On the spectrum for Km+2 Km designs

On the Spectrum for Km+ \ KmDesigns

Peter Adams* and Elizabeth J. Billington*Centre for Combinatorics, Department of Mathematics, The University of Queens-land, Queensland 4072, Australia

D. G. Hoffman†Department of Discrete and Statistical Sciences, Math Annex, Auburn University,Auburn, Alabama, U.S.A. 36849-5307

ABSTRACT

The spectrum problem for the decomposition of Kn into copies of the graph Km+2 \ Km issolved for n ≡ 0 or 1 (mod 2m + 1). c© 1997 John Wiley & Sons, Inc.

1. INTRODUCTION AND PRELIMINARY RESULTS

A G-design of order n is an edge-disjoint decomposition of the complete undirected graphKn into isomorphic copies ofG,which covers precisely all of the edges ofKn.For example,if G is a cycle of length m, then such a G-design is usually called an m-cycle system, whileif G is K3, then we usually refer to a K3-design as a Steiner triple system. A great deal ofwork has been done on G-designs for particular small G. If G is the graph K4 \K2, thatis, K4 − e (where e is any edge), the spectrum problem (for which orders n do these G-designs exist?) and 2-colouring problems have been dealt with; see [3, 6]. If G is K5 \K3,the spectrum problem was solved in [2]. More generally, a small embedding for partialKm \ Km−2 designs is given in [7]. However, the spectrum problem for Km+2 \ Km

designs (using our preferred terminology) has not been considered for m ≥ 4, and thisarticle addresses that problem.

*Research supported by Australian Research Council Grants A49532750 and A695104.†Research supported by ONR Grant N00014-95-1-0769.

c© 1997 John Wiley & Sons, Inc. CCC 1063-8539/97/010049-12Journal of Combinatorial Designs, Vol. 5, No. 1 (1997)

49

50 ADAMS, BILLINGTON, AND HOFFMAN

FIG. 1. The Graph Km+2 \Km.

NowKm+2\Km is a graph onm+2 points with 2m+1 edges, so if there is aKm+2\Km

design of ordern then it must contain n(n−1)2(2m+1) isomorphic copies ofKm+2\Km.Henceforth

we shall refer to such a copy of Km+2 \Km in a decomposition of Kn as a block, and wedenote the Km+2 \Km graph shown in Figure 1 by (x1, x2;x3, x4, . . . , xm+2), or also by(x1α, x2α;x3β, x4β, . . . , xm+2β), where α is any permutation on {x1, x2} and β is anypermutation on {x3, x4, . . . , xm+2}. So we list the two vertices of degree m+1 first, in anyorder, followed by the m vertices of degree 2, in any order. Clearly, a necessary conditionfor existence of a Km+2 \Km design of order n is that n(n−1)

2(2m+1) is an integer. So we shallconsider the cases

n ≡ 0, 1, 2m + 1, 2m + 2 (mod 4m + 2).

Note that if 2m + 1 is a prime power, then this settles the entire spectrum.When m is odd, every vertex of Km+2 \Km has even degree, and so a further necessary

requirement then is that n should be odd. Hence we have

Lemma 1.1. There is no Km+2 \Km design of order n when m is odd and n is even.

Of course, for certain composite values of 2m + 1, we have other plausible values forn (mod 2m + 1) (see the boldface entries in Table I), but in this article we concentrate onn ≡ 0, 1 (mod 2m + 1).

Our aim in this article is to prove the following theorem:

Theorem 1.1. If n ≡ 0 or 1 (mod 2m+ 1) when m is even, and n ≡ 1 or 2m+ 1 (mod4m + 2) when m is odd, then there exists a Km+2 \Km design of order n for all m ≥ 1except when (m,n) = (2M, 4M + 1),M ≥ 1.

The case m = 1 in the above theorem is, of course, that of a Steiner triple system.

2. EXAMPLES

In this section we present some Km+2 \ Km decompositions of various graphs. Thesesmall examples are useful in the general constructions given in Section 3.

SPECTRUM FOR Km+2 \Km DESIGNS 51

TABLE I. Necessary Conditions for Existence of a Km+2 \Km Design of Order n for m ≤ 20

m 2m + 1 n (mod 4m + 2)

2 5 0, 1, 5, 63 7 1, 74 9 0, 1, 9, 105 11 1, 116 13 0, 1, 13, 147 15 1, 15, 21, 258 17 0, 1, 17, 189 19 1, 19

10 21 0, 1, 7, 15, 21, 22, 28, 3611 23 1, 2312 25 0, 1, 25, 2613 27 1, 2714 29 0, 1, 29, 3015 31 1, 3116 33 0, 1, 12, 22, 33, 34, 45, 5517 35 1, 15, 21, 2518 37 0, 1, 37, 3819 39 1, 13, 27, 3920 41 0, 1, 41, 42

Example 2.1. There exist Km+2 \Km designs of order 4m + 3 for all m ≥ 1.

Proof. Let the vertex set of K4m+3 be Z4m+3. Then a Km+2 \ Km decomposition ofK4m+3 can be obtained by cycling the following starter block modulo 4m + 3:

(0, 1; 3, 5, 7, . . . , 2m + 1).

Example 2.2. There exist Km+2 \Km designs of order 8m + 5 for all m ≥ 1.

Proof. Let the vertex set of K8m+5 be Z8m+5. Then a Km+2 \ Km decomposition ofK8m+5 can be obtained by cycling the following two starter blocks modulo 8m + 5:

(0, 1; 3, 5, 7, . . . , 2m− 1, 2m + 2),

(0, 2m; 4m + 3, 4m + 4, . . . , 5m + 2).

Example 2.3. There exists a Km+2 \Km decomposition of the complete tripartite graphK2m+1,2m+1,2m+1.

Proof. Let the vertex set of K2m+1,2m+1,2m+1 be

{01, 11, . . . , (2m)1} ∪ {02, 12, . . . , (2m)2} ∪ {03, 13, . . . , (2m)3}.Then a Km+2 \Km decomposition of K2m+1,2m+1,2m+1 can be obtained by cycling thefollowing three starter blocks modulo 2m + 1:

(01, 02; 03, 13, . . . , (m− 1)3),


(01,m3; (m + 1)2, (m + 2)2, . . . , (2m)2),

(02,m3; (m + 1)1, (m + 2)1, . . . , (2m)1).

Lemma 2.1. When m is even, there is no Km+2 \Km design of order 2m + 1.

Proof. Note that K2m+1 is a 2m-regular graph, while Km+2 \ Km has two vertices ofodd degree m + 1. Now the number of edges in K2m+1 is m(2m + 1), and the number ofexpectedKm+2\Km blocks in a decomposition ofK2m+1 is thereforem, sinceKm+2\Km

contains 2m + 1 edges.Assuming that a Km+2 \Km design of order 2m+1 exists, when we remove one block

from K2m+1, two vertices will have degree (2m − (m + 1)) = m − 1, which is odd.However, each block has two vertices of degree m + 1 and m of degree 2. So clearly nosuch design is possible.

Lemma 2.2. When m is odd, a Km+2 \Km design of order 2m + 1 exists.

Proof. A suitable design is given in Lemma 3.2 of [7]; we give an alternative one here.Since m is odd, let m = 2M + 1, and let the vertex set of K2m+1 = K4M+3 be

{∞} ∪ {ij | 0 ≤ i ≤ 2M, j = 1, 2}.Then a Km+2 \Km design of order 4M +3 can be obtained by cycling the following starterblock modulo 2M + 1:

(01, 02;∞, 11, 12, 21, 22, . . . ,M1,M2).

Lemma 2.3. When m is even, a Km+2 \Km design of order 2m + 2 exists.

Proof. Again, a suitable design is given in Lemma 3.2 of [7]; we give an alternative onehere. Since m is even, let m = 2M, and let the vertex set of K2m = K4M+2 be

{ij | 0 ≤ i ≤ 2M, j = 1, 2}.Then a Km+2 \Km design of order 4M +2 can be obtained by cycling the following starterblock modulo 2M + 1:

(01, 02; 11, 12, 21, 22, . . . ,M1,M2).

We remark that the case m = 2 in Lemma 2.1 gives the nonexistence of a K4 \K2 (thatis, K4 − e) design of order 5, as detailed in [3] and [6].

Next we give some more useful examples.

Example 2.4. There exist Km+2 \Km designs of order 10m + 5 for all odd m.

Proof. Since m is odd, let m = 2M + 1, so K10m+5 = K20M+15. Let the vertex set ofK20M+15 be

{∞} ∪ {ij | 0 ≤ i ≤ 10M + 6, j = 1, 2}.


Then a Km+2 \Km decomposition of K20M+15 can be obtained by cycling the followingfive starter blocks modulo 10M + 7:

(01, 02;∞, (4M + 4)1, (4M + 5)1, . . . , (5M + 3)1, (4M + 4)2, . . . , (5M + 3)2),

(01, 11; 31, 51, . . . , (4M + 3)1),

(01, 12; 22, 32, . . . , (2M + 2)2),

(01, (10M + 6)2; (6M + 4)2, (6M + 5)2, . . . , (8M + 4)2),

(02, (4M + 3)2; 21, 31, 41, . . . , (2M)1, (6M + 4)1, (6M + 5)1).

Example 2.5. There exist Km+2 \Km designs of order 6m + 3 for all even m.

Proof. Sincem is even, letm = 2M, soK6m+3 = K12M+3.Let the vertex set ofK12M+3

be

{∞} ∪ {ij | 0 ≤ i ≤ 6M, j = 1, 2}.Then a Km+2 \Km decomposition of K12M+3 can be obtained by cycling the following

three starter blocks modulo 6M + 1:

(01, (4M)2;∞, (2M + 1)1, (2M + 2)1, . . . , (3M)1, 12, 22, . . . , (M − 1)2),

(02, 12; 32, 52, 72, . . . , (2M + 1)2, (2M + 3)1, (2M + 5)1, . . . , (4M + 1)1),

(01, 02; 11, 21, 31, . . . , (2M)1).

3. THE CONSTRUCTIONS

For our constructions we shall employ some group divisible designs (GDDs) given in thenext lemma; these all have index 1. For a proof, see for example Lemma 2.1 of [1] and thereferences therein.

Lemma 3.1. There exists

• a GDD on 6s + 4 points with blocks of size 3 and with one group of size 4 and 3sgroups of size 2, for s ≥ 1. Such a GDD will be denoted GDD[3, 1, {2, 4∗}; 6s+ 4];

• a GDD on 6s + δ points, δ = 0 or 2, with blocks of size 3 and groups of size 2, fors ≥ 1. Such a GDD will be denoted GDD[3, 1, 2; 6s + δ].

The examples given in Section 2, together with the existence of these GDDs, now provideall that is necessary in order to prove the following:

Theorem 3.1. If n ≡ 1 (mod 4m+2), then there exists aKm+2\Km design of order n.

Proof. Let n = (4m+ 2)t+ 1, so n = (2m+ 1)(2t) + 1, and let the vertex set of Kn be

{∞} ∪ {(i, j) | 1 ≤ i ≤ 2t, 1 ≤ j ≤ 2m + 1}.


Then on the set {i | 1 ≤ i ≤ 2t} we take a GDD[3, 1, 2; 2t] if 2t ≡ 0 or 2 (mod 6), or aGDD[3, 1, {2, 4∗}; 2t] if 2t ≡ 4 (mod 6); see Lemma 3.1.

When 2t ≡ 0 or 2 (mod 6), for each group {x, y} of the GDD we take a Km+2 \Km

design of order 4m + 3 on the set

{∞} ∪ {(x, j), (y, j) | 1 ≤ j ≤ 2m + 1},and for each block {a, b, c} of the GDD, on the set

{(a, j) | 1 ≤ j ≤ 2m + 1} ∪ {(b, j) | 1 ≤ j ≤ 2m + 1} ∪ {(c, j) | 1 ≤ j ≤ 2m + 1}we place a Km+2 \Km decomposition of K2m+1,2m+1,2m+1.

When 2t ≡ 4 (mod 6), the only difference in our construction is that for the one groupof size 4, say {x1, x2, x3, x4}, of the GDD, we take a Km+2 \Km design of order 8m+ 5on the set

{∞} ∪ {(xi, j) | 1 ≤ i ≤ 4, 1 ≤ j ≤ 2m + 1}.These component designs all exist; see Examples 2.1, 2.2 and 2.3.

Theorem 3.2. If n ≡ 2m + 1(mod 4m + 2), then there exists a Km+2 \Km design oforder n, provided n /= 2m + 1 when m is even.

Proof. Let n = (4m + 2)t + (2m + 1), so n = (2m + 1)(2t + 1), and let the vertex setof Kn be

{(i, j) | 1 ≤ i ≤ 2t + 1, 1 ≤ j ≤ 2m + 1}.First, suppose that m is odd; then by Lemma 2.2 there exists a Km+2 \Km design of

order 2m + 1. For 2t + 1 ≡ 1 or 3 (mod 6), take a Steiner triple system (STS) on the set{i | 1 ≤ i ≤ 2t + 1}. Then on each set {(i, j) | 1 ≤ j ≤ 2m + 1}, for i = 1, 2, . . . , 2t + 1,place a Km+2 \Km design of order 2m + 1, and for each triple {x, y, z} of the STS, onthe set

{(x, j) | 1 ≤ j ≤ 2m + 1} ∪ {(y, j) | 1 ≤ j ≤ 2m + 1} ∪ {(z, j) | 1 ≤ j ≤ 2m + 1}(∗)place a decomposition of K2m+1,2m+1,2m+1 (see Example 2.3).

When 2t + 1 ≡ 5 (mod 6), we use a GDD on the set {i | 1 ≤ i ≤ 2t + 1} with blocksof size 3 and one group of size 5, and 2t − 4 groups of size 1. This GDD exists; see forexample [5]. For the group of size 5, say {xi | 1 ≤ i ≤ 5}, on the set

{(xi, j) | 1 ≤ i ≤ 5, 1 ≤ j ≤ 2m + 1}place a decomposition of K10m+5 (see Example 2.4), and for each group {x} of size oneof the GDD, place on the set

{(x, j) | 1 ≤ j ≤ 2m + 1}a decomposition of K2m+1. Then for each block {x, y, z} of the GDD, repeat the pro-cedure described above on the set (∗), again using the Km+2 \ Km decomposition ofK2m+1,2m+1,2m+1.

Secondly, suppose m is even; so by Lemma 2.1 we know there is no Km+2 \Km designof order 2m + 1. However, we do have a design of order 6m + 3; see Example 2.5. Weconsider three cases, depending on the value of 2t + 1 (modulo 6).

When 2t + 1 ≡ 1(mod 6), for t ≥ 9 [and necessarily t ≡ 0 (mod 3)] we use a GDDon the set {i | 1 ≤ i ≤ 2t + 1} with blocks of size 3, and all groups of size 3 except one


of size 7 (see [5]). Let a typical block be {a, b, c}, a typical group be {x, y, z}, and let thegroup of size 7 be {xi | 1 ≤ i ≤ 7}. Then on each set

{(i, j) | i = x, y, z, 1 ≤ j ≤ 2m + 1}we place a decomposition of K6m+3 (see Example 2.5); on the set

{(xi, j) | 1 ≤ i ≤ 7, 1 ≤ j ≤ 2m + 1}we place a decomposition of K14m+7 (see the next section for this decomposition); and onthe set

{(a, j) | 1 ≤ j ≤ 2m + 1} ∪ {(b, j) | 1 ≤ j ≤ 2m + 1} ∪ {(c, j) | 1 ≤ j ≤ 2m + 1}we place a decomposition of K2m+1,2m+1,2m+1 (see Example 2.3).

For t = 3, we have the case K14m+7, and for t = 6 we have the case K26m+13. We dealwith these in the next section.

When 2t+1 ≡ 3 (mod 6), for all relevant t there exists a Kirkman triple system of order2t + 1 on the set {i | 1 ≤ i ≤ 2t + 1}. Take one parallel class, and on each block {x, y, z}of this class, place a suitable decomposition of K6m+3 (see Example 2.5) on the vertex set{(i, j) | i = x, y, z, 1 ≤ j ≤ 2m+ 1}. Then for each block {a, b, c} not in this one parallelclass, on the set

{(a, j) | 1 ≤ j ≤ 2m + 1} ∪ {(b, j) | 1 ≤ j ≤ 2m + 1} ∪ {(c, j) | 1 ≤ j ≤ 2m + 1}we place a decomposition of K2m+1,2m+1,2m+1 (see Example 2.3).

Finally, when 2t + 1 ≡ 5 (mod 6), [so t ≡ 2 (mod 3)], for t ≥ 8 we place a GDD withblock size 3, one group of size 5, and all other groups of size 3 (see [5]). We repeat theconstruction given when 2t+1 ≡ 1 (mod 6), except that the block of size 7 in the previouscase is replaced by the block of size 5 here, so a decomposition of K10m+5 is used (see thenext section).

When t = 5, the case K22m+11 needs separate treatment; see the next section. Also,when t = 2, the case K10m+5 is dealt with in the next section.

This completes the proof of Theorem 3.2.

The next result deals with the case m even and order n ≡ 0 (mod 4m+ 2). It also usesdesigns of orders 4m + 2 and 8m + 4 (see the next section).

Theorem 3.3. If m is even, there is a Km+2 \Km design of order congruent to 0 (mod4m + 2).

Proof. Let n = (4m + 2)t, so n = 2t(2m + 1), and let the vertex set of Kn be

{(i, j) | 1 ≤ i ≤ 2t, 1 ≤ j ≤ 2m + 1}.When 2t ≡ 0 or 2 (mod 6) we use a GDD with blocks of size 3 and groups of size 2 on theset {i | 1 ≤ i ≤ 2t} (see Lemma 3.1). Then for each block {a, b, c} of the GDD, on the set

{(a, j) | 1 ≤ j ≤ 2m + 1} ∪ {(b, j) | 1 ≤ j ≤ 2m + 1} ∪ {(c, j) | 1 ≤ j ≤ 2m + 1}we place a decomposition of K2m+1,2m+1,2m+1 (see Example 2.3), and for each group{x, y} of the GDD we take a design of order 4m + 2 (see the next section) on the set{(x, j), (y, j) | 1 ≤ j ≤ 2m + 1}.


When 2t ≡ 4 (mod 6), we take a GDD on the set {i | i ≤ i ≤ 2t} with blocks of size 3,one group of size 4 and the rest of size 2. The one change is that the group of size 4 requiresthe use of a design of order 8m + 4 (see the next section).

Finally we prove

Theorem 3.4. If m is even, there is a Km+2 \Km design of order 2m+2 (mod 4m+2).

Proof. Let n = (4m + 2)t + (2m + 2), so n = (2m + 1)(2t + 1) + 1. Let the vertex setof Kn be

{∞} ∪ {(i, j) | 1 ≤ i ≤ 2t + 1, 1 ≤ j ≤ 2m + 1}.The construction parallels that given in Theorem 3.2 above when m is odd; the only dif-ference is that we use a design of order 2m + 2 here (see Lemma 2.3), where one oforder 2m + 1 was used in Theorem 3.2, and one of order 10m + 6 (see the next section)here, where one of order 10m + 5 was used in Theorem 3.2, and with the element ∞appropriately used.

4. SOME DESIGNS WITH HOLES

The constructions in the previous section require seven more ingredients: designs of order4m + 2, 8m + 4, 10m + 5, 10m + 6, 14m + 7, 22m + 11 and 26m + 13, all for m even.We construct these here by enlarging designs previously constructed.

Specifically, we define a Km+2 \ Km design of order v with a hole of size h to be apartition of the edges of the graph Kv \ Kh into copies of Km+2 \ Km. The followingresult clearly holds.

Lemma 4.1. If there is a Km+2 \ Km design of order v with a hole of size h, and aKm+2 \Km design of order h, then there is a Km+2 \Km design of order v.

Next, we give some definitions. If a and b are integers with a ≤ b + 1, we denote by[a, b] the set of all integers x in the interval a ≤ x ≤ b. (Note that when a = b + 1, then[a, b] = ∅.)

Let d be a positive integer. For each integer x, define |x | d as follows: find the one andonly integer y with

−d/2 < y ≤ d/2, and y ≡ x (mod d);

then |x | d is defined to be the usual absolute value of y.Let C = [0, d − 1], and let D = [1, bd/2c]. We consider certain subgraphs of the

complete graph K2d with vertex set C × {L,R}. (Here L and R refer to Left and Right,helpful for a pictorial representation.)

Let e = ab be an edge of K2d. If a = (x, L) and b = (y, L), then we call e an edgeof left pure difference | y − x | d. If a = (x, L) and b = (y,R), then we call e an edge ofmixed difference r if y − x ≡ r (mod d), where r ∈ C. If a = (x,R) and b = (y,R), thenwe call e an edge of right pure difference | y − x | d. So mixed differences are in C, whilepure differences are in D.

If S,U ⊆ D and T ⊆ C, let 〈S, T, U〉 denote the spanning subgraph of K2d withexactly the following edges: those of left pure differences in S, together with those ofmixed differences in T, together with those of right pure differences in U.


A collection of edges of a graph G is called a one-factor if every vertex of G is met byexactly one edge of the collection. Furthermore, G is said to be one-factored if the edgesof G can be partitioned into one-factors.

Lemma 4.2. Let S ⊆ D,T ⊆ C with T /= ∅. Then 〈S, T, S〉 can be one-factored.

Proof. We use an idea due to Stern and Lenz; see [8]. Let a ∈ T. For each b ∈ T \{a}, theedges of 〈∅, {b}, ∅〉 form a one-factor, so it remains to one-factor the graphG = 〈S, {a}, S〉.Let H = 〈S, ∅, ∅〉, and let ∆ be the common degree in H of every vertex in the left part ofH,C × {L}. By Vizing's theorem (see [9]), the edges of H can be coloured with ∆ + 1colours in such a way that incident edges receive different colours. If (x, L)(y, L) is anedge of H, colour the edge (x + a,R)(y + a,R) with the colour used on (x, L)(y, L).

For each x ∈ C, there is at this point exactly one colour that does not appear at (x, L),and the same colour does not appear at (x + a,R). Colour the edge (x, L)(x + a,R) withthis missing colour.

For each of the ∆ + 1 colours, the edges of that colour form a one-factor.

Theorem 4.1. Let 0 ≤ 2t < d, and let 2t ≤ m. Then there is a Km+2 \Km design oforder v = 2m(d− 2t− 1)+m− 2t+ 2dwith a hole of sizeh = 2m(d− 2t− 1) +m− 2t.

Proof. The hole points will be (F ×M) ∪ S, where F,M and S are sets of cardinality2(d− 2t− 1),m andm− 2t, respectively. The points outside the hole will beC × {L,R},where C = [0, d − 1]. Consider first the base block

((0, L), (0, R); ([1, t] × {L,R}) ∪ S).

This is to be developed cyclically modulo d.On the points C × {L,R}, there remains a graph with the same left differences as right

differences, namely [t+1, bd/2c]. So, by Lemma 4.2, this graph can be one-factored; therewill be 2(d− 2t− 1) one-factors.

Let φ be a bijection from F onto the set of one-factors. For each j ∈ F, and for eachedge ab of φ(j), take the block (a, b; {j} ×M).

Corollary 4.1. There are Km+2 \Km designs of orders 4m+ 2, 8m+ 4 and 10m+ 6if m is even.

Proof. We shall use Lemma 4.1 and Theorem 4.1.For v = 4m + 2, take t = m/2 − 1 and d = m in Theorem 4.1. The hole of size

h = 2m + 2 can be filled by Lemma 2.3.For v = 8m + 4, take t = m/2 − 2 and d = m in Theorem 4.1. The hole of size

h = 6m + 4 can be filled by Theorem 3.4.For v = 10m + 6, take t = m/2 − 2 and d = m + 1 in Theorem 4.1. The hole of size

h = 8m + 4 can be filled by the previous case.It now remains to show existence of designs of orders (2s+1)(2m+1) for s = 2, 3, 5, 6

when m is even. We may assume that m ≥ 4, for if m = 2,K4 − e designs of these ordersall exist (see [3], [4] or [6]).

Theorem 4.2. Let m ≥ 4 be even, and let 2 ≤ s ≤ m. Then there is a Km+2 \ Km

design of order v = (2s + 1)(2m + 1) with a hole of size h = 4(s− 1)m + (2s− 1).


Proof. The hole points will be (F ×M) ∪ S, where F,M and S are sets of cardinality4s− 4,m and 2s− 1, respectively. The points outside the hole will be C ×{L,R}, whereC = [0, 3m], so C is really a set of integers, but we shall also think of C as the residuesmodulo 3m + 1. For the construction, we shall first carefully choose sets

X,YL, YR, ZL, ZR,W ⊆ C,

and a set P ⊆ D = [1, 3m/2]. We do this as follows:First suppose s < m. Choose any integer zL satisfying

1m/2 − s + 2

}≤ zL ≤

{m/2 + s− 1m− s

This is possible because each of the two lower bounds on zL is less than or equal to eachof the two upper bounds. Choose any integer x satisfying

0m− s− zL + 3

m− 2s + 1

≤ x ≤

{mm− s + 1 + zL

This is possible, again because each of the lower bounds is less than or equal to each of theupper bounds.

Let

X = [2, x + 1],

YL = [2,m− s− zL + 1],

YR = [2m + s− zL, 3m− x] and

ZL = {m− s− zL + 1 + 2i | 1 ≤ i ≤ zL}.

Now it is possible to partition the set

[x + 2, 2m + s− zL − 1] ∪ [3m + 1 − x, 2m + s + zL − 1] ∪ {3m}into m− zL pairs of the form {a− 1, a}, and a set W of size 2s−m + 2zL − 2. (This ispossible because 2s−m + 2zL − 2 ≥ 2.) Let ZR be the set of a's that occur in the pairs{a− 1, a}. Finally, let

P = [m− s + 2 + zL, 3m/2].

We now construct three base blocks of our design; these are to be developed modulo3m + 1:

((0, L), (1, R); (X × {R}) ∪ S1),

((0, L), (0, R); (YL × {L}) ∪ (YR × {R}) ∪ S2),

((0, L), (1, L); (ZL × {L}) ∪ (ZR × {R})).Here, S1 and S2 are sets which partition S, with |S1 | = m − x and |S2 | =

2s−m + x− 1.What remains on the vertices C × {L,R} is a graph with pure differences on C × {L}

the same as those on C ×{R}, namely P, and mixed differences (computed as R−L) theset W. By Lemma 4.2, this graph can be one-factored into 4(s− 1) one-factors; let φ be a


bijection from F onto the set of one-factors. For each j ∈ F, and for each edge ab of φ(j),take the block (a, b; {j} ×M).

A slight variation allows us to handle the case s = m. We just describe the appropriatesets in this case:

X = ∅,YL = ∅,YR = {3m},ZL = ∅,ZR = {2i + 1 | 1 ≤ i ≤ m},W = [2m + 2, 3m− 1] and

P = [2, 3m/2].

Corollary 4.2. If m is even, there exist Km+2 \Km designs of orders 10m + 5, 14m +7, 22m + 11 and 26m + 13.

Proof. We shall use Theorem 4.2 with s = 2, 3, 5 and 6, and then apply Lemma 4.1. Foreach case, the size of the hole is congruent to 1 (mod 4m + 2), so Theorem 3.1 can beused to fill the hole. The only problem is the condition s ≤ m in Theorem 4.2; this fails ifm = 4, and v = 99 or 117. We give specific designs in these cases; they both have holesizes congruent to 1 (mod 4m + 2 = 18), so again Theorem 3.1 fills the hole.

v = 99 (hole of size 55).Let the hole vertices be {∞1,∞2, . . . ,∞55}, and let the vertices outside the hole be

Z22 × {L,R}. Then four base blocks (mod 22) are:

(0L, 1L; 9R, 11R, 13R, 15R),

(0L, 0R; 3L, 4L, 6R, 7R),

(0L, 1R; 6L, 7L, 4R, 5R),

(2L, 0R; 1R,∞1,∞2,∞3).

Left over outside the hole is

〈{2, 5, 8, 9, 10, 11}, {2, 3}, {2, 5, 8, 9, 10, 11}〉,which by Lemma 4.2 is 13 one-factors for the remaining 13 · 4 points in the hole.

v = 117 (hole of size 73).Let the hole vertices be {∞1,∞2, . . . ,∞73}, and let the vertices outside the hole be

Z22 × {L,R}. Then three base blocks (mod 22) are:

(0L, 1L; 7R, 9R, 11R, 13R),

(0L, 1R; 2L, 2R, 5R,∞1),

(0L, 0R; 3L, 4L, 20R, 3R).

Left over outside the hole is

〈{5, 6, . . . , 11}, {4, 14, 15, 16, 17}, {5, 6, . . . , 11}〉,


which by Lemma 4.2 is 18 one-factors for the remaining 18 · 4 points in the hole.The designs constructed in this section now complete the proof of our main Theorem 1.1.

REFERENCES

[1] P. Adams, E. J. Billington, and C. C. Lindner, k-perfect 3k-cycle systems, J. Combin. Math.Combin. Comput. 15 (1994), 141–154.

[2] J.-C. Bermond, C. Huang, A. Rosa, and D. Sotteau, Decompositions of complete graphs intoisomorphic subgraphs with f ive vertices, Ars Combin. 10 (1980), 211–254.

[3] J. C. Bermond and J. Schonheim, G-decomposition of Kn, where G has four vertices or less,Discrete Math. 19 (1977), 113–120.

[4] E. J. Billington, M. Gionfriddo, and C. C. Lindner, The intersection problem forK4 −e designs,J. Statistical Plann. Inference (to appear).

[5] C. J. Colbourn, D. G. Hoffman, and R. Rees, A new class of group divisible designs with blocksize three, J. Combin. Theory Ser. A 59 (1992), 73–89.

[6] M. Gionfriddo, C. C. Lindner, and C. A. Rodger, 2-colouring K4 − e designs, Australasian J.Combinatorics 3 (1991), 211–229.

[7] C. C. Lindner and C. A. Rodger, A small embedding for partial Km \ Km−2 designs, ArsCombin. 35A (1993), 193–201.

[8] G. Stern and H. Lenz, Steiner triple systems with given subspaces: another proof of the Doyen–Wilson theorem, Boll. Un. Math. Ital. A(5) 17 (1980), 109–114.

[9] V. G. Vizing, On an estimate of the chromatic class of a p-graph, Discrete Analiz. 3 (1964),25–30.

Received October 5, 1995Accepted February 12, 1996

On the spectrum for Km+2 Km designs

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