Given a hypercube $\mathcal{Q}^{n} := \{0,1\}^{n}$ in $\mathbb{R}^{n}$ and $k \in \{0, \dots, n\}$, the $k$-th layer $\mathcal{Q}^{n}_{k}$ of $\mathcal{Q}^{n}$ denotes the set of all points in $\mathcal{Q}^{n}$ whose coordinates contain exactly $k$ many ones. For a fixed $t \in \mathbb{N}$ and $k \in \{0, \dots, n\}$, let $P \in \mathbb{R}\left[x_{1}, \dots, x_{n}\right]$ be a polynomial that has zeroes of multiplicity at least $t$ at all points of $\mathcal{Q}^{n} \setminus \mathcal{Q}^{n}_{k}$, and $P$ has zeros of multiplicity exactly $t-1$ at all points of $\mathcal{Q}^{n}_{k}$. In this short note, we show that $$deg(P) \geq \max\left\{ k, n-k\right\}+2t-2.$$Matching the above lower bound we give an explicit construction of a family of hyperplanes $H_{1}, \dots, H_{m}$ in $\mathbb{R}^{n}$, where $m = \max\left\{ k, n-k\right\}+2t-2$, such that every point of $\mathcal{Q}^{n}_{k}$ will be covered exactly $t-1$ times, and every other point of $\mathcal{Q}^{n}$ will be covered at least $t$ times. Note that putting $k = 0$ and $t=1$, we recover the much celebrated covering result of Alon and Füredi (European Journal of Combinatorics, 1993). Using the above family of hyperplanes we disprove a conjecture of Venkitesh (The Electronic Journal of Combinatorics, 2022) on exactly covering symmetric subsets of hypercube $\mathcal{Q}^{n}$ with hyperplanes. To prove the above results we have introduced a new measure of complexity of a subset of the hypercube called index complexity which we believe will be of independent interest. We also study a new interesting variant of the restricted sumset problem motivated by the ideas behind the proof of the above result.

翻译：给定 $\mathbb{R}^{n}$ 中的超立方体 $\mathcal{Q}^{n} := \{0,1\}^{n}$ 及 $k \in \{0, \dots, n\}$，$\mathcal{Q}^{n}$ 的第 $k$ 层 $\mathcal{Q}^{n}_{k}$ 表示 $\mathcal{Q}^{n}$ 中坐标恰好包含 $k$ 个1的所有点的集合。对于固定的 $t \in \mathbb{N}$ 和 $k \in \{0, \dots, n\}$，设 $P \in \mathbb{R}\left[x_{1}, \dots, x_{n}\right]$ 是一个多项式，它在 $\mathcal{Q}^{n} \setminus \mathcal{Q}^{n}_{k}$ 的所有点上至少具有 $t$ 阶零点，而在 $\mathcal{Q}^{n}_{k}$ 的所有点上恰好具有 $t-1$ 阶零点。在这篇短注中，我们证明 $$\deg(P) \geq \max\left\{ k, n-k\right\}+2t-2.$$ 匹配上述下界，我们显式构造了 $\mathbb{R}^{n}$ 中的一族超平面 $H_{1}, \dots, H_{m}$，其中 $m = \max\left\{ k, n-k\right\}+2t-2$，使得 $\mathcal{Q}^{n}_{k}$ 的每个点被恰好覆盖 $t-1$ 次，而 $\mathcal{Q}^{n}$ 中所有其他点被至少覆盖 $t$ 次。注意，当 $k=0$ 且 $t=1$ 时，我们恢复了 Alon 和 Füredi 著名的覆盖结果（European Journal of Combinatorics, 1993）。利用上述超平面族，我们否定了 Venkitesh（The Electronic Journal of Combinatorics, 2022）中关于用超平面精确覆盖超立方体 $\mathcal{Q}^{n}$ 对称子集的一个猜想。为证明上述结果，我们引入了超立方体子集的一种新的复杂度度量，称为指标复杂度，我们认为该度量将具有独立意义。我们还研究了一个由上述结果证明思想所启发的受限和集问题的新变体。