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Also available at http://amc-journal.eu ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 10 (2016) 19-29
ars mathematica contemporanea
Squashing maximum packings of 6-cycles into maximum packings of triples
Curt C. Lindner
Department of Mathematics and Statistics, Auburn University, Auburn, AL 36849-5307, U.S.A.
Giovanni Lo Faro *
Dipartimento di Matematica e Informatica, Universita di Messina, 98166 Messina, Italia
Antoinette Tripodif
Dipartimento di Matematica e Informatica, Universita di Messina, 98166 Messina, Italia
Received 21 July 2014, accepted 28 September 2014, published online 24 January 2015
Abstract
A 6-cycle is said to be squashed if we identify a pair of opposite vertices and name one of them with the other (and thereby turning the 6-cycle into a pair of triples with a common vertex). The squashing problem for 6-cycle systems was introduced by C. C. Lindner, M. Meszka and A. Rosa and completely solved by determining the spectrum. In this paper, by employing PBD and GDD-constructions and filling techniques, we extend this result by squashing maximum packings of Kn with 6-cycles into maximum packings of Kn with triples. More specifically, we establish that for each n > 6, there is a max packing of Kn with 6-cycles that can be squashed into a maximum packing of Kn with triples.
Keywords: Maximum packing with triples, maximum packing with 6-cycles. Math. Subj. Class.: 05B07
* Supported by P.R.I.N., P.R.A. and I.N.D.A.M.(G.N.S.A.G.A.) tSupportedby P.R.I.N., P.R.A. and I.N.D.A.M.(G.N.S.A.G.A.)
E-mail addresses: lindncc@auburn.edu (Curt C. Lindner), lofaro@unime.it (Giovanni Lo Faro), atripodi@unime.it (Antoinette Tripodi)
©® This work is licensed under http://creativecommons.org/licenses/by/3.0/
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Ars Math. Contemp. 10 (2016) 31-44
1 Introduction
Let G be a graph. A G-design of order n is a pair (S, B) where B is a collection of subgraphs (blocks), each isomorphic to G, which partitions the edge set of the complete undirected graph Kn with vertex set S. After determining the spectrum for G-designs for different graphs G, many problems have been studied also recently (for example, see
A Steiner triple system (more simply, triple system) of order n is a G-design of order n where G is the graph K3. It is well known that the spectrum for triple systems is precisely the set of all n = 1 or 3 (mod 6) [9], and that if (S, T) is a triple system of order n, then |T| = n(n - 1)/6. Similarly, a 6-cycle system of order n is a G-design of order n where G is 6-cycle. The spectrum for 6-cycle systems is precisely the set of all n = 1 or 9 (mod 12) [15], and if (X, C) is a 6-cycle system of order n, then |C| = n(n - 1)/12. It is worth noting that if (S, T) and (X, C) have order n, then |T| = 2|C|.
Given the fact that triple systems and 6-cycle systems coexist for all n = 1 or 9 (mod 12), an obvious question to ask is: are there any connections between the two when n = 1 or 9 (mod 12)? The answer, of course, is yes. One much studied connection is that of 2-perfect 6-cycle systems. A 6-cycle system is 2-perfect provided the collection of triples obtained by replacing each 6-cycle (a, b, c, d, e, f) with the two triples (a, c, e) and (b, d, f) is a Steiner triple system. Such systems exist for all n = 1 or 9 (mod 12) > 13 [15].
Quite recently a new connection between triple systems and 6-cycle systems has been introduced: the squashing of a 6-cycle system into a Steiner triple system. A definition is in order. Let (a, b, c, d, e, f) be a 6-cycle and form the following six bowties (a pair of triples with a common vertex).
If B is any one of the six bowties in Figure 1, we say that we have squashed (a, b, c, d, e, f) into B. So there are six different ways to squash a 6-cycle into a bowtie. If (X, C) is a 6-cycle system, 2|C| = 2n(n — 1)/12 = n(n — 1)/6 is the number of triples in a Steiner triple system. Therefore it makes sense to ask the following question: what is the spectrum for 6-cycle systems that can be squashed into Steiner triple systems? In [11], a complete solution is given to this problem by constructing for every n = 1 or 9 (mod 12) a 6-cycle system that can be squashed into a Steiner triple system.
Example 1.1. (A 6-cycle system of order 9 squashed into a triple system [11].)
Now if n = 3 or 7 (mod 12) there does not exist a 6-cycle system of order n. However, there does exist a maximum packing (max packing) of Kn with 6-cycles with leave a triple (i.e., a pair (X, C) and a set L, the leave, where C is a collection of edge disjoint 6-cycles with verteces in X, L is the set of the edges of Kn not belonging to any 6-cycle of C and |L| is as small as possible) and so the following question makes sense. Does there exist for each n = 3 or 7 (mod 12) a max packing of Kn with 6-cycles which can be squashed into bowties so that the bowties plus the leave (a triple) form a Steiner triple system?
[1]-[7]).
(0,1,2,3,4,5) (3,6,0,2,4,1) (2,8,4,0,3,7) (7,0,8,6,5,1) (6,1,8,5,7,4) (5,2,6,7,8,3)
SQUASH
(0,1,2)(0,4,5) (3,6,0)(3,4,1) (2,8,4)(2,3,7) (7,0,8)(7,5,1) (6,1,8)(6,7,4) (5,2,6)(5,8,3)
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c	e
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'd
Figure 1
Example 1.2. (A max packing of K7 squashed into a triple system [11].)
(2,3,4,5,0,1)	SQUASH	(2,1,0)(2,3,4)
(4,6,0,2,5,1)	—>	(4,1,5X4,0,6)
(5,6,2,4,0,3)	(5,3,0)(5,6,2)
(1,3,6) leave	—>	(1,3,6)
The following theorem is proved in [11].
Theorem 1.3. [11] There exists a 6-cycle system of every order n = 1 or 9 (mod 12) that can be squashed into a triple system and a 6-cycle maximum packing that can be squashed into a triple system for every n = 3 or 7 (mod 12), n > 7.	□
The object of this paper is to finish off the problem of squashing maximum packings of Kn with 6-cycles into maximum packings of Kn with triples. We need to be a bit more precise.
Let (X, C) be a maximum packing of Kn with 6-cycles with leave L. In what follows, to keep the vernacular from getting out of hand, to say that C has been squashed means that the resulting collection S(C) of bowties is a partial triple system.
Further, if t is a triple belonging to L and S(C) U {t} is a maximum packing of Kn with triples (or a triple system), we will say that we have squashed (X, C) into a maximum
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Ars Math. Contemp. 10 (2016) 31-44
packing of Kn with triples. So, for example, Example 1.2 is the squashing of a maximum packing of K7 with 6-cycles into a triple system of order 7.
The following easy to read table gives the leaves for max packings for both 6-cycles and triples not covered by Theorem 1.3. (See [8] and [13].)
Kn	6-cycles leave	triples leave
n = 0, 2, 6, 8 (mod 12)	1-factor	1-factor
n = 5 (mod 12)	4-cycle	4-cycle
n = 11 (mod 12)	4 leaves are possible	•-• •-•
n = 4 or 10 (mod 12)	AIII-1 22 leaves are possible for n > 16	<11 I" I tripole [13]
We remark that if n = 0, 2, 6, 8 or 5 (mod 12) and a 6-cycle maximum packing can be squashed, there are no triples to be added; i.e., the resulting collection of bowties is a maximum packing of Kn with triples. If n = 4, 10 or 11 (mod 12) and a 6-cycle maximum packing can be squashed, then a triple is taken from the 6-cycle leave in order to obtain a maximum packing of Kn with triples.
2 Preliminaries
From now on to say that the 6-cycle (a, b, c, d, e, f) is squashed we will always mean that it has been squashed into the bowtie (a, b, c)(a, e, f); see Figure 2.
Figure 2
So, for example, in Example 1.1 we can simply list the 6-cycles (without listing the bowties they have been squashed into) and say they can be squashed into a triple system. The following three examples are used repeatedly in what follows.
Example 2.1. (A max packing of K6 with 6-cycles squashed into a max packing of K6 with triples.)
C = {(5,0,1,2,4, 3), (2,3,1, 5,4,0)}, leave L = {(0, 3), (1,4), (2,5)}. (There are no triples in the leave.)
f
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Example 2.2. (A max packing of K8 with 6-cycles squashed into a max packing of K8 with triples.)
X = Z4 x Z2; C = {(0o, 3o, 1i, 2o, 3i, 0i), (1o, 0o, 2i, 3o, 0i, 1i), (2o, 1q, 3i, 0o, 1i, 2i), (3o, 2q, 0i, 1q, 2i, 3i)}, leave L = {(0q, 2q), (1q, 3q), (0i, 2i), (1i, 3i)}. (There are no triples in the leave.)
Example 2.3. (Decomposition of K4 4 4 into 6-cycles squashed into triples.) (An obvious definition.)
X = Z4 x {1, 2, 3}; C = {(12,13,0i, 02, 03,1i), (02, 23, 0i, 12,03, 2i), (1i, 02, 33,0i, 22, 13), (0i, 32, 33,1i, 22,03), (32,1i, 23,12, 3i, 03), (12, 2i, 23, 32, 3i, 33), (3i, 02,13, 2i, 22, 23), (2i, 32,13, 3i, 22,33)}. (There is no leave.)
3 Basic Lemmas
With the examples of Section 2 in hand we can go to the general constructions, where we shall make use of GDDs. Let H be a set of integers and X be a set of size n; a GDD(n, H, k) is a triple (X, G, B) where G is a partition of X into subsets called groups of size in H, B is a set of subsets of X (called blocks) of size k, such that a group and a block contain at most one common point and every pair of points from distinct groups occurs in exactly one block. A PBD is a GDD(n, {1}, k).
We break the constructions into the eight cases: 2, 6, 8; 0; 11; 4, 10; 5 (mod 12).
3.1	n = 2, 6 and 8 (mod 12)
These are the easiest cases, so a good place to start.
n = 2 (mod 12) Write 12k + 2 = 2(6k + 1) and let (X, T) be a Steiner triple system of order 6k + 1. Let S = X x {1, 2} and define a collection C of 6-cycles as follows: For each triple t = {a, b, c} G T define a copy of Example 2.1 on {a, b, c} x {1, 2} with leave Lt = {(ai, a2), (bi, b2), (ci, c2)} and put these 6-cycles in C. Then C is a max packing of Ki2k+2 with 6-cycles with leave L = {Lt| t G T}. Trivially, C can be squashed into a max packing of Ki2k+2 with triples with leave L.
n = 6 (mod 12) The case for n = 6 is handled with Example 2.1. So now write 12k+6 = 2(6k + 3) and proceed exactly as in the case n = 2 (mod 12).
n = 8 (mod 12) Write 12k + 8 = 2(6k + 4). The case n = 8 is handled by Example 2.2. So let 12k + 8 > 20. It is well kown that there is a PBD with block sizes 3 and 4 for every n = 4 (mod 6) [13]. Let (X, B) be such a PBD, |X| = 4 (mod 6), and proceed exactly as in the cases for n = 2 or 6 (mod 12), using Example 2.2 as well as Example 2.1.
Lemma 3.1. There exists a max packing of Kn with 6-cycles that can be squashed into a max packing of Kn with triples for all n = 2, 6, 8 (mod 12) > 6.	□
3.2	n = 0 (mod 12) We begin with an example. Example 3.2. (n = 12)
Let X = {toi, to2} U Zi0 and define a collection of 6-cycles C as follows:
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Ars Math. Contemp. 10 (2016) 31-44
(0, TOI, 2,1, 3, 6), (2, TO2, 4, 3, 5, 8), (4, TO^ 6, 5, 7,0), (6, TO2, 8, 7, 9, 2) (8, TOI, 9,0,1, 4), (l, TOI, 3,0, 2, 7), (3, TO2, 5, 2,4, 9), (5, TOI, 7, 4, 6, l), (7, to2, 9, 6, 8, 3), (0, TO2, 1, 8, 9, 5),
with leave L = {(0, 8), (1,9), (2,3), (4, 5), (6, 7), (tox, to2)}. Then (X, C) is a max packing of K12 with 6-cycles and can be squashed into a max packing of K12 with triples with leave L.
We will need two constructions for 12k > 24: one when k is even and one when k is odd.
12k, k even Write 12k = 4(3k) and let (P,G,B) be a GDD(3k, {2}, 3), set X = P x {1,2,3,4} and define a collection of 6-cycles C as follows:
(i)	For each group g G G place Example 2.2 on g x {1,2,3,4} with leave Lg = {g x
{1}, g x {2}, g x {3}, g x {4}} and place these 6-cycles in C.
(ii)	For each triple t = {a, b, c} G B place a copy of Example 2.3 on K4j4j4 with parts
{a} x {1,2, 3,4}, {b} x {1, 2, 3,4}, and {c} x {1,2,3,4} and place these 6-cycles in C.
Then (X, C) is a max packing of K12k with 6-cycles with leave L = {g x{1},g x {2}, g x {3}, g x {4} | g G G}. It is straightforward to see that the 6-cycles in (i) and (ii) can be squashed into a max packing of K12k with triples with leave L.
12k, k odd Write 12k = 4(3k). Since k is odd, 3k is the order of a Kirkman triple system (P, T). Let X = P x {1, 2, 3,4}, n a parallel class in T, and define a collection of 6-cycles C as follows:
(i)	For each triple t = {a, b, c} G n, place a copy of Example 3.2 on {a, b, c} x {1, 2, 3,4} with leave Lt and place these 6-cycles in C.
(ii)	For each triple {a, b, c} G T \ n, place a copy of Example 2.3 on K4j4j4 with parts
{a} x {1,2,3,4}, {b} x {1,2, 3,4}, and {c} x {1, 2, 3,4}, and place these 6-cycles in C.
Then (X, C) is amax packing of K12k with 6-cycles with leave L = {Lt| t G n}. Squashing these 6-cycles produces a max packing of K12k with triples with leave L.
Lemma 3.3. There exists a max packing of Kn with 6-cycles that can be squashed into a max packing of Kn with triples for all n = 0 (mod 12).	□
3.3 n = 11 (mod 12)
We begin with an example.
Example 3.4. (n = 11)
Let X = Z9 U {to1, to2} and define a collection of 6-cycles C as follows:
(4, 8, to2, 7, TO 1, 0), (5,0, TO2, 4, TO1, 1), (6,1, TO2, 5, TO1, 2), (7, 2, TO2, 6, TO1, 3) (3, 2, 5, 7, 0,1), (7,4, 5, 3, 0, 6), (4, 3, 6, 8,1, 2), (8, 5, 6,4,1, 7)
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with leave L = {(to^ to2, 3, 8), (0, 2,8)}. Then (X, C) is a max packing of Kn with 6-cycles with leave L. Squashing these 6-cycles and adding (0, 2, 8) from the leave L gives a max packing of Kn with triples with leave the 4-cycle (to^ to2, 3, 8). We can now give a general construction for 11 (mod 12) > 23.
12k +11 > 23 Write 12k +11 = 2(6k + 4) + 3. Let (P, G, B) be a GDD(6k + 4, {4*, 2}, 3) [13], set X = {toi, TO2, TO3} U (P x {1, 2}), and define a collection of 6-cycles C as follows:
(i)	Let b* be the unique group of size 4 and define a copy of Example 3.4 on { TOi, TO>2, TO3}U(b* x{1, 2}) with leave L = {(to1, to2, to3), (x, y, z, w)}, where {x, y, z, w} C b* x {1,2} and place these 6-cycles in C.
(ii)	For each group g G G of size 2, define a copy of a max packing of K7 with 6-cycles, with vertex set {to1, to2, to3}U (b x {1,2}), that can be squashed into 6-triples with leave (to1, to2, to3) [11]. Add these 6-cycles to C.
(iii)	For each triple t = {a, b, c} G B, place a copy of Example 2.1 on t x {1, 2} with leave {a} x {1,2}, {b} x {1, 2}, and {c} x {1,2} and place these 6-cycles in C.
Then (X, C) is a max packing of K12k+11 with 6-cycles with leave L in (i). If we squash these 6-cycles and add the triple (to1, to2, to3) from the leave L in (i) we have a max packing of K12k+11 with triples with leave (x, y, z, w) in (i).
Lemma 3.5. There exists a max packing of Kn with 6-cycles that can be squashed into a max packing of Kn with triples for all n = 11 (mod 12).	□
3.4 n = 4 or 10 (mod 12)
The following three examples are necessary for the constructions in this section. Example 3.6. (n = 10)
Let X = {to} U (Z3 x Z3) and define a collection of 6-cycles C as follows: (01,11,00,12, 02, TO), (12, 2i, 0o, 0i, 02,1o), (1i, 2i, 1o, 22,12, to), (22, 0i, 1o, 1i, 12, 2o), (2i, 0i, 2o, 02, 22, to), (02,1i, 2o, 2i, 22,0o) with leave L = {{to, 2o, 1o, 0o}, (0i, 12), (1i, 22), (21,02)}. (We remark that {to, 20,10,00} is a copy of K4 and not a 4-cycle.) Then (X, C) is a max packing of K10 with 6-cycles with leave L. If we squash these 6-cycles and remove a triple from {to, 20,10,00}, the result is a max packing of K10 with triples with leave the tripole Ki,3 U {(0i, 12), (1i, 22), (2i, 02)}.
Example 3.7. (n = 16)
Let X = {to1, to2, to3, to4} U {ij | i G Z6, j G {0,1}}. Further, for each i G Z6, define a(i) = to1 if i is odd and to2 if i is even. For each i G Z6 define a collection of 6-cycles C as follows: (ii,io, (4 + i)i, (2 + i)i, (1 + i)i,a(i)), (io, (1 + i)i, (4 + i)o, (2 + i)o, (1 + i)o, a(i)), and (io, (2 + i)i, TO3, (1 + i)o, TO4, (5 + i)i) with leave L = {to 1, to2, to3, to4} U {(ij, (3 + i)j) | i G {0,1,2},j G {0,1}}. (Once again we remark that {to1, to2, to3, to4} is a copy of K4.) Then (X, C) is a max packing of K16 with 6-cycles with leave L. If we squash these 6-cycles and remove a triple from {to1, to2, to3, to4}, the result is a max packing of K16 with triples, with leave the tripole Ki,3 U {(ij, (3 + i)j) | i G {0,1, 2}, j G {0, 1}}.
Example 3.8. (n = 28)
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Ars Math. Contemp. 10 (2016) 31-44
Let X = {toi, to2, to3, to4} U (Z12 x {1, 2}), and let (P,G,B) beaGDD(12, {3}, 4) (equivalent to a pair of orthogonal quasigroups of order 3) and define a collection of 6-cycles C as follows:
(i)	For each group g G G, place a copy of Example 3.6 on {to^ to2, to3, to4} U (g x
{1, 2}) with leave {toi, to2, to3, to4} U {(x^x2) | x G g} (with K4 based on
{TOi, TO2, TO3, TO4}).
(ii)	For each block b G B place a copy of Example 2.2 on b x {1,2} with leave
{{xi, X2} | x G b}.
Then (X, C) is a max packing of K28 with 6-cycles with the leave in (i). Now squashing the 6-cycles in (i) and (ii) and removing a triple from K4 gives a max packing of K28 with triples with leave a tripole.
We can now go to the general constructions for n = 10 (mod 12), n > 22 and n = 4 (mod 12), n > 40.
n = 10 (mod 12), n > 22 Write 12k + 10 = 2(6k + 5) and let (P, B) be a PBD(6k + 5, {5*, 3}) [13]. Set X = P x {1,2} and define a collection C of 6-cycles as follows:
(i)	Let b* be the unique block of size 5 and define a copy of Example 3.6 on b* x {1,2} and place these 6-cycles in C. (The leave is K4 U {1-factor}.)
(ii)	For each triple t = {a, b, c} G B, define a copy of Example 2.1 on t x {1, 2} with leave {(a1, a2), (b1, b2), (c1, c2)} and place these 6-cycles in C.
Then (X, C) is a max packing of K12fc+10 with 6-cycles with leave K4 U {1-factor}. Squashing the 6-cycles in C and removing a triple from the leave in (i) produces a max packing of K12k+10 with triples with leave a tripole.
n = 4 (mod 12), n > 40 Write 12k + 4 = 4 + 2(6k) and let (P, G, B) be aGDD(6k, {6}, 3) [13]. Set X = {to1, to2, to3, to4} U (P x {1, 2}) and define a collection of 6-cycles as follows:
(i)	For each group g G G define a copy of Example 3.7 on{TO1, to2, to3, to4} U (g x {1, 2}) with leave K4 U {(x1, x2) | x G G} (K4 is based on {to1, to2, to3, to4}).
(ii)	For each triple t = {a, b, c} G B, define a copy of Example 2.1 on t x {1, 2} with leave {(a1, a2), (b1, b2), (c1, c2)} and place these 6-cycles in C.
Then (X, C) is a max packing of K12k+4 with 6-cycles. Squashing the 6-cycles in C and removing a triple from the leave K4 in (i) produces a max packing of K12k+4 with triples with leave a tripole.
Lemma 3.9. There exists a max packing of Kn with 6-cycles that can be squashed into a max packing of Kn with triples for all n = 4 or 10 (mod 12) > 6.	□
3.5 n = 5 (mod 12)
This case requires three examples.
Example 3.10. (n = 17)
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Let X = {toi, to2} U Z15 and define a collection of 6-cycles C as follows:
{(0, 9,4, 5,1, 3) + i | i € Z15}U{(7,14, to2, 13, 0), (8,0, to2, 7, to1 ,1), (9,1, to2, 8, 2), (10, 2, to2, 9, 3), (11, 3, to2, 10, , 4), (12, 4, to2, 11, 5), (13, 5, to2, 12, 6)} with leave the 4-cycle (to1, to2, 6,14). Squashing all of the 6-cycles in C produces a max packing of K17 with triples with leave the 4-cycle (to1, to2, 6,14).
Example 3.11. (n = 29)
Let X = {to1, to2} U Z27 and define a collection of 6-cycles C as follows:
{(0, 3,1, 5, 4, 9) + i, (0,16,10,17, 7,15) + i | i € Z27} U {(14, 0, to2, 13, 1) + j € {0,1, 2,..., 11}, (13, 26, to2, 25, 0)} with leave the 4- cycle (to1, to2, 12, 26). Squashing the 6-cycles in C gives a max packing of K29 with triples with leave the 4-cycle
(to1, to2, 12, 26).
Example 3.12. (n = 53)
Let X = {to1, to2} U Z51 and define a collection of 6-cycles C as follows: {(0, 3,1, 5, 4, 21) + i, (0,19, 9, 20,11, 23) + i, (0, 24, 8,15, 7, 22) + i, (0, 20, 6,11, 5,18) + i | i € Z51}u{(26, 0, to2, 25, 1) + j | j € {0,1, 2,..., 23}}u{(25, 50, to2, 49, 0)} with leave the 4-cycle (to 1, to2 , 24,50). Squashing these 6-cycles gives a max packing of K53 with triples with leave the 4-cycle (to1, to2, 24, 50).
We can now give two general constructions to finish off the case n = 5 (mod 12). 12k + 5, k odd Write 12k+5 = 1+4(3k+1). Since k is odd, 1+4(3k+1) = 1+4(6t+4). Let (P, G, B) be a GDD(6t, {4*, 2}, 3), set X = {to} U (P x {1,2,3,4}) and define a collection of 6-cycles as follows:
(i)	For the unique group b* of size 4, define a copy of Example 3.10 on {to} U (b* x {1, 2, 3,4}) (the leave is a 4-cycle) and place these 6-cycles in C.
(ii)	For each group g of size 2, define a copy of Example 1.1 on {to} U (g x {1,2}) and place these 6-cycles in C. (There is no leave.)
(iii)	For each triple {a, b, c} € B, place a copy of Example 2.3 on {a, b, c} x {1, 2,3,4} with parts {a} x {1, 2, 3,4}, {b} x {1, 2, 3,4}, {c} x {1, 2, 3,4}, and place these 6-cycles in C. (There is no leave.)
Then (X, C) is a max packing of K12fc+5 with 6-cycles with leave a 4-cycle. If we squash the 6-cycles in (i), (ii) and (iii), we have a max packing of K12fc+5 with triples with leave a 4-cycle.
12k + 5, k even Write 12k + 5 = 1+ 4(3k + 1). Since k is even, 1 + 4(3k + 1) = 1 + 4(6t + 1). Since 12k + 5 > 77, 6t + 1 > 19, and there exists a GDD(6t + 1, {7*, 3}, 3) [11] (P, G, B). Define a collection C of 6-cycles on X = {to} U (P x {1, 2, 3,4}) as follows:
(i)	For the unique group b* of size 7, define a copy of Example 3.11 on {to} U (b x {1, 2, 3,4}) (with leave a 4-cycle) and place these 6-cycles in C.
(ii)	For each group g of size 3, place a copy of a 6-cycle system of order 13 which can be squashed into a triple system [11] (no leave) on {to}U (g x{1, 2,3,4}) and place these 6-cycles in C.
(iii)	For each triple {a, b, c} € B, place a copy of Example 2.3 on {a, b, c} x {1, 2,3,4} with parts {a} x {1, 2,3,4}, {b} x {1, 2, 3,4}, {c} x {1,2, 3,4}, and place these 6-cycles in C. (There is no leave.)
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Ars Math. Contemp. 10 (2016) 31-44
Then (X, C) is a max packing of Ki2k+5 with 6-cycles with leave the 4-cycle in (i). Squashing the 6-cycles in (i), (ii) and (iii) produces a max packing of K12k+5 with triples with leave the 4-cycle in (i).
Lemma 3.13. There exists a max packing of Kn with 6-cycles that can be squashed into a max packing of Kn with triples for all n = 5 (mod 12) > 17.	□
4 Main result and further developments
Putting together the results in Section 3 gives the following theorem.
Theorem 4.1. For each n = 0, 2,4,5,6,8,10,11 (mod 12), n > 6, there exists a max packing of Kn with 6-cycles that can be squashed into a max packing of Kn with triples. There are no exceptions.	□
Since a complete solution is also a max packing, we can combine Theorem 1.3 and Theorem 8.1 into the following corollary (giving a complete solution to the squashing of max packings of 6-cycles into max packings with triples).
Corollary 4.2. For each n > 6, there is a max packing of Kn with 6-cycles that can be squashed into a max packing of Kn with triples.	□
In this paper we give a complete solution to the problem of squashing maximum packings of Kn with 6-cycles into maximum packings of Kn with triples. An open problem is to solve the general case, i.e. squashing a maximum packing of Kn with 2m-cycles into a maximum packing of Kn with m-cycles; the case m = 4 is completely solved in [10] .
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