Sorting a permutation by reversals is a famous problem in genome rearrangements. Since 1997, quite some biological evidence were found that in many genomes the reversed regions are usually flanked by a pair of inverted repeats. This type of reversals are called symmetric reversals, which, unfortunately, were largely ignored until recently. In this paper, we investigate the problem of sorting by symmetric reversals, which requires a series of symmetric reversals to transform one chromosome $A$ into the another chromosome $B$. The decision problem of sorting by symmetric reversals is referred to as {\em SSR} (when the input chromosomes $A$ and $B$ are given, we use {\em SSR(A,B)}) and the corresponding optimization version (i.e., when the answer for {\em SSR(A,B)} is yes, using the minimum number of symmetric reversals to convert $A$ to $B$), is referred to as {\em SMSR(A,B)}. The main results of this paper are summarized as follows, where the input is a pair of chromosomes $A$ and $B$ with $n$ repeats. (1) We present an $O(n^2)$ time algorithm to solve the decision problem {\em SSR(A,B)}, i.e., determine whether a chromosome $A$ can be transformed into $B$ by a series of symmetric reversals. (2) We design an $O(n^2)$ time algorithm for a special 2-balanced case of {\em SMSR(A,B)}, where chromosomes $A$ and $B$ both have duplication number 2 and every repeat appears twice in different orientations in $A$ and $B$. (3) We show that SMSR is NP-hard even if the duplication number of the input chromosomes are at most 2, hence showing that the above positive optimization result is the best possible. As a by-product, we show that the \emph{minimum Steiner tree} problem on \emph{circle graphs} is NP-hard, settling the complexity status of a 38-year old open problem.
翻译:在基因组重新布局中,通过反转排序是一个著名的问题。自1997年以来,已经发现一些生物证据,在许多基因组中,被反转的区域通常被一对反转的重复。这种反转被称为对称反转,不幸的是,直到最近,这种反转在很大程度上被忽略。在本文中,我们调查了通过对称反转排序的问题,这需要一系列对称反转将一个染色体($A$)转换成另一个染色体($B$)。在许多基因组中,通过对称反转的调整进行分解的问题通常被一对称为“反转 ” (A,B) 反转, 当输入的染色体($B) 和相应的优化版本(例如,当对 emSR(A,B) 的解析(美元) 的解析(美元) 问题出现两次。使用最起码的对数(A) 和对价(美元) 的反转数(B) 被指为“SSR(A,B) 美元) 最重的变数(美元) (美元) (O) 的解算结果,我们的变数显示的是“O) 。