论文标题
关于通过换位的某些变化的复杂性
On the Complexity of Some Variations of Sorting by Transpositions
论文作者
论文摘要
计算生物学的主要挑战之一是找到两个生物之间的进化距离。在比较基因组学领域,估计这种距离的一种方法是找到将一个基因组转化为另一个基因组所需的最低成本序列(大规模突变),这称为重排距离。在过去的几十年中,研究了这些问题,考虑了许多类型的重排(例如逆转,转换,跨性别和RevRevs),并考虑所有重排的重量相同,或者根据重排类型的不同重量而考虑相同的重量。即使最近给出了逆转和换位的问题的硬度证明,涉及逆转,转置和两个重排的问题的复杂性是已知的。在本文中,我们通过证明涉及逆转,跨反向和RevRevs并列的模型是NP-HARD,从而增强了这些问题的知识,考虑到逆转的权重W1,而对于其他重新排列的权重W1,W2/W1的重量是W2的重量,以至于W2/W1的权重小于或等于或等于1.5。此外,我们解决了与重排引起的碎片数量相关的成本函数,证明了寻找最低成本排序顺序的问题,即考虑片段化的成本函数具有一些限制,是转置的NP,是转置的NP,以及逆转和转换的组合。
One of the main challenges in Computational Biology is to find the evolutionary distance between two organisms. In the field of comparative genomics, one way to estimate such distance is to find a minimum cost sequence of rearrangements (large scale mutations) needed to transform one genome into another, which is called the rearrangement distance. In the past decades, these problems were studied considering many types of rearrangements (such as reversals, transpositions, transreversals, and revrevs) and considering the same weight for all rearrangements, or different weights depending on the types of rearrangements. The complexity of the problems involving reversals, transpositions, and both rearrangements is known, even though the hardness proof for the problem combining reversals and transpositions was recently given. In this paper, we enhance the knowledge for these problems by proving that models involving transpositions alongside reversals, transreversals, and revrevs are NP-hard, considering weights w1 for reversals and w2 for the other rearrangements such that w2/w1 is less than or equal to 1.5. In addition, we address a cost function related to the number of fragmentations caused by a rearrangement, proving that the problem of finding a minimum cost sorting sequence, considering the fragmentation cost function with some restrictions, is NP-hard for transpositions and the combination of reversals and transpositions.