论文标题
通过雷曼·罗恩（Lehman-Ron）结果的概括，确切的抗抗饱和数
Exact antichain saturation numbers via a generalisation of a result of Lehman-Ron
论文作者
论文摘要
For given positive integers $k$ and $n$, a family $\mathcal{F}$ of subsets of $\{1,\dots,n\}$ is $k$-antichain saturated if it does not contain an antichain of size $k$, but adding any set to $\mathcal{F}$ creates an antichain of size $k$.我们使用sat $^*（n，k）$来表示这样一个家庭的最小规模。对于所有$ k $和足够大的$ n $，我们确定SAT $^*（n，k）$的确切值。我们的结果意味着SAT $^*（n，k）= n（k-1）-θ（k \ log k）$，这证实了对抗季饱和度的几种猜想。以前，SAT $^*（N，K）$的精确值仅以$ K $高达$ 6 $而闻名。 我们还证明了雷曼·罗恩（Lehman-Ron）的结果可能具有独立利益。我们表明，给定布尔晶格中的$ m $分离链，我们可以创建$ m $脱节的无连接链，以覆盖相同的元素（如果连续两个连续的元素的尺寸完全不同，我们将其称为无链条，恰好是一个）。
For given positive integers $k$ and $n$, a family $\mathcal{F}$ of subsets of $\{1,\dots,n\}$ is $k$-antichain saturated if it does not contain an antichain of size $k$, but adding any set to $\mathcal{F}$ creates an antichain of size $k$. We use sat$^*(n, k)$ to denote the smallest size of such a family. For all $k$ and sufficiently large $n$, we determine the exact value of sat$^*(n, k)$. Our result implies that sat$^*(n, k)=n(k-1)-Θ(k\log k)$, which confirms several conjectures on antichain saturation. Previously, exact values for sat$^*(n,k)$ were only known for $k$ up to $6$. We also prove a generalisation of a result of Lehman-Ron which may be of independent interest. We show that given $m$ disjoint chains in the Boolean lattice, we can create $m$ disjoint skipless chains that cover the same elements (where we call a chain skipless if any two consecutive elements differ in size by exactly one).
