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Succinct Data Structures for Representing Equivalence Classes

机译:用于代表等效类的简洁数据结构

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Given a partition of an n element set into equivalence classes, we consider time-space tradeoffs for representing it to support the query that asks whether two given elements are in the same equivalence class. This has various applications including for testing whether two vertices are in the same connected component in an undirected graph or in the same strongly connected component in a directed graph. We consider the problem in several models. - Concerning labeling schemes where we assign labels to elements and the query is to be answered just by examining the labels of the queried elements (without any extra space): if each vertex is required to have a unique label, then we show that a label space of Σ_(i=1)~n「n/i」 is necessary and sufficient. In other words, lg n+lg lg n+O(1) bits of space are necessary and sufficient for representing each of the labels. This slightly strengthens the known lower bound and is in contrast to the known necessary and sufficient bound of 「lg n」 for the label length, if each vertex need not get a unique label. - Concerning succinct data structures for the problem when the n elements are to be uniquely assigned labels from label set {1, . . ., n}, we first show that Θ(n(1/2)) bits are necessary and sufficient to represent the equivalence class information. This space includes the space for implicitly encoding the vertex labels. We can support the query in such a structure in O(lg n) time in the standard word RAM model. We then develop structures where the queries can be answered ? in O(lg lg n) time using O(n(1/2) lg n/ lg lg n) bits, and ? in O(1) time using O(n(1/2) lg n) bits of space. We also develop a dynamic structure that uses O(n(1/2) lg n) bits to support equivalence queries and unions in O(lg n/ lg lg n) worst case time or O(α(n)) expected amortized time where α(n) is the inverse Ackermann function.
机译:鉴于将N个元素的分区设置为等价类,我们考虑时空折衷,以表示支持该查询,询问两个给定元素是否在同一等价类中。这具有包括用于测试两个顶点是否在无向图中的相同连接分量中或在定向图中的相同强连接的组件中测试的各种应用。我们考虑了几种模型中的问题。 - 关于标签方案,我们将标签分配给元素和查询是通过检查查询元素的标签(没有任何额外的空间):如果每个顶点都需要唯一标签,那么我们就会显示一个标签Σ_(i = 1)〜n「n / i」的空间是必要的并且充分。换句话说,LG N + LG LG N + O(1)空间是必要的并且足以表示每个标签。这略微增强了已知的下限,与标签长度为标签长度的已知必要和足够的界限相反,如果每个顶点不需要获得唯一的标签。 - 关于n个元素是从标签集{1唯一分配标签时的问题的简洁数据结构。 。 。,n},我们首先表明θ(n(1/2))比特是必要的并且足以表示等同类信息。此空间包括隐式编码顶点标签的空间。我们可以在标准字RAM模型中以O(LG N)时间的这种结构支持查询。然后我们开发可以回答查询的结构?在O(lg lg n)时间使用O(n(1/2)lg n / lg n)位,并且?在O(1)时间使用O(n(1/2)lg n)空间位。我们还开发了一种动态结构,它使用O(n(1/2)lg n)位来支持O(lg n / lg lg n)最坏情况时间或o(α(n))预期摊销时间的等效查询和联合其中α(n)是逆Ackermann函数。

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