All graphs with maximum degree three whose complements have 4-cycle decompositions

www.elsevier.com/locate/disc

All graphs with maximum degree three whose complements have

4-cycle decompositions

Chin-Mei Fu

, Hung-Lin Fu

, C.A. Rodger

, Todd Smith

Received 30 May 2003; received in revised form 23 November 2004; accepted 29 September 2006 Available online 4 September 2007

In honor of Jennie Seberry on the occasion of her 60th birthday

Abstract

LetG be the set that contains precisely the graphs on n vertices with maximum degree 3 for which there exists a 4-cycle system of their complement in Kn. In this paperG is completely characterized.

© 2007 Elsevier B.V. All rights reserved.

MSC: 05B30; 05C38

Keywords: 4-cycle; Specified leave; 3-regular; Graph decomposition

1. Introduction

Ak-cycle system of a graph G is a partition of the edges of G into sets, each of which induces a cycle of length k. Over the past 30 years there has been considerable interest in the problem of finding k-cycle systems for various

families of graphs. The classic result in this area is that of Sotteau[17], who found necessary and sufficient conditions for the existence of a k-cycle system of Ka,b, the complete bipartite graph. Probably the best result follows after years

of activity in the literature (see for example[12,14]) from a recent pair of papers that completely settle the existence problem for k-cycle systems for complete graphs, and for complete graphs with the edges in a 1-factor removed[1,16]. More recently, attention has focused on fixing k and allowing G to vary in non-symmetric ways. For example, it is now known when it is possible to find k-cycle systems of G when G is formed from Knby removing the edges of any tree

whenever k is 4 or 6 (see[9,2], respectively), and by removing the edges of any 2-regular graph whenever k is 3[8], 4

[10], 6[3], or n[4,6,15]. In this paper we solve the existence problem for 4-cycle systems of G for all the myriad of graphs one can form from Knby removing the edges of any subgraph of maximum degree 3.

This problem has also been of some statistical interest. Balanced sampling designs excluding contiguous units were first introduced by Hedayat et al.[11]. In terms of this paper, they were interested in a 1-dimensional problem that is equivalent to a 3-cycle decomposition of the complete graph with the edges in a 2-factor removed. Since then,

E-mail addresses:cmfu@mail.tku.edu.tw(C.-M. Fu),hlfu@math.nctu.edu.tw(H.-L. Fu),rodgec1@auburn.edu(C.A. Rodger), tbsmith@mail.ucf.edu(T. Smith).

0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2007.07.082

2-dimensional variants have been considered, where the graph removed from Knis a 4-factor[5]. One can also define

similar variants of the existence problem for neighbor designs, a statistical experimental design that is equivalent to a decomposition of the complete graph into cycles of a fixed length[13], corresponding to antigens being placed around the rim of petri dishes, surrounding an antiserum. Results in this paper relate directly to this generalization, allowing for the situation where we do not need to see how some pairs of antigens react together. Another version allows one antigen to be placed at the center of the petri dish, so the design corresponds to a decomposition of Kninto wheels[7].

A graph G is said to be a3-graph if(G)3. G is said to be odd if all vertices have odd degree. A graph G on n

vertices is said to be n-admissible if:

(a) all vertices in Kn− E(G) have even degree and

(b) 4 dividesn₂− |E(G)|.

If the value of n is irrelevant then we can also refer to G as being simply admissible. LetG(n)= {G|G is an odd

3-graph of order n for which there exists a 4-cycle system of K_n − E(G)}. Clearly if G ∈ G(n) then G is n-admissible. LetG =_n1G(n). In this paper we completely determine G (see Theorem 1), showing that there are exactly two non-admissible graphs that are not inG.

Let G denote the complement of G. Let G ∨ H denote the graph formed from two vertex-disjoint graphs G and H by joining each vertex in G to each vertex in H with one edge. Let G[A] denote the subgraph of G induced by A. It is easy to see that the edges of Ka,bwith vertex sets A and B can be partitioned into sets of size 4, each of which induces

a 4-cycle, if and only if both a and b are even; we let C(A, B) denote such a set of 4-cycles. 2. Some helpful results

The following result makes it easy to check if a given graph is admissible.

Lemma 1. If G is an odd3-graph, then G is admissible if and only if the number of vertices of degree 1 is congruent

to|V (G)| modulo 4.

Proof. Suppose G is admissible. By (a), let|V (G)| = 4x + 2y with y ∈ {0, 1}. Let the number of vertices of degree 1 in G be z. Then 2|E(G)| = z + 3(4x + 2y − z) = 12x + 6y − 2z. By (b), 4 divides (4x + 2y)(4x + 2y − 1)/2 −

(6x + 3y − z) = 4(2x2+ 2xy − 2x − y) + 2y2− z, so z ≡ 2y (modulo 4) as required. Similarly, if z ≡ 2y (modulo

4) then z ≡ 2y2(modulo 4), so reversing the above argument shows that (b) is true. Of course (a) is true since every graph that is odd must have an even number of vertices. 

It will be useful to defineS = {n| there exists a 4-cycle system of Kn− E(G) for all admissible odd 3-graphs G

with|V (G)| = n}. An odd 3-graph of order n, G, is said to be maximal if there is no odd3-graph of order n, say G, with the property that Gcan be formed from G by adding 4 new edges which induce a 4-cycle. Clearly the following is true.

Lemma 2. If each maximal odd3-graph on n vertices is inG(n), then n ∈ S.

In particular, since the complement of any subgraph of G induced by four vertices of degree 1 must contain a 4-cycle, it follows that to prove that n ∈ S we need only consider the admissible odd 3-graphs that have at most 2 vertices of degree 1. Therefore, throughout this paper, when we consider graphs with at least 10 vertices we assume that they have at most 2 vertices of degree 1 (since 8 /∈ S, all graphs on 8 vertices need to be considered).

The main result is proved inductively. One common tool used to effect such a proof is to remove four vertices of degree 3 in G that induce a path, P. Such a subgraph can arise in several ways, depending on the neighborhood of P. In the following result we consider four such cases. Let N(P ) denote the set of vertices that are each adjacent to at least one vertex in P.

Lemma 3. Suppose that G is an admissible odd3-graph on n vertices that contains an induced path P =(a1, a2, a3, a4)

of a4be c1and c2(possibly these vertices are not all distinct). Suppose that one of the following conditions is satisfied

by G:

(1) bi = cjfor 1i, j 3 and

(a) |{b1, b2, b3}| = |{c1, c2, c3}| = 3 and {{b1, c3}, {b2, c2}, {b3, c1}} ∩ E(G) = ∅ or (b) b2= b3, c2= c3, and{{b1, c2}, {b2, c2}, {b2, c1}} ∩ E(G) = ∅ or

(c) b2= b3, |{c1, c2, c3}| = 3 and {{b1, c1}, {b2, c2}, {b2, c3}} ∩ E(G) = ∅; (2) b1= c1,|{b1, b2, b3, c1, c2, c3}| = 5, and {{b1, b3}, {b1, c3}, {b2, c2}} ∩ E(G) = ∅.

If n − 4 ∈ S then G ∈ G.

Proof. We consider the four cases in turn.

Case (1a): Suppose that b3∈ {b1/ , b2} and c3∈ {c1/ , c2}. Form G1from G−V (P ) by adding the edges {b1, c3}, {b2, c2}, and{b3, c1}. Then dG1(v) = dG(v) for each v ∈ V (G1). So since n − 4 ∈ S, by Lemma 1 there exists a 4-cycle system (V (G)−V (P ), C1) of K_n−4−E(G1). Let C2=C1∪{(a1, a3, b1, c3), (a1, a4, b2, c2), (a2, a4, b3, c1), (a1, b3, a3, c1),

(a2, b1, a4, c3), (a2, b2, a3, c2)} ∪ C(V (P ), V (G) − N(P )). Then (V (G), C2) is a 4-cycle system of K_n− E(G).

Case (1b): Suppose b2= b3and c2= c3. Form a graph G1from G − V (P ) by adding the edges {b2, c2}, {b1, c2}, and{b2, c1}. Then, as before, there exists a 4-cycle system (V (G) − V (P ), C1) of Kn−4− E(G1). Define C2= C1∪ {(a2, b1, a3, c1), (a1, a4, b2, c2), (a1, a3, b2, c1), (a2, a4, b1, c2)} ∪ C(V (P ), V (G) − N(P )). Then (V (G), C2) is a 4-cycle system of Kn− E(G).

Case (1c): Suppose b2= b3and c3∈ {c1/ , c2}. Since |V (G)| is even, G contains another vertex z. Form G1from G −

V (P ) by adding the edges {b1, c1}, {b2, c2}, and {b2, c3}. Then, as before, let (V (G)−V (P ), C1) be a 4-cycle system of

Kn−4−E(G1). Let C2=C1∪{(a1, a4, b2, c3), (a1, a3, b2, c2), (a2, a4, b1, c1), (a2, b1, a3, c2), (a1, c1, a3, z), (a2, c3,

a4, z)} ∪ C(V (P ), V (G) − (N(P ) ∪ {z})). Then (V (G), C2) is a 4-cycle system of Kn− E(G).

Case 2: Suppose b1= c1. Since|V (G)| is even, G contains another vertex z. Form G1from G − V (P ) by adding the edges{b1, b3}, {b1, c3}, and {b2, c2}. As before, there exists a 4-cycle system (V (G) − V (P ), C1) of Kn−4− E(G1).

Define C2 = C1∪ {(a1, b3, a3, z), (a1, c3, a2, a4), (a1, c2, b2, a3), (a2, b2, a4, z), (a2, b1, a3, c2), (a4, b3, b1, c3)} ∪

C(V (P ), V (G) − (N(P ) ∪ {z})). Then (V (G), C2) is a 4-cycle system of Kn− E(G).  3. The main result

The following result handles the small cases, of which there are many! For example, if G is connected, cubic and has 12 vertices then it is one of the 85 such graphs—the proof handles these in a neat way. Lemma 2 is useful throughout this section.

Proposition 1. If G is a cubic graph with at most 12 vertices, then G ∈ G if and only if G /∈ {G1, G2} (seeFig. 1).

Proof. The result is trivial if n = 4.

If n = 6 then G must have either 2 or 6 vertices of degree 1. So G is a 1-factor, or consists of the 2 components

K2and K4, or is induced by the edge set{{1, 2}, {2, 3}, {2, 4}, {3, 4}, {3, 5}, {4, 5}, {5, 6}}. In each case the required

decomposition is easy to find.

Suppose that n = 8. Then G has 0, 4, or 8 vertices of degree 1. If G contains a component isomorphic to K2 then the result follows by deleting this component, applying the solution when n = 6, then adding three 4-cycles between the six vertices and the replaced K2. If G consists of two components isomorphic to K4 then the result follows from a 4-cycle system of K4,4. If G is connected and has four vertices of degree 1 then G has edge set {{1, 3}, {2, 3}, {3, 4}, {4, 5}, {4, 6}, {5, 6}, {5, 7}, {6, 8}}, or {{1, 5}, {2, 6}, {3, 7}, {4, 8}, {5, 6}, {6, 7}, {7, 8}, {5, 8}}. In either case, G contains two vertices u and v of degree 1 whose neighbors w and x, respectively, are non-adjacent. Delete

u and v and join w and x to form a graph on six vertices that by the case n = 6 has a 4-cycle system; to this system

add the 4-cycle (u, v, w, x) and then the 4-cycles in a 4-cycle system of K2,4. Finally we consider the case where G is connected and has 0 vertices of degree 1. All connected non-isomorphic cubic graphs on at most 12 vertices are listed on Gordon Royle’s website:http://www.cs.uwa.edu.au/gordon/remote/cubics/index.html. There are five such graphs on eightvertices. It is easy to establish that three of them are inG(8) and the two depicted inFig. 1(graphs 2 and 5 in the Royle list) are not.

Suppose that n=10. As in the previous case, if G is disconnected then one component must be either K2or K4, so the

result follows by using previous cases unless one of the components is one of the graphs inFig. 1. If it is G1then the

4-cycles in {(1, 7, 3, 6), (2, 5, 4, 8), (1, 3, 8, 9), (8, 6, 2, 10), (2, 4, 7, 9), (7, 5, 1, 10), (10, 3, 9, 4), (10, 5, 9, 6)} provide the required system. If it is G2, then the 4-cycles in {(1, 7, 2, 6), (3, 8, 5, 9), (3, 10, 4, 6), (1, 3, 5, 10), (2, 4, 9, 8), (4, 7, 10, 8), (2, 10, 6, 9), (1, 9, 7, 5)} provide the required system.

If n = 10 and G is connected then proceed as follows. If G contains two vertices say 9 and 10 of degree 1 with

dG(9, 10)4 then delete 9 and 10 and join their neighbors u and v, respectively, to form G. If G= G1, G2then adding

the 4-cycle (9, 10, u, v) to 4-cycle systems of K8−E(G) and of K2,6(with bipartition{9, 10} and {1, . . . , 8}−{u, v})

provides the result. If G=G1then note that removing 8-cycle (a1, a2, . . . , a8)=(1, 3, 7, 5, 2, 4, 8, 6) from K8−E(G)

leaves two 4-cycles c1 and c2. Since for some i we have u = ai+1 and v = ai+3, ai+4, or ai+5 (reducing the

sum modulo 8), to c1and c2add the six 4-cycles in {(ai+5, 9, ai+7, 10), (9, ai+2, ai+1, ai+8), (9, ai+4, ai+5, ai+6),

(10, ai+2, ai+3, ai+4), (10, ai+6, ai+7, ai+8), (9, 10, ai+1, ai+3)}, {(ai+2, 9, ai+8, 10),(ai+3, 9, ai+6, 10) (9, ai+5,

ai+6, ai+7), (10, ai+1, ai+8, ai+7),(ai+1, ai+2, ai+3, ai+4),(9, 10, ai+5, ai+4)}, or {(ai+3, 9, ai+7, 10), (9, ai+2, ai+1,

ai+8), (9, ai+4, ai+5, ai+6),(10, ai+2, ai+3, ai+4), (10, ai+6, ai+7, ai+8), (9, 10, ai+1, ai+5)}, respectively, to form

the required 4-cycle system. If G = G2 then this approach does not work. Fortunately, each of the two

vertex-permutations (1, 5)(2, 4)(3)(6, 8)(7) and (1)(2)(3, 5)(4)(6, 7)(8) are automorphisms of G2, so there exist

automor-phisms of G2acting transitively on the edges in G2[{1, . . . , 5}], on the edges in {{1, 8}, {3, 7}, {5, 6}}, and on the edges

in G2[{6, 7, 8}]. So we need only consider the three cases: where (u, v) = (1, 2), where (u, v) = (1, 8), and where (u, v) = (6, 8). In these cases the result follows: from Lemma 2, Case 1(c) with (a1, a2, a3, a4, c1) = (8, 7, 3, 2, 5)

when (u, v) = (1, 2); by Lemma 2, Case 1(c) with (a1, a2, a3, a4, c1) = (8, 7, 3, 2, 5) when (u, v) = (1, 8); and by the

4-cycles in{(9, 1, 6, 8), (9, 2, 6, 4), (9, 3, 1, 5), (9, 7, 1, 10), (10, 2, 4, 7), (10, 3, 8, 4), (10, 5, 3, 6), (2, 8, 5, 7)} when

(u, v) = (6, 8).

If n= 10 and dG(9, 10)3, then G is one of the graphs inFig. 2. In each case except when G is G3 or G4,

the vertices have been labeled so that the result follows from Lemma 2, case 1(a) or (c) (as indicated below the graph). If G = G3or G4then the 4-cycles in {(1, 2, 7, 8), (1, 4, 7, 6), (1, 5, 3, 7), (2, 4, 8, 5), (2, 6, 3, 8), (1, 9, 2, 10), (3, 9, 4, 10), (5, 9, 6, 10)} or {(1, 2, 6, 5), (1, 4, 9, 6), (1, 7, 2, 8), (1, 9, 2, 10), (2, 4, 10, 5), (3, 6, 7, 9), (3, 5, 8, 10), (3, 7, 4, 8)}, respectively, provide the result.

Now, suppose that n = 12 and G is connected. According to the Royle list, there are 85 possibilities for G. For exam-ple, the 4-cycles in S1= {(1, 4, 2, 5), (3, 7, 4, 8), (5, 11, 6, 12), (1, 8, 6, 10), (4, 10, 9, 11), (2, 6, 3, 9), (2, 10, 3, 11), (1, 3, 5, 9), (2, 7, 5, 8), (1, 11, 8, 12), (4, 6, 7, 12), (7, 9, 12, 10)} form a 4-cycle system of K12−G where G happens to

be graph 15 in the list, containing the edges in{{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 1}, {1, 7}, {2, 12}, {3, 12}, {4, 9}, {5, 10}, {6, 9}, {7, 8}, {7, 11}, {8, 9}, {8, 10}, {10, 11}, {11, 12}}. This set S1of cycles has been carefully selected. Each

of the first six cycles (a, b, c, d) listed have the property that the edges {a, c} and {b, d} are in E(G). So we could replace this 4-cycle with (a, c, b, d) to form another 4-cycle system of K12− G1for some other graph G1, and then

replace it with (a, c, d, b) to form another 4-cycle system of K12 − G2 for some third graph G2. So we can get

36_{graphs by making all such switches. Of course there will be many repetitions, so we test for isomorphism using}

NAUTY, but hopefully most of the 85 graphs can be obtained in this way. In fact, doing this produces 4-cycle systems for all such graphs except for those numbered 1, 2, 3, 4, 5, 6, 9, 11, 12, 16, 17, 20, 23, 32, 39, 46, 56, and 67 in the Royle list. Similarly one can switch cycles using the first 4 and the 7th cycle in S1 to get graphs 11, 17, and 56. If

Fig. 2. Graphs on 10 vertices.

(1, 7, 12, 10), (2, 6, 8, 12), (2, 10, 4, 11), (3, 6, 4, 8), (3, 7, 9, 11), (3, 9, 4, 12)} then all edges occur in a 4-cycle

ex-cept for those in{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {7, 8}, {8, 9}, {9, 10}, {10, 11}, {11, 12}, {12, 1}, {1, 4}, {2, 7}, {3, 10}, {5, 8}, {6, 11}, {9, 12}; these edges induce graph 5 in the Royle list. In S2the first four cycles can be switched to produce up to 34possible graphs for which there exists a 4-cycle system of the complement in K12. Among these, graphs 1, 2, 3, 4, 5 and 12 in the Royle list appear. Next let S3={(2, 10, 3, 11), (5, 11, 6, 12), (4, 7, 10, 8), (1, 3, 12, 9),

(1, 7, 2, 8), (1, 4, 12, 10), (1, 5, 9, 11), (2, 4, 6, 9), (2, 6, 8, 12), (3, 5, 10, 6), (3, 7, 5, 8), (4, 9, 7, 11)}, thus avoiding

the edges in {1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {7, 8}, {8, 9}, {9, 10}, {10, 11}, {11, 12}, {12, 1}, {1, 6}, {2, 5}, {3, 9}, {4, 10}, {7, 12}, {8, 11} which induce graph 6. The first four cycles can be switched, thereby producing graphs 6, 9, 16, 20, 23, and 39 in the Royle list. We also get graph 67 by replacing the first, fourth and fifth cycles in S3with

(2, 3, 11, 10), (1, 12, 9, 3), and (1, 2, 7, 8). The edges in the 4-cycles in {(1, 3, 8, 9), (1, 6, 4, 12), (1, 7, 3, 11), (1, 8,

2, 10), (2, 3, 10, 7), (2, 6, 5, 11), (2, 9, 3, 12), (4, 5, 12, 11), (4, 7, 5, 8), (4, 9, 5, 10), (6, 7, 12, 10), (6, 8, 11, 9)} av-oid the edges in graph 32. Finally, the edges in the 4-cycles in {(1, 3, 8, 4), (1, 5, 10, 7), (1, 8, 12, 10), (1, 11, 2, 12),

(2, 4, 7, 9), (2, 5, 8, 10), (2, 6, 3, 7), (3, 5, 7, 11), (3, 9, 5, 12), (4, 6, 8, 11), (4, 9, 6, 10), (6, 11, 9, 12)} avoid the

edges in graph 46.

It remains to suppose that n = 12 and G is disconnected. Then G has 0, 4, 8, or 12 vertices of degree 1. First, if G has a component K2, then the result follows by deleting this component, using the solution when n = 10 and adding a 4-cycle system of K2,10. Second, if G has a component of order 4, then it must be K4. If the rest of G is G = G1, G2(see

Fig. 1) then this can be handled by combining 4-cycle systems of Gand K4,8. Otherwise, Gis either G1or G2. Observe

that both G1and G2, can be decomposed into the two Hamiltonian cycles in:{(a1, a2, . . . , a8), (b1, b2, . . . , b8)} =

{(1, 7, 3, 6, 8, 2, 4, 5), (1, 3, 8, 4, 7, 5, 2, 6)} or {(1, 7, 2, 8, 4, 6, 3, 5), (1, 3, 8, 5, 7, 4, 2, 6)} for G1or G2, respectively.

Now, let V (K4)={9, 10, 11, 12}. Then the 4-cycle system of K4∨Gis{(9, a1, a2, a3), (10, a3, a4, a5), (9, a5, a6, a7), (10, a7, a8, a1), (9, a2, 10, a4), (9, a6, 10, a8), (11, b1, b2, b3), (12, b3, b4, b5), (11, b5, b6, b7), (12, b7, b8, b1), (11, b2, 12, b4), (11, b6, 12, b8)}.

Finally, we consider the case that G has two components of order 6. Let the two components be H1and H2. If H1

system of H1∨ H2can be obtained by combining 4-cycle systems of H1, H2, and K6,6, respectively. Therefore, it is left to consider the case when xi ≡ 0 (mod 4). Clearly, if H1(or H2) has four vertices of degree 1, then we can join

these four vertices by a 4-cycle. That is, it suffices to consider the case when H1and H2are cubic graphs of order 6. This implies that each of H1and H2is either a 6-cycle or a vertex-disjoint union of 3-cycles. Thus, the proof will be concluded by verifying C4|C6∨ 2C3, C4|C6∨ C6, and C4|2C3∨ 2C3.

Starting with C6∨ 2C3, we let V (C6∪ 2C3) = {ai, bi|i ∈ Z6}. Then the 4-cycle system of C6∨ 2C3 can be

obtained as follows:{(a_i, bi, ai+3, bi+1), (ai, ai+1, bi, bi+2)|i ∈ Z6}. (Note that the 6-cycle is (a0, a1, a2, a3, a4, a5) and 2C3= (b0, b2, b4) ∪ (b1, b3, b5).)

Next, let V (C6∪C6)={ai, bi|i ∈ Z6}. The 4-cycle system of C6∨C6is obtained as follows: {(b0, a0, a1, a2), (b2, a2,

a3, a4), (b4, a4, a5, a0), (a1, b0, b1, b2), (a3, b2, b3, b4), (a5, b4, b5, b0), (b0, a3, b1, a4), (b2, a0, b1, a5), (b4, a1, b1,

a2), (b3, a0, b5, a1), (b3, a2, b5, a3), (b3, a4, b5, a5)}.

Finally, let V (2C3∪2C3)={ai, bi, ci, di|i ∈ Z3}. Then the 4-cycle system of 2C3∨2C3is the following collection of

4-cycles: {(a1, a2, b2, b1), (a1, a3, d3, d1), (a1, b2, c1, b3), (a1, d2, c1, d3), (c1, c2, d2, d1), (c1, c3, b3, b1),(a2, a3, b2,

b3), (a2, d3, c3, b1), (b3, a3, b1, c2), (b2, c2, d1, c3), (d2, d3, c2, c3), (d1, a2, d2, a3)}.  Moving on to the general setting, we begin with the disconnected case.

Proposition 2. Suppose that G is an admissible, odd3-graph that is disconnected. If n − 4 ∈ S then G ∈ G. Proof. If one component of G is a copy of K4defined on the vertex set A, then since n − 4 ∈ S there exists a 4-cycle system (V (G)\A, C1) of Kn−4− E(G1), where G1= G[V (G)\A]. Then (V (G), C1∪ C(V (G)\A, A)) is a 4-cycle system of Kn− E(G).

Otherwise, each vertex in v is incident with an edge that occurs in no 3-cycle. Since we can assume that G has at most two vertices of degree 1, such an edge exists in each component that joins two vertices of degree 3; let{a1, a2}

and{a3, a4} be two such edges that occur in different components of G. For 1i 4, let bi,1and bi,2be the other two

neighbors of ai. Let A = {a1, a2, a3, a4} and B = {bi,j | 1i 4, 1j 2}.

Since n−4 ∈ S, by Lemma 1 there exists a 4-cycle system (V (G)\A, C1) of Kn−4−E(G), where G1is formed from G−A by adding the edges {bi,j, bi+2,j} for 1i, j 2. Since n−4 ∈ S, there exists a 4-cycle system (V (G)\A, C1) of Kn−4− E(G). Let C2= C1∪ {(a1, b3,1, b1,1, a3), (a1, b4,1, b2,1, a4), (a2, b3,2, b1,2, a4), (a2, b4,2, b2,2, a3), (b1,1, a2,

b3,1, a4), (b1,2, a2, b4,1, a3), (b2,1, a1, b4,2, a3), (b2,2, a1, b3,2, a4)} ∪ C(A, V (G)\(A ∪ B)). Then (V (G), C2) is a 4-cycle system of Kn− E(G). 

Now we provide the final piece of the puzzle.

Proposition 3. Let n14. Suppose that G is a connected, admissible, odd 3-graph on n vertices. If n − 4 ∈ S then

G ∈ G.

Proof. We consider several cases in turn. Recall that we are assuming that G contains at most two vertices of degree 1.

Case 1: Suppose G contains no 3-cycle and no 4-cycle.

Let a be a vertex of degree 3 with neighbors b1, b2, and b3that also have degree 3. For 1i 3 let ci,jfor 1j 2 be

the neighbors of bi. Being in Case 1, all 10 vertices defined so far must be distinct, and the only possible edges joining

those vertices in G must be of the form{c_i,j, ck,l} where i = k. Let V1be the set of these 10 vertices. Form a graph G1

from G[V (G)\{a, b1, b2, b3}] by adding the edges {ci,1, ci,2} for 1i 3. Then by Lemma 1, G1is an admissible odd 3-graph, on n − 4 vertices, so since n − 4 ∈ S we know G1∈ G. Let (V (G)−{a, b1, b2, b3}, C1) be a 4-cycle system of Kn−4−E(G1). Define C2=C1∪{(a, c1,1, b3, c2,1), (a, c1,2, b2, c3,1), (a, c2,2, b1, c3,2), (b1, b2, c3,2, c3,1), (b2, b3,

c1,2, c1,1), (b1, b3, c2,2, c2,1)} ∪C({a, b1, b2, b3}, V (G)\V1).Then (V (G), C2) is a 4-cycle system of K_n− E(G).

Case 2: Suppose G contains a 3-cycle but no 4-cycle.

Let = (a1, a2, a3) be a 3-cycle in G chosen so that the neighbor of a3outside has degree 3; let this neighbor of a3be b. Note that G[{a1, a2, a3, b}] has exactly four edges since G has no 4-cycles. Naming further vertices, let

c1, c2, c3,1, and c3,2 be neighbors of a1, a2, b, and b, respectively. Being in Case 1, all eight vertices defined so far are distinct (for example, c1 = c2; for otherwise (a1, a3, a2, c1) is a 4-cycle in G). Let V1 be the set of these eight vertices.

Suppose that c1is not adjacent to one of c3,1and c3,2, say c3,1, and that c2is not adjacent to the other, c3,2. Then let

G1be formed from G − {a1, a2, a3, b} by adding the edges {c1, c3,1} and {c2, c3,2}. By Lemma 1, G1is an admissible odd3-graph, so since n − 4 ∈ S, there exists a 4-cycle system (V (G)\{a1, a2, a3, b}, C1) of Kn−4− E(G). Define C2= C1∪ {(a1, b, c1, c3,1), (a2, b, c2, c3,2), (a1, c2, a3, c3,2), (a2, c1, a3, c3,1)} ∪ C({a1, a2, a3, b}, V (G)\V1). Then

(V (G), C2) is a 4-cycle system of Kn− E(G).

Next notice that since G contains no 4-cycles, it follows that neither c1nor c2can be joined to both of c3,1and c3,2, nor can they have a common neighbor. Therefore, it now suffices to finish this case by considering the possibility that {{c1, c3,1}, {c2, c3,1}} ⊆ E(G) and {{c1, c3,2}, {c2, c3,2}} ∩ E(G) = ∅. Then P = (a3, b, c3,1, c2) is an induced path in

G that satisfies condition (1c) in Lemma 2. So G ∈ G. Case 3: Suppose G contains a 4-cycle.

Let (a1, a2, a3, a4) be a 4-cycle in G. We consider several subcases in turn. Let A = {a1, a2, a3, a4}.

(1) Suppose{{a1, a3}, {a2, a4}} ⊆ E(G). Then G[A] is a component in G, contradicting that G is connected.

(2) Suppose{a1, a3} ∈ E(G) and {a2, a4} /∈ E(G). Let b2and b4be the third neighbors of a2and a4, respectively. If b2= b4and{b2, b4} /∈ E(G) then let G1be formed from G − A by adding the edge {b2, b4}. Since n − 4 ∈ S, let (V (G)\A, C1) be a 4-cycle system of K_n−4− E(G1), and let C2= C1 ∪ {(a1, b2, a3, b4), (a2, a4, b2, b4)} ∪

C(V (G)\(A ∪ {b2, b4}), A). Then (V (G), C2) is a 4-cycle system of K_n− E(G).

If b2= b4and{b2, b4} ∈ E(G), then let c2and c4, be the third neighbors of b2and b4, respectively. If c2= c4then let d be the third neighbor of c2, and let e1and e2be neighbors of d; apply Lemma 2 with P = (e1, d, c2, b2) using condition (1b) or (1c) depending on whether{e1, e2} ∈ E(G) or not. If one of b2 and b4 has degree 1, then let the other, bx(so x ∈ 2, 4) be adjacent to c which is adjacent to d1and d2; apply Lemma 2 to P = (d1, c, bx, ax) using

condition (1c) or (1a) depending on whether or not{d1, d2} ∈ E(G). So now suppose that c2 = c4. If c2or c4has degree 1, then proceed as above. If either c2or c4, say cx (so x ∈ {2, 4}) has a neighbor d /∈ {c2, c4} that is not also

adjacent to the other c vertex, then apply Lemma 2 with P = (d, cx, bx, ax) using condition (1a). Otherwise c2and c4 have a common neighbor d that has a third neighbor e that has two further neighbors f1and f2that are different from all the vertices defined so far (since n14 and G is connected); apply Lemma 2 to P = (f1, e, d, c2) using condition (1a–c), depending on whether{c2, c4} ∈ E(G) and/or {f1, f2} ∈ E(G).

If b2= b4then let c be the third neighbor of b2. If c has degree 1 in G then we have just defined a component in

G, contradicting the assumption that G is connected. If c has degree 3, then let the neighbors of c be d1and d2. Then

P = (a2, b2, c, d1) satisfies condition (1a) or (1c) of Lemma 2, depending on whether d1and d2are adjacent or not. So again the result follows.

(3) We can now assume that A induces a 4-cycle in G. Let B be the set of vertices in V (G)\A that are adjacent to vertices in A. We consider the values of|B| in turn. For 1i 4 let bi be adjacent to ai; so B = {bi|1i 4} where b1, . . . , b4need not all be distinct.

(3i) Suppose that|B| = 4. First suppose that the complement of G[B] contains a 1-factor. Without loss of gener-ality we may assume that either both{b1, b3} and {b2, b4} are not edges in G, or both {b1, b4} and {b2, b3} are not

edges in G.

If both the edges{b1, b3} and {b2, b4} do not occur in G then form G1 by adding them to G[V (G)\A]. Since

n − 4 ∈ S, let (V (G)\A, C1) be a 4-cycle system of Kn−4− E(G1). Let C2= C1∪ {(a1, a3, b1, b3), (a2, a4, b2, b4),

(a1, b2, a3, b4), (a2, b1, a4, b3)} ∪C(V (G)\(A ∪ B), A). Then (V (G), C2) is a 4-cycle system of Kn− E(G).

If both the edges {b1, b4} and {b2, b3} do not occur in G then form G1 by adding them to G[V (G)\A]. Since

n − 4 ∈ S, there exists a 4-cycle system (V (G)\A, C1) of Kn−4− E(G1). Let {z1, z2} ⊂ V (G)\(A ∪ B). Let C2=

C1∪{(a1, b4, b1, a3), (a2, b3, b2, a4), (b2, a3, b4, z1), (b4, a2, b1, z2), (b1, a4, b3, z1), (b3, a1, b2, z2)}∪C(V (G)\(A∪

B), A), where z1and z2are any two vertices in V (G)\(A ∪ B). Then (V (G), C2) is a 4-cycle system of Kn− E(G).

If the complement of G[B] contains no 1-factor then since G[B] has maximum degree at most 2 it must contain a 3-cycle, say (b2, b3, b4). Since G is connected, let c1and c2be the two other neighbors of b1. Then P = (c1, b1, a1, a2) satisfies condition (1c) or (1a) of Lemma 2, depending on whether c1and c2are adjacent or not. So again the result follows.

(3ii) Suppose that|B| = 3. We can assume that either b3= b4or that b2= b4.

First suppose that b3= b4. If neither{b1, b3} nor {b2, b3} are edges in G then let G1be formed by adding them to G[V (G)\A]. Since n − 4 ∈ S there exists a 4-cycle system (V (G)\A, C1) of Kn−4− E(G1). Let C2= C1∪ {(a1, a3, b2, b3), (a2, a4, b1, b3), (b1, a2, z, a3), (b2, a1, z, a4)} ∪ C(V (G)\(A ∪ B ∪ {z}), A), where z is any vertex in

Otherwise, if b3= b4then it now follows that one of{b1, b3} or {b2, b3} must be an edge in G (we can assume it is {b2, b3} by symmetry). Either b1has degree 3 or it has degree 1 in G. If b1has degree 3, then let its two other neighbors be c1and c2(possibly c2= b2). If dG(b1) = 3 and c2= b2then P = (c1, b1, a1, a4) satisfies condition (1a) or (1c) of Lemma 2 depending on whether c1is adjacent to c2or not. If dG(b1) = 3 and c2= b2then since G is connected, c1 must have two neighbors d1and d2; then P = (d1, c1, b1, a1) satisfies condition (1c) or (1a) of Lemma 2 depending on whether d1is adjacent to d2or not. If b1has degree 1, then let c be the remaining neighbor of b2, and let d1and d2 be the other two neighbors of c (all vertices defined in this situation are necessarily distinct because for each vertex all three incident edges have been defined). Then P = (d1, c, b2, a2) satisfies condition (1c) or (1a) of Lemma 2 depending on whether d1is adjacent to d2or not, so the result follows.

Next suppose that b2= b4. If either{b1, b2} or {b2, b3} is in G (by symmetry we assume {b2, b3} ∈ E(G)) then (a2, a3, a4, b2) is a 4-cycle in G where N(a2, a3, a4, b2) contains only two further vertices, namely a1and b3; so this case is handled when we consider|B| = 2. Therefore, we can assume that neither {b1, b2} nor {b2, b3} is an edge in G.

If dG(b1) = 1 = dG(b3) then let c be the remaining neighbor of b2, and let d1and d2be the other neighbors of c. Then

P = (d1, c, b2, a4) satisfies condition (1c) or (1a) of Lemma 2 depending on whether d1and d2are adjacent or not. So suppose dG(b1) = 3. Let c1and c2be neighbors of b1, and let c3be the other neighbor of b2, where possibly c2= c3. Then P = (c1, b1, a1, a2) satisfies condition (1c) or (1a) of Lemma 2, depending on whether c1and c2are adjacent or not.

(3iii) Suppose that|B| = 2. Then we have three cases to consider.

Suppose that{a1, b1}, {a2, b1}, {a3, b3}, and {a4, b3} are edges in G. Since G is connected, {b1, b3} /∈ E(G), so let

the remaining neighbors of b1and b3be c1and c3, respectively. If c1= c3then P = (a1, b1, c1, d) is a path satisfying condition (1c) of Lemma 2, where d is the third neighbor of c1. If c1= c3then P = (a1, b1, c1, d1) satisfies condition (1b) or (1c) of Lemma 2, where d1∈ {b1/ , c3} is a neighbor of c1.

Next, suppose that{a1, b1}, {a2, b2}, {a3, b1}, and {a4, b2} are edges in G. Since G is connected, {b1, b2} /∈ E(G).

Let the remaining neighbors of b1and b2and c1and c2, respectively. Then the rest of this case follows the previous case, using P = (a1, b1, c1, d1) and Lemma 2 to obtain the result.

Finally, suppose that b1 is adjacent to a1, a2, and a3, and{a4, b4} ∈ E(G). Then P = (c4, b4, a4, a3) satisfies condition (1c) or (1a) of Lemma 2, where c4= a4is a neighbor of b4. 

We can now collect all the pieces.

Theorem 1. Let G be a graph on n vertices, where n is even and(G)3. Then there exists a 4-cycle system of

Kn− E(G) if and only if

1. all vertices in G have odd degree, 2. 4 divides n(n − 1)/2 − |E(G)|, and

3. G is not one of the two graphs of order 8 described in Proposition 1.

Proof. Since the necessity is clear, we prove the sufficiency. So let G satisfy conditions (1–3) of this theorem. If n12 then the result follows from Proposition 1. So suppose that n14. Using induction, suppose that m ∈ S for all m satisfying 10mn − 2. If G is disconnected then the result follows from Proposition 2. If G is connected then the result follows from Proposition 3. 

Acknowledgment

The first two authors would like to express their thanks to the Discrete and Statistical Sciences Department at Auburn University for their support of the visit during which this research was completed. The third author wishes to thank the Department of Mathematics at Tamkang University for their kind support for a visit during which this research was begun.

References

[1]B. Alspach, H. Gavlas, Cycle decompositions of Knand Kn−I, J. Combin. Theory B 81 (2001) 77–99.

[3]D.J. Ashe, H.L. Fu, C.A. Rodger, All 2-regular leaves of partial 6-cycle systems, Ars Combin. 76 (2005) 129–150. [4]D. Bryant, Hamilton cycle rich two-factorizations of complete graphs, J.Combin. Design 12 (2) (2004) 147–155.

[5]D. Bryant, Y. Chang, C.A. Rodger, R. Wei, Two-dimensional balanced sampling plans excluding contiguous units, Comm. Statist. Theory Methods 31 (2002) 1441–1455.

[6]H. Buchanan II, Graph factors and hamiltonian decompositions, Ph.D. Dissertation, West Virginia University, 1997.

[7]C.J. Colbourn, C.C. Lindner, C.A. Rodger, Neighbor designs and m-wheel systems, J. Statist. Plann. Inference 27 (1991) 335–340. [8]C.J. Colbourn, A. Rosa, Quadratic leaves of maximal partial triple systems, Graphs Combin. 2 (1986) 317–337.

[9]H.L. Fu, C.A. Rodger, Forest leaves and four-cycles, J. Graph Theory 33 (2000) 161–166.

[10]H.L. Fu, C.A. Rodger, Four-cycle systems with two-regular leaves, Graphs Combin. 17 (2001) 457–461.

[11]A.S. Hedayat, C.R. Rao, J. Stufken, Sampling plans excluding contiguous units, J. Statist. Plann. Inference 19 (1988) 159–170.

[12]C.C. Lindner, C.A. Rodger, Decomposition into Cycles II: Cycle Systems, Contemporary Design Theory, Wiley-Interscience Series Discrete Mathematics Optimization, Wiley, New York, 1992 pp. 325–369.

[13]D.H. Rees, Some designs of use in serology, Biometrics 23 (1967) 779–791.

[14]C.A. Rodger, in: C.J. Colbourn, J.H. Dinitz (Eds.), Cycle Systems, The CRC Handbook of Combinatorial Designs, CRC Press, Boca Raton, FL, 1996, pp. 266–269.

[15]C.A. Rodger, Hamilton decomposable graphs with specified leaves, Graphs Combin. 20 (2004) 541–543.

[16]M. Sajna, Cycle decompositions. III. Complete graphs and fixed length cycles, J. Combin. Design 10 (2002) 27–78. [17]D. Sotteau, Decomposition of Km,n(Km,n∗ ) into cycles (circuits) of length 2k, J. Combin. Theory B 30 (1981) 75–81.