Two genomes over the same set of gene families form a canonical pair when each of them has exactly one gene from each family. Different distances of canonical genomes can be derived from a structure called breakpoint graph, which represents the relation between the two given genomes as a collection of cycles of even length and paths. Then, the breakpoint distance is equal to n - (c_2 + p_0/2), where n is the number of genes, c_2 is the number of cycles of length 2 and p_0 is the number of paths of length 0. Similarly, when the considered rearrangements are those modeled by the double-cut-and-join (DCJ) operation, the rearrangement distance is n - (c + p_e/2), where c is the total number of cycles and p_e is the total number of even paths. The distance formulation is a basic unit for several other combinatorial problems related to genome evolution and ancestral reconstruction, such as median or double distance. Interestingly, both median and double distance problems can be solved in polynomial time for the breakpoint distance, while they are NP-hard for the rearrangement distance. One way of exploring the complexity space between these two extremes is to consider the {\sigma}_k distance, defined to be n - [c_2 + c_4 + ... + c_k + (p_0 + p_2 + ... +p_k)/2], and increasingly investigate the complexities of median and double distance for the {\sigma}_4 distance, then the {\sigma}_6 distance, and so on. While for the median much effort was done in our and in other research groups but no progress was obtained even for the {\sigma}_4 distance, for solving the double distance under {\sigma}_4 and {\sigma}_6 distances we could devise linear time algorithms, which we present here.

翻译：两个具有相同基因家族集合的基因组，当每个基因组在每个家族中恰好包含一个基因时，构成规范对。规范基因组的不同距离可通过称为断裂点图的结构推导得出，该图将两个给定基因组之间的关系表示为偶数长度环和路径的集合。此时，断裂点距离等于 n - (c_2 + p_0/2)，其中 n 为基因数量，c_2 为长度为 2 的环数量，p_0 为长度为 0 的路径数量。类似地，当考虑由双切割与连接（DCJ）操作模拟的重排时，重排距离为 n - (c + p_e/2)，其中 c 为环的总数，p_e 为偶数路径的总数。该距离公式是基因组进化和祖先重建相关的若干其他组合问题（如中位数或双重距离）的基本单元。有趣的是，对于断裂点距离，中位数和双重距离问题均可在多项式时间内求解，而对于重排距离，它们属于NP难问题。探索这两个极端之间复杂性空间的一种方法是考虑 k 距离，定义为 n - [c_2 + c_4 + ... + c_k + (p_0 + p_2 + ... + p_k)/2]，并逐步研究 4 距离、6 距离等下的中位数和双重距离的复杂性。尽管我们和其他研究团队在中位数问题上投入了大量努力，但即使在 4 距离下也未取得进展，而对于 4 和 6 距离下的双重距离，我们设计了线性时间算法，并在本文中加以呈现。