ISRN Discrete MathematicsVolume 2012 (2012), Article ID 308595, 8 pagesdoi:10.5402/2012/308595

Research Article

Bipancyclic Properties of Faulty Hypercubes

Chun-Nan Hung and Min-Kun Hsiao

Department of Computer Science and Information Engineering, Da-Yeh University, Changhua 51591, Taiwan

Received 24 July 2012; Accepted 5 September 2012

Academic Editors: L. Ji, U. Vaccaro, W. Wallis, and X. Yong

Copyright © 2012 Chun-Nan Hung and Min-Kun Hsiao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

A bipartite graph is bipancyclic if it contains cycles of every even length from 4 to and edge bipancyclic if every edge lies on a cycle of every even length from 4 to . Let denote the -dimensional hypercube. Let be a subset of such that can be decomposed into two parts and , where is a union of disjoint adjacent pairs of , and consists of edges. We prove that is bipancyclic if . Moreover, is edge bipancyclic if with .

1. Introduction

Interconnection networks play an important role in parallel computing/communication systems. The graph embedding problem, which is a central issue in evaluating a network, asks if the guest graph is a subgraph of a host graph. A benefit of graph embedding is that we can apply an existing algorithm for guest graphs to host graphs. This problem has attracted a burst of studies in recent years. Note that cycle networks are useful for designing simple algorithms with low communication costs. Thus, there are many studies on the cycle embedding problem. The cycle embedding problem deals with identifying all possible lengths of the cycles in a given graph. For the graph definition and notation, we follow [1].

Let be -bit binary strings. The Hamming weight of , denoted by , is the number of such that . Let and be two -bit binary strings. The Hamming distance between two vertices and is the number of different bits in the corresponding strings of both vertices. The -dimensional hypercube, denoted by , has all -bit binary strings as its vertices; two vertices and are adjacent if and only if . Obviously, is a bipartite graph with bipartition and . A vertex of is white if is odd, otherwise is black. It is known that the distance between and is . For , , let denote the subgraph of induced by . Obviously, is isomorphic to . For any vertex , we use to denote the bit . Moreover, we use to denote the vertex with for and . An edge is of dimension if .

The hypercube is one of the most popular interconnection networks for parallel computers/communication systems [2]. This is partly due to its attractive properties, such as regularity, recursive structure, vertex and edge symmetry, maximum connectivity as well as effective routing and broadcasting algorithms.

Note that the hypercube is a bipartite graph for every integer . The corresponding cycle embedding problem on bipartite graphs is called the bipancyclic property. A bipartite graph is bipancyclic if it contains cycles of every even length from 4 to , inclusive.

There are some variations of the bipancyclic property. A bipartite graph is edge bipancyclic if every edge lies on a cycle of every even length from 4 to , inclusive. A bipartite graph is -edge fault tolerant bipancyclic if is bipancyclic for any with . Moreover, a bipartite graph is -edge fault tolerant edge bipancyclic if is edge bipancyclic for any with . The following theorem is proved.

Theorem 1.1 (see [3]). is -edge fault tolerant edge bipancyclic if .

In this paper, we improve Theorem 1.1 by considering both edge faults and vertex faults. However, we restrict the faults on the vertex set to those occurring only on disjoint adjacent pairs. Let be a subset of such that can be decomposed into two parts and where is a union of disjoint adjacent pairs of , and consists of edges. More precisely, , where and for . Without loss of generality, we assume that is a set of black vertices, and is a set of white vertices. We will prove that is bipancyclic if . Moreover, is edge bipancyclic if with .

2. Preliminary

We need the following lemmas.

Lemma 2.1 (see [4]). Let be any edge of for . There are cycles of length four that contain in common.

Lemma 2.2 (see [5]). Assume that is any positive integer with , and is a subset of with . Then there exists a Hamiltonian path of joining any two vertices from different bipartite sets. Moreover, there exists a Hamiltonian path joining to of for in some partite set and , in the other partite set for .

Lemma 2.3 (see [6]). Assume that is any positive integer with . Let and be two distinct white vertices of and and be two distinct black vertices of . There are two disjoint paths and such that (1) joins to , (2) joins to , and (3) spans .

We extend the above lemma by considering the occurrence of edge faults.

Lemma 2.4. Assume that is any positive integer with , and is a subset of with . Let and be two distinct white vertices of and and be two distinct black vertices of . There are two disjoint paths and of such that (1) joins to , (2) joins to , and (3) spans .

Proof. We prove this lemma by induction on . By Lemma 2.3, this lemma is true for . Thus, we assume and . For , let denote the set of -dimensional edges in . Thus, . Without loss of generality, we assume that . For , 1, we use to denote the set . Obviously, . Let for , 1. Without loss of generality, we can assume that . We have the following cases.Case  1  . By induction, there exist two spanning disjoint paths and of such that joins to , and joins to . Since , there exists an edge of or such that . Without loss of generality, we can assume that is in . Thus, can be written as . By Lemma 2.2, there exists a Hamiltonian path joining to of . We set as and set as .Case  2  . Without loss of generality, we can assume that . Since there are black vertices in and for , there exists a black vertex of such that and . By induction, there exist two spanning disjoint paths and of such that joins to , and joins to . By Lemma 2.2, there exists a Hamiltonian path joining to of . We set as . Thus, and form the required paths.Case  3  . Without loss of generality, we can assume that . By Lemma 2.2, there exists a Hamiltonian path of joining to , and there exists a Hamiltonian path of joining to . Obviously, and form the required paths.Case  4  . Without loss of generality, we can assume that . Since there are black vertices in and for , there exist two black vertices and of such that . By induction, there exist two spanning disjoint paths and of such that joins to and joins to . Similarly, there exist two spanning disjoint paths and of such that joins to , and joins to . We set as and set as .

Lemma 2.5 (see [7]). Assume that . Let be a subset of such that can be decomposed into two parts and where is a union of disjoint adjacent pairs of and consists of edges. Then there exists a Hamiltonian cycle of if .

Lemma 2.6 (see [7]). Assume that . Let be a subset of such that can be decomposed into two parts and , where is a union of disjoint adjacent pairs of , and consists of edges. Then there exists a Hamiltonian path of between any two vertices from different partite sets of if .

We improve Lemma 2.6 into the following lemma.

Lemma 2.7. Assume that . Let be a subset of such that can be decomposed into two parts and , where is a union of disjoint adjacent pairs of , and consists of edges. Then there exists a Hamiltonian path of between any two vertices from different partite sets of if with .

Proof. Let be any fault-free white vertex and be any fault-free black vertex. We need to construct a Hamiltonian path of joining to by induction on . Let . For , let denote the set of -dimensional edges in . Thus, . Without loss of generality, we assume that .By brute force, we can check that the required paths exist for . By Lemmas 2.2 and 2.6, the required paths exist when or . Therefore, we only need to consider the case and for . Thus, . Let , , and for , 1. Let and be a pair of where is a black vertex.Case  1   We first consider the case that and are in the same subcube. Without loss of generality, we can assume that both and are in . By induction, there exists a Hamiltonian path of joining to . Note that . We can write as for some and such that . By induction, there exists a Hamiltonian path of joining to . Thus, is a desired path.Thus we consider the case that and are in different subcubes. Without loss of generality, we can assume that and . Since there are black vertices in and for , there exists a black vertex in such that . By induction, there exists a Hamiltonian path of joining to and there exists a Hamiltonian path of joining to . Thus, is a desired path.Case  2 or for some . Without loss of generality, we can assume that or . Thus, .Assume that both and are in . By induction, there exists a Hamiltonian path of joining to . Suppose that the edge is in . Without loss of generality, we can write as . By Lemma 2.2, there exists a Hamiltonian path of joining to . Obviously, is a desired Hamiltonian path. Thus, we consider the case that is not in . Without loss of generality, can be written as . By Lemma 2.4, there are two disjoint spanning paths and of such that joins to and joins to of . Obviously, is a desired path.Now, we consider the case that and are in different subcubes. Without loss of generality, we can assume that and . By induction, there exists a Hamiltonian path of joining and . Suppose that the edge is in . We can write as . By Lemma 2.2, there exists a Hamiltonian path of joining to . Obviously, is a desired path. Thus, we consider the case that is not in . We can write as . Assume that . By Lemma 2.4, there are two disjoint spanning paths and of such that joins to and joins to . Obviously, is a desired path. Assume that . By Lemma 2.2, there exists a Hamiltonian path of joining to . Obviously, is a desired path.Finally, we consider the case that and are in . Let be any faulty edge of , where is a white vertex. By Lemma 2.5, there exists a Hamiltonian cycle of .Suppose that is not an edge of . Since the length of is at least , we can write as such that and is a white vertex. By Lemma 2.4, there exist two disjoint spanning paths and of such that joins to and joins to . Obviously, is a desired path.Thus, we consider the case that is in . We can write as . Suppose that . By Lemma 2.4, there exist two disjoint spanning paths and of such that joins to and joins to . Obviously, is a desired path. Suppose that . Without loss of generality, we assume that . By Lemma 2.2, there exists a Hamiltonian path of joining to . Thus, is a Hamiltonian path of . Suppose that . Obviously, can be written as for some black vertex . By induction, there exists a Hamiltonian path of joining to . Obviously, is a desired path.

3. Bipancyclic Properties

Theorem 3.1. Assume that . Let be a subset of such that can be decomposed into two parts and , where is a union of disjoint adjacent pairs of , and consists of edges. Then is bipancyclic if .

Proof. To prove this theorem, we will construct a cycle in of length for every even integer with by induction. Let . For , let denote the set of -dimensional edges in . Thus, . Without loss of generality, we assume that . Let ,  , and for , .By brute force, we can check that the required cycles exist for , 4. By Theorem 1.1, the required cycles exist if . Therefore, we consider the case and . Assume is an even integer from 4 to . Let for , 1. Without loss of generality, we can assume .Suppose that . We first consider the case that . By induction, the desired cycle exists in . Suppose that . Let be a cycle of length in . Let . Note that there are sections in of length depending on choice of the beginning and terminating vertices. Since , one such section, say , joins vertex to vertex such that . By Lemma 2.6, there exists a Hamiltonian path of joining to . Obviously, is a desired cycle.Thus, we consider . Therefore, . Assume that . By Theorem 1.1, the desired cycle exists in . Thus, we consider . Let be a pair of adjacent vertices of with being a black vertex. By Lemma 2.5, there exists a Hamiltonian cycle in . Obviously, can be written as . Without loss of generality, we assume that . Thus, . We can write as .Suppose that . Let . Let be the section of joining vertex to of length . By Lemma 2.2, there exists a Hamiltonian path of joining to . Obviously, is a desired cycle.Suppose that . Since , . We can write as . Let . Let be the section of joining vertex to of length . By Lemma 2.4, there exist two spanning disjoint paths and of such that joins to and joins to . Thus, forms a desired cycle. The theorem is proved.

Theorem 3.2. Assume that . Let be a subset of such that can be decomposed into two parts and where is a union of disjoint adjacent pairs of and consists of edges. Then is edge bipancyclic if and .

Proof. Let be any fault-free edge of where is a black vertex. For , we will construct a cycle containing of length in by induction to prove this theorem. Suppose , the desired cycle exists by Lemma 2.1. Thus, we consider .Let . For , let denote the set of -dimensional edges in . Since , we can assume without loss of generality that . Thus, are vertices in for some . Let and for .By Theorem 1.1, the desired cycles exist if . Thus, the desired cycles exist for . By brute force, we can check that the desired cycles exist for . Thus, we only consider the case with . Without loss of generality, we can assume that .Suppose that and . Without loss of generality, we can assume that both and are in . Suppose that . By induction, the desired cycle exists in . Thus, we consider that . Let . By induction, there exists a Hamiltonian cycle of containing the edge . Since , there exists an edge in such that . Thus, can be written as . By induction, there exist cycles of length of . Obviously, forms the desired cycle.Thus, we consider or . We have the following two cases.Case  1 . Suppose that . By Theorem 1.1, the desired cycle exists. Thus, we consider . Let and be a pair of adjacent vertices in . By induction, there exists a Hamiltonian cycle in containing the edge . Thus, we can write as .Suppose that . Let . Obviously, there exists a section of joining and of length such that is a black vertex. Suppose that . By Lemma 2.4, there exist two spanning disjoint paths and of such that joins to and joins to . Obviously, forms the desired cycle. Suppose that . Without loss of generality, we assume that . By Lemma 2.2, there exists a Hamiltonian path of joining to . Obviously, , forms the desired cycle. Suppose that . Thus, and . We can write as for some and . By induction, there exists a Hamiltonian cycle of containing the edge . Thus, can be written as . Obviously, forms the desired cycle.Case  2 . Suppose that . Let . By Theorem 1.1, there exists a cycle of length in containing the edge . We can write as . Obviously, forms the desired cycle.Now, we consider . Let and be a pair of adjacent vertices in . By induction, there exists a Hamiltonian cycle in containing .Suppose that . Obviously, can be written as . Let . Obviously, there exists a section of joining to of length containing . By Lemma 2.2, there exists a Hamiltonian path joining to of . Obviously, forms the desired cycle.Suppose that . The cycle can be written as . Without loss of generality, we can assume that is on the path . Suppose that . Let . There exists a section of such that (1) joins to , (2) contains , and (3) is of length . By Lemma 2.2, there exists a Hamiltonian path of joining to . Obviously, forms the desired cycle. Suppose that . Let . There exists a section of joining to of length . By Lemma 2.4, there exist two spanning disjoint paths   and of such that joins to and joins to . Obviously, forms the desired cycle.

4. Conclusion

In this paper, we study the bipancyclic property of faulty hypercubes. We improve previous results by considering both edge faults and vertex faults. Let be a subset of such that can be decomposed into two parts and where is a union of disjoint adjacent pairs of and consists of edges. We prove that is bipancyclic if . This result is optimal. Let be any vertex of . Assume that forms a set of faulty edges. Obviously, . is not bipancyclic.

We also prove that is edge bipancyclic if with . Again, this result is optimal. Assume that forms a set of adjacent faulty vertices. Obviously, . Thus, any Hamiltonian cycle of contains the path . Therefore, there is no Hamiltonian cycle containing the edge for . Thus, is not edge bipancyclic.

Two interesting observations, Lemmas 2.4 and 2.7, are used in this paper. In the following, we claim that these two lemmas are also optimal.

Let , , and be three distinct white vertices of . Assume that for forms a set of faulty edges. It is observed that are the neighbors of in . Moreover, and are black vertices. Obviously, cannot be any vertex in for any two disjoint paths and of such that joins to and joins to . Therefore, there do not exist two disjoint paths and of such that joins to and joins to . Thus, the number of faulty edges of Lemma 2.4 is optimal.

Assume that . Let be any vertex of and be a set of adjacent faulty vertices. Obviously, and are vertices in different partite sets. Assume that there is a Hamiltonian path joining to . Since , must include the section . Obviously, is not a Hamiltonian path and we get a contradiction. Thus, the number of adjacent faulty vertices of Lemma 2.7 is optimal.

Acknowledgment

This work was supported in part by the National Science Council of Republic of China under NSC 100-2221-E-212-014.

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