Independent Spanning Trees on Crossed Cubes

碩士 === 國立臺北商業技術學院 === 資訊與決策科學研究所 === 102 === A set of spanning trees in a graph is said to be independent (ISTs for short) if all the trees are rooted at the same node r and for any other node v(≠r), the paths from v to r in any two trees are node-disjoint except the two end nodes v and r. For an n-...

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Bibliographic Details
Main Authors: Jhen-Ding Wang, 王溱鼎
Other Authors: Jou-Ming Chang
Format: Others
Language:en_US
Published: 2014
Online Access:http://ndltd.ncl.edu.tw/handle/59244316344879651711
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Summary:碩士 === 國立臺北商業技術學院 === 資訊與決策科學研究所 === 102 === A set of spanning trees in a graph is said to be independent (ISTs for short) if all the trees are rooted at the same node r and for any other node v(≠r), the paths from v to r in any two trees are node-disjoint except the two end nodes v and r. For an n-connected graph, the independent spanning trees problem asks to construct n ISTs rooted at an arbitrary node of the graph. Recently, Zhang et al. [Y.-H. Zhang, W. Hao, and T. Xiang, Independent spanning trees in crossed cubes, Information Processing Letters, 113 (2013) 653–658] proposed an algorithm to construct n ISTs with a common root at node 0 in an n-dimensional crossed cube CQn. However, it has been proved by Kulasinghe and Bettayeb [P.D. Kulasinghe and S. Bettayeb, Multiply-twisted hypercube with 5 or more dimensions is not vertex transitive, Information Processing Letters, 53 (1995) 33–36] that the CQn (a synonym called multiply-twisted hypercube in that paper) fails to be node-transitive for n≥5. Thus, the result of Zhang et al. does not really solve the ISTs problem in CQn. In this thesis, we revisit the problem of constructing n ISTs rooted at an arbitrary node in CQn. As a consequence, we show that the proposed algorithm can be parallelized to run in O(log N) time using N = 2^n nodes of CQn as processors.