首页> 外文期刊>The Journal of Combinatorial Mathematics and Combinatorial Computing >AN ALGEBRAIC APPROACH FOR FINDING DISJOINT PATHS IN THE ALTERNATING GROUP GRAPH
【24h】

AN ALGEBRAIC APPROACH FOR FINDING DISJOINT PATHS IN THE ALTERNATING GROUP GRAPH

机译:在交替群图中寻找不连续路径的一种代数方法

获取原文
获取原文并翻译 | 示例

摘要

For the purpose of large scale computing,we are interested in linking computers into large interconnection networks. In order for these networks to be useful,the underlying graph must possess desirable properties such as a large number of vertices, high connectivity and small diameter. In this paper, we are interested in the alternating group graph, as an interconnection network, and the k-Disjoint Path Problem. In 2005, Cheng, Kikas and Kruk showed that the alternating group graph, AG_n has the (n - 2)-Disjoint Path Property. However, their proof was an existence proof only. They did not show how to actually construct the (n - 2) disjoint paths. In 2006, Boats, Kikas and Oleksik developed an algorithm for constructing the three disjoint paths in the graph AG_5. Their algorithm exploited the hierarchal structure of AG_n to construct the paths. In this paper we develop a purely algebraic algorithm that constructs the (n - 2) disjoint paths from scratch. This algebraic approach can be used for other Cayley graphs such as the split-star and the star graphs. Indeed, we believe that our approach can be used for any Cayley graph. We close with remarks on possible research directions stemming from this work.
机译:为了进行大规模计算,我们对将计算机链接到大型互连网络感兴趣。为了使这些网络有用,基础图必须具有所需的属性,例如大量的顶点,高连通性和较小的直径。在本文中,我们对作为互通网络的交替群图和k-不相交路径问题感兴趣。在2005年,Cheng,Kikas和Kruk证明了交替群图AG_n具有(n-2)-不相交路径性质。但是,他们的证明仅仅是存在证明。他们没有显示如何实际构造(n-2)个不相交的路径。 2006年,Boats,Kikas和Oleksik开发了一种算法,用于构造图AG_5中的三个不相交的路径。他们的算法利用AG_n的层次结构来构造路径。在本文中,我们开发了一种纯代数算法,该算法从头开始构造(n-2)个不相交的路径。此代数方法可用于其他Cayley图,例如裂星图和星图。确实,我们相信我们的方法可以用于任何Cayley图。最后,我们将对这项工作可能产生的研究方向进行评论。

著录项

相似文献

  • 外文文献
  • 中文文献
  • 专利
获取原文

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号