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Recognizing Treelike k-Dissimilarities

机译:识别艰难的K-异化

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A k-dissimilarity D on a finite set X, {pipe}X{pipe} ≥ k, is a map from the set of size k subsets of X to the real numbers. Such maps naturally arise from edgeweighted trees T with leaf-set X: Given a subset Y of X of size k, D(Y) is defined to be the total length of the smallest subtree of T with leaf-set Y. In case k = 2, it is well-known that 2-dissimilarities arising in this way can be characterized by the so-called "4-point condition". However, in case k > 2 Pachter and Speyer (2004) recently posed the following question: Given an arbitrary k-dissimilarity, how do we test whether this map comes from a tree? In this paper, we provide an answer to this question, showing that for k ≥ 3 a k-dissimilarity on a set X arises from a tree if and only if its restriction to every 2 k-element subset of X arises from some tree, and that 2 k is the least possible subset size to ensure that this is the case. As a corollary, we show that there exists a polynomial-time algorithm to determine when a k-dissimilarity arises from a tree. We also give a 6-point condition for determining when a 3-dissimilarity arises from a tree, that is similar to the aforementioned 4-point condition.
机译:有限组X,{管道x {管道}≥k上的k-异化d是来自X的大小k子集集的映射到实数。这种地图自然地由叶柄X的边缘重量x:给定尺寸K的子集Y,D(y)被定义为T的最小子树的总长度,其中叶片y。在k = 2,众所周知,以这种方式引起的2个异化可以通过所谓的“4点状况”来表征。但是,如果k> 2个pachter和speyer(2004)最近提出了以下问题:给定任意K-异化,我们如何测试这张地图是否来自树?在本文中,我们提供了对这个问题的答案,表明对于k≥3,对于只有在从某些树中出现的每2 k个元素子集的限制时,k≥3k≥3k-siviliary会从树中出现。并且,2 k是至少可能的子集大小,以确保这种情况。作为推论,我们表明存在多项式时间算法,以确定何时从树中产生k-异化。我们还给出了6点状况,以确定何时从树上产生3异化,类似于上述4点状况。

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