Cube Factorizations of Complete Graphs

Peter Adams, Darryn Bryant, Barbara MaenhautDepartment of Mathematics, The University of Queensland, QLD 4072, Australia,E-mail: bmm@maths.uq.edu.au

Received November 4, 2003; revised January 22, 2004

Published online 16 April 2004 in Wiley InterScience (www.interscience.wiley.com).

DOI 10.1002/jcd.20015

Abstract: A cube factorization of the complete graph on n vertices, Kn, is a 3-factorization of

Kn in which the components of each factor are cubes. We show that there exists a cube

factorization of Kn if and only if n � 16 ðmod 24Þ, thus providing a new family of uniform

3-factorizations as well as a partial solution to an open problem posed by Kotzig in 1979.

# 2004 Wiley Periodicals, Inc. J Combin Designs 12: 381–388, 2004

Keywords: factorization; cube decomposition; uniform 3-factorization

1. INTRODUCTION

A k-factor of a graph G is a k-regular spanning subgraph of G. A k-factorization of Gis a set of k-factors of G whose edge sets partition the edge set of G. A k-factorizationin which all of the k-factors are isomorphic is called uniform. It is easy to see that ifthere exists a 3-factorization of the complete graph on n vertices, Kn, thenn � 4 ðmod 6Þ, and for these values of n it is straightforward to construct a uniform3-factorization of Kn from a 1-factorization of Kn that has a cyclic automorphism oforder n� 1. Such 1-factorizations are well known, see [14]. However, for uniform 3-factorizations in which the structure of the 3-factors is specified in advance, very littleis known. For n ¼ 10, the smallest non-trivial value of n for which there exists a 3-factorization of Kn, a complete enumeration of 3-factorizations of Kn is given in [10].Resolvable block designs of order v with block size k ¼ 4 and index � ¼ 1 areequivalent to 3-factorizations of Kv in which each 3-factor consists of v=4 vertexdisjoint copies of K4, and are known to exist for all v � 4 ðmod 12Þ; see [9]. Inthis paper (see Theorem 4.1) we give a second complete family of uniform 3-factorizations of Kn, where each component of each 3-factor is the 3-cube.

Contract grant sponsor: Australian Research Council

# 2004 Wiley Periodicals, Inc.

381

The d-cube is a graph whose vertex set can be labelled with the set of all binaryd-tuples, so that its edge set consists of all pairs of vertices which differ in exactlyone coordinate. It is clear that the d-cube has 2d vertices, d � 2d�1 edges, and isd-regular and bipartite. A d-cube factorization of a graph G is a d-factorization ofG in which each component of each factor is a d-cube. The factors in a d-cubefactorization are called d-cube factors. Related to the concept of a factorization isthat of a decomposition. A d-cube decomposition of G is a set of d-cubes whose edgesets partition the edge set of G.

In 1979, two problems posed by Kotzig, Problems 15 and 16 in [12], sparkedinterest in d-cube decompositions and d-cube factorizations of complete graphs. Thetwo main open problems are:

Cube Decomposition Problem. For which values of n and d does there exist ad-cube decomposition of Kn?

Cube Factorization Problem. For which values of n and d does there exist ad-cube factorization of Kn?

Since these problems were introduced, progress on the cube factorization problemhas been restricted to sporadic values of n or the case where n is a power of two,see [5] and [6]. The d-cube has 2d vertices and is d-regular, so necessary conditionsfor the existence of a d-cube factorization of the complete graph Kn are that

n � 0 ðmod 2dÞ and n � 1 ðmod dÞ:

The first condition implies that n must be even and the second condition implies thatn must have opposite parity to d. Hence, if a d-cube factorization of Kn exists, then dmust be odd. In this paper, we settle the cube factorization problem for d ¼ 3 byshowing that these necessary conditions are also sufficient. (Note that d ¼ 3 is thefirst non-trivial value of d. A 1-cube factorization is simply a 1-factorization.)

In order to settle the 3-cube factorization problem, we apply a constructiontechnique that uses frames. The crucial insight which enabled this technique to beapplied was the use of a 3-cube factorization of a particular 24-regular graph on40 vertices (see Section 3). This factorization was constructed by combining severalsmaller factorizations, one of which was found using a computational search.

The cube decomposition problem and its variations have been investigated bymany people and several results have been obtained, although the cube decomposi-tion problem itself is far from being completely solved. If we ask only for a d-cubedecomposition of Kn rather than a d-cube factorization, then the necessary conditionsare less stringent. The d-cube is d-regular and has d � 2d�1 edges, so we require thatd j n� 1 and d � 2d j nðn� 1Þ. These conditions imply that if there exists a d-cubedecomposition of Kn then

(a) if d is even then n � 1 ðmod d � 2dÞ;(b) if d is odd then either

(i) n � 1 ðmod d � 2dÞ, or(ii) n � 0 ðmod 2dÞ and n � 1 ðmod dÞ.

Kotzig [11] established sufficiency of these conditions in the cases when d is evenand when d is odd and n is odd (cases (a) and (b) (i) above). Sufficiency in the case

382 ADAMS, BRYANT, AND MAENHAUT

where d ¼ 3 follows from Kotzig’s results along with a result of Maheo [13] thatsettles the case d ¼ 3 and n � 16 ðmod 24Þ. The case d odd and n even remains openfor d � 5. It is interesting to note that the open case for the cube decompositionproblem involves precisely those values of n for which a d-cube factorization of Kn

may exist, so solving the cube factorization problem would also solve the cubedecomposition problem.

Cube decompositions of graphs other than Kn have also been studied. Decomposi-tions of complete bipartite graphs into d-cubes were investigated by several authorsin [3,6–8], decompositions of complete multigraphs and complete bipartite multi-graphs into 3-cubes were investigated in [1], and maximum packings and minimumcoverings of Kn by 3-cubes were investigated in [2].

2. PRELIMINARIES AND EXISTING RESULTS

The remainder of this paper deals exclusively with 3-cubes, so the term cube will beused for 3-cube. Let KmðxÞ denote the complete multipartite graph with x parts of sizem. The complete bipartite graph Kmð2Þ may be written as Km;m. All graphs in this paperare simple. The vertex set and edge set of a graph G will be denoted by VðGÞ andEðGÞ, respectively.

A useful tool in the construction of factorizations is a frame. A k-frame of type nx

is a decomposition of KnðxÞ into edge-disjoint copies of Kk, called blocks, such that theset of blocks can be partitioned into subsets, called partial factors, which satisfy thefollowing two conditions:

(1) each partial factor is a set of blocks in which each vertex of KnðxÞ occurs eitherone or zero times;

(2) in each partial factor, the vertices that do not occur are precisely the vertices ina part (of size n) of KnðxÞ .

Related to a k-frame, we define a cube frame of type nx to be a cube decomposition ofKnðxÞ , such that the set of cubes can be partitioned into partial factors satisfyingconditions (1) and (2) above.

We require some additional notation to describe our graphs. The cube with vertexset fv1; v2; v3; v4; v5; v6; v7; v8g and edge set

ffv1; v2g; fv2; v3g; fv3; v4g; fv1; v4g; fv5; v6g; fv6; v7g; fv7; v8g; fv5; v8g;fv1; v5g; fv2; v6g; fv3; v7g; fv4; v8gg

will be denoted by ½v1; v2; v3; v4; v5; v6; v7; v8�. The cycle of length m with verticesfv1; v2; . . . ; vmg and edges ffvi; vjgj j � iþ 1 ðmod mÞg will be denoted byðv1; v2; . . . ; vmÞ. The letter I will be used to denote a 1-factor of a graph. If H is asubgraph of a graph G, then the graph obtained by removing the edges of H from Gwill be denoted by G� H. If G and H are edge-disjoint graphs, then the union of Gand H, denoted G [ H, is the graph with vertex set VðG [ HÞ ¼ VðGÞ [ VðHÞ andedge set EðG [ HÞ ¼ EðGÞ [ EðHÞ. If G and H are vertex-disjoint graphs, then the

CUBE FACTORIZATIONS 383

join of G and H, denoted G _ H, is the graph with vertex set VðG _ HÞ ¼VðGÞ [ VðHÞ and edge set EðG _ HÞ ¼ EðGÞ [ EðHÞ [ ffx; ygjx 2 VðGÞ;y 2 VðHÞg. The complement of a graph G will be denoted Gc.

Our constructions make use of the following results.

Lemma 2.1. There exists a cube factorization of K8 � I.

Proof. The existence of such a factorization is well known, see for instance [2]. Acube factorization of K8 � I, with vertex set Z8, is given by the following cubefactors.

Factor 1 : ½0; 1; 2; 3; 4; 5; 6; 7� Factor 2 : ½0; 2; 7; 5; 6; 4; 1; 3� &

Lemma 2.2. Let T be any 2-regular graph of order 16 such that each component ofT is a cycle of length 4; 8; 12 or 16. Then there exists a cube factorization ofK8;8 � T .

Proof. Let

T ffi ð0; 1; . . . ;m1 � 1Þ [ ðm1;m1 þ 1; . . . ;m2 � 1Þ [ � � � [ ðmt;mt þ 1; . . . ; 15Þ:

Note that m1;m2; . . . ;mt � 0 ðmod 4Þ. Let X be the graph obtained by removingthe edges of T from K8;8 with bipartition f0; 2; 4; . . . ; 14g; f1; 3; 5; . . . ; 15g. Fori; j 2 f0; 1; 2; 3g, let Xij ¼ fx 2 VðXÞ : x � i ðmod 4Þ or x � j ðmod 4Þg. The sub-sets X01; X03; X12, and X23 of VðXÞ each induce a cube in X. To see this note that thecube is isomorphic to K4;4 � I. Pairing the cube induced by X01 with the cube inducedby X23, and the cube induced by X03 with the cube induced by X12, gives the requiredcube factorization. &

Lemma 2.3 [6]. Let d be odd and let s be the order of 2 ðmod dÞ. If r is any integerwith r � d=s, then there exists a d-cube factorization of K2rs .

3. CUBE FACTORIZATIONS OF CERTAIN GRAPHS

Lemma 3.1. There exist cube factorizations of K16 and K64.

Proof. This result follows from Lemma 2.3. &

Lemma 3.2. There exists a cube factorization of K12;12.

Proof. A cube factorization of K12;12 has four cube factors, each containing threecubes. Label the vertices of K12;12 with the elements of Z24, with the bipartitionf0; 1; . . . ; 11g; f12; 13; . . . ; 23g. A cube factorization is given by the following cubefactors.

Factor 1: ½0; 12; 1; 13; 14; 2; 15; 3�; ½4; 16; 5; 17; 18; 6; 19; 7�;½8; 20; 9; 21; 22; 10; 23; 11�:

Factor 2: ½0; 15; 4; 19; 20; 5; 12; 3�; ½1; 14; 11; 18; 21; 10; 13; 2�;½6; 17; 9; 22; 23; 8; 16; 7�:

384 ADAMS, BRYANT, AND MAENHAUT

Factor 3: ½0; 16; 3; 17; 22; 1; 23; 2�; ½4; 13; 5; 14; 20; 6; 21; 7�;½8; 12; 10; 18; 19; 11; 15; 9�:

Factor 4: ½0; 18; 3; 21; 23; 5; 22; 4�; ½1; 17; 10; 19; 20; 11; 16; 2�;½6; 12; 9; 14; 15; 7; 13; 8�: &

The essential ingredient of our cube factorization construction is a cube factoriza-tion of a 24-regular graph on 40 vertices that contains a set of 16 independentvertices, and whose complement also has a cube factorization. Finding such a graphusing a purely computational approach is a sizeable problem. To avoid this, therequired graph was found by carefully arranging the union of six copies of K8;8 � T(with T as defined in Lemma 2.2), three copies of K8 � I, and a six-regular graph. Wenow construct this 24-regular graph and show that it has a cube factorization.

Let G be the following 8-regular graph on 24 vertices. Let VðGÞ ¼ Z24 andlet fx; yg 2 EðGÞ if and only if x; y 2 f0; 1; . . . ; 7g; x; y 2 f8; 9; . . . ; 15g; x; y 2f16; 17; . . . ; 23g or

fx; yg 2 ff0; 8g; f1; 17g; f2; 10g; f3; 19g; f4; 12g; f5; 21g;f6; 14g; f7; 23g; f9; 16g; f11; 18g; f13; 20g; f15; 22gg:

Thus G consists of three copies of K8 with each pair of copies of K8 joined by fourvertex-disjoint edges.

Lemma 3.3. There exists a cube factorization of Kc16 _ G.

Proof. Consider the graph H defined as follows. Let VðHÞ ¼ Z40 and EðHÞ ¼ffx; ygjfx�16; y� 16g2 EðGÞg [ ffx; ygjx 2 f0; 1; . . . ; 15g; y 2 f16; 17; . . . ; 39gg.Thus H is isomorphic to Kc

16 _ G. Note that H has 480 edges and 40 vertices, so acube factorization of H necessarily has eight cube factors, each containing five cubes.Such a cube factorization is given by the following cube factors.

Factor 1: ½0; 16; 24; 25; 33; 17; 8; 32�; ½1; 18; 26; 27; 35; 19; 9; 34�;½2; 28; 5; 36; 37; 12; 20; 10�; ½3; 30; 7; 38; 39; 14; 22; 11�;½4; 21; 6; 23; 29; 13; 31; 15�:

Factor 2: ½0; 20; 28; 29; 37; 21; 8; 36�; ½1; 22; 30; 31; 39; 23; 9; 38�;½2; 17; 3; 19; 25; 10; 27; 11�; ½4; 24; 5; 32; 33; 12; 16; 13�;½6; 26; 7; 34; 35; 14; 18; 15�:

Factor 3: ½0; 18; 4; 17; 21; 2; 16; 1�; ½5; 22; 6; 19; 23; 3; 20; 7�;½8; 26; 13; 25; 29; 10; 24; 9�; ½12; 30; 15; 27; 31; 11; 28; 14�;½32; 34; 38; 36; 37; 39; 35; 33�:

Factor 4: ½0; 22; 4; 19; 23; 2; 20; 1�; ½5; 18; 6; 17; 21; 3; 16; 7�;½8; 30; 13; 27; 31; 10; 28; 9�; ½12; 26; 15; 25; 29; 11; 24; 14�;½32; 38; 33; 39; 35; 37; 34; 36�:

Factor 5: ½0; 24; 7; 31; 30; 1; 25; 5�; ½4; 28; 6; 27; 26; 3; 29; 2�;½8; 37; 14; 34; 39; 9; 32; 12�; ½10; 33; 15; 38; 35; 11; 36; 13�;½16; 18; 22; 20; 21; 23; 19; 17�:

Factor 6: ½0; 28; 7; 27; 26; 1; 29; 5�; ½4; 25; 6; 30; 31; 3; 24; 2�;½8; 33; 14; 38; 35; 9; 36; 12�; ½10; 32; 15; 39; 34; 11; 37; 13�;½16; 22; 17; 23; 19; 21; 18; 20�:

CUBE FACTORIZATIONS 385

Factor 7: ½0; 35; 3; 36; 38; 5; 33; 6�; ½4; 37; 7; 39; 34; 1; 32; 2�;½8; 19; 14; 16; 18; 12; 17; 9�; ½10; 22; 15; 21; 23; 13; 20; 11�;½24; 26; 30; 28; 29; 31; 27; 25�:

Factor 8: ½0; 34; 3; 32; 39; 5; 37; 6�; ½4; 36; 7; 35; 38; 1; 33; 2�;½8; 23; 14; 20; 22; 12; 21; 9�; ½10; 19; 15; 16; 18; 13; 17; 11�;½24; 30; 25; 31; 27; 29; 26; 28�:

&

We make the following remarks on the construction of the above factorization.The graph Kc

16 _ G can be regarded as the union of four spanning subgraphs each ofdegree six: three copies of ðK8;8 � TÞ [ ðK8;8 � TÞ [ ðK8 � IÞ (with T as definedin Lemma 2.2) and one 6-regular spanning subgraph which we will call H0. The cubesin Factors 1 and 2 are a cube factorization of H0. These cubes (and hence thesubgraph H0) were found computationally. It has been shown that K8 � I has a cubefactorization containing two cubes (Lemma 2.1), and K8;8 � T has a cube factoriza-tion with two cube factors each containing two cubes (Lemma 2.2). Thus each copyof ðK8;8 � TÞ [ ðK8;8 � TÞ [ ðK8 � IÞ has a cube factorization with two cube factors.These cube factors make up Factors 3 to 8.

Lemma 3.4. There exists a cube factorization of the graph Gc.

Proof. Notice that Gc is isomorphic to K8;8;8 � I, so Gc has 180 edges and 24vertices. Thus a cube factorization of Gc has five cube factors, each containing threecubes. Such a cube factorization is given by the following cube factors.

Factor 1: ½0; 11; 3; 16; 20; 7; 9; 2�; ½4; 23; 6; 21; 18; 14; 22; 12�;½5; 8; 17; 13; 10; 1; 15; 19�:

Factor 2: ½0; 9; 4; 14; 12; 6; 8; 2�; ½3; 23; 5; 17; 21; 1; 19; 7�;½11; 22; 13; 16; 20; 10; 18; 15�:

Factor 3: ½0; 13; 7; 15; 10; 4; 16; 6�; ½2; 23; 12; 17; 21; 8; 19; 11�;½3; 14; 5; 22; 18; 1; 20; 9�:

Factor 4: ½0; 23; 10; 21; 19; 9; 17; 14�; ½2; 11; 6; 13; 15; 4; 20; 3�;½5; 12; 1; 16; 18; 7; 22; 8�:

Factor 5: ½0; 22; 4; 17; 18; 2; 19; 6�; ½3; 12; 20; 8; 10; 16; 14; 7�;½5; 9; 1; 11; 15; 21; 13; 23�: &

Lemma 3.5. There exists a cube factorization of K40.

Proof. Let V0 ¼ fv1; v2; . . . ; v16g. Define the graphs H1; H2, and H3 as follows. LetH1 ffi K16 with VðH1Þ ¼ V0. Let H2 ffi Kc

16 _ G with VðH2Þ ¼ V0 [ Z24. Let H3 ffi Gc

with VðH3Þ ¼ Z24. Then H1 [ H2 [ H3 ffi K40. Let R be a cube factorization of H1

(Lemma 3.1), with cube factors R1;R2; . . . ;R5. Let Q be a cube factorization of H2

(Lemma 3.3), with cube factors Q1;Q2; . . . ;Q8. Let S be a cube factorization ofH3 (Lemma 3.4), with cube factors S1;S2; . . . ;S5. Then the following 13 cubefactors give a cube factorization of H1 [ H2 [ H3.

fQ1;Q2; . . . ;Q8;R1 [ S1;R2 [ S2; . . . ;R5 [ S5g: &

386 ADAMS, BRYANT, AND MAENHAUT

4. CUBE FACTORIZATIONS OF Kn

We first construct a family of cube frames.

Lemma 4.1. For all integers x � 3, there exists a cube frame of type 24x.

Proof. A 2-frame of type 2x exists for all x � 3 [4]. Let M be a 2-frame of type 2x

with the underlying graph K2ðxÞ . Replace each vertex of K2ðxÞ by a set of 12 vertices,and each edge of K2ðxÞ by a copy of K12;12. Replace each block in M by acube factorization of K12;12, which exists by Lemma 3.2. This gives a cube frame oftype 24x. &

In the above construction of a cube frame, for each group of two vertices in K2ðxÞ ,there are two partial factors of the 2-frame that miss that pair of vertices. Hence, inthe cube frame of type 24x, for each group of 24 vertices, there are eight partialfactors that miss that group.

Theorem 4.1. Let n be a positive integer. There exists a cube factorization of Kn ifand only if n � 16 ðmod 24Þ.Proof. The requirement that n � 16 ðmod 24Þ is clearly necessary. Let n ¼24xþ 16: Cube factorizations of K16; K40, and K64 are given by Lemmas 3.1 and3.5. Thus it remains to consider x � 3.

Let V0 ¼ f0; 1; . . . ; 15g and let Vi ¼ ff0i; 1i; . . . ; 23igj1 � i � xg. For x � 3, letthe vertex set of K24xþ16 be [x

i¼0Vi. Let P be a cube frame of order 24x on the vertexset [x

i¼1Vi, with partial factors Pi1;Pi2; . . . ;Pi8 missing the vertices in Vi.For 1 � i � x, let Qi be a subgraph of K24xþ16 with vertex set V0 [ Vi, which is

isomorphic to Kc16 _ G, where if fx; yg 2 EðGÞ, then fxi; yig 2 EðQiÞ. Let Qi be a

cube factorization of Qi. This cube factorization has eight cube factors, call theseQi1;Qi2; . . . ;Qi8.

Let R be a complete subgraph of K24xþ16 with vertex set V0. Let R be a cubefactorization of R and call the five cube factors R1;R2; . . . ;R5.

For 1 � i � x, let Si be a subgraph of K24xþ16 with vertex set Vi, which is iso-morphic to Gc, where if fx; yg 2 EðGcÞ, then fxi; yig 2 EðSiÞ. Let Si be a cubefactorization of Si. This cube factorization has five cube factors, call these Si1;Si2; . . . ;Si5:

A cube factorization of K24xþ16 is given by the following 8xþ 5 cube factors:

fPij [ Qijj1 � i � x; 1 � j � 8g;fRj [x

i¼1 Sijj1 � j � 5g: &

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CUBE FACTORIZATIONS 387

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388 ADAMS, BRYANT, AND MAENHAUT