2}d\log d) rows, uses one Hadamard pass with a padded internal dimension N=O~(n+ε−2d3)N=\widetilde{O}(n+ε^{-2}d^3), and is efficient to apply: the sketched pair (SA,Sb)(SA, Sb) can be computed in O(Ndlog⁡N)=O~(nd+ε−2d4)O(Nd\log N)=\widetilde{O}(nd+ε^{-2}d^4) time.

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