O(log(N))算法视角:基于h-extra边连通性的折叠交叉超立方体可靠性评估

Hengji Qiao, Mingzu Zhang, Wenhuan Ma, Xing Yang
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引用次数: 1

摘要

互连网络可以建模为连通图[公式:见正文]。互连网络的可靠性对多处理器系统至关重要。为了准确地反映各种现实网络情况,过去已经引入了几种条件边连接,其中[公式:见文本]-额外边连接就是这样一种条件边连接。[公式:见文]的[公式:见文]-额外边-连通性,用[公式:见文]表示,是错误边的最小基数,这些错误边的删除将图[公式:见文]与每个包含至少[公式:见文]处理器的结果组件断开连接。一般来说,对于连通图[公式:见文],确定图是否存在[公式:见文]-额外的边缘切割是[公式:见文]-很难。折叠交叉超立方体[公式:见文]是带有[公式:见文]处理器的交叉超立方体[公式:见文]的变体。本文在挖掘了折叠交叉超立方体的层结构后,研究了[公式:见文]的一些递归性质,在这些递归性质的基础上,设计了[公式:见文]-折叠交叉超立方体额外边连通性的有效[公式:见文]算法,该算法可以确定每个正整数[公式:见文]的精确值和[公式:见文]的最优性[公式:见文]。我们的研究结果彻底解决了这个问题。
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An O(log(N)) Algorithm View: Reliability Evaluation of Folded-crossed Hypercube in Terms of h-extra Edge-connectivity
An interconnection network can be modelled as a connected graph [Formula: see text]. The reliability of interconnection networks is critical for multiprocessor systems. Several conditional edge-connectivities have been introduced in the past for accurately reflecting various realistic network situations, with the [Formula: see text]-extra edge-connectivity being one such conditional edge-connectivity. The [Formula: see text]-extra edge-connectivity of [Formula: see text], denoted by [Formula: see text], is the minimum cardinality of faulty edges whose deletion disconnects the graph [Formula: see text] with each resulting component containing at least [Formula: see text] processors. In general, for a connected graph [Formula: see text], determining whether the graph exists an [Formula: see text]-extra edge-cut is [Formula: see text]-hard. The folded-crossed hypercube [Formula: see text] is a variation of the crossed hypercube [Formula: see text] with [Formula: see text] processors. In this paper, after excavating the layer structure of folded-crossed hypercube, we investigate some recursive properties of [Formula: see text], based on some recursive properties, an effective [Formula: see text] algorithm of [Formula: see text]-extra edge-connectivity of folded-crossed hypercube is designed, which can determine the exact value and the [Formula: see text]-optimality of [Formula: see text] for each positive integer [Formula: see text]. Our results solve this problem thoroughly.
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