Research Article | Open Access
On 3-Regular Bipancyclic Subgraphs of Hypercubes
The -dimensional hypercube is bipancyclic; that is, it contains a cycle of every even length from 4 to . In this paper, we prove that contains a 3-regular, 3-connected, bipancyclic subgraph with vertices for every even from 8 to except 10.
The cartesian product of two graphs and is a graph with the vertex set , and any two vertices and are adjacent in if and only if either and is adjacent to in or and is adjacent to in . A graph with even number of vertices is bipancyclic if it contains a cycle of every even length from 4 to . The hypercube of dimension is a graph obtained by taking cartesian product of the complete graph on two vertices with itself times; that is, ( times). The hypercube is an -regular, -connected, bipartite, and bipancyclic graph with vertices. It is one of the most popular interconnection network topologies . The bipancyclicity of a given network is an important factor in determining whether the network topology can simulate rings of various lengths. The connectivity of a network gives the minimum cost to disrupt the network. Regular subgraphs, bipancyclicity, and connectivity properties of hypercubes are well studied in the literature [2–6].
Since () is bipancyclic, it contains a 2-regular, 2-connected subgraph (cycle) with vertices for every even integer from 4 to . Suppose . Mane and Waphare  proved that contains a spanning -regular, -connected, bipancyclic subgraph. So the natural question arises; what are the other possible orders existing for -regular, -connected and bipancyclic subgraphs of As , can be regarded as a subgraph of . Hence has a -regular, -connected, bipancyclic subgraph with vertices. In this paper, we answer the question for . We prove that () contains a -regular, -connected, and bipancyclic subgraph with vertices for every even integer from 8 to except 10.
The cartesian product of a nontrivial path with the complete graph is a ladder graph. Let be the graph obtained from a path () by adding one extra edge . We call the graph a ladder type graph on vertices (see Figure 1).
Lemma 1. A ladder graph is bipancyclic.
Proof. Let be a ladder graph with vertices. Label the vertices of by ’s and ’s so that is the union of the paths and and the edges for . Suppose . Let be the subpath of from to and let be the subpath of from to . Then is a cycle of length in . Hence has a cycle of every even length from 4 to .
The vertices of the hypercube can be labeled by the binary strings of length so that two vertices are adjacent in if and only if their binary strings differ in exactly one coordinate. Denote by the subgraph of induced by the set of all vertices of each having first coordinate for . Then and are vertex-disjoint and each of them is isomorphic to . We can express as , where . Note that is a perfect matching in .
Lemma 2. For every with , there exists a ladder type subgraph in () with vertices.
Proof. We first prove that contains a Hamiltonian cycle with a chord which forms a 4-cycle with three edges of . This is obvious for . Suppose . Write as . By induction, there exists a Hamiltonian cycle in with a chord which forms a 4-cycle with three edges of . Let be the corresponding Hamiltonian cycle in . Let be any edge on which is not on and let be the corresponding edge on . Then and belong to . Let . Then is a Hamiltonian cycle in such that is its chord which forms the 4-cycle with three edges of .
Now, we prove that contains a ladder type graph with vertices. Obviously, itself is a ladder type graph on 8 vertices. Suppose . By the above part, contains a Hamiltonian cycle with a chord which forms a 4-cycle with three edges of . Label the vertices of by ’s so that and . Let be the subgraph of obtained by taking the union of the subpath of and the edge . Then is a ladder type subgraph of with vertices.
As a consequence of a result of , we get the following lemma.
Lemma 3. Let be an -regular, -connected graph for . Then the graph is -regular, -connected.
It is well known that the hypercube does not contain the complete bipartite graph as a subgraph. The following result is the main theorem of this paper.
Theorem 4. Let be an integer such that . Then there exists a -regular, -connected, and bipancyclic subgraph of on vertices if and only if is an even integer with and .
Proof. Suppose contains a 3-regular subgraph with vertices. By Handshaking Lemma, the sum of the degrees of all vertices of a graph is even. Hence is even. Consequently, is even. The minimum degree of is three. Therefore contains an even cycle. Since is simple, . If , then contains a triangle, a contradiction. Thus . Suppose . Then must contain a cycle of length four. A vertex of outside has at least two neighbours in giving a triangle or a in , which is a contradiction. Suppose . Let be an edge of . Without loss of generality, we may assume that the end vertices of differ in the first coordinate. Write as . Then . Therefore intersects with both and . Let be a component of for . Then is a subgraph of with minimum degree two and hence it contains a cycle. As is simple bipartite, has at least four vertices. Since , is the only component of in . We may assume that . Then or . Let be an even cycle in . Then . If has vertices, then the vertex of which is not on is adjacent to at least two vertices of giving a triangle or a in , a contradiction. Consequently, has vertices. Thus . Let be the cycle in corresponding to . Since is -regular, each vertex of has one neighbour in along an edge of . Therefore all vertices of belong to . As has six vertices, it has a vertex which is not on . Then has no neighbour in . Thus has three neighbours in . Therefore has at least two neighbours in the 4-cycle giving a triangle or a in , a contradiction. Hence . Thus is an even integer with and .
Now, we construct a 3-regular, 3-connected, bipancyclic subgraph of with vertices for every even integer with and . Suppose for some integer with . Write as . Since is a bipancyclic graph and is even, there is a cycle of length in . By Lemma 3, is a -regular, -connected subgraph of with vertices. Let be an edge of . Then is a ladder graph which spans . By Lemma 1, is bipancyclic.
Suppose with . Write as . As , there exists a ladder type subgraph in on vertices by Lemma 2. Label the vertices of by ’s and ’s so that and are paths and is an edge of for . Let be the ladder type subgraph of on vertices corresponding to . Label the vertices of by and , where the vertex corresponds to , and the vertex corresponds to for every . Let be the graph obtained from by deleting the edges and . Let be the graph obtained from by deleting two vertices and . Then is a subgraph of with vertices and is a ladder subgraph of with vertices.
Let , where (see Figure 2). Then is a 3-regular subgraph of with vertices. We claim that is bipancyclic and 3-connected.
Claim 1. is bipancyclic.
Clearly, , is a Hamiltonian cycle in . By deleting two vertices and and then adding the edge to , we get a cycle of length in . Similarly, we obtain a cycle of length in from by deleting four vertices and then adding the edge . Now, by deleting six vertices from adding the edges and gives a cycle of length in . Suppose . Then has vertices. We get a cycle of length 4 and a cycle of length 6 in the ladder as, by Lemma 1, it is a bipancyclic graph on six vertices. Thus contains a cycle of every even length from 4 to 14. Suppose . Then has at least 10 vertices. Let be the ladder in formed by two paths and , and the matching and for and . By Lemma 1, is bipancyclic. Hence contains a cycle of every even length from 4 to . Thus contains a cycle of every even length from 4 to . Therefore is bipancyclic.
Claim 2. is 3-connected.
Since contains a Hamiltonian cycle, it is 2-connected. It suffices to prove that deletion of any two vertices from leaves a connected graph. Let with . We prove that is connected. Let . Suppose intersects both and . We may assume that and . Being Hamiltonian graphs, both and are 2-connected. Hence and are connected. There are at least two edges from the set which connects to in . Therefore is connected.
Suppose . Then and . Obviously, is connected. Suppose is connected. Then it is joined to through at least two edges from the set . This implies that is connected. Suppose is not connected. Then one vertex of belongs the path and the other vertex belongs to the path . Let be a Hamiltonian cycle of . Then has exactly two components, say, and with vertex set and . Note that or may have a single vertex. Therefore has two components one with vertex set and the other with vertex set . It is easy to see that contains a vertex from the set and hence has a neighbour in along an edge of the set for . Consequently, each component of has a neighbour in in the graph . This implies that is connected.
Suppose . Then is connected. Let . Then and . If each component of contains a vertex of the set , then all the components of are connected to by the edges of the set giving connected. Therefore it suffices to prove that each component of contains a vertex of the set . If is connected, then we are done. Suppose is not connected. Consider the case when . Then is the union of the two 4-cycles , , , , and , , , , , and the two edges , . Each of the vertices , , , has degree two in . If , then is connected. Therefore . Thus , , , , or . In any case, each component of contains a vertex of the set . Suppose . Then is a Hamiltonian cycle in . Therefore has only two components. It follows that one component of contains a vertex from and the other component contains a vertex from the set . Hence the vertex set of each component of intersects . Consequently, is connected. Therefore is 3-connected.
Thus, from Claims 1 and 2, is a 3-regular, -connected, bipancyclic subgraph of with vertices.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
The authors would like to thank anonymous referees for their valuable suggestions. The first author is supported by the Department of Science and Technology, Government of India via Project no. SR/S4/MS: 750/12.
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