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Unique Lowest Common Ancestors in Dags Are Almost as Easy as Matrix Multiplication

机译:DAG中的独特最低的常见祖先几乎与矩阵乘法一样简单

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We consider the problem of determining for each pair of vertices of a directed acyclic graph (dag) on n vertices whether or not it has a unique lowest common ancestor, and if so, finding such an ancestor. We show that this problem can be solved in time O(n{sup}ω log n), where ω < 2.376 is the exponent of the fastest known algorithm for multiplication of two n × n matrices. We show also that the problem of determining a lowest common ancestor for each pair of vertices of an arbitrary dag on n vertices is solvable in time O{top}~(n{sup}2p+n{sup}ω), where p is the minimum number of directed paths covering the vertices of the dag. With the help of random bits, we can solve the latter problem in time O(n{sup}2p).
机译:我们考虑在n顶点上确定针对一对周角的每对顶点的问题,无论是独特的最低共同祖先,还是如此,寻找这样的祖先。我们表明该问题可以在时间o(n {sup}ωlog n)中解决,其中ω<2.376是用于乘法的最快已知算法的指数,用于两个n×n矩阵。我们还表明,在N个顶点上确定任意DAG的每对顶点的最低公共祖先的问题是可在时间o {top}〜(n {sup} 2p + n {sup}ω)中的可溶解的覆盖DAG顶点的最小定向路径数。在随机位的帮助下,我们可以在时间o(n {sup} 2p)中解决后一个问题。

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