An $n$-Venn diagram is a diagram in the plane consisting of $n$ simple closed curves that intersect only finitely many times such that each of the $2^n$ possible intersections is represented by a single connected region. An $n$-Venn diagram has at most $2^n-2$ crossings, and if this maximum number of crossings is attained, then only two curves intersect in every crossing. To complement this, Bultena and Ruskey considered $n$-Venn diagrams that minimize the number of crossings, which implies that many curves intersect in every crossing. Specifically, they proved that the total number of crossings in any $n$-Venn diagram is at least $L_n:=\lceil\frac{2^n-2}{n-1}\rceil$, and if this lower bound is attained then essentially all $n$ curves intersect in every crossing. Diagrams achieving this bound are called minimum Venn diagrams, and are known only for $n\leq 7$. Bultena and Ruskey conjectured that they exist for all $n\geq 8$. In this work, we establish an asympototic version of their conjecture. For $n=8$ we construct a diagram with 40 crossings, only 3 more than the lower bound $L_8=37$. Furthermore, for every $n$ of the form $n=2^k$ for some integer $k\geq 4$, we construct an $n$-Venn diagram with at most $(1+\frac{33}{8n})L_n=(1+o(1))L_n$ many crossings. Via a doubling trick this also gives $(n+m)$-Venn diagrams for all $0\leq m<n$ with at most $40\cdot 2^m$ crossings for $n=8$ and at most $(1+\frac{33}{8n})\frac{n+m}{n}L_{n+m}=(2+o(1))L_{n+m}$ many crossings for $k\geq 4$. In particular, we obtain $n$-Venn diagrams with the smallest known number of crossings for all $n\geq 8$. Our constructions are based on partitions of the hypercube into isometric paths and cycles, using a result of Ramras.
翻译:一个$n$-维恩图是由$n$条简单闭曲线在平面上相交有限次构成的图形,使得$2^n$种可能的交集各由一个连通区域表示。$n$-维恩图最多有$2^n-2$个交叉点,若达到此最大交叉数,则每个交叉点仅由两条曲线相交。作为补充,Bultena和Ruskey研究了最小化交叉点数量的$n$-维恩图,这意味着每个交叉点有多条曲线相交。具体而言,他们证明了任意$n$-维恩图的总交叉点数至少为$L_n:=\lceil\frac{2^n-2}{n-1}\rceil$,若达到此下界,则本质上所有$n$条曲线在每个交叉点处相交。达到此界限的图称为最小维恩图,目前仅知对$n\leq 7$存在。Bultena和Ruskey猜想对所有$n\geq 8$均存在。本文中,我们建立了该猜想的渐近版本。对$n=8$,我们构造了一个具有40个交叉点的图,仅比下界$L_8=37$多3个。此外,对每个形如$n=2^k$(整数$k\geq 4$)的$n$,我们构造了一个交叉点数不超过$(1+\frac{33}{8n})L_n=(1+o(1))L_n$的$n$-维恩图。通过倍增技巧,这也为所有$0\leq m<n$给出了$(n+m)$-维恩图:对$n=8$交叉点数不超过$40\cdot 2^m$,对$k\geq 4$交叉点数不超过$(1+\frac{33}{8n})\frac{n+m}{n}L_{n+m}=(2+o(1))L_{n+m}$。特别地,我们得到了对所有$n\geq 8$具有目前已知最小交叉点数的$n$-维恩图。我们的构造基于将超立方体分割为等距路径和环,使用了Ramras的一个结果。