Abstract

Let stand for the sunlet graph which is a graph that consists of a cycle and an edge terminating in a vertex of degree one attached to each vertex of cycle . The necessary condition for the equipartite graph to be decomposed into for is that the order of must divide , the order of . In this work, we show that this condition is sufficient for the decomposition. The proofs are constructive using graph theory techniques.

1. Introduction

Let ,  ,   denote cycle of length , complete graph on vertices, and complement of complete graph on vertices. For even, denotes the multigraph obtained by adding the edges of a 1-factor to , thus duplicating edges. The total number of edges in is . The lexicographic product, , of graphs and , is the graph obtained by replacing every vertex of by a copy of and every edge of by the complete bipartite graph .

For a graph , an -decomposition of a graph , , is a set of subgraphs of , each isomorphic to , whose edge set partitions the edge set of . Note that for any graph and and any positive integer , if then .

Let be a graph of order and any graph. The corona (crown) of with , denoted by , is the graph obtained by taking one copy of and copies of and joining the th vertex of with an edge to every vertex in the th copy of . A special corona graph is , that is, a cycle with pendant points which has vertices. This is called sunlet graph and denoted by , .

Obvious necessary condition for the existence of a -cycle decomposition of a simple connected graph is that has at least vertices (or trivially, just one vertex), the degree of every vertex in is even, and the total number of edges in is a multiple of the cycle length . These conditions have been shown to be sufficient in the case that is the complete graph , the complete graph minus a -factor [1, 2], and the complete graph plus a -factor [3].

The study of cycle decomposition of was initiated by Hoffman et al. [4]. The necessary and sufficient conditions for the existence of a -decomposition of , where ( is prime) that (i) is even and (ii) divides , were obtained by Manikandan and Paulraja [5, 6]. Similarly, when is a prime, the necessary and sufficient conditions for the existence of a -decomposition of were given by Smith [7]. For a prime number , Smith [8] showed that -decomposition of exists if the obvious necessary conditions are satisfied. In [9], Anitha and Lekshmi proved that the complete graph and the complete bipartite graph for even have decompositions into sunlet graph . Similarly, in [10], it was shown that the complete equipartite graph has a decomposition into sunlet graph of length , for a prime .

We extend these results by considering the decomposition of into sunlet graphs and prove the following result.

Let , , and be even integers. The graph can be decomposed into sunlet graph of length if and only if divides , the number of edges in .

2. Proof of the Result

To prove the result, we need the following.

Lemma 1 (see [10]). For , decomposes .

Lemma 2. For any integer and a positive even integer , the graph has a decomposition into sunlet graph , for .

Proof   
Case 1 ( is even). First observe that can be decomposed into sunlet graphs with vertices. Now, set and decompose into cycles . To decompose into -cycles , denote vertices in th part of by for , and create base cycles . Next, combine these base cycles into one cycle by replacing each edge with . To create the remaining cycles , we apply mappings for defined on the vertices as follows.
Subcase  1.1 ( odd). Consider This is the desired decomposition into cycles .
Subcase  1.2 ( even). Consider This is the desired decomposition into cycles .
Now take each cycle , and make it back into . Each decomposes into sunlet graphs (by Lemma 1), and we have decomposing into sunlet graphs with length for even. Note that Case 2 ( is odd)
Subcase 2.1 (). Set . First create cycles in as in Case . Then, take complete tripartite graph with partite sets for and and decompose it into triangles using well-known construction via Latin square, that is, construct Latin square and consider each element in the form where denotes the row, denotes the column, and denotes the entry with . Each cycle is of the form ,  ,  . Then, for every triangle , replace the edge in each , by the edges and to obtain cycles . Therefore, . Now take each cycle , make it into , and by Lemma 1, has a decomposition into sunlet graphs .
Subcase 2.2 (). Set . The graph decomposes into Hamilton cycle by [11]. Next, make each cycle into . Each graph decomposes into sunlet graph by Lemma 1.

Theorem 3. Let ,   be positive integers satisfying , then decomposes .

Proof. Let the partite sets (layers) of the -partite graph be . Set . Obtain a new graph from as follows.
Identify the subsets of vertices , for and into new vertices , and identify the subset of vertices for and into new vertices and two of these vertices , where , are adjacent if and only if the corresponding subsets of vertices in induce . The resulting graph is isomorphic to . Next, decompose into cycles as follows: where ,   are calculated modulo .
To construct the remaining cycles, apply mapping defined on the vertices.
Subcase 1.1 ( odd in each cycle). Consider This is the desired decomposition of into cycles .
Subcase 1.2 ( even in each cycle). Consider This is the desired decomposition of into cycles .
By lifting back these cycles of to , we get edge-disjoint subgraphs isomorphic to . Obtain a new graph again from as follows.
For each , identify the subsets of vertices , where into new vertices , and two of these vertices are adjacent if and only if the corresponding subsets of vertices in induce . The resulting graph is isomorphic to . Then, decompose into cycles . Each decomposes into cycles by [12]. By lifting back these cycles of to , we get edge-disjoint subgraph isomorphic to . Finally, each decomposes into two sunlet graphs (by Lemma 1), and we have decomposing into sunlet graphs as required.

Theorem 4 (see [12]). The cycle decomposes for every even .

Theorem 5 (see [12]). If and are odd integers, then decomposes .

Theorem 6. The sunlet graph decomposes if and only if either one of the following conditions is satisfied. (1)is a positive odd integer, and is a positive even integer. (2),   are positive even integers with .

Proof. (1) Set , where is a positive integer. Let the partite sets (layers) of the -partite graph be . For each , where , identify the subsets of vertices , for into new vertices , and two of these vertices are adjacent if and only if the corresponding subsets of vertices in induce . The resulting graph is isomorphic to . Then, decompose into cycles , where is a positive integer.
Now, by Theorems 4 and 5.
By lifting back these -cycles of to , we get edge-disjoint subgraphs isomorphic to . Each copy of decomposes into sunlet graphs of length (by Lemma 1), and we have decomposing into sunlet graphs of length as required.
(2) Set , where is an even integer since .
Obtain a new graph from the graph as in Case . By Theorem 4, . By lifting back these -cycles of to , we get edge-disjoint subgraphs isomorphic to . Each copy of decomposes into sunlet graph of length (by Lemma 1). Therefore, as required.

Remark 7. In [10], it was shown that This, coupled with Lemma 1, gives the following.

Theorem 8 (see [10]). The graph decomposes into sunlet graphs for any positive integer .

Lemma 9 (see [3]). Let be an even integer. Then, is -decomposable.

Lemma 10 (see [3]). Let and be integers with odd, ,  , and . Then, is -decomposable.

Lemma 11 (see [3]). Let and be integers with odd, ,  , and . Then, is -decomposable.

We can now prove the major result.

Theorem 12. For any even integers ,  , and , the sunlet graph decomposes if and only if .

Proof. The necessity of the condition is obvious, and so we need only to prove its sufficiency. We split the problem into the following two cases.
Case 1 ()
Subcase 1.1 (). Cycle decomposes by Lemma 9, and we have Each graph decomposes into sunlet graph , where by Lemma 2, and we have decomposing into sunlet graph , where .
Subcase 1.2 (). First, consider .
Cycle decomposes by Lemma 9, and we have Now, sunlet graph by Theorem 3, and hence sunlet graph decomposes .
Also, consider .
Suppose . Cycle decomposes by Lemma 10, and we have Now, sunlet graph decomposes by Theorem 8, and we have decomposing into sunlet graph of length .
Case 2 ()
Subcase 2.1 (). Suppose , and by Lemma 9, cycle decomposes , and we have Also, sunlet graph decomposes each by Theorem 6, and we have sunlet graph decomposing .
Subcase 2.2 (). Let and an odd integer. Cycle decomposes , by Lemmas 9, 10, and 11, and we have Now, each decomposes into sunlet graph by Theorem 6, and we have decomposing into sunlet graph as required.
Subcase 2.3 (). Set , where is any positive integer, then by Subcases 2.1 and 2.2, we have Each graph decomposes into sunlet graph by Remark 7, and we have decomposing into sunlet graph .