Constructing Node-Disjoint Routes in K-Ary N-Cubes

: In this paper, a method for constructing node-disjoint (parallel) paths in k -ary n -cube interconnection networks is described. We start by showing in general how to construct parallel paths in any Cartesian product of two graphs based on known paths in the factor graphs. Then we apply the general result to build a complete set of parallel paths (i.e., as many paths as the degree of the network) between any two nodes of a k -ary n -cube which can be viewed as the Cartesian product of complete graphs. Each of the constructed paths is of length at most 2 plus the minimum distance between the two nodes. These parallel paths are useful in speeding-up the transfer of large amounts of data between two nodes and in offering alternate routes in cases of faulty nodes.

There is a confusion in the literature about which graph is called the k-ary n-cube.For example, what is called the torus network in (Dally and Seitz) and (Gravano et al, 1994) is called the k-ary n-cube in (Linder and Harden, 1991) and (Bose et al, 1995) while in (Graham and Seidel, 1993) the k-ary n-cube refers to a different topology.As defined later, the k-ary n-cube considered in this paper is the same as in (Graham and Seidel, 1993).Graham and Seidel (1993) have shown that the k-ary ncube performs better in terms of one-to-all broadcasting and complete broadcasting than the star graph with a comparable number of nodes for practical network sizes.In general, the criteria used in evaluating interconnection networks relate to their topological properties of symmetry, scalability, low degree and diameter, efficient distributed routing algorithms, recursive structure, fault tolerance, low-cost embedding of other topologies, support of efficient broadcasting, and existence of parallel paths.This paper contributes to the study of k-ary n-cubes by presenting a method for constructing complete sets of node-disjoint (parallel) paths between arbitrary nodes.
These sets are complete in the sense that we obtain as many parallel paths between any two nodes as the degree of the network.Furthermore, the obtained paths are of optimal lengths plus at most 2 independently of the network size and the distance between the two nodes.Many research works have addressed such constructions of parallel paths on various interconnection networks such as the hypercube (Saad and Schultz, 1988), the star graph (Day and Tripathi, 1994), and the arrangement graph (Day and Tripathi, 1998).

PRELIMINARIES
In the following we present a number of definitions and notations used in the paper.
Definition 1 (Graham and Seidel, 1993): The k-ary n-cube Qis formed of N = k n nodes labeled by the base-k integers of the form a n-1 a n-2 ...a 0 , where 0 ≤ a i < k for 0 ≤ i < n.Two nodes are connected if and only if they differ by exactly one of their n digits.
It has been shown that Qhas degree (k-1)n and diameter n (Agrawal and Bhuyan, 1984).
Definition 2 (Leighton, 1992): and , where V and E are given by: Let N 1 , δ 1 , ∆ 1 be respectively the size (number of nodes), degree, and diameter of G 1 ; and let N 2 , δ 2 , ∆ 2 be respectively the size, degree, and diameter of G 2 .The size N, degree δ , and diameter ∆ of G 1 ⊗ G 2 are given by: N = N 1 .N 2 , δ = δ 1 + δ 2 , ∆ = ∆ 1 + ∆ 2 .The size and degree expressions are fairly obvious.As for the diameter expression it can be justified by noticing that a path between any two vertices u = <x u ,y u > and v = <x v ,y v > of G 1 ⊗ G 2 is composed of two types of edges: G 1edges (affecting the G 1 -component) and G 2 -edges (affecting the G 2 -component).If all the G 1edges (resp.the G 2 -edges) in the path from u to v are extracted and listed maintaining their relative order, we obtain a path from x u to x v in G 1 (resp.from y u to y v in G 2 ).Therefore u would be at a maximum distance from v in G 1 ⊗ G 2 if, and only if, x u is at maximum distance from x v in G 1 and y u is at maximum distance from y v in G 2 .
Let K k denote the complete graph with k nodes.If we perform the Cartesian product of K k by K k for n times, the k-ary n-cube Qwill be obtained.
Lemma 1: Qis isomorphic to Proof: This can be shown by induction on n.First, notice that Qis isomorphic to K k (trivial).
Next, we can obtain an isomorphism between Qand K k ⊗Qby mapping the node a n-1 a n-2 ...a 0 of Qto the node <a n-1 , a n-2 a n-3 ...a 0 > of K k ⊗Q.By definition 2, the nodes connected to the node a = Therefore a and b are connected if for exactly one position i, In this section we show that if G1 and G 2 are two regular graphs such that for each one of these graphs there exists a complete family of parallel paths (i.e. as many paths as the degree of the graph) between any two of its nodes, then the same property holds for G1⊗G2.
We start by introducing some notations used in the construction of parallel paths.A path π in some graph from a node u 1 to a node u m going through the intermediate nodes u 2 , u 3 ... u m-1 will be denoted by: π = u 1 u 2 ... u m .The length m of such a path π is denoted by | π |.We will also denote π -1 the path from u 1 to u m-1 obtained from π by removing its last edge.
Let x be a node in G 1 and let π = y 1 y 2 ... y m be a path in G 2 .We denote by <x, π > the path <x, .. x m is a path in G 1 and y is a node in G2, then < π , y> denotes the path <x 1 , y> <x 2 ,y> ... <x m , y> in G1 ⊗ G2.
Definition 2: A regular graph G of degree δ is said to have complete parallel paths with maximum length increase r if, and only if, for any two distinct nodes u and v of G, there exists a set of δ node-disjoint paths between u and v such that each of these paths is of length at most dist(u,v)+r, and at least one of these paths is of length dist (u,v), where dist (u,v) is the minimum distance between u and v in G.We use the abbreviation: G has CPP/r-MLI.
Theorem 1: Let π , π ,... π be δ 1 parallel paths between x u and x v in G 1 .These paths exist since x u ≠ x v , G 1 is regular of degree δ 1 and G 1 has CPP/r 1 -MLI.Each of these paths is of length at most dist(x u ,x v )+r 1 and at least one of them (say π ) is of length dist(x u ,x v ).Similarly, there exists δ 2 parallel paths π , π , ... π between y u and y v in G 2 .Each of these paths is of length at most dist(y u ,y v )+r 2 and at least one of them (say π ) is of length dist(y u ,y v ).We must therefore have: Let x i denote the last intermediate node on the path π for all i, 2 ≤ i ≤ δ 1 .In other words, is the last edge of π .Let also y i denote the last intermediate node on the path π for all i, 2 ≤ i ≤ δ 2 .Therefore, y i is such that (y i ,y v ) is the last edge of π .We construct the following δ 1 + δ 2 paths between u = <x u ,y u > and v = <x v ,y v >: The path π 1 is obtained by moving along the edges of π transforming x u into x v followed by the edges of π transforming y u into y v Therefore π 1 is of length dist(x u ,x v )+dist(y u ,y v ) which is equal to dist (u,v).The path π i , 2 ≤ i ≤ δ 1 , is obtained by following the edges of π except the last edge, followed by all the edges of π and finally the last edge of π .Since π is of minimum length and π is of length at most r 1 plus the minimum length, therefore π i is of length at most dist(u,v)+r 1 .
A similar argument is used to infer that the path πδ 1+1 is of length dist(u,v) and that each path πδ 1+i , 2 ≤ i ≤ δ 2 , is of length at most dist(u,v)+r 2 .
To show that the π i paths, 1 ≤ i ≤ δ 1 + δ 2 are node-disjoint we introduce the following notations.
Let S A be the set of all the intermediate (i.e.other than the source and the destination) nodes along all the paths π i , 1 ≤ i ≤ δ 1 , which have y u as G 2 -component.Notice that all these nodes appear first (at the left) in π i , for each i, 1 ≤ i ≤ δ 1 .Let S B be the set of all the remaining intermediate nodes along these paths.Similarly, let S C be the set of all the intermediate nodes along all the paths πδ 1+i , 1 ≤ i ≤ δ 2 which have x u as G 1 -component.These nodes appear first (at the left) in πδ 1+i .foreach i, 1 ≤ i ≤ δ 2 Finally, let S D be the set of the remaining intermediate nodes along these πδ 1+i paths.
None of the sets S A , S B , S C , and S D has a node that appears in it more than once.This is justified by the fact that the π paths are node-disjoint in G 1 and the π paths are node-disjoint in G 2 .The set S A is disjoint with each of S B , S C , and S D since all the nodes in S A have y u as G 2component which is not the case for any node in any of the sets S B , S C , and S D .Similarly, the set S C is disjoint with each of S B and S D since all the nodes in S C have x u as G 1 -component which is not the case for any node in any of the sets S B and S D .
It remains to show that S B and S D are disjoint.Let Sbe the subset of nodes of S B that appear in π 1 and let Sdenote the set S B -S. Define Sas the subset of nodes of S D that appear in πδ 1+1 and let Sdenote the set S D -S.Sand Sare disjoint since all the nodes of Shave x v as G 1 -component which is not the case for any node of S. Sand Sare disjoint since each node of Shas some intermediate node of π as G 2 -component which is not true for any node in S. Sand Sare disjoint since each node of Shas some intermediate node of π as G 1 -component which is not true for any node in S.
Sand Sare disjoint since each node of Shas an intermediate node of π or y v as G 2 -component which is not true for any node in S. Therefore the π i paths, 1 ≤ i ≤ δ 1 + δ 2 , form a maximum-size family of δ 1 + δ 2 node-disjoint paths between u and v and each is of length at most dist(u,v)+max(r 1 ,r 2 ).Notice that π 1 is of minimum length.
Case 2: If x u ≠ x v and y u = y v (the case x u = x v and y u ≠ y v is similar) Let π , π ,... π be δ 1 parallel paths between x u and x v in G 1 .These paths exist since x u ≠ x v , G 1 is regular of degree δ 1 and G 1 has CPP/r 1 -MLI.Each of these paths is of length at most dist(x u ,x v )+r 1 and at least one of them (say π ) is of length dist(x u ,x v ).Let y, 1 ≤ i ≤ δ 2 , be the δ 2 distinct adjacent nodes to y u in G 2 .We construct the following δ 1 + δ 2 paths between u and v: The paths π i , 1 ≤ i ≤ δ 1 , are node-disjoint among each other since the π , 1 ≤ i ≤ δ 1 , are node- disjoint in G 1 .Each of these paths is of length at most dist(u,v)+r 1 .Notice that π 1 is of minimum length.The paths πδ 1+i , for all i, 1 ≤ i ≤ δ 2 are node-disjoint among each other since each path has a different fixed G 2 -component yat its intermediate nodes.Each πδ 1+i path is of length dist(u,v)+2.Finally, every π i path, 1 ≤ i ≤ δ 1 , is node-disjoint with every πδ 1+j path, 1 ≤ j ≤ δ 2 , since all intermediate nodes of π i have y u as G 2 -component which is not the case for any of the intermediate nodes of πδ 1+j .QED.

PARALLEL PATHS IN THE k-ARY n-CUBE
We start by showing how to construct a complete family of parallel paths between any two nodes of a complete graph.
Lemma 2: The complete graph on k nodes K k has CPP/1-MLI.
Proof: Let 0, 1, ... , k-1 be the k nodes of K k .Let x and y be any two distinct nodes in K k .Hence, 0 ≤ x, y < k and x ≠ y.Consider the following k-1 paths between x and y: π = xy, π = xiy, for all i, 0 ≤ i < k, i ≠ x, and i ≠ y.
π is of minimum length equal to 1.Each π , 0 ≤ i < k, i ≠ x, and i ≠ y is of length 2 and has only one intermediate node i which is different from the intermediate node j of any other π .
Therefore, K k has CPP/1-MLI.QED The following theorem is a direct derivation from Lemma 1, Theorem 1, and Lemma 2.
Example: Let us follow the method described in the proof of Theorem 1 to construct step by step the parallel paths between the two nodes 0000 and 0120 of the 3-ary 4-cube Q.By Lemma 1, Qis isomorphic to K 3 ⊗Q.The node 0000 of Qis mapped to the node <0, 000> of K 3 ⊗Qand 0120 to <0, 120>.We apply the construction of Case 2 of Theorem 1.
Step 1: Find the parallel paths between 000 and 120 in Q.By Lemma 1, Qis isomorphic to K 3 ⊗Q.The node 000 is mapped to <0, 00> and 120 is mapped to <1, 20>.We need to apply the construction of Case 1 of Theorem 1.
Step 1.1: Find the parallel paths between nodes 0 and 1 in K 3 .These are given by Lemma 2 as follows (we use the symbol ' → ' in path descriptions to denote edges in the path): Step 1.2: Find the parallel paths between 00 and 20 in Q.But, Qis isomorphic to K 3 ⊗K 3 (by Lemma 1).The node 00 is mapped to <0, 0> and 20 is mapped to <2, 0>.We need to apply the construction of Case 2 of Theorem 1 using the parallel paths 0 → 2 and 0 → 1 → 2 between nodes 0 and 2 in K 3 (by Lemma 2).This results in the following paths between 00 and 20: Let π 1 be a minimum length path in S 1 S 2 = ∅ for each path π in S 1 do S 2 =S 2 ∪ {< a n-1 , π _>} Let π 1 be a minimum length path in S 1 S 2 = {a n-1 ...a 0 → b n-1 a n-2 ...a 0 || <b n-1 , π 1 >} for each i, 0 ≤ i < k and i ≠ a n-1 and i ≠ b n-1 do S 2 =S 2 ∪ {a n-1 ...a 0 → ia n-2 ...a 0 || <i, π 1 > || ib n-2 ...b 0 → b n-1 ...b 0 } S 2 = S 2 ∪ {<a n-1 , π 1 > || a n-1 b n-2 ...b 0 → b n-1 ...b 0 } for each path π in S 1 such that π ≠ π 1 do S 2 =S 2 ∪ {<a n-1 , π -1> || In this paper, we have contributed to the study of the topological properties of the k-ary n-cube by presenting a simple algorithm (easy to implement) for constructing node-disjoint (parallel) paths between arbitrary nodes of the k-ary n-cube.In fact, we show how to construct effectively a complete set of parallel paths (i.e. as many paths as the degree of the network) between any two nodes of a k-ary n-cube.Furthermore, each of the constructed paths is shown to be of length at most 2 plus the minimum distance between the two nodes.