The {\em discrepancy} of a matrix $M \in \mathbb{R}^{d \times n}$ is given by $\mathrm{DISC}(M) := \min_{\boldsymbol{x} \in \{-1,1\}^n} \|M\boldsymbol{x}\|_\infty$. An outstanding conjecture, attributed to Koml\'os, stipulates that $\mathrm{DISC}(M) = O(1)$, whenever $M$ is a Koml\'os matrix, that is, whenever every column of $M$ lies within the unit sphere. Our main result asserts that $\mathrm{DISC}(M + R/\sqrt{d}) = O(d^{-1/2})$ holds asymptotically almost surely, whenever $M \in \mathbb{R}^{d \times n}$ is Koml\'os, $R \in \mathbb{R}^{d \times n}$ is a Rademacher random matrix, $d = \omega(1)$, and $n = \omega(d \log d)$. The factor $d^{-1/2}$ normalising $R$ is essentially best possible and the dependency between $n$ and $d$ is asymptotically best possible. Our main source of inspiration is a result by Bansal, Jiang, Meka, Singla, and Sinha (ICALP 2022). They obtained an assertion similar to the one above in the case that the smoothing matrix is Gaussian. They asked whether their result can be attained with the optimal dependency $n = \omega(d \log d)$ in the case of Bernoulli random noise or any other types of discretely distributed noise; the latter types being more conducive for Smoothed Analysis in other discrepancy settings such as the Beck-Fiala problem. For Bernoulli noise, their method works if $n = \omega(d^2)$. In the case of Rademacher noise, we answer the question posed by Bansal, Jiang, Meka, Singla, and Sinha. Our proof builds upon their approach in a strong way and provides a discrete version of the latter. Breaking the $n = \omega(d^2)$ barrier and reaching the optimal dependency $n = \omega(d \log d)$ for Rademacher noise requires additional ideas expressed through a rather meticulous counting argument, incurred by the need to maintain a high level of precision all throughout the discretisation process.

翻译：矩阵 $M \in \mathbb{R}^{d \times n}$ 的{\em 差异度}定义为 $\mathrm{DISC}(M) := \min_{\boldsymbol{x} \in \{-1,1\}^n} \|M\boldsymbol{x}\|_\infty$。一个由科姆洛什提出的著名猜想断言：当 $M$ 是科姆洛什矩阵（即 $M$ 的每一列均位于单位球内）时，有 $\mathrm{DISC}(M) = O(1)$。本文的主要结果表明，对于科姆洛什矩阵 $M \in \mathbb{R}^{d \times n}$ 和拉德马赫随机矩阵 $R \in \mathbb{R}^{d \times n}$，当 $d = \omega(1)$ 且 $n = \omega(d \log d)$ 时，$\mathrm{DISC}(M + R/\sqrt{d}) = O(d^{-1/2})$ 渐近几乎必然成立。其中归一化因子 $d^{-1/2}$ 本质上是最优的，且 $n$ 与 $d$ 之间的依赖关系是渐近最优的。我们的主要灵感来源于 Bansal、Jiang、Meka、Singla 和 Sinha（ICALP 2022）的结果，他们在平滑矩阵为高斯分布的情形下得到了类似结论。他们提出：在伯努利随机噪声或任何其他离散分布噪声（这类噪声在如贝克-菲亚拉问题等其他差异度场景的平滑分析中更具适用性）的情形下，能否以最优依赖关系 $n = \omega(d \log d)$ 实现相应结论？对于伯努利噪声，他们的方法在 $n = \omega(d^2)$ 时有效。本文针对拉德马赫噪声的情形，回答了 Bansal 等人提出的问题。我们的证明在很大程度上继承并发展了他们的方法，给出了其离散版本。为突破 $n = \omega(d^2)$ 的限制并在拉德马赫噪声下达到最优依赖 $n = \omega(d \log d)$，需要引入额外的思想，这些思想通过相当精细的计数论证实现，以确保在整个离散化过程中始终保持高精度。