Abstract
Embedding cycles into a network topology is crucial for a network simulation. In particular, embedding Hamiltonian cycles is a major requirement for designing good interconnection networks. A graph G is called r-spanning cyclable if, for any r distinct vertices of G, there exist r cycles in G such that is on for every i, and every vertex of G is on exactly one cycle . If , this is the classical Hamiltonian problem. In this paper, we focus on the problem of embedding spanning disjoint cycles in bipartite k-ary n-cubes. Let be even and . It is shown that the n-dimensional bipartite k-ary n-cube is m-spanning cyclable with . Considering the degree of , the result is optimal.
1. Introduction
An interconnection network’s design is typically represented by a graph G, with vertices representing processors and edges representing links between processors. Due to its appealing characteristics, including its symmetry, small diameter, regularity, high connectivity, relatively cheap link complexity, and recursive construction, the hypercube is one of the most widely used network topologies [1,2]. Among various generalizations of the hypercube, the k-ary n-cube may be the most important one that can be used to build networks on chips, data center networks, and parallel computing systems [3]. Also, it has many attractive properties, such as regularity, recursive structure, vertex symmetry and edge symmetry, reduced message latency, and ease of implementation [4].
Two basic network topologies that can be utilized to create straightforward algorithms with low communication costs are cycles (rings) and paths (linear arrays) [5]. For a network simulation, it is essential to incorporate cycles (paths) into the network architecture [5,6]. Numerous parallel methods based on cycles and paths have been developed to address a wide range of algebraic, graph, picture, and signal processing issues [1]. Embedding cycles or paths can also save a significant amount of transmission time. Numerous studies have been conducted on these issues; see, for instance, [6,7,8,9,10].
A path that contains all vertices of G is referred to as a Hamiltonian cycle. A graph G is Hamiltonian if it has such a cycle and Hamiltonian-connected if there is a Hamiltonian path between any two vertices . Hamiltonicity is a widely studied and essential subject. Problems involving the embedding of two or more paths in networks have appeared recently. Let S and T be two disjoint vertex subsets of size t in a graph G. A set of t vertex-disjoint paths between S and T whose unions cover every vertex of G forms a many-to-many t-disjoint path cover of G connecting S and T. If every vertex in S needs to be connected to a specific vertex in T, then it is paired; if not, it is unpaired. There are many papers on Hamiltonian paths or disjoint path covers in various networks. Let us now turn to k-ary n-cubes. Refs. [11,12,13,14,15,16] investigated Hamiltonian paths under various conditions of k-ary n-cubes. For disjoint path covers of k-ary n-cubes, we refer to [17,18,19,20] and the references therein.
There are several essential rules that we should follow when designing an interconnection network. Two fundamental principles are extendability and embeddability for various topologies [5]. This implies that building huge networks from small ones should be achievable. Thus, a large network can often be decomposed into a set of small sub-networks. It is well known that embedding Hamiltonian cycles is a major requirement in designing good interconnection networks [21]. Furthermore, embedding Hamiltonian cycles into these sub-networks is also important and essential. This feature can be used to identify failing parts in a multiprocessor system [22].
Considering the above research motivations, this paper will study the spanning cyclability of k-ary n-cubes. A graph G is called r-spanning cyclable if, for any r distinct vertices of G, there exist r cycles in G such that is on for every i, and every vertex of G is on exactly one cycle . The spanning cyclability is closely related to another graph notation named a 2-factor. A 2-factor in a graph G is a spanning subgraph of G such that every vertex has degree two. Therefore, a graph G is r-spanning cyclable if, for any r distinct vertices of G, it has a 2-factor with each cycle containing exactly one prescribed vertex.
Lin et al. [22] considered the spanning cyclability of hypercubes and showed that an n-dimensional hypercube is r-spanning cyclable with . Yang et al. [23] considered the 2-spanning cyclability of the generalized Petersen graph with . Qiao et al. [24] considered the spanning cyclability of enhanced hypercubes and showed that an n-dimensional enhanced hypercube is r-spanning cyclable with . Also, they [25] considered the spanning cyclability of the Cayley graph generated by a transposition tree and showed that is r-spanning cyclable with . Wu et al. [26] considered the spanning cyclability of augmented cubes and showed that an n-dimensional augmented cube is r-spanning cyclable with . Also, Wu and Sabir [27] studied the spanning cyclability of hypercubes with prescribed edges in each cycle and showed that an n-dimensional hypercube is spanning r-edge-cyclable with .
In this paper, we focus on the problem of embedding spanning disjoint cycles in k-ary n-cubes, with each cycle containing a prescribed vertex. Let be even and ; it is shown in this paper that the n-dimensional bipartite k-ary n-cube is m-spanning cyclable with . The result is optimal in view of the degree of .
2. Preliminaries
We refer the reader to [28] for fundamental graph definitions that are not covered here. Let H and G be two graphs. If and , then H is a spanning subgraph of G. Assume that a path in G is . The inverse path of P is represented by . A similar cycle, , can be defined from a cycle of G.
The k-ary n-cube, denoted by , is a graph with the vertex set , where . Two vertices and are adjacent in if and only if there exists an integer such that and for every . Such an edge is referred to in this instance as an i-dimensional edge for . And, we use , or for short, to represent the set of all i-dimensional edges of .
It is clear that has vertices and is a bipartite graph when k is even. For any vertex , it has neighbors. Moreover, the neighbor of u joining by an i-dimensional edge, denoted by (resp. ), is the vertex , where (resp. ). If (resp. ), then we define u to be an even vertex (resp. odd vertex). If any two vertices are both even or odd, then the parity indicator ; otherwise, .
Let i be a specific dimension in . Then, along its i-dimensional edges, can be divided into k disjoint subgraphs: (written for short for every . It is clear that each subgraph is isomorphic to , where . For , define (or to be the subgraph induced by the vertex set . In particular, we have if . And, we have if by taking modulo k. Illustrations for are shown in Figure 1, where .
Before proceeding to the main results, we first present two important lemmas on the existence of Hamiltonian paths or disjoint path covers of .
Lemma 1
([19]). Given and its k subcubes for , let j be an integer with . When k is odd, let u and v be any pair of vertices in . There exists a path between u and v that covers all vertices of . On the other hand, when k is even, let w be a white vertex and b be a black vertex in . There exists a path between w and b that covers all vertices of .
Lemma 2
([17]). Let be a bipartite k-ary n-cube, where is even and . Let F be a set of faulty edges with . Given any four vertices , and , such that each partite set contains two vertices, the graph has a paired 2-disjoint path cover consisting of an -path and an -path.
Bose et al. [3] show that is Hamiltonian, so it is 1-spanning cyclable. In what follows, we will investigate the r-spanning cyclability of the bipartite k-ary n-cube for .
3. Spanning Cyclability of Bipartite -ary -cubes
Lemma 3.
The k-ary n-cube is m-spanning cyclable with .
Proof.
It suffices to consider the cases in which . Let and be two distinct vertices of , where for . Thus, or . Without loss of generality, together with the fact that the k-ary n-cube is vertex-transitive, we may assume that and . Let , . Then, is a 2-factor such that contains u and contains v.
Now, suppose that . Let , , and be three distinct vertices of , where for . For the same reason, we may let . If and both of them are not equal to 0, without loss of generality, let and , where and . Then, it is easy to see that the subgraphs , , and form a desired 2-factor containing u, v, and w, respectively. The case in which and both of them are not equal to 0 can be solved similarly.
In what follows, suppose that both the sets and have a cardinality of two. Without loss of generality, suppose that , , and . Let . It is easy to find a 2-factor in the residual graph such that contains v and contains w. Thus, forms a desired 2-factor of . The case in which , , and can be solved similarly. This completes the proof of Lemma 3. □
Theorem 1.
Let be even, and let . Then, the bipartite k-ary n-cube is m-spanning cyclable with .
Proof.
It proceeds by induction on n. The result follows from Lemma 3 when . Suppose that is m-spanning cyclable with and . Divide into k disjoint subgraphs , , …, along its n-dimensional edges. Then, each with is isomorphic to . Let S be any prescribed vertices of , and let with . Moreover, we use to denote the cardinality of , and then . Since is vertex-transitive, we may let . Thus, there exists at least one prescribed vertex belonging to . According to the cardinality of the prescribed vertices in each , we have the following cases.
Case 1. For every , we have . Let , , …, be subcubes so that each of them contains at least one prescribed vertex, where and . By the induction hypothesis, is -spanning cyclable for . That is, there exist disjoint cycles in such that each cycle contains exactly one prescribed vertex and the union of all cycles spans , and . Therefore, we have obtained disjoint cycles spanning , where . If , then form a desired 2-factor of . So, suppose that ; we shall add vertices of all other subcubes that contain no prescribed vertices to these cycles.
Let be a cycle in ; then, is an even cycle because is bipartite for even k. Let be an edge on ; then, we have . Moreover, it is easy to see that . By Lemma 1, there is a Hamiltonian path P joining and in . Removing the edge and adding will result in a new cycle that contains all vertices of . With a similar method, the vertices of can be added to a cycle of , which results in a new cycle . Continuing the above process recursively, the vertices of can be added to a cycle of , which results in a new cycle . Thus, the new cycles, together with other cycles in , form a desired 2-factor of .
Case 2. There exists some such that . Without loss of generality, we may assume that . We distinguish several subcases as follows.
Case 2.1. For every with , we have . Let and be two distinct prescribed vertices in . Then, there exists some such that . Divide into k disjoint subgraphs , , …, along its i-dimensional edges. Thus, the prescribed vertices u and v belong to different subcubes of , and each subcube contains at most prescribed vertices, which can be reduced to Case 1.
Case 2.2. For some with , we have . Let and . Since the subgraph is Hamiltonian, we obtain a cycle containing the vertex that spans . By the induction hypothesis, there are disjoint cycles of such that is in for , and spans . Thus, the prescribed vertex must be on some cycle , where . Without loss of generality, suppose that is on . We have the following subcases.
Case 2.2.1. Let . Furthermore, if is adjacent to on , one may assume that . Note that is bipartite with even k, and we have . Since , it follows that there exist such x and y satisfying , , and . By Lemma 1, there is a Hamiltonian path joining to passing through every vertex of . Meanwhile, there is a Hamiltonian path joining to that spans . Let and . Then, is a desired 2-factor of ; see Figure 2a for an illustration.
If is not adjacent to on and , we may assume that . It is easy to see that and . Thus, Lemma 1 results in a Hamiltonian path joining to that spans and a Hamiltonian path joining to that spans . Let and . Then, is a desired 2-factor of .
Now, assume that is not adjacent to on and ; then, . Let . Such vertices z and x can be guaranteed because is not adjacent to on . Moreover, we have and . Lemma 1 gives a Hamiltonian path joining to that spans and a Hamiltonian path joining to that spans . Let and . Then, is a desired 2-factor of ; see Figure 2b for an illustration.
Case 2.2.2. or . We may assume that because the case can be solved with a similar method. Also, we distinguish several situations according to the locations of and on . If is adjacent to on , assume that . Note that is bipartite with even k, and we have . Moreover, it follows that there exist such x and y satisfying , , and . By Lemma 2, there is a paired 2-disjoint path cover in such that is a path joining to , and is a path joining to . Moreover, let be an edge of the path cover. Without loss of generality, assume that is on , and . By Lemma 1, there is a Hamiltonian path P joining to that spans . Let and . Then, is a desired 2-factor of ; see Figure 3a for an illustration.
If is not adjacent to on and , we may assume that . It is easy to see that and . By Lemma 2, there is a paired 2-disjoint path cover in such that is a path joining to , and is a path joining to . Moreover, let be an edge of the path cover; that is, is on . Suppose that . By Lemma 1, there is a Hamiltonian path P joining to that spans . Let and . Then, is a desired 2-factor of .
Now, assume that is not adjacent to on and . It is easy to see that . Let . Thus, we have and . Lemma 2 gives a paired 2-disjoint path cover in such that is a path joining to , and is a path joining to . Moreover, let be an edge of the path cover. Without loss of generality, assume that is on , and . Lemma 1 gives a Hamiltonian path P joining to that spans . Let and . Then, is a desired 2-factor of ; see Figure 3b for an illustration.
Case 3. There exists some such that . Without loss of generality, we may assume that . Let and be two distinct prescribed vertices in . Then, there exists some such that . Divide into k disjoint subgraphs , , …, along its i-dimensional edges. Thus, the prescribed vertices u and v belong to different subcubes of , and each subcube contains at most prescribed vertices, which can be reduced to Case 2.
This completes the proof of Theorem 1. □
We provide an example to show that is not -spanning cyclable. Let v be any vertex of , and let its neighbors be . Then, cannot be -spanning cyclable. Otherwise, we select the prescribed vertices as . Since v must be on some cycle, such a cycle must contain two of , which is a contradiction.
4. Experimental Results and Analysis
In this section, we provide some experimental examples to demonstrate the performance of the main theorem. We focus on the 4-ary 3-cube , which contains 64 vertices. The spanning disjoint cycles of are listed in Table 1. In order to make the experimental results easier to follow, we give some explanations. The left column of the table represents prescribed vertices of , while the right column of the table represents vertex-disjoint cycles of such that every cycle contains a prescribed vertex and the union of all cycles spans . Let ; it is not difficult to see that cannot be 6-spanning cyclable such that every cycle contains exactly one vertex of S. Therefore, the main theorem is optimal with respect to the degree of .
5. Concluding Remarks
In a network simulation, it is essential to embed cycles into the network. In particular, having Hamiltonian cycles is a crucial requirement for developing high-quality interconnection networks. This paper studies the spanning cyclability of k-ary n-cubes and shows that an n-dimensional bipartite k-ary n-cube is m-spanning cyclable with . Considering the degree of , the result is optimal. A natural question arises: Why do our results need to exclude the non-bipartite k-ary n-cube when is odd? In fact, our original findings do not exclude the non-bipartite k-ary n-cube when k is odd. However, there are some difficulties we have not overcome. As shown in this paper, the proofs of several subcases in Case 2 of Theorem 1 depend on every cycle having a length of at least 4, and the methods are not appropriate for the case where . Let be odd. The spanning cyclability for the non-bipartite k-ary n-cube will be addressed in future research.
Author Contributions
Conceptualization, H.Q. and W.Z.; methodology, H.Q.; writing original draft preparation, H.Q.; writing review and editing, H.Q.; supervision, W.Z.; funding acquisition, H.Q. All authors have read and agreed to the published version of the manuscript.
Funding
This work was supported by the Xinjiang Tianchi Doctoral Project, China (No. 5105240150c), and the Foundation of China University of Petroleum, Beijing, at Karamay (No. QZX20240031).
Data Availability Statement
No new data were created or analyzed in this study. Data sharing is not applicable to this article.
Acknowledgments
The authors would like to express their gratitude to the editor and anonymous reviewers for their valuable comments and constructive suggestions on the original manuscript.
Conflicts of Interest
The authors declare no conflict of interest.
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