Abstract

The construction of vertex-disjoint paths (disjoint paths) is an important research topic in various kinds of interconnection networks, which can improve the transmission rate and reliability. The -ary -cube is a family of popular networks. In this paper, we determine that there are disjoint paths in 3-ary -cube covering from to (many-to-many) with and from to (one-to-many) with where is in a fault-free cycle of length three.

1. Introduction

Vertex-disjoint paths (disjoint paths for short) are a set of paths in a graph that they do not share any vertices. A disjoint path cover of a graph is a disjoint path between two different vertices, and it covers all vertices of the graph. The disjoint path cover problem has been applied in many fields such as software testing, database design, and code optimization. Actually, vertex-disjoint paths can accelerate data transmission by providing parallel communication paths to avoid communication congestion. Besides, disjoint paths enhance the robustness of vertex failure and load balancing capability [1] and accelerate the transmission of large amounts of data by splitting data into disjoint communication paths of multiple vertices. The problem of many-to-many disjoint paths is an important problem in networks. It has attracted much attention because of its application in fault-tolerant routings for high-performance interconnection network.

Depending on the number of sources or sinks, there are one-to-one, one-to-many, and many-to-many disjoint paths problems. Among them, the one-to-one disjoint path is what we usually call the Hamilton path problem. A nonbipartite graph is one-to-one -disjoint path coverable if there is an -disjoint path cover between any two vertices of . A bipartite graph is one-to-one -disjoint path coverable if there is an -disjoint path cover between any two vertices in different partite sets. Shih and Kao [1] constructed the one-to-one -disjoint path cover of -ary -cube for , and . Li et al. [2] studied many-to-many -disjoint paths in hypercubes with , and each fault-free vertex has at least two fault-free neighbors. Chen in [3] considered many-to-many disjoint -paths in the hypercube with , and any two sets and of fault-free vertices in different parts. Zhang and Wang [4] proved that many-to-many disjoint paths cover in -ary -cubes with even and . So, we consider some disjoint path problems about 3-ary -cubes. There is also research on other graphs. Such as one-to-many disjoint paths of -star graph [5] and hyper-star networks [6] and one-to-one disjoint path covers on alternating group graphs [7].

For a set of sources and a set of sinks in with , the many-to-many -disjoint path problem is to determine whether there exist disjoint paths each joining a source and a sink. In other words, we need to find disjoint paths in joining and , which cover all the vertices in the graph; that is, and for and . In addition, this problem is also subclassified into paired and unpaired types. In paired type, each source should be joined to a specific sink; that is, should be joined to . Otherwise, each source can be joined to an arbitrary sink; that is, could be joined to for . We study the unpaired type in this paper. Specially, if , then it is called one-to-many disjoint paths; that is, for all and .

With the development of science and the expansion of the scale of computer system, people have more higher requirements for the reliability of network. However, computer failure is inevitable in a system. So, the disjoint path cover problem is also extended to a graph with some faulty elements (vertices and/or edges). A graph is called -fault hamiltonian-connected if each pair of vertices are joined by a hamiltonian path in for any faulty elements .

In this paper, we study disjoint paths of 3-ary -cube covering with faulty vertices from to for in Theorem 1. Besides, we also consider disjoint paths of 3-ary -cube covering with faulty vertices from to where and is in a fault-free cycle of length three in Theorem 2.

2. Preliminaries

2.1. Notations

An interconnection network can be represented by a graph , where represents the vertex set and represents the edge set. Let be a simple undirected graph. For a vertex , is the set of vertices adjacent to in . The path is a sequence of adjacent vertices where all the vertices are distinct, which is denoted by . If the path satisfies , it is called a cycle. In addition, . For , denotes the section from to in . For , . A path or cycle which contains every vertex of a graph is called a Hamilton path (hamiltonian path) or Hamilton cycle (hamiltonian cycle) of the graph. A graph is hamiltonian if it contains a Hamilton cycle. A graph is hamiltonian-connected if there exists a Hamilton path between any two distinct vertices of the graph. For graph-theoretical terminology and notation not defined here, we follow [8].

2.2. The -Ary -Cube

Definition 1. The -ary -cube is a graph with vertices, each of which has the form , where for every . Two vertices and in are adjacent if and only if there exists an integer , , such that and for every . For clarity of presentation, we omit writing “” in similar expressions for the remainder of the paper.
We discuss the 3-ary -cube with regular. Obviously, is a cycle of length 3, and is a wrap-around mesh. We can partition along -dimension, , by deleting all the -dimension edges, into three disjoint subcubes, , and (abbreviated as , and , if there are no ambiguities). is isomorphic to for . For and , the edge set between and is a perfect matching.

Lemma 1 (see [9]). The -ary -cube is hamiltonian-connected when is odd.

Lemma 2 (see [10]). Assume is odd with . Let be a set of faulty vertices and/or edges in a -ary -cube .(1)If , then the graph is hamiltonian.(2)If , then the graph is hamiltonian-connected.

Lemma 3 (see [11]). Let , , . If are four vertices in , then there are two vertex-disjoint paths and such that .

3. Many-to-Many Disjoint Paths

Theorem 1. Let be a 3-ary -cube for . Let be a vertex faulty set of . Let with . If , then there are disjoint paths from to covering .

Proof. We prove the theorem by induction on . When , and . By Lemma 3, the theorem holds. Assume the theorem holds on . We prove the theorem holds on for .
Suppose there is for . Note . By the induction hypothesis, there are disjoint paths from to covering . Let . Then, the theorem holds. Thus, we suppose for all .
We decompose into three isomorphic subgraphs along one dimension. Let , and for . Let and . If for and , then let and for . Case 1. for all . Case 1.1. for some . Without loss of generality, let . Then, . Let and . Case 1.1.1. . Then, . Let and . Then, . By the induction hypothesis, there are disjoint paths from to covering . Then, . Case 1.1.1.1. for some . Without loss of generality, let . Suppose is before in . Then, . Let (resp. ) be the before vertex (resp. later vertex) of (resp. ) in . Let and . By Lemma 1, there is a Hamilton path in . Let , and . By Lemma 1, there is a Hamilton path in . Let , and . Similarly, is before in . Case 1.1.1.2. for and . Without loss of generality, let and . Let and be the before and later vertex of in , respectively. Let and . Let and be the before and later vertex of in , respectively. Let and . Note for . By Lemma 3, there are two disjoint paths and covering . By Lemma 1, there is a Hamilton path in . Let , , and . Case 1.1.2. . Then, . Without loss of generality, let and . Let and . Then, . Note . By the induction hypothesis, there are disjoint paths from to covering . Then, for some . Without loss of generality, let . Then, . Let and be the before and later vertex of in , respectively. Let and for . If or , then and . By Lemma 2, there is a path and covering and , respectively. Let , and . Suppose and . By Lemma 2, there is a path and covering and , respectively. Let , and . Case 1.1.3. . In this case, . We get for . We have . Without loss of generality, let . Since , there are two vertices for some . Without loss of generality, let . Let and . Then, . Note . By the induction hypothesis, there are disjoint paths from to covering . Since for , there are two edges and such that for . Note for . By Lemma 3, there are two disjoint paths and covering . Note for . By Lemma 2, there is a path covering . Let . Case 1.1.4. . Let and for and . Since for , there is an edge such that and for . Without loss of generality, let be satisfied. Let and for and . Since for , there is an edge such that and for . Without loss of generality, let be satisfied. Note . By the induction hypothesis, there are disjoint paths covering. Since for , there are two edges , such that for . Note for . By Lemma 3, there are two disjoint paths and covering . Note for . By Lemma 2, there is a path covering . Let . Case 1.2. for some . Without loss of generality, let , and . Let , , and . By the induction hypothesis, there are disjoint paths covering . Since for , there are two edges , such that for . Note for . By Lemma 3, there are two disjoint paths and covering . By Lemma 2, there is a path covering . Let . Case 1.3. for all . Let for . Then, . Note . By the induction hypothesis, there are disjoint paths covering for . Case 1.4. for only one . Let . If there is or for some , then it is similar to Case 1.2. Thus, suppose for . Let . Then, . Let , , and . Note . By the induction hypothesis, there are disjoint paths covering . Since for , there are two edges such that for . Note . By the induction hypothesis, there are disjoint paths , and covering . Note for . By Lemma 2, there is a path covering . Let . Case 2. for only one . Without loss of generality, let . Then, , and . Without loss of generality, let . Then, and . Since for , there are cycles of length three for such that for . Case 2.1. . Then, . Case 2.1.1. and . Then, and . Let , . Case 2.1.1.1. . Let , . Then, . By the induction hypothesis, there are disjoint paths from to covering . Let . Let and . Note and for . By Lemma 2, there is a path covering . Similarly, there is a path covering . Let . Suppose . Then, . Let and . By Lemma 2, there are two paths and (resp. and ) covering (resp. ). Let for . Then, for . Suppose . Let . Let and be the before and later vertex of and , respectively. Let and . If , then let . Let , and . Note for . By Lemma 2, there is a path covering . Let . Suppose . Note . By Lemma 1, there is a Hamilton path in . Let and . Suppose and . Let and be the before and later vertex of and in and , respectively. Let and . If , then let . Let , and . Note for . By Lemma 2, there is a path covering . Let and . Suppose . Note . By Lemma 1, there is a Hamilton path in . Let , and . Case 2.1.1.2. . There is at most one vertex or for . Without loss of generality, let for all . Let , . Then, . Note . By the induction hypothesis, there are disjoint paths from to covering . Let . Let and . Note and for . By Lemma 3, there are two disjoint paths and covering . Similarly, there are two disjoint paths and covering . Let and . Case 2.1.1.3. . Then, for . Let and for . Since for , there is an edge such that and . Without loss of generality, let be satisfied. Let and for . Since for , there is an edge such that and . Without loss of generality, let be satisfied. Let and . Then, . Let and . Note and . By the induction hypothesis, there are disjoint paths (resp. ) from to (resp. from to ) covering (resp. ). Note for . By Lemma 2, there is a path covering . Let and . Case 2.1.2. and . Let . Let . Then, . Let and . Let . Note . By the induction hypothesis, there are disjoint paths covering . Case 2.1.2.1. . Let for , for . Then, . Note . By the induction hypothesis, there are disjoint paths and covering . Let , . Then, for some . We get . Let and be the before and later vertex of in , respectively. Let , . Let . Suppose and . Note . By Lemma 3, there are two disjoint paths and covering . Let , and . Suppose . Let be the before vertex of in and . Note for . By Lemma 2, there is a path covering . Let , and . Similar to . Case 2.1.2.2. . Then, for . Since , there is a vertex such that for . Without loss of generality, let . Let for , for . Then, . Note and . By the induction hypothesis, there are disjoint paths and covering . Let and for . Let . Note for . By Lemma 2, there is a path covering . Let . Case 2.1.2.3. . Then, for . Let and for . Since for , there is an edge such that and . Without loss of generality, let be satisfied. Let for , for . Then, . Note and . By the induction hypothesis, there are disjoint paths and covering . Let and . Let . Note for . By Lemma 2, there is a path covering . Let . Case 2.1.3. . Let . Then, . Let and . If or , then it is similar to Case 2.1.2. Thus, for . Let . Then, . Let and . Let for and for where . Then, . Since for , there are two edges , such that for . Note for . By Lemma 2, there is a path covering . Note . By the induction hypothesis, there are disjoint paths , and from to covering . Note . By the induction hypothesis, there are disjoint paths , , from to covering . Let , , and . Case 2.2. . By the induction hypothesis, there are disjoint paths from to covering . Then, . Let for . Then, and . By the induction hypothesis, there are (resp. ) disjoint paths from to (resp. to ) covering (resp. ). Case 3. for all . By Case 2, there are cycles of length three for such that for . Case 3.1. for only one . Without loss of generality, let . Then, , . We get . Case 3.1.1. and . Then, and . Let and . Let , and . Without loss of generality, let . Then, . We get and . Let . By the induction hypothesis, there are disjoint paths from to covering . Case 3.1.1.1. . Let . Then, . Let . Note . By the induction hypothesis, there are disjoint paths from to covering . Then, for some . Without loss of generality, let . Let be the before vertex of in and . Suppose . Without loss of generality, let . Let . Note and . By the induction hypothesis, there are disjoint paths covering . Let , , and . Suppose . Note . By the induction hypothesis, there are disjoint paths , from to covering . Let , , and . Case 3.1.1.2. . Then, . Since , there is a vertex such that for . Without loss of generality, let . Let . Note and . By the induction hypothesis, there are disjoint paths from to covering . Let . Let . Note . By the induction hypothesis, there are disjoint paths , from to covering . Let and . Case 3.1.1.3. . If , then it is similar to Case 3.1.1.2. Suppose . Let and for . Since for , there is a vertex such that . Without loss of generality, let be satisfied. Let and . Then, . By the induction hypothesis, there are disjoint paths from to covering . Let . Let . Note . By the induction hypothesis, there are disjoint paths , from to covering . Let and . Case 3.1.2. and . Let . Since for , for . Let , and . Let , and . Case 3.1.2.1. . Then, . Let and . Note and . By the induction hypothesis, there are (resp. ) disjoint paths (resp. ) from to (resp. to covering (resp. ). Let and . Let . Note and for . By the induction hypothesis, there are disjoint paths from to covering . Case 3.1.2.2. . Since , there is a vertex such that for . Without loss of generality, let . Let for . Note and . By the induction hypothesis, there are disjoint paths and from to covering . Let . Let . Note and . By the induction hypothesis, there are disjoint paths from to covering . Let . Let for and for . Since , . Note and . By the induction hypothesis, there are disjoint paths and from to covering . Let and . Case 3.1.2.3. . Let and for . Since for , there is a vertex such that . Without loss of generality, let be satisfied. Let and . Note and . By the induction hypothesis, there are disjoint paths and from to covering . Let . Let . Note . By the induction hypothesis, there are disjoint paths from to covering . Let . Let for and for . Note . By the induction hypothesis, there are disjoint paths and from to covering . Let and . Case 3.1.3. . If for some , then it is similar to Case 3.1.1 or 3.1.2. Since for , for . Let , and . Thus, . Note , and . By the induction hypothesis, there are , and disjoint paths from to , from to and from to covering , and , respectively. Case 3.2. for two . There is only one such that for . It is similar to Case 3.1.