Fault Tolerant Routing in Star Graphs Using Fault Vector

Author(s):  
Rajib K. Das
Keyword(s):  
1996 ◽  
Vol 06 (01) ◽  
pp. 127-136 ◽  
Author(s):  
QIAN-PING GU ◽  
SHIETUNG PENG

In this paper, we give two linear time algorithms for node-to-node fault tolerant routing problem in n-dimensional hypercubes Hn and star graphs Gn. The first algorithm, given at most n−1 arbitrary fault nodes and two non-fault nodes s and t in Hn, finds a fault-free path s→t of length at most [Formula: see text] in O(n) time, where d(s, t) is the distance between s and t. Our second algorithm, given at most n−2 fault nodes and two non-fault nodes s and t in Gn, finds a fault-free path s→t of length at most d(Gn)+3 in O(n) time, where [Formula: see text] is the diameter of Gn. When the time efficiency of finding the routing path is more important than the length of the path, the algorithms in this paper are better than the previous ones.


2015 ◽  
Vol 93 (10) ◽  
pp. 1695-1707 ◽  
Author(s):  
Yali Lv ◽  
Yonghong Xiang ◽  
Jianxi Fan
Keyword(s):  

2006 ◽  
Vol 183 (2) ◽  
pp. 972-979 ◽  
Author(s):  
Min Xu ◽  
Xiao-Dong Hu ◽  
Qiang Zhu
Keyword(s):  

1997 ◽  
Vol 08 (02) ◽  
pp. 127-142 ◽  
Author(s):  
Shuo-Cheng Hu ◽  
Chang-Biau Yang

The capability of fault tolerance is one of the advantages of multiprocessor systems. In this paper, we prove that the fault tolerance of an n-star graph is 2n-5 with restriction to the forbidden faulty set. And we propose an algorithm for examining the connectivity of an n-star graph when there exist at most 2n - 4 faults. The algorithm requires O(n2 log n) time. Besides, we improve the fault-tolerant routing algorithm proposed by Bagherzadeh et al. by calculating the cycle structure of a permutation and the avoidance of routing message to a node without any nonfaulty neighbor. This calculation needs only constant time. And then, we propose an efficient fault-tolerant broadcasting algorithm. When there is no fault, our broadcasting algorithm remains optimal. The penalty is O(n) if there exists only one fault, and the penalty is O(n2) if there exist at most n - 2 faults.


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