The Conditional Fault-Diameter of the K-ary n-Cube

We obtain the conditional fault diameter of the k-ary n-cube interconnection network. It has been previously shown that under the condition of forbidden faulty sets (i.e. assuming each non-faulty node has at least one non-faulty neighbor), the k-ary n-cube, whose connectivity is 2n, can tolerate up to 4n-3 faulty nodes without becoming disconnected. We extend this result by showing that the conditional fault-diameter of the k-ary n-cube is equal to the fault-free diameter plus two. This means that if there are at most 4n-3 faulty nodes in the k-ary n-cube and if every non-faulty node has at least one non-faulty neighbor, then there exists a fault-free path of length at most the diameter plus two between any two non faulty nodes. We also show how to construct these fault-free paths. With this result the k-ary n-cube joins a group of interconnection networks (including the hypercube and the star-graph) whose conditional fault diameter has been shown to be only two units over the fault-free diameter.


ABSTRACT:
We obtain the conditional fault diameter of the k-ary n-cube interconnection network.It has been previously shown that under the condition of forbidden faulty sets (i.e.assuming each non-faulty node has at least one non-faulty neighbor), the k-ary n-cube, whose connectivity is 2n, can tolerate up to 4n-3 faulty nodes without becoming disconnected.We extend this result by showing that the conditional fault-diameter of the k-ary n-cube is equal to the fault-free diameter plus two.This means that if there are at most 4n-3 faulty nodes in the k-ary n-cube and if every non-faulty node has at least one non-faulty neighbor, then there exists a fault-free path of length at most the diameter plus two between any two non faulty nodes.We also show how to construct these fault-free paths.With this result the k-ary n-cube joins a group of interconnection networks (including the hypercube and the star-graph) whose conditional fault diameter has been shown to be only two units over the fault-free diameter.KEYWORDS: Fault-tolerance, multiprocessor systems, interconnection architectures, kary n-cube, torus.

Introduction
he node connectivity and the fault diameter have been used as measures of the fault-tolerance of interconnection networks.These measures however do not reflect the real resilience of these networks.It is true that when the number of faulty nodes is equal to the connectivity the network may become disconnected.However, this is very unlikely to happen since only very special fault distributions of these faults cause disconnection.For instance, in the k-ary n-cube, the nodeconnectivity is 2n and 2n faulty nodes may cause disconnection.But the network becomes disconnected only when all the 2n faults are adjacent to the same node which is very improbable.
In an attempt to better quantify the fault resilience of a network, the concept of forbidden faulty sets has been introduced by Esfahanian (1989).The idea is to assume that each node has at least one non-faulty neighbor.Under this forbidden faulty set condition, the number of tolerable faulty nodes is significantly larger with a slight increase in the fault diameter.Esfahanian (1989) has proven that for the binary n-cube, whose connectivity is n, 2n-3 nodes can fail (under the forbidden faulty set condition) without disconnecting the network.Latifi (1993) has then showed that the corresponding conditional fault diameter increases only by 2 over the fault-free diameter.In Rouskov et al. (1996) similar results for the star graph network have been established.Latifi et al. (1994) have generalized this idea by assuming that each node has at least k non-faulty neighbors.Similar results for the m-ary generalized n-cube network have also been obtained in Wu (1998).
It has been previously shown (Day, 2004) that for k ≥ 4 and n ≥ 2, the k-ary n-cube, whose connectivity is 2n, can tolerate up to 4n-3 faulty nodes without becoming disconnected.The corresponding conditional node connectivity is therefore 4n-2.The result for the remaining small values of k and n has also been obtained in Day (2004).We extend these results in this paper by showing that the conditional fault-diameter of the k-ary n-cube is equal to / 2 2 n k ⎢ ⎥ ⎣ ⎦ + .We therefore establish that the k-ary n-cube, like the hypercube and the star-graph, has conditional fault diameter equal to two plus the fault-free diameter.
This paper is organized as follows: section 2 presents some notations; section 3 obtains some preliminary results useful for the derivation of the conditional fault diameter in section 4. Section 5 concludes the paper.

Notations
The k-ary n-cube  H (X,Y) is not the length of a shortest path between X and Y.The length of a shortest path between X and Y is equal to the Lee distance (Bose et al. 1995) given by: d Q nodes whose addresses are of the form Q we denote by p A , 0 ≤ p < k, the node of the form p A = p a n-2 ... a 0 obtained from A by replacing its leftmost digit a n-1 by p. Definition 1: The node-connectivity C(G) (or point-connectivity) of a graph G is the minimum number of nodes of G whose removal results in a disconnected or trivial graph.It has been shown in Day and Al-Ayyoub (1997 Definition 2: The fault-diameter FD(G) of a graph G is the maximum distance between any two nodes of G in the presence of at most C(G)-1 faulty nodes.It has been shown in Day and Al-Ayyoub (1997 Definition 3: The conditional node connectivity CC(G) of a graph G is the minimum number of nodes of G whose removal results in a disconnected or trivial graph, provided that each of the remaining nodes has at least one adjacent node in G that is not removed.
Definition 4: The conditional fault diameter CFD(G) of a graph G is the maximum distance between any two nodes of G in the presence of at most CC(G)-1 faulty nodes, provided that each of the nonfaulty nodes has at least one non-faulty adjacent node in G.
We use the abbreviation FFSC for denoting the forbidden faulty set condition which corresponds to the requirement that each non-faulty node must have at least one non faulty neighbor.

Preliminary Results
In this section we present some preliminary results that will be used in the next section for the derivation of the conditional fault diameter of the k-ary n-cube.We start by the following result which has been proven in by Day (2004): Theorem 1.The conditional node connectivity of the k-ary n-cube for n ≥ 2 and k ≥ 4 is: We will denote by Q .Additional results about node disjoint paths in the k-ary n-cube can be found in Day and Al-Ayyoub (1997).
There are a total of 2n node-disjoint paths between X and Y of which: 1) h paths have length l, 2) 2n-2h paths have length l+2, and 3) for each i such that w i > 0, there is a path of length l+k-2w i (h paths).
In preparation for the proof of the conditional fault-diameter of the k-ary n-cube given in the next section, we present the corresponding result for the special case of the k-ary 2-cube (also called the k-torus).The k-torus is a wrap-around mesh which consists of k rows and k columns.Each row and each column consists of a cycle of k nodes.There are four node-disjoint paths connecting any two nodes of the k-torus.The connectivity of the k-torus is therefore equal to 4. Any two nodes on the same cycle (row or column) are connected by two paths along the cycle.One is called the shortest path on the cycle and the other is called the longest path.We show that under the FFSC, the k-torus, whose connectivity is 4, can tolerate up to 5 faulty nodes without becoming disconnected.The conditional node connectivity of the k-torus is therefore 6.We also show that the conditional faultdiameter of the k-torus is equal to its fault-free diameter plus two.
The proof of the following theorem consists of a lengthy manual construction of fault-free paths between any two non faulty nodes of the k-torus considering all possible relative locations of the source and destination nodes and those of the 5 faulty nodes.For brevity, we omit this lengthy construction here.Interested readers can find it in Touzene and Day (2005).Theorem 3. The conditional fault diameter of the k-torus under the FFSC is 2 /2 2 k The next result states that it is always possible to construct between any two non-faulty nodes of k n Q a fault-free path of length at most the diameter if the number of faults does not exceed two.This result will be used in the next section to prove the conditional fault diameter of , n ≥ 2 and if there are two or less faulty nodes, then there exists between any two non-faulty nodes at least one fault-free path of length at most / 2 n k Proof.Let X and Y be any two distinct non-faulty nodes in and k≥4).
Hence we have at least 3 paths each of length at most / 2 n k ⎢ ⎥ ⎣ ⎦ and at least one of these paths must be fault-free.
• If h = 2 and n = 2, the result is derived from the proof of the fault-diameter of the 2-ary n-cube (k-torus) available in Touzene and Day (2005).
and k≥4).Hence we have at least 4 paths each of length at most / 2 n k ⎢ ⎥ ⎣ ⎦ , one of which must be fault-free.
• If h ≥ 3 then at least one of the first set of h paths each of length at most / 2 / 2 h k n k

The Conditional Fault Diameter of the k-Ary n-Cube
The following result establishes a lower bound on the conditional fault-diameter of Proof.Let X and Y be two nodes in Z be a neighbor of Y and let T be a neighbor of Z other than Y (see Figure 1).Assume that all 2n-2 neighbors of Z other than Y and T are faulty and that all 2n-1 neighbors of T other than Z are faulty.Notice that the former set of 2n-2 faults and the latter set of 2n -1 faults must be disjoint since k ≥ 4 (otherwise, if F were a faulty node appearing in both sets then Z F T would form a cycle of length 3 which is impossible for k ≥ 4).Assume there are no other faults.The total number of faults is therefore 4n-3.A path from T to X must first go through Z then through Y then from Y to X in at least / 2 n k ⎢ ⎥ ⎣ ⎦ moves and, hence, will be of length of at least / 2 2 n k ⎦ .Now we establish an upper bound on the conditional fault-diameter of Proof .We proceed by induction on n.The induction basis (n = 2) is given by Theorem 3. Now we prove the result for n ≥ 3 assuming it is true for smaller n values.Consider two arbitrary non-faulty Assume the number of faulty nodes is at most 4n-3 and that the FFSC is satisfied.Our aim is to show that it is always possible to find a fault-free path between X and Y of length at most / 2 2 n k ⎢ ⎥ ⎣ ⎦ + .We distinguish the following cases: Case 1. there exists m, 0 ≤ m < n, such that x m = y m = i (assume without loss of generality m = n-1), then both nodes X and Y belong to the sub-graph , 1 Q − has at most 4(n-1)-3 = 4n-7 faults.By induction hypothesis there must exist a fault- Case 1.2.
, 1 Q − has more than 4n-7 faults.This means that there are at most 3 faults outside , 1 Case 1.2.1.each of X and Y has all its 2n-2 neighbors inside , 1 There must exist at most one faulty node outside , 1 ). Similarly for Y i-1 and Y i+1 (see Figure 2).

Since either
, 1 1 is fault-free, at least one of the two paths: is fault-free and is of length at most ( 1) / 2 2 n k graph except for two edges).The symbol || is used to denote path concatenation.
Case 1.2.2.X or Y has a non-faulty neighbor inside , 1 has at most one faulty node (the other case is similar).
Case 1.2.2.1.X i-1 is faulty (hence X i-1 is the only faulty node in , 1 1 3).By Lemma 1, there exists at least one fault-free path π (X' i-1 ,Y i-1 ) of length at most ( 1) /2 Since there is at most one fault in 1 , 1 Case 1.2.2.3.X i-1 non-faulty and Y i-1 faulty (the only fault in 1 , 1 4).By Lemma 1, there exists at least one fault-free path π (X i-1 ,Y' i-1 ) of length at most ( 1) /2 n k In this case Y i+1 must be non-faulty (by the FFSC).We know that 1 , 1 If it has only one faulty node then it is possible to enter 1 , 1 or from X to X' then to X' i+1 requiring in both cases at most two moves (see Figure 5).We can then use Lemma 1 in 1 , 1 X' i+1 to Y i+1 and then make a final move from Y i+1 to Y. Hence we have a fault-free path from X to Y of length at most ( 1) /2 3  5), otherwise the total number of faulty nodes would include 2n-2 faulty neighbors of Y inside for a total of 4n-2 which exceeds 4n-3.Let X" i+1 denote the neighbor of X" located in 1 , 1 . At least one of the three nodes X i+1 , X' i+1 or X" i+1 must be non-faulty since there are only two faults in 1 , 1 starting from X in at most two moves.We can then use Lemma 1 in 1 , 1 and then make a final move from Y i+1 to Y. Hence we have in total a fault-free path from X to Y of length at most ( 1) /2 3 n k Assume without loss of generality that i, i+1, … j is the shortest path between i and j on the cycle 0, 1, 2, … k- Π be the set of 2n-1 minimum-distance paths each joining either X to j X or any of the 2n- 6).The notations x π (X, j X ) and x π (X', j X ' ) will be used to denote these 2n-1 paths of x Π .Similarly, let y Π be the set of 2n-1 minimum-distance paths each joining either ⎦ and they are all mutually node-disjoint.Figure 7 illustrates the 4n-2 paths of x y Π ∪ Π for n = 3, k = 7, X = 000, and Y = 333.These 4n-2 paths are mutually disjoint and therefore at least one of them must be completely fault-free (since the total number of faults is at most 4n-3).Let π be this fault-free path and assume that π ∈ x Π (the case π ∈ y Π is symmetric).Let X* denote the end node of * X located in e. X* is either j X or one of the 2n-2 nodes denoted j X ' ).From the previous discussion we have: Notice also that at most one of the sub-graphs can possibly contain more than 4n-7 faulty nodes otherwise the number of faults would exceed 4n-3 since n ≥ 3 (by assumption of the induction step).
− has more than 4n-7 faults (hence at most 3 faults outside In this case we can build a path: − .Notice here that the induction hypothesis applies in − ) is at most 3 which is less than 4(n-1)-2 (since n ≥ 3, by induction step assumption).Furthermore, the FFSC is satisfied in − to make all the neighbors of one of its nodes faulty which is not possible since 2n-2 > 3 (because n ≥ 3) and there are at most 3 faults in has at most two faults, Lemma 1 can be used to go from one of the nodes X since the initial move is a minimum distance move) and then from which is not faulty (by FFSC).We can, therefore, build a fault-free

Q
(this is justified by the fact that the paths in y Π are minimum distance paths).Therefore, at least one of the following two paths must be fault-free and is of length at most / 2 2 n k for a total of 4n-2 faults which exceeds the number of faults 4n-3.In this case the path:

Conclusion
We have contributed to the study of the fault-tolerance of the k-ary n-cube interconnection network by establishing its conditional fault-diameter under the FFSC (i.e., assuming that each nonfaulty processor has at least one non-faulty neighbor).We have shown that under this condition and for k ≥ 4 and n ≥ 2, the conditional fault-diameter of the k-ary n-cube is / 2 2 n k ⎢ ⎥ ⎣ ⎦ + .This means that if there are less than 4n-2 faults in the k-ary n-cube and if every non faulty node has at least one nonfaulty neighbor, then there is a fault-free path of length at most / 2 2 n k ⎢ ⎥ ⎣ ⎦ + between any two non- Y) a minimum-length path between nodes X and Y in k n Q .The following theorem established in by Bose et al. (1995) about the existence of a complete set of node disjoint paths between any two nodes of the k-ary n-cube k n Q will be used in the next section for establishing the conditional fault diameter of k n
Figure 7.An Example of the 4n-2 node-disjoint paths in for 3 and 7. n k x y Π ∪ Π = = minimum distance moves from X to Y along two different dimensions (these neighbors exist since n ≥ 3).Consider the 3 paths joining X to the sequence of moves i → i-1 → i-2 → … → j+1 in dimension n-1.These paths are node-disjoint and each is of length at most / moves (in fact if we had started with the edge X→

−
that X and Y differ in all dimensions and that contains at most one faulty node).In fact it is possible to find a non faulty neighbor i at dimension n-1 along the opposite direction of that followed along yπ ( i Y ,Y) (i.e. in the direction i → i-1 → i-2 … → j+1).Going in this direction along dimension n-1 requires at most one extra move beyond the minimum distance since X and Y are diametrically opposite along dimension n-1.The alternative path can be completed by the two all moves are along minimum distance paths except possibly for the one extra move on the path from The conditional fault-diameter of the k-ary n-cube is equal to / k ≥ 4 and n ≥ 2. Proof.Derived from combining the results of Lemma 2 and Lemma 3. .
In this case we can build a path: π that there are no faults in

2.2.4.
where π ( (i.e.X and Y are diametrically opposite along dimension n-1) then a different path is needed.An alternative path can be built going first from X to a non faulty neighbor