Stack Solution for Finding Optimal One

P. Kumar, G. Sahoo
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Abstract

In comparative genomics, genome rearrangement evolution is an important effort. Genome conversion is the major problem in this field using different sorting process. Transforming one sequence into another and finding an optimal solution is a useful tool for analyzing real evolutionary scenario but it will be much better if we find all possible solution for that. In order to obtain more accurate result, some solution should be taken into consideration as there is large number of different optimal sorting sequence. Reversal and translocation are the two common genome sorting process used in development of mammalian species. The problem of genome sorting using reversal and translocation is to find the shortest sequence that transforms any source genome A into some target genome B. Currently the question is resolved by lessening of sorting by reversal and sorting by translocation problem separately, but here we are applying both the sorting process together at the same time. By this paper we present an algorithm for the two sorting process that explicitly treats them as two distinct operations, along with that finding the various solutions which is a better hypothetical and real-world solution than just finding a solo one. If we have single solution for any problem then we cannot decide whether this solution is the perfect one or not but if we have more solution indeed we can find the best one among them and say this is the perfect solution. We also present an example which proves that this solution is more prominent than previous one.
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寻找最优解的堆栈解
在比较基因组学中,基因组重排进化是一个重要的研究方向。基因组转换是该领域的主要问题,采用不同的分选工艺。将一个序列转换为另一个序列并找到最优解是分析真实进化场景的有用工具,但如果我们能找到所有可能的解决方案,那就更好了。由于存在大量不同的最优排序顺序,为了获得更准确的结果,需要考虑一定的解。逆转和易位是哺乳动物物种发育中常见的两种基因组分选过程。利用反转和易位进行基因组排序的问题是找到将任意源基因组A转化为目标基因组b的最短序列,目前这个问题是分别通过减少反转排序和易位排序来解决的,但这里我们同时应用了这两种排序过程。通过本文,我们提出了一种算法,用于两个排序过程,明确地将它们视为两个不同的操作,以及寻找各种解决方案,这是一个更好的假设和现实世界的解决方案,而不仅仅是寻找一个单独的解决方案。如果我们对任何问题都有单一的解决方案,那么我们无法决定这个解决方案是否是完美的解决方案,但如果我们有更多的解决方案,我们可以找到其中最好的一个,并说这是完美的解决方案。最后给出了一个算例,证明了该解比之前的解更加突出。
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