AI 中文总结
本文确定了裁剪超立方体的Lasserre秩，给出了其精确递推公式、多项式时间算法及渐近表达式，并推广至任意固定裁剪距离。
AI 中文摘要
在一个$n$维的\emph{裁剪超立方体}中，$2^n$个裁剪不等式中的每一个都通过一个$\ell_1$距离$\rho$裁剪掉$0$--$1$超立方体的一个角。$\rho=1/2$的情形已在文献中被广泛研究。本文证明，当$\rho=1/2$且$n\geq 2$时，$n$维裁剪超立方体的Lasserre秩是满足递推关系$\Delta_{-1}=1$，$\Delta_{0}=n-1$，$\Delta_t=(n-1)\Delta_{t-1}-t(n-t+1)\Delta_{t-2}$中$\Delta_t<0$的最小整数$0\leq t\leq n$。由此可知，Lasserre秩可以在$O(n^2 \log^2 n)$时间内计算。渐近地，该秩为$\frac{n}{2}+c_{1/2}\sqrt{n}+o(\sqrt{n})$，其中$c_{1/2}$是某个给定函数的唯一零点。数值上，$c_{1/2}\approx 0.3825$。实际上，我们证明了对于任意固定的$0<\rho<1$，此类结果均成立。
英文摘要
In an $n$-dimensional cropped hypercube each of the $2^n$ cropping inequalities chops off a single corner of the $0$--$1$ hypercube by an $\ell_1$-distance $ρ$. The case $ρ= 1/2$ has been extensively studied in the literature. This paper shows that the Lasserre rank of the $n$-dimensional cropped hypercube where $ρ= 1/2$, $n \geq 2$, is the smallest integer $0\leq t \leq n$ such that $Δ_t < 0$ in the recurrence $Δ_{-1} = 1$, $Δ_{0} = n-1$, $Δ_t = (n-1)Δ_{t-1} - t(n-t+1)Δ_{t-2}$. It follows that the Lasserre rank can be computed in time $O(n^2 \log^2 n)$. Asymptotically, the rank is $\frac{n}{2} + c_{1/2}\sqrt{n} + o(\sqrt{n})$, where $c_{1/2}$ is the unique zero of a given function. Numerically, $c_{1/2} \approx 0.3825$. In fact, we prove such results for any fixed $0 < ρ< 1$.
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