Linear arrangements of graphs are a well-known type of graph labeling and are found in many important computational problems, such as the Minimum Linear Arrangement Problem ($\texttt{minLA}$). A linear arrangement is usually defined as a permutation of the $n$ vertices of a graph. An intuitive geometric setting is that of vertices lying on consecutive integer positions in the real line, starting at 1; edges are often drawn as semicircles above the real line. In this paper we study the Maximum Linear Arrangement problem ($\texttt{MaxLA}$), the maximization variant of $\texttt{minLA}$. We devise a new characterization of maximum arrangements of general graphs, and prove that $\texttt{MaxLA}$ can be solved for cycle graphs in constant time, and for $k$-linear trees ($k\le2$) in time $O(n)$. We present two constrained variants of $\texttt{MaxLA}$ we call $\texttt{bipartite MaxLA}$ and $\texttt{1-thistle MaxLA}$. We prove that the former can be solved in time $O(n)$ for any bipartite graph; the latter, by an algorithm that typically runs in time $O(n^4)$ on unlabelled trees. The combination of the two variants has two promising characteristics. First, it solves $\texttt{MaxLA}$ for almost all trees consisting of a few tenths of nodes. Second, we prove that it constitutes a $3/2$-approximation algorithm for $\texttt{MaxLA}$ for trees. Furthermore, we conjecture that $\texttt{bipartite MaxLA}$ solves $\texttt{MaxLA}$ for at least $50\%$ of all free trees.
翻译:图的线性排列是一种广为人知的图标记方式,并存在于许多重要的计算问题中,例如最小线性排列问题($\texttt{minLA}$)。线性排列通常定义为图 $n$ 个顶点的一个排列。一个直观的几何设定是顶点位于实数轴上从1开始的连续整数位置;边通常绘制为实数轴上方的半圆。本文研究最大线性排列问题($\texttt{MaxLA}$),即 $\texttt{minLA}$ 的最大化变体。我们提出了通用图最大排列的新刻画,并证明了对于环图可在常数时间内求解 $\texttt{MaxLA}$,对于 $k$-线性树($k\le2$)可在 $O(n)$ 时间内求解。我们提出了 $\texttt{MaxLA}$ 的两个约束变体,分别称为 $\texttt{二分图 MaxLA}$ 和 $\texttt{1-蓟 MaxLA}$。我们证明了前者对于任何二分图可在 $O(n)$ 时间内求解;后者通过一个算法求解,该算法在未标记树上通常运行时间为 $O(n^4)$。这两种变体的组合具有两个有前景的特性。首先,它能为几乎所有包含几十个节点的树求解 $\texttt{MaxLA}$。其次,我们证明它构成了树 $\texttt{MaxLA}$ 的一个 $3/2$-近似算法。此外,我们推测 $\texttt{二分图 MaxLA}$ 能为至少 $50\%$ 的所有自由树求解 $\texttt{MaxLA}$。