arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新,休息一下吧,周末愉快~~
arXiv 0709.0170cs.CGcs.DM
收藏

解平面图

Untangling a Planar Graph

Xavier Goaoc, Jan Kratochvil, Yoshio Okamoto, Chan-Su Shin, Andreas Spillner, Alexander Wolff

摘要:

平面图$G$的直线图$δ$不必是平面的,但可以通过\emph{untangling}使其成为平面,即通过移动$G$的一些顶点。令shift $(G,δ)$表示需要移动以解开$δ$的顶点的最小数目。我们表明,移位$(G,δ)$是NP难计算和近似。我们的硬度结果扩展到一个版本的\textsc{1BendPointSetEmbeddability},一个著名的图形绘制问题。此外,我们定义固定$(G,δ)=n-shift(G,δ)$的平面$n$ -顶点图$G$的顶点的最大数量,可以固定时解开$δ$。我们给出了一个算法,固定至少$\sqrt{((\log n)-1)/\log \log n}$顶点时解开一个$n$ -顶点图$G$的绘图。如果$G$是外平面的,则相同的算法至少固定$\sqrt{n/2}$顶点。另一方面,对于任意大的$n$,我们构造了一个$n$ -顶点平面图$n$和一个$G$的绘图$δ_G$,其中$(G,δ_G) \le \sqrt{n-2}+1$是固定的,以及一个$n$ -顶点外平面图$H$和一个$H$的绘图$δ_H$,其中$(H,δ_H) \le 2 \sqrt{n-1}+1$是固定的.因此,我们的算法是渐近最坏情况下最优的外平面图。

英文摘要:

A straight-line drawing $δ$ of a planar graph $G$ need not be plane, but can be made so by \emph{untangling} it, that is, by moving some of the vertices of $G$. Let shift$(G,δ)$ denote the minimum number of vertices that need to be moved to untangle $δ$. We show that shift$(G,δ)$ is NP-hard to compute and to approximate. Our hardness results extend to a version of \textsc{1BendPointSetEmbeddability}, a well-known graph-drawing problem. Further we define fix$(G,δ)=n-shift(G,δ)$ to be the maximum number of vertices of a planar $n$-vertex graph $G$ that can be fixed when untangling $δ$. We give an algorithm that fixes at least $\sqrt{((\log n)-1)/\log \log n}$ vertices when untangling a drawing of an $n$-vertex graph $G$. If $G$ is outerplanar, the same algorithm fixes at least $\sqrt{n/2}$ vertices. On the other hand we construct, for arbitrarily large $n$, an $n$-vertex planar graph $G$ and a drawing $δ_G$ of $G$ with fix$(G,δ_G) \le \sqrt{n-2}+1$ and an $n$-vertex outerplanar graph $H$ and a drawing $δ_H$ of $H$ with fix$(H,δ_H) \le 2 \sqrt{n-1}+1$. Thus our algorithm is asymptotically worst-case optimal for outerplanar graphs.

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。

↑