Decompositions of Complete Graphs into Bipartite 2-regular Subgraphs

It is shown that if G is any bipartite 2-regular graph of order at most n 2 or at least n − 2, then the obvious necessary conditions are sufficient for the existence of a decomposition of the complete graph of order n into a perfect matching and edge-disjoint copies of G.

If G is a fixed graph and D = {G 1 , G 2 , . . ., G t } is a decomposition such that G i is isomorphic to G for i = 1, 2, . . ., t, then D is called a G-decomposition.See [9] for a survey on G-decompositions, and see [5,21] for general asymptotic existence results.
This paper concerns G-decompositions of complete graphs in the case where G is a 2-regular graph.See [1] for a survey of results on G-decompositions of complete graphs.The complete graph of order n is denoted by K n , the cycle of order n is denoted by C n , and the path of order n is denoted by P n (so P n has n − 1 edges).If G is 2-regular and n is even, then there is no G-decomposition of K n , and it is common to instead consider decompositions of K n − I, where K n − I denotes the graph obtained from K n by deleting the edges of a perfect matching.For each positive integer n, define K * n to be K n if n is odd and K n − I if n is even.The number of edges in K * n is given by n n−1

2
. If G is a 2-regular graph of order k and there exists a G-decomposition of K * n (n 3), then it is obvious that If G is a 2-regular graph of order k, then the conditions given in (1) are called the obvious necessary conditions for the existence of a G-decomposition of K * n .The following problem presents itself.
Problem 1: For each 2-regular graph G and each positive integer n satisfying the obvious necessary conditions, determine whether there exists a G-decomposition of K * n .It is known that if G is a cycle, then the obvious necessary conditions are sufficient for the existence of a G-decomposition of K * n [4,18].However, when G is 2-regular but is not a cycle, there are cases where the obvious necessary condition are satisfied but no G-decomposition of K * n exists.There is no G-decomposition of K * n in each of the following cases (see [9] and [10]).If G has order n, then Problem 1 is precisely the well-known Oberwolfach Problem.See [10,11,20] for more information on the Oberwolfach Problem, and see [12] for a generalisation of the problem.Problem 1 has been solved for every 2-regular graph of order at most 10 when n is odd [2], and various results on Problem 1 have been obtained via graph labellings.For example, in [3] it is shown that if G has order k and is 2-regular with at most three components, then there exists a G-decomposition of K 2k+1 , and in [6] it is shown that if G is bipartite and 2-regular of order k, then there exists a G-decomposition of K 2kx+1 for each positive integer x.Several strong results have also been obtained on Problem 1 for the case where G consists of disjoint 3-cycles [13,14].These results relate to Kirkman signal sets which are are used in devising codes for unipolar communication, see [15].
In [16], a simple but powerful idea is used to show that if both n and n n−1 2 /k are even, then there is a G-decomposition of K * n for every bipartite 2-regular graph G of order k.Our main result, see Theorem 10, extends this result to the case n n−1 2 /k is the electronic journal of combinatorics 23(2) (2016), #P2.1 odd, except when n 2 < k < n − 2. The special case of this extension where k = n (that is, the case corresponding to the Oberwolfach Problem) is the main result in [8].

Notation and Preliminary Results
For a given graph K, we define the graph K (2) by V (K (2) t }.Observe that if F is a decomposition of K, then F (2) is a decomposition of K (2) .
The following result of Häggkvist [16] is a critical ingredient in many of our constructions.
Parker [17] has completely settled the problem of decomposing complete bipartite graphs into paths of uniform length, and we need the following special case of her result.
Theorem 2. (Parker [17]) If r and a are even with r 2a − 2, r 2b, and r dividing ab, then there is a P r+1 -decomposition of K a,b .
We also need the following result of Tarsi on decompositions of complete graphs into isomorphic paths [19].
For each even r 2, let Y r denote any graph isomorphic to the graph with vertex set {v 1 , v 2 , . . ., v r+1 } and edge set Lemma 4. For each even r 2, there exists a decomposition of K r+1 into r−2 2 Hamilton paths and a copy of Y r .
Proof.Let r 2 be even and for i = 0, 1, . . ., r let M i be the matching with edge set {{x, y} : x = y, x + y = i} in the complete graph with vertex set Z r+1 .Then is the required decomposition.Proof.By Lemma 4, there is a P r+1 -decomposition of K r+1 − Y r , so it suffices to show that there is a P r+1 -decomposition of K m − K r+1 .But K m − K r+1 can be decomposed into K r,m−r−1 and K m−r , so it suffices to prove that K r,m−r−1 and K m−r each have P r+1decompositions.The former has a P r+1 -decomposition by Theorem 2, and the latter by Theorem 3. It is routine to check that the hypotheses of these two theorems are satisfied when r is even, 2 r m−1 2 and r divides 1  2 m(m − 1) − 3r 2 .For each even r 2 we define the graph J 2r (see Figure 1) to be the graph with vertex set V (J 2r ) = {u 1 , u 2 , . . ., u r+2 } ∪ {v 1 , v 2 , . . ., v r+2 } and edge set Figure 1: The graph J 2r The following result is proved in [8], see Lemma 10 and the proof of Lemma 11.
Lemma 6.If G is a bipartite 2-regular graph of order 2r where r 4 is even, then there is a decomposition

Main Results
Lemma 8.If n 6 is even and G is any bipartite 2-regular graph of order n − 2, then there is a G-decomposition of K n − I.
If m is even, then let D be a decomposition of K m into m 2 Hamilton paths, and if m is odd, then let D be a decomposition of K m into m−3

2
Hamilton paths and a copy of Y m−1 .The first of these decompositions exists by Theorem 3 and the second exists by Lemma 4. In either case let the vertex set of K m be Z m and let I be the 1-regular graph with m and the perfect matching I when m is even, and is a decomposition of m−1 , and the perfect matching I when m is odd.Since the union of the copy of Y (2) m−1 and I is a copy of X n−2 , the result follows by Theorem 1 and Lemma 7.
Lemma 9. Let r 2. If there is a P r+1 -decomposition of K m or if r is even and there is a P r+1 -decomposition of K m − Y r , then there is a G-decomposition of K 2m − I for every bipartite 2-regular graph of order 2r.
Proof.Let G be a bipartite 2-regular graph of order 2r, let the vertex set of K m be Z m and let I be the 1-regular graph with If there is a P r+1 -decomposition D of K m , then D (2) is a P r+1 -decomposition of K 2m −I.By Theorem 1, we can decompose each copy of r+1 in D (2) into two copies of G, thereby obtaining a G-decomposition of K 2m − I.
Thus, we can assume r is even and there is a P r+1 -decomposition of K m − Y r , and hence a decomposition D of K m into one copy of Y r and ( m 2 − 3r 2 )/r copies of r+1 , and a perfect matching.There are r + 1 edges of I which form a 1-regular graph on the vertex set of the copy of Y r+1 , and a matching M with m − (r + 1) edges (such that M and the copy of X 2r are vertex-disjoint).
By Theorem 1, we can decompose each copy of P r+1 in D (2) into two copies of G. Let D P be the union of all of these decompositions.By Lemma 7, there is a decomposition D X ∪ {M } of the copy of X 2r where D X contains three copies of G and M is a perfect matching in the copy of X 2r .This means that the union of M and M is a perfect matching in K 2m .It follows that D P ∪ D X is a G-decomposition of K 2m − I. 2 so there is nothing to prove.Thus, it remains only to show that there is a G-decomposition of K n − I when 3 k n 2 and k divides n(n−2)

2
. Let m = n 2 and let r = k 2 (since G is bipartite, k is even and r 2 is an integer).By Lemma 9, it suffices to show that there is a P r+1 -decomposition of K m or that r is even and there is a P r+1 -decomposition of K m − Y r .If 2m(m − 1)/k is even, then r divides m(m − 1)/2 and so by Theorem 3, there is a P r+1 -decomposition of K m .If 2m(m − 1)/k is odd, then it follows that r is even, r divides It is worth remarking that the constructions used to prove Theorem 10 can be easily generalised as follows.In the proof of Lemma 9, each copy of P (2) r+1 can be decomposed independently, resulting in decompositions of K n − I into 2-regular graphs which are not all isomorphic.Although each copy of P (2) r+1 produces two isomorphic 2-regular graphs in the final decomposition, and the copy of X 2r , when it is present, produces three isomorphic 2-regular graphs in the final decomposition, this construction can produce a wide variety of different combinations of 2-regular graphs in the final decomposition.
The 2-regular graphs given by the construction of the preceding paragraph will all have the same order, namely k = 2r, but it is also possible to get around this constraint.Instead of using a P r+1 -decomposition of K m or K m − Y r , one may use a decomposition of K m or K m − Y r into paths which are not necessarily all isomorphic.In [7] it is shown that the obvious necessary conditions are sufficient for the existence of a decomposition of K m into paths of any specified lengths.This facilitates the construction of decompositions of K n − I into many combinations of 2-regular graphs of many different orders.

( 2 )
r , and the union of this 1-regular graph with the copy of Y(2) r is a copy of X 2r .Thus, we have a decomposition of K 2m into one copy the electronic journal of combinatorics 23(2) (2016), #P2.1 of X 2r , ( m 2 − 3r 2 )/r copies of P

Theorem 10 . 2 ,
Let G be a bipartite 2-regular graph, let k be the order of G, and let n 4 be even.There exists a G-decomposition of K n − I if and only if 3 k n and k divides n(n−2) except possibly when n 2 < k < n − 2 and n(n−2) 2k is odd both hold.Proof.The conditions 3 k n and k divides n(n−2) 2 are clearly necessary for the existence of a G-decomposition of K n − I.The case k = n is covered by the main theorem in [8] and the case k = n − 2 is covered by Lemma 8.If k = n − 1, then k does not divide n(n−2)
each of H 1 , H 2 and H 3 is isomorphic to G, If r 2 is even and G is any bipartite 2-regular graph of order 2r, then there is a decomposition of X 2r into three copies of G and a 1-factor.Proof.If r = 2, then G is a 4-cycle, X 2r is isomorphic to K 6 and the result holds.So assume r 4. Observe that if the edges {u 1 , u 4 }, {u 1 , v 4 }, {v 1 , u 4 }, {v 1 , v 4 } of J 2r are replaced with {u 2 , u 4 }, {u 2 , v 4 }, {v 2 , u 4 }, {v 2 , v 4 }, the vertices u 1 and v 1 are deleted, and the edge {u 2 , v 2 } is added, then the resulting graph is X 2r .Let {H 1 , H 2 , H 3 , H 4 } be the decomposition of J 2r given by Lemma 6, let H 1 be the graph obtained from H 1 by replacing the edges {u 1 , u 4 } and {u 1 , v 4 } with {v 2 , u 4 } and {v 2 , v 4 }, let H 2 be the graph obtained from H 2 by replacing the edges {v 1 , u 4 } and {v 1 , v 4 } with {u 2 , u 4 } and {u 2 , v 4 }, let H 3 = H 3 and let H 4 be the graph obtained from H 4 by adding the edge {u 2 , v 2 } (and the vertices u 2 and u 4 ).It is easy to see that {H 1 , H 2 , H 3 , H 4 } is the required decomposition of X 2r .