Fault-tolerant Routing in Metacube - CiteSeerX

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Table 1. Multi-channel examplesExample 1 Example 2 Example 3 Example 4(0000000000,0001000000) (0000000000,0001011100) (0000000000,0100001111) (0000000001,1101000000)P 0 ( j = 0) P 1 ( j = 1) P 0 ( j = 0) P 1 ( j = 1) P 0 ( j = 0) P 1 ( j = 1) P 0 ( j = 0) P 1 ( j = 1)0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000001 00000000010100000000 1000000000 0100000000 1000000000 0100000000 1000000000 0100000001 10000000010100000100 1000010000 0100000100 1000100000 0100000100 1000010000 0100000101 10000100011100000100 1100010000 1100000100 1100100000 0100001100 1100010000 1100000101 11000100011101000100 1101010000 1101000100 1101100000 1100001100 0100010000 1101000101 11010100011001000100 0101010000 1001000100 0101100000 1000001100 0100010100 1001000101 01010100010001000100 0001010000 1001010100 0101100100 0000001100 0100011100 0001000101 00010100010101000100 1001010000 0001010100 0101101100 0000001101 0000011100 0001000100 00010100000101000000 1001000000 0101010100 0001101100 0000001111 0000011101 0101000100 10010100000001000000 0001000000 0101011100 1001101100 0100001111 0000011111 0101000000 10010000000001011100 1001111100 1000011111 1101000000 11010000001001011100 10000011110001011100 11000011110100001111end.v i ) is constructed by the cross-edges specifiedby Z j .Example 1: Assume m = k = 2. c u = c v = 0, C u ={0,1,2,3} and C v = {64,65,66,67}. According to the algorithm,the four node pairs are (0,64), (1,65), (2,66),(3,67). We only show the two paths for (0,64), so thesimplified notations u = 0 and v = 64 are used (ignored ifor i = 0,1,2,3). Since M u [1] = M v [1] = 00 and M u [2] =M v [2] = 00, we choose y = 2 for both j = 0 and 1 as akey-bit of type 1. The two paths from node 0 to node64 of class 0 are shown in Table 1. The key-bit is shownwith boldface. |P 0 | = |P 1 | = d(u,v) + 4 = 5 + 4 = 9, whered(0,64) = H(0,64) + 2 2 = 1 + 4 = 5.Example 2: Assume m = k = 2. c u = c v = 0, C u ={0,1,2,3} and C v = {92,93,94,95}. (0,92) a pair. Let u =0 and v = 92. Since M u [1] = 00 and M v [1] = 11, we choosey = 2 for j = 0 as a key-bit of type 2. Since M u [2] = 00 andM v [2] = 01, we choose y = 3 for j = 1 as a key-bit of type 1.The two paths from node 0 to node 92 are shown in Table 1.|P 0 | = d(u,v) + 2 = 10 and |P 1 | = d(u,v) + 4 = 12, whered(0,92) = H(0,92) + 2 2 = 4 + 4 = 8.Example 3: Assume m = k = 2. c u = 0, c v = 1, C u ={0,1,2,3} and C v = {3,7,11,15}. The four pairs of thetwo clusters are (0,15), (1,11), (2,7), (3,3). To constructthe two paths for pair (0,15), we notice that |c u ⊕ c v | = 1and c (0)u = c v . Therefore, we construct path P 0 by Case 2.1.We choose y = 2 for j = 1. The two paths from node u =0000000000 to node v = 0100001111 are shown in Table 1.|P 0 | = d(u,v)+1 = 9, |P 1 | = d(u,v)+4+|Z 1 | = 8+4+1 =13, where d(0,15) = H(0,15) + 2 2 = 4 + 4 = 8.Example 4: Assume m = k = 2. c u = 0, c v = 3, C u ={0,1,2,3} and C v = {0,64,128,192}. The four pairs of thetwo clusters are (0,0), (1,64), (2,128), (3,192). Since|c u ⊕ c v | = 2, we construct path P 0 and P 1 for pair (1,64)by Case 2.2. We choose y = 2 for both j = 0 and 1. Sincec (0)u = c (1)v and c (1)u = c (0)v , we have |Z 0 | = |Z 1 | = 1. The twopaths from node u = 0000000001 to node v = 1101000000are shown in Table 1. |P 0 | = |P 1 | = d(u,v) + 4 + |Z 1 | = 5 +4 + 1 = 10, where d(1,64) = H(1,64) + 2 2 = 1 + 4 = 5.We first prove that the k paths (u : u ( j) → v ( j) : v), 0 ≤j ≤ k − 1, constructed in Case 1, are disjoint. For any twopaths P j1 and P j2 , if one of them has a key-bit of type 1 thenit is clear that they cannot intersect each other since no otherpath will change the value of that key-bit. If the key-bits ofthe two paths are of type 2, from the definition of type 2key-bit, we know that the two paths cannot intersect eachother when we use the nodes in the hamiltonian cycle that∈ H k , path P j2 updatesthe values in field M[c ( j1) ] and the values in the field M[c j2) ]was partially updated only (the key-bit changed and otherbits unchanged), however, the values in the field M[c ( j2) ]for P j1 is either unchanged or fully updated. Therefore, twopaths cannot meet at the nodes of class c ( j1) . Similarly, theyare not c ( j 1)u nor c ( j 2)u . At node c ( j 1)ucannot meet at the nodes of class c ( j 2)u . We conclude thatthe k paths are disjoint. Next, since each pair (u i ,v i ) has aunique node_id and this unique node_id will become partof the cluster_id of the nodes in path (u (p) → v (q) ), differentpaths cannot pass through the same cluster. Therefore, allk2 m paths for connecting C u and C v are disjoint for Case 1.ii348
To show the k paths for each node pair constructedin Case 2 are disjoint, we divide each path into to parts,→ w i, j ) and (w i, j → v i ). Since the first part of the pathincludes a signature (except P q in Case 2.1), they shouldbe disjoint following the same argument as in Case 1 (theargument is true even one of the paths does not carry signature).Moreover, it is also disjoint with the second part(u ( j)iof other paths because of its unique signature. The secondparts of the paths P j , 0 ≤ j ≤ k − 1, contain onlythe cross-edges that are identical to the disjoint class pathsZ j , they are also disjoint. Therefore, the k paths constructedby the algorithm are disjoint. To prove that the hsets of paths (u i → v i ), 0 ≤ i ≤ 2 m − 1 are disjoint also,we argue as follows: First for each u i , consider the paths(u i → u ( j)i→ w i → w ′ i ), where c w ≠ c v and c next(wi ) = c v ,c w ′i≠ c v and c prev(w ′i ) = c v . From the algorithm, we knowthat the h sets of paths (u i → w i ) are disjoint since the valueM ui [c u ] is different for every u i ∈ C u and M ui [c u ] is part ofthe cluster_id of every node in path (u ( j)i→ w ′ i ). In path(w i → w ′ i ), the value M u i[c v ] is changed to M vi [c v ]. SinceM ui [c u ]⊕M vi [c v ] = M ui [c v ]⊕M vi [c u ], M vi [c v ] takes differentvalues for every u i . That is, node w ′ i is in distinct clustersfor every u i . Since M vi [c v ] is part of the cluster_id of everynode in path (w ′ i → v i), the h sets of paths (w ′ i → v i)are also disjoint. Therefore, we conclude that all paths forconnecting the clusters C u and C v are disjoint for Case 2.From the path construction in Case 1, the length of thelongest path (u i → v i ) is |(u i → v i )| = d(u i ,v i ) + 4, whered(u i ,v i ) = H(u i ,v i )+2 k , H(u i ,v i ) is Hamming distance betweenu i and v i . From the path construction in Case 2,the longest length of the paths (u i → v i ) is |(u i → v i )| =d(u i ,v i ) + 4 + max 0≤ j≤k−1 {|Z j |} ≤ d(u i ,v i ) + m + 5. Wesummarize the results in the following theorem.Theorem 1. Given any two clusters C u and C v in MC(k,m),we can find a multi-channel cube C u ∪C v ∪ E, where E isa set of k2 m disjoint paths connecting C u and C v , k pathsfor each node pair (u i ,v i ), u i ∈ C u , v i ∈ C v , and the lengthof the paths in E is at most max(d(u i ,v i )) + m + 5, for i =0,1,...,2 m − 1.5. Fault-tolerant routingWith the multi-channel cube structure, the fault-tolerantrouting problem can be solved efficiently in an MC(k,m) assumingthat there are up to k+m−1 faulty nodes. The faulttolerantrouting from source s to destination t in hypercubehas been solved in [4]. The hypercube algorithm is summarizedbelow. First, we partition the n-cube into two (n − 1)-dimensional subcubes along a dimension k such that s andt are in the distinct subcubes. If the subcube containing thas less faulty nodes then we route s to the subcube containingt by a fault-free path of length at most 2: (s → s (k) )or (s → s ( j,k) ) for some j, 0 ≤ j ≤ n − 1, j ≠ k. Then theproblem is reduced to the sub-problem in the subcube containingt. Repeat this process at most logn times until thesubcube contains no faulty node. The path (s → s ( j,k) ) is abad candidate if the jth bit of s is the same as that of t (abad candidate will increase the length of the routing path).The algorithm guarantees that the bad candidate is used byat most one partition. It was shown in [4] that given twononfaulty nodes s and t, and up to n − 1 faulty nodes inan n-cube, we can find a fault-free path of length at mostd(s,t) + 2 in O(n) time.Under the conditions k ≤ m and the number of faultynodes in C s ∪C t is at most m, we show that an efficient algorithmfor fault-tolerant routing from s to t in MC(k,m)can be found. For most of the interesting MC(k,m), we havek ≤ m as indicated in the previous sections. Statistically, thecondition that there are at most m faculty nodes in the twoclusters is held with very high probability. Assuming uniformdistribution, the probability of at most m fault nodes in) (/2 m2 k +kC s ∪C t is m ∑( m+k−1ii=0)( 2 m2 k +k −(m+k−1)2 m+1 −i2 m+1 )≈ 1. For example,the probability is 3186303/3186304 = 0.99999969in the case of k = m = 2. An algorithm for fault-tolerantrouting under the conditions above in an MC(k,m) is givenbelow.Algorithm 3 (Fault_Tolerant_Routing(k,m,s,t))beginCase 1: C s = C t . If C s contains at most m − 1 faulty nodesthen we apply the hypercube algorithm. Otherwise,at least m faulty nodes are in C s . We find k disjointpaths (s → t) outside the cluster C s as follows: P j =(s : s ( j) : s ( j,k) : s ( j,k, j) → t ( j,k, j) : t ( j,k) : t ( j) : t), 0 ≤j ≤ k − 1, where s ( j,k, j) and t ( j,k, j) are in the samecluster ≠ C s . Since there at most k − 1 faulty nodesoutside the cluster C s , there exists a fault-free path P j .Case 2: C s ≠ C t . We create a multi-channel cube with C sand C t as described in the previous subsection. Sincethere are at most m fault nodes in C s ∪C t , we can finda fault-free path of length at most d(s,t) + 2 in O(m)time in the (m + 1)-cube formed by C s ∪C t . If the kpaths (u → v) of the multi-channel cube, which correspondto a single (m + 1)th dimensional cube-edge,are all faulty, then we mark v as a faulty node. Ifthe number of faulty nodes in C s ∪ C t is m, then nomarked faulty node will be created. Otherwise, sincea marked faulty node consumes k faulty nodes outsideclusters C s and C t , the total number of faulty andmarked faulty nodes will be at most m + 1 − 1 = m.Therefore, we can apply the hypercube algorithm,and find a fault-free path in the (m + 1)-cube formedby C s ∪C t . Replacing the (m+1)th dimensional cubeedge(u,v) in the fault-free path (s → t) by a fault-free349
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