The Complexity of Intersection Graphs of Lines in Space and Circle Orders

Jean Cardinal
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Abstract

We consider the complexity of the recognition problem for two families of combinatorial structures. A graph $G=(V,E)$ is said to be an intersection graph of lines in space if every $v\in V$ can be mapped to a straight line $\ell (v)$ in $\mathbb{R}^3$ so that $vw$ is an edge in $E$ if and only if $\ell(v)$ and $\ell(w)$ intersect. A partially ordered set $(X,\prec)$ is said to be a circle order, or a 2-space-time order, if every $x\in X$ can be mapped to a closed circular disk $C(x)$ so that $y\prec x$ if and only if $C(y)$ is contained in $C(x)$. We prove that the recognition problems for intersection graphs of lines and circle orders are both $\exists\mathbb{R}$-complete, hence polynomial-time equivalent to deciding whether a system of polynomial equalities and inequalities has a solution over the reals. The second result addresses an open problem posed by Brightwell and Luczak.
空间直线与圆阶相交图的复杂性
我们考虑了两类交叉结构的识别问题的复杂性。如果 V$ 中的每一条 $v\ 都可以映射到 $\mathbb{R}^3$ 中的一条直线 $\ell (v)$,从而当且仅当 $\ell(v)$ 和 $\ell(w)$ 相交时,$vw$ 才是 E$ 中的一条边,那么我们就可以说一个图 $G=(V,E)$ 是空间中的线段交集图。如果 X$ 中的每一个 $x\ 都可以映射到一个封闭的圆盘 $C(x)$ 中,从而当且仅当 $C(y)$ 包含在 $C(x)$ 中时,$y\prec x$ 才是一个部分有序集合 $(X,/prec)$,那么这个部分有序集合 $(X,\prec)$ 可以说是一个圆阶,或者说是一个 2 时空阶。我们证明了线和圆阶交点图的识别问题都是 $\exists\mathbb{R}$ 完全的,因此多项式时间等价于决定一个多项式等式和inequalities 系统在实数上是否有解。第二个结果解决了布莱特韦尔和卢卡克提出的一个开放性问题。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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