arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 1909.08680math.CO
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布尔格的偏序集Ramsey数

Poset Ramsey Numbers for Boolean Lattices

Linyuan Lu, Joshua C. Thompson

摘要:

偏序集$Q$的子集$Q'$是\textit{copy of a poset} $P$,如果在$P$和$Q'$的元素之间存在双射$f$,使得$x \le y$ in $P$当且仅当$f(x) \le f(y)$ in $Q'$。对于偏序集$P, P'$,令\textit{poset Ramsey number} $R(P,P')$是最小的$N$,使得无论布尔格$Q_N$的元素如何被着色为红色和蓝色,都有一个$P$的副本,其中所有元素都是红色的,或者$P'$的副本,其中所有元素都是蓝色的。Axenovich和Walzer在\textit{Order}(2017)中引入了这个概念,在那里他们证明了$R(Q_2, Q_n) \le 2n + 2$和$R(Q_n, Q_m) \le mn + n + m$,其中$Q_n$是维度$n$的布尔格。他们后来证明了$2n \le R(Q_n, Q_n) \le n^2 + 2n$。沃尔泽后来证明了$R(Q_n, Q_n) \le n^2 + 1$。我们提供了一些改进的界限$R(Q_n, Q_m)$各种$n,m \in \mathbb{N}$。特别是,我们证明了$R(Q_n, Q_n) \le n^2 - n + 2$,$R(Q_2, Q_n) \le \frac{5}{3}n + 2$,和$R(Q_3, Q_n) \le \frac{37}{16}n + \frac{39}{16}$。我们还证明了$R(Q_2,Q_3) = 5$,和$R(Q_m, Q_n) \le (m - 2 + \frac{9m - 9}{(2m - 3)(m + 1)})n + m + 3$的所有$n \ge m \ge 4$。

英文摘要:

A subposet $Q'$ of a poset $Q$ is a \textit{copy of a poset} $P$ if there is a bijection $f$ between elements of $P$ and $Q'$ such that $x \le y$ in $P$ iff $f(x) \le f(y)$ in $Q'$. For posets $P, P'$, let the \textit{poset Ramsey number} $R(P,P')$ be the smallest $N$ such that no matter how the elements of the Boolean lattice $Q_N$ are colored red and blue, there is a copy of $P$ with all red elements or a copy of $P'$ with all blue elements. Axenovich and Walzer introduced this concept in \textit{Order} (2017), where they proved $R(Q_2, Q_n) \le 2n + 2$ and $R(Q_n, Q_m) \le mn + n + m$, where $Q_n$ is the Boolean lattice of dimension $n$. They later proved $2n \le R(Q_n, Q_n) \le n^2 + 2n$. Walzer later proved $R(Q_n, Q_n) \le n^2 + 1$. We provide some improved bounds for $R(Q_n, Q_m)$ for various $n,m \in \mathbb{N}$. In particular, we prove that $R(Q_n, Q_n) \le n^2 - n + 2$, $R(Q_2, Q_n) \le \frac{5}{3}n + 2$, and $R(Q_3, Q_n) \le \frac{37}{16}n + \frac{39}{16}$. We also prove that $R(Q_2,Q_3) = 5$, and $R(Q_m, Q_n) \le (m - 2 + \frac{9m - 9}{(2m - 3)(m + 1)})n + m + 3$ for all $n \ge m \ge 4$.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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