The Metamorphosis of Maximum Packings of 2Kn with Triples into Maximum
Packings of 2Kn with 4-cycles
by
PJ Couch
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
August 4, 2012
Keywords: graph decompositions, design theory
Copyright 2012 by PJ Couch
Approved by
C.C. Lindner, Chair, Professor, Mathematics & Statistics
D.G. Ho man, Professor, Mathematics & Statistics
P.D. Johnson, Professor, Mathematics & Statistics
Abstract
In this work, the problem of constructing a maximum packing of 2Kn with triples having
a metamorphosis into a maximum packing of 2Kn with 4-cycles is concluded by solving the
problem for every n 11 such that n 2;5;8; or 11 (mod 12).
ii
Acknowledgments
This work is dedicated to Curtis White, November 9, 1959 - April 4, 2012.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 The 12n + 11 Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3 The 12n + 5 Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4 The 12n + 2 Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
5 The 12n + 8 Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
iv
Chapter 1
Introduction
A Steiner triple system (more simply, triple system) of order n is a pair (S;T), where T
is a collection of edge-disjoint triangles (or triples) which partitions the edge set of Kn with
vertex set S.
Example 1.1 (STS(7)).
0 1
2
3
4
5
6
0 1
3
1
2
4
2
35
3
4
6
45
0
1
5
62
6
0
K7
It is well-known that the spectrum for Steiner triple systems is precisely the set of all
n 1 or 3 (mod 6) and that if (S;T) is a Steiner triple system of order n, jTj= n(n 1)=6
[2].
A 2-fold triple system of order n is a pair (X;T), where T is a collection of edge-disjoint
triples which partitions the edge set of 2Kn (every pair of vertices is joined by 2 edges) with
vertex set X.
1
Example 1.2 (2-fold triple system of order 9).
2K9
0 1
2
3
4
5
6
7
8
1 1
2 2
65
1 1
33
5 80 1 1
4 4
0 7 2 2
33
0
0
2 2
44
8
3 3
44
6 7
5 5
6 6
40 4
5 5
7 7
2
3
5 5
88
0 0
66
77
1
6 6
88
2 3 0 7
7
88
1
Just as for triple systems it is well-known that the spectrum for 2-fold triple systems
is precisely the set of all n 0 or 1 (mod 3) and that if (X;T) is a 2-fold triple system of
order n, jTj= n(n 1)=3 [3].
Finally, a 2-fold 4-cycle system of order n is a pair (X;C), where C is a collection of
edge-disjoint 4-cycles which partitions the edge set of 2Kn with vertex set X. The spectrum
for 2-fold 4-cycle systems is precisely the set of all n 0 or 1 (mod 4) [4]. Since the spectra
for 2-fold triple systems and 2-fold 4-cycle systems agree when n 0;1;4; or 9 (mod 12), an
obvious question arises: Are there any connections between the two systems when they have
the same order? The answer is yes!
In what follows, we will call the graph
a hinge. Notice that if (X;T) is a 2-fold triple system, then jTj is even. This yields the
following problem for 2-fold triple systems.
2
Suppose n  0;1;4; or 9 (mod 12). Does there exist a 2-fold triple system (X;C) of
order n such that:
(i) the triples can be paired into a collection of hinges, H, and
(ii) the collection of double edges, D, can be organized into a collection of 4-cycles D ?
If so, (X;(HnD)[D ) is a 2-fold 4-cycle system called a metamorphosis of the 2-fold
triple system (X;C).
2Kn T
H
H\D D
?
3
H\D D?
?
Example 1.3 (metamorphosis of Example 1.2 into a 2-fold 4-cycle system).
1 1
2 2
65
1 1
33
50
0
2 2
44
8
3 3
44
6 7
3
5 5
88
0 0
66
77
1
8
1 1
4 4
0 7
2 2
33
0
5 5
6 6
4
0 4
5 5
7 7
2
6 6
88
2 3
0 7 7
88
1
1 1
2 3
2
4
3
4
5
8
6
7
1
4
2
3
5
6
5
7
6
8
7
8
5
0
3
8
0
2
6
8
0
0
4
3
0
6
1
7
4
0
5
7
0
0
2
1
T H
4
H\D D
?
1 1
2 3
2
4
3
4
5
8
6
7
1
4
2
3
5
6
5
7
6
8
7
8
5
0
3
8
0
2
6
8
0
0
4
3
0
6
1
7
4
0
5
7
0
0
2
1
1 2
34
7
5 6
8
D?
?
1 2
1 2
1 2
34
34
34
5 6
5 6
5 6
78
78
78H\D
1 1
2 3
2
4
3
4
5
8
6
7
1
4
2
3
5
6
5
7
6
8
7
8
5
0
3
8
0
2
6
8
0
0
4
3
0
6
1
7
4
0
5
7
0
0
2
1
In [5], a complete solution is given of this problem.
Theorem 1.4 (M. Gionfriddo and C.C. Lindner [5]). There exists a 2-fold triple system
of every order n  0;1;4; or 9 (mod 12) having a metamorphosis into a 2-fold 4-cycle
system.  
Now, when n 3;6;7; or 10 (mod 12), a metamorphosis of a 2-fold triple system into
a 4-cycle system is impossible, since 2-fold 4-cycle systems of these orders do not exist.
However, a maximum packing of 2Kn with 4-cycles is possible [1]. In this case, the leave is
always a double edge.
5
Example 1.5 (metamorphosis of a 2-fold triple system of order 10 into a maximum packing
of 2K10 with 4-cycles).
T
0
2
1 3 0
4
5 6
0
8
7 9
2
4
1 7
2
8
3 6
4
8
6 7
1
5
7 8
3
5
7 8
3
9
4 6
1
9
6 8
5
6
0 2 6
7
1 3
7
9
0 2
5
9
2 4
1
3
0 4
0
2
0
4
0
8
2
4
2
8
4
8
1
5
3
5
3
9
1
9
5
6
6
7
7
9
5
9
1
3
H
0
2
1 3
0
4
5 6
0
8
7 9
2
4
1 7
2
8
3 6
4
8
6 7
1
5
7 8
3
5
7 8
3
9
4 6
1
9
6 8
5
6
0 2
6
7
1 3
7
9
0 2
5
9
2 4
1
3
0 4
6
H\D
0 2
84
1 5
39
5 6
79 1
3 D
?
0
2
1 3
0
4
5 6
0
8
7 9
2
4
1 7
2
8
3 6
4
8
6 7
1
5
7 8
3
5
7 8
3
9
4 6
1
9
6 8
5
6
0 2
6
7
1 3
7
9
0 2
5
9
2 4
1
3
0 4
?
1 5
39
1 5
39
5 6
79
5 6
79
0
4 8
2 0
4 8
2 0
4 8
2
1 3
D?
H\D
0
2
1 3
0
4
5 6
0
8
7 9
2
4
1 7
2
8
3 6
4
8
6 7
1
5
7 8
3
5
7 8
3
9
4 6
1
9
6 8
5
6
0 2
6
7
1 3
7
9
0 2
5
9
2 4
1
3
0 4
7
The following theorem is proved in [1].
Theorem 1.6 (S.E. McClanahan). The spectrum for 2-fold triple systems having a meta-
morphosis into a maximum packing of 2Kn with 4-cycles is precisely the set of all n 3;6;7;
or 10 (mod 12) 10.  
The object of this thesis is the completion of the metamorphosis trilogy by constructing
for every n  2;5;8; or 11 (mod 12) a maximum packing of 2Kn with triples having a
metamorphosis into a maximum packing of 2Kn with 4-cycles. A maximum packing of 2Kn
with triples will always have an even number of triples with leave a double edge. (However,
for n 5 or 8 (mod 12), a maximum packing of 2Kn with 4-cycles is a 2-fold 4-cycle system.
In this case, the double edge leave in the maximum packing of 2Kn with triples is used in
the metamorphosis into 4-cycles.)
Example 1.7 (metamorphosis of a maximum packing of 2K11 with triples into a maximum
packing of 2K11 with 4-cycles).
T
?1
?2
11 11
12 12
?1 ?2 ?1 ?211 11
13 13
12 12
13 13
?1 ?2
21 21
22 22
21 21 22 22
23 23 23 23
?1 ?2 ?1 ?2 ?1 ?2
31 31 31 31
32 32
32 32
33 33 33 33
?1 ?2 ?1 ?2 ?1 ?2
11 11
21 21
32 33
12 12
22 22
31 33
13 13
23 23
31 32
21 21
31 31
12 13
22 22
32 32
11 13
23 23
33 33
11 12
11 11
31 31
12 12
32 32
13 13
33 33
22 23 21 23 21 22
8
H
?1
?2
11
12
11
13
12
13
21
22
21
23
22
23
31 31 32
32 33 33
?1
?1
?1
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?2
?2
?2
11
21
21
31
11
31
12
22
22
32
12
32
13
23
23
33
13
33
32
12
22
33
13
23
31
11
21
33
13
23
31
11
21
32
12
22
H\D
?1
?2
11 12 13
21 22 23
31 32 33
D
?
11
12
11
13
12
13
21
22
21
23
22
23
31 31 32
32 33 33
?1
?1
?1
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?2
?2
?2
11
21
21
31
11
31
12
22
22
32
12
32
13
23
23
33
13
33
32
12
22
33
13
23
31
11
21
33
13
23
31
11
21
32
12
22
9
?1
?2
D?
?
11 21
2212
11 21
2313
11 31
3313
11 31
3212
21 22
3231
21 31
3323
12 22
2313
12 32
3313
22 32
3323
H\D
11
12
11
13
12
13
21
22
21
23
22
23
31 31 32
32 33 33
?1
?1
?1
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?2
?2
?2
11
21
21
31
11
31
12
22
22
32
12
32
13
23
23
33
13
33
32
12
22
33
13
23
31
11
21
33
13
23
31
11
21
32
12
22
The following theorem gives a complete solution of the problem of constructing max-
imum packings of 2Kn with triples having metamorphoses into maximum packings of 2Kn
with 4-cycles. As mentioned above, the object of this thesis is the proof of this theorem.
Theorem 1.8. The spectrum for maximum packings of 2Kn with triples having a meta-
morphosis into a maximum packing of 2Kn with 4-cycles is the set of all n  2;5;8; or
11 (mod 12) 8, except n = 5 and n = 8.  
10
In what follows, we will use the following notation:
x y edge (x;y) or (y;x)
x y double edge < x;y > or < y;x >
x
y
z 3-cycle or triple Any cyclic shift of (x;y;z)
x
y
z
w
4-cycle
Any cyclic shift of
(x;y;z;w)
x
y
z
w
hinge
< x;z;y;w >,
< x;z;w;y >,
< z;x;y;w >, or
< z;x;w;y >
We will organize our results into 6 chapters: an introduction (this chapter), the 12n+11
Construction, the 12n+5 Construction, the 12n+2 Construction, the 12n+8 Construction,
and a summary.
11
Chapter 2
The 12n + 11 Construction
This is the easist construction out of the four basic constructions in this thesis; so, a
perfect place to begin.
To begin with, Example 1.7 gives a metamorphosis of a maximum packing of 2K11 with
triples into a maximum packing of 2K11 with 4-cycles. So, it is only necessary to give a
construction for 12n + 11 23.
We begin with a necessary example.
Example 2.1 (metamorphosis of a maximum packing of 2K11 n2K5 with triples into a
maximum packing of 2K11n2K5 with 4-cycles (the leave is a double edge)).
T
1 1
3 3
?1 ?2 ?1 ?51 1
6 6
1 1
5 5
?2 ?3
2 2
6 6
2 2 2 2
5 5 4 4
?1 ?2 ?1 ?5 ?2 ?3
4 4 3 3
5 5
3 3
4 4 6 6
?1 ?2 ?1 ?5 ?2 ?3
1 1
2 2
?3 ?4 1 1
4 4
?4 ?5
3 3
5 5
?3 ?4 2 2
3 3
?4 ?5
4 4
6 6
5 5
6 6
?3 ?4 ?4 ?5
12
H
1
3
1
6
1
5
2
6
2
5
2
4
4 3 3
5 4 6
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?5
?5
?5
?3
?3
?3
1
2
3
5
4
6
1
4
2
3
5
6
?3
?3
?3
?4
?4
?4
?4
?4
?4
?5
?5
?5
H\D
1
3
1
6
1
5
2
6
2
5
2
4
4 3 3
5 4 6
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?5
?5
?5
?3
?3
?3
1
2
3
5
4
6
1
4
2
3
5
6
?3
?3
?3
?4
?4
?4
?4
?4
?4
?5
?5
?5
13
?
654
31 2
Dis2K6 on{1,2,3,4,5,6}
H\D
1
3
1
6
1
5
2
6
2
5
2
4
4 3 3
5 4 6
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?5
?5
?5
?3
?3
?3
1
2
3
5
4
6
1
4
2
3
5
6
?3
?3
?3
?4
?4
?4
?4
?4
?4
?5
?5
?5
14
1 4 D?
?
2 3
65
2 3
56
2 5
36
1 2
45
1 3
46
1 3
46
1 2
45
The 12n+11 23 Construction
With the above example in hand, we can proceed to the 12n + 11  23 Construction.
Write 12n + 11 = 3(4n + 2) + 5. Since 12n + 11  23, 4n + 2  6. This is important! Let
1=f11;12;13;14;15gand Q =f1;2;3;:::;4n+ 2g. Let H(Q) =fh0;h1;:::;h2ngbe
a partition of Q into pairwise disjoint sets of size 2 (called holes), where hi =f2i+1;2i+2g.
Let (Q; ) be a commutative quasigroup of order 4n + 2 with holes H(Q) (see [6]) and set
X =1[(Q f1;2;3g). For 0 i 2n, let Bi = hi f1;2;3g and Ai =1[Bi. De ne a
collection of hinges as follows:
(1) Use the n = 11 example (Example 1.7) to  nd a hinge system (obvious de nition) with
our desired metamorphosis (A0;H0), where the leave is <11;12 >.
...?
?1 ?2
A0 = B
0leave
(1,1) (2,1)
15
(2) For 1  i 2n, use the n = 11 with a hole of size 5 (Example 2.1) Construction to
 nd a hinge system with our desired metamorphosis (Ai;Hi), where the hole is1
and the leave is < (1;2i + 1);(1;2i + 2) >.
...? B
1 B2n
(3,1) (4,1) (4n+1,1)(4n+2,1)
leave
(3) We now need to use the edges between vertices in Bi and Bj for i 6= j. De ne a
collection of hinges H? as follows. For x2hi, y2hj, i6= j, place the hinges
< (x;1);(y;1);(x y;2);(x y;3) >; < (x;2);(y;2);(x y;1);(x y;3) >;
and < (x;3);(y;3);(x y;1);(x y;2) > in H?:
...? (x,1)
(y,1)
(x?y,2)
(x?y,3)
...?
(x,2) (y,2)
(x?y,3)
(x?y,1)
...?
(x,3)
(y,3)
(x?y,1)
(x?y,2)
16
It is straightforward to see that (X =
2n[
i=0
Ai;H?[
2n[
i=0
Hi) is a hinge system of order
12n+11 with leave <11;12 >. To proceed with our metamorphosis,  rst use the prescribed
metamorphoses in (1) and (2) and place these 4-cycles in C. After removing the double edges
from our hinges in H? (from (3)), we still have the following edges remaining to use in our
metamorphosis: <11;12 >, edges of the type < (2i + 1;1);(2i + 2;1) >, 1 i 2n, and
edges of the type < (x;k);(y;k) >, where k2f1;2;3g and x2hi, y2hj, i6= j.
?1 ?2
Edgesleftfrom(1)and(2)
...(3,1) (4,1) (5,1) (6,1)
(4n+1,1)
(4n+2,1)
Edgesleftfrom(3)
...
...
2K2,2,...,2 on{1,2,...,4n+2}?{k}
fork?{1,2,3}
(3,k) (5,k) (4n+1,k)
(2,k) (4,k) (6,k) (4n+2,k)
(1,k)
(4) For k = 1, we have 2K4n+2 on f(1;1);(2;1);:::;(4n + 2;1)g minus < (1;1);(2;1) >.
For 1 i n, on f(4i 1;1);(4i;1);(4i + 1;1);(4i + 2;1)g, we have:
f((4i 1;1);(4i;1);(4i + 1;1);(4i + 2;1));
((4i 1;1);(4i + 1;1);(4i + 2;1);(4i;1));
((4i 1;1);(4i + 1;1);(4i;1);(4i + 2;1))g
 C
17
(4i,1)
(4i+2,1) (4i+1,1)
(4i-1,1) (4i,1)
(4i+2,1) (4i+1,1)
(4i-1,1) (4i,1)
(4i+2,1) (4i+1,1)
(4i-1,1) (4i,1)
(4i+2,1) (4i+1,1)
(4i-1,1)
? ?
(5) For 0 i < j 2n, j i > 1 or i = 0, onf(2i+1;1);(2i+2;1);(2j +1;1);(2j +2;1)g,
we have:
f((2i + 1;1);(2j + 2;1);(2i + 2;1);(2j + 1;1));
((2i + 1;1);(2j + 2;1);(2i + 2;1);(2j + 1;1))g
 C.
(2,1)(1,1) (2j+2,1)(2j+1,1)
(2,1)(1,1) (2j+2,1)(2j+1,1)
i=0and1?j?2n
(2i+2,1)(2i+1,1) (2j+2,1)(2j+1,1)
(2i+2,1)(2i+1,1) (2j+2,1)(2j+1,1)
1?i<j?2nandj?i>1
(6) For k = 2 and k = 3, we have 2 copies of the complete (2n + 1)-partite graph with all
partite sets having size 2. Between each pair of partite sets fx;yg and fz;wg in these
2K2;2;:::;2, we have f(x;w;y;z);(x;w;y;z)g C.
yx wz
yx wz
{x,y}=hi?{k},{z,w}=hj?{k}, inegationslash=j, k?{2,3}
18
The result is a maximum packing with 4-cycles of 2K12n+11 with vertex set X with leave
<11;12 >.
Theorem 2.2. There exists a maximum packing of 2K12n+11 with triples having a metamor-
phosis into a maximum packing of 2K12n+11 with 4-cycles of every order 12n + 11.  
We illustrate the 12n + 11 Construction with the following example.
Example 2.3 (metamorphosis of a maximum packing of 2K23 with triples into a maximum
packing of 2K23 with 4-cycles).
(1) Rename the vertices of the hinge system in Example 1.7 to get (A0;H0) with leave
<11;12 >.
H0 =
f< (1;1);(2;1);11;12 >; < (1;1);13;11;12 >;
< (2;1);13;11;12 >; < (1;2);(2;2);11;12 >;
< (1;2);14;11;12 >; < (2;2);14;11;12 >;
< (1;3);(2;3);11;12 >; < (1;3);15;11;12 >;
< (2;3);15;11;12 >; < (1;1);(1;2);(2;3);15 >;
< (2;1);(2;2);(1;3);15 >; <13;14;(1;3);(2;3) >;
< (1;2);(1;3);(2;1);13 >; < (2;2);(2;3);(1;1);13 >;
<14;15;(1;1);(2;1) >; < (1;1);(1;3);(2;2);14 >;
< (2;1);(2;3);(1;2);14 >; <13;15;(1;2);(2;2) >g
H0
?1
?2
(1,1)
(2,1)
(1,1)
?3
(2,1)
?3
(1,2)
(2,2)
(1,2)
?4
(2,2)
?4
(1,3) (1,3) (2,3)
(2,3) ?5 ?5
?1
?1
?1
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?2
?2
?2
(1,1)
(1,2)
(1,2)
(1,3)
(1,1)
(1,3)
(2,1)
(2,2)
(2,2)
(2,3)
(2,1)
(2,3)
?3
?4
?4
?5
?3
?5
(2,3)
(2,1)
(2,2)
?5
?3
?4
(1,3)
(1,1)
(1,2)
?5
?3
?4
(1,3)
(1,1)
(1,2)
(2,3)
(2,1)
(2,2)
19
f((1;1);11;(2;1);12); ((1;1);11;13;12); ((2;1);11;13;12);
((1;2);11;(2;2);12); ((1;2);11;14;12); ((2;2);11;14;12);
((1;3);11;(2;3);12); ((1;3);11;15;12); ((2;3);11;15;12);
((1;1);(2;3);(1;2);15); ((2;1);(1;3);(2;2);15); (13;(1;3);14;(2;3));
((1;2);(2;1);(1;3);13); ((2;2);(1;1);(2;3);13); (14;(1;1);15;(2;1));
((1;1);(2;2);(1;3);14); ((2;1);(1;2);(2;3);14); (13;(1;2);15;(2;2));
((1;1);(1;2);(2;2);(2;1)); ((1;1);(1;2);14;13); ((1;1);(1;3);15;13);
((1;1);(1;3);(2;3);(2;1)); ((1;2);(2;2);(2;3);(1;3)); ((1;2);(1;3);15;14);
((2;1);(2;2);14;13); ((2;1);(2;3);15;13); ((2;2);(2;3);15;14)g
 C
(2) For 1 i 2, rename the vertices of the hinge system in Example 2.1 to get (Ai;Hi),
where the hole is on1and the leave is < (2i + 1;1);(2i + 2;1) >.
H1 =
f< (3;1);(3;3);11;12 >; < (3;2);(4;3);11;12 >;
< (4;1);(4;2);11;12 >; < (3;1);(4;3);11;15 >;
< (3;2);(4;2);11;15 >; < (3;3);(4;1);11;15 >;
< (3;1);(4;2);12;13 >; < (3;2);(4;1);12;13 >;
< (3;3);(4;3);12;13 >; < (3;1);(3;2);13;14 >;
< (3;3);(4;2);13;14 >; < (4;1);(4;3);13;14 >;
< (3;1);(4;1);14;15 >; < (3;2);(3;3);14;15 >;
< (4;2);(4;3);14;15 >g
H1
(3,1)
(3,3)
(3,1)
(4,3)
(3,1)
(4,2)
(3,2)
(4,3)
(3,2)
(4,2)
(3,2)
(4,1)
(4,1) (3,3) (3,3)
(4,2) (4,1) (4,3)
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?5
?5
?5
?3
?3
?3
(3,1)
(3,2)
(3,3)
(4,2)
(4,1)
(4,3)
(3,1)
(4,1)
(3,2)
(3,3)
(4,2)
(4,3)
?3
?3
?3
?4
?4
?4
?4
?4
?4
?5
?5
?5
20
f(11;(3;1);12;(3;3)); (11;(3;2);12;(4;3)); (11;(4;1);12;(4;2));
(11;(3;1);15;(4;3)); (11;(3;2);15;(4;2)); (11;(3;3);15;(4;1));
(12;(3;1);13;(4;2)); (12;(3;2);13;(4;1)); (12;(3;3);13;(4;3));
(13;(3;1);14;(3;2)); (13;(3;3);14;(4;2)); (13;(4;1);14;(4;3));
(14;(3;1);15;(4;1)); (14;(3;2);15;(3;3)); (14;(4;2);15;(4;3));
((3;2);(3;3);(4;2);(4;3)); ((3;2);(4;3);(3;3);(4;2));
((3;2);(3;3);(4;3);(4;2)); ((3;1);(3;2);(4;1);(4;2));
((3;1);(3;2);(4;1);(4;2)); ((3;1);(3;3);(4;1);(4;3));
((3;1);(3;3);(4;1);(4;3))g  C
H2 =
f< (5;1);(5;3);11;12 >; < (5;2);(6;3);11;12 >;
< (6;1);(6;2);11;12 >; < (5;1);(6;3);11;15 >;
< (5;2);(6;2);11;15 >; < (5;3);(6;1);11;15 >;
< (5;1);(6;2);12;13 >; < (5;2);(6;1);12;13 >;
< (5;3);(6;3);12;13 >; < (5;1);(5;2);13;14 >;
< (5;3);(6;2);13;14 >; < (6;1);(6;3);13;14 >;
< (5;1);(6;1);14;15 >; < (5;2);(5;3);14;15 >;
< (6;2);(6;3);14;15 >g
H2
(5,1)
(5,3)
(5,1)
(6,3)
(5,1)
(6,2)
(5,2)
(6,3)
(5,2)
(6,2)
(5,2)
(6,1)
(6,1) (5,3) (5,3)
(6,2) (6,1) (6,3)
?1
?1
?1
?1
?1
?1
?2
?2
?2
?2
?2
?2
?5
?5
?5
?3
?3
?3
(5,1)
(5,2)
(5,3)
(6,2)
(6,1)
(6,3)
(5,1)
(6,1)
(5,2)
(5,3)
(6,2)
(6,3)
?3
?3
?3
?4
?4
?4
?4
?4
?4
?5
?5
?5
21
f(11;(5;1);12;(5;3)); (11;(5;2);12;(6;3)); (11;(6;1);12;(6;2));
(11;(5;1);15;(6;3)); (11;(5;2);15;(6;2)); (11;(5;3);15;(6;1));
(12;(5;1);13;(6;2)); (12;(5;2);13;(6;1)); (12;(5;3);13;(6;3));
(13;(5;1);14;(5;2)); (13;(5;3);14;(6;2)); (13;(6;1);14;(6;3));
(14;(5;1);15;(6;1)); (14;(5;2);15;(5;3)); (14;(6;2);15;(6;3));
((5;2);(5;3);(6;2);(6;3)); ((5;2);(6;3);(5;3);(6;2));
((5;2);(5;3);(6;3);(6;2)); ((5;1);(5;2);(6;1);(6;2));
((5;1);(5;2);(6;1);(6;2)); ((5;1);(5;3);(6;1);(6;3));
((5;1);(5;3);(6;1);(6;3))g  C
(3) Let Q = f1;2;3;4;5;6g. Let H(Q) = fh0;h1;h2g, where hi = f2i + 1;2i + 2g for
0  i  2. De ne (Q; ) to be the following symmetric quasigroup of order 6 with
holes H(Q):
 1 2 3 4 5 6
1 5 6 3 4
2 6 5 4 3
3 5 6 1 2
4 6 5 2 1
5 3 4 1 2
6 4 3 2 1
We now need to use the edges between vertices in Bi and Bj for i6= j. This is done as
follows.
22
H? =
f< (1;1);(3;1);(5;2);(5;3) >; < (1;2);(3;2);(5;1);(5;3) >;
< (1;3);(3;3);(5;1);(5;2) >; < (1;1);(4;1);(6;2);(6;3) >;
< (1;2);(4;2);(6;1);(6;3) >; < (1;3);(4;3);(6;1);(6;2) >;
< (1;1);(5;1);(3;2);(3;3) >; < (1;2);(5;2);(3;1);(3;3) >;
< (1;3);(5;3);(3;1);(3;2) >; < (1;1);(6;1);(4;2);(4;3) >;
< (1;2);(6;2);(4;1);(4;3) >; < (1;3);(6;3);(4;1);(4;2) >;
< (2;1);(3;1);(6;2);(6;3) >; < (2;2);(3;2);(6;1);(6;3) >;
< (2;3);(3;3);(6;1);(6;2) >; < (2;1);(4;1);(5;2);(5;3) >;
< (2;2);(4;2);(5;1);(5;3) >; < (2;3);(4;3);(5;1);(5;2) >;
< (2;1);(5;1);(4;2);(4;3) >; < (2;2);(5;2);(4;1);(4;3) >;
< (2;3);(5;3);(4;1);(4;2) >; < (2;1);(6;1);(3;2);(3;3) >;
< (2;2);(6;2);(3;1);(3;3) >; < (2;3);(6;3);(3;1);(3;2) >;
< (3;1);(5;1);(1;2);(1;3) >; < (3;2);(5;2);(1;1);(1;3) >;
< (3;3);(5;3);(1;1);(1;2) >; < (3;1);(6;1);(2;2);(2;3) >;
< (3;2);(6;2);(2;1);(2;3) >; < (3;3);(6;3);(2;1);(2;2) >;
< (4;1);(5;1);(2;2);(2;3) >; < (4;2);(5;2);(2;1);(2;3) >;
< (4;3);(5;3);(2;1);(2;2) >; < (4;1);(6;1);(1;2);(1;3) >;
< (4;2);(6;2);(1;1);(1;3) >; < (4;3);(6;3);(1;1);(1;2) >g
23
Hstar
(1,1)
(3,1)
(1,1)
(4,1)
(1,1)
(5,1)
(1,2)
(3,2)
(1,2)
(4,2)
(1,2)
(5,2)
(1,3) (1,3) (1,3)
(3,3) (4,3) (5,3)
(5,2)
(5,1)
(5,1)
(6,2)
(6,1)
(6,1)
(3,2)
(3,1)
(3,1)
(5,3)
(5,3)
(5,2)
(6,3)
(6,3)
(6,2)
(3,3)
(3,3)
(3,2)
(1,1)
(6,1)
(1,2)
(6,2)
(1,3)
(6,3)
(2,1)
(3,1)
(2,2)
(3,2)
(2,3)
(3,3)
(2,1)
(4,1)
(2,2)
(4,2)
(2,3)
(4,3)
(4,2)
(4,1)
(4,1)
(4,3)
(4,3)
(4,2)
(6,2)
(6,1)
(6,1)
(6,3)
(6,3)
(6,2)
(5,2)
(5,1)
(5,1)
(5,3)
(5,3)
(5,2)
(2,1)
(5,1)
(2,1)
(6,1)
(3,1)
(5,1)
(2,2)
(5,2)
(2,2)
(6,2)
(3,2)
(5,2)
(2,3) (2,3) (3,3)
(5,3) (6,3) (5,3)
(4,2)
(4,1)
(4,1)
(3,2)
(3,1)
(3,1)
(1,2)
(1,1)
(1,1)
(4,3)
(4,3)
(4,2)
(3,3)
(3,3)
(3,2)
(1,3)
(1,3)
(1,2)
(3,1)
(6,1)
(3,2)
(6,2)
(3,3)
(6,3)
(4,1)
(5,1)
(4,2)
(5,2)
(4,3)
(5,3)
(4,1)
(6,1)
(4,2)
(6,2)
(4,3)
(6,3)
(2,2)
(2,1)
(2,1)
(2,3)
(2,3)
(2,2)
(2,2)
(2,1)
(2,1)
(2,3)
(2,3)
(2,2)
(1,2)
(1,1)
(1,1)
(1,3)
(1,3)
(1,2)
24
f((1;1);(5;2);(3;1);(5;3)); ((1;2);(5;1);(3;2);(5;3));
((1;3);(5;1);(3;3);(5;2)); ((1;1);(6;2);(4;1);(6;3));
((1;2);(6;1);(4;2);(6;3)); ((1;3);(6;1);(4;3);(6;2));
((1;1);(3;2);(5;1);(3;3)); ((1;2);(3;1);(5;2);(3;3));
((1;3);(3;1);(5;3);(3;2)); ((1;1);(4;2);(6;1);(4;3));
((1;2);(4;1);(6;2);(4;3)); ((1;3);(4;1);(6;3);(4;2));
((2;1);(6;2);(3;1);(6;3)); ((2;2);(6;1);(3;2);(6;3));
((2;3);(6;1);(3;3);(6;2)); ((2;1);(5;2);(4;1);(5;3));
((2;2);(5;1);(4;2);(5;3)); ((2;3);(5;1);(4;3);(5;2));
((2;1);(4;2);(5;1);(4;3)); ((2;2);(4;1);(5;2);(4;3));
((2;3);(4;1);(5;3);(4;2)); ((2;1);(3;2);(6;1);(3;3));
((2;2);(3;1);(6;2);(3;3)); ((2;3);(3;1);(6;3);(3;2));
((3;1);(1;2);(5;1);(1;3)); ((3;2);(1;1);(5;2);(1;3));
((3;3);(1;1);(5;3);(1;2)); ((3;1);(2;2);(6;1);(2;3));
((3;2);(2;1);(6;2);(2;3)); ((3;3);(2;1);(6;3);(2;2));
((4;1);(2;2);(5;1);(2;3)); ((4;2);(2;1);(5;2);(2;3));
((4;3);(2;1);(5;3);(2;2)); ((4;1);(1;2);(6;1);(1;3));
((4;2);(1;1);(6;2);(1;3)); ((4;3);(1;1);(6;3);(1;2))g
 C
(4) For k = 1, we have 2K6 on f(1;1);(2;1);:::;(6;1)g minus < (1;1);(2;1) >. Since
n = 1 on f(3;1);(4;1);(5;1);(6;1)g we have:
f((3;1);(4;1);(5;1);(6;1));((3;1);(5;1);(6;1);(4;1));((3;1);(5;1);(4;1);(6;1))g C
(4,1)
(6,1) (5,1)
(3,1) (4,1)
(6,1) (5,1)
(3,1) (4,1)
(6,1) (5,1)
(3,1) (4,1)
(6,1) (5,1)
(3,1)
? ?
(5) For 1 j 2, on f(1;1);(2;1);(2j + 1;1);(2j + 2;1)g, we have:
25
f((1;1);(2j + 2;1);(2;1);(2j + 1;1));((1;1);(2j + 2;1);(2;1);(2j + 1;1))g C.
(2,1)(1,1) (4,1)(3,1)
(2,1)(1,1) (4,1)(3,1)
i =0andj =1
(2,1)(1,1) (6,1)(5,1)
(2,1)(1,1) (6,1)(5,1)
i =0andj =2
(6) For k = 2 and k = 3, we have 2 copies of the complete tripartite graph with partite
sets h0 fkg, h1 fkg, and h2 fkg. Between each pair of partite sets fx;yg and
fz;wg in these 2K2;2;2, we have f(x;w;y;z);(x;w;y;z)g C.
(2,2)(1,2) (4,2)(3,2)
(2,2)(1,2) (4,2)(3,2)
{x,y}=h0?{2},{z,w}=h1?{2}
(2,2)(1,2) (6,2)(5,2)
(2,2)(1,2) (6,2)(5,2)
{x,y}=h0?{2},{z,w}=h2?{2}
(4,2)(3,2) (6,2)(5,2)
(4,2)(3,2) (6,2)(5,2)
{x,y}=h1?{2},{z,w}=h2?{2}
(2,3)(1,3) (4,3)(3,3)
(2,3)(1,3) (4,3)(3,3)
{x,y}=h0?{3},{z,w}=h1?{3}
26
(2,3)(1,3) (6,3)(5,3)
(2,3)(1,3) (6,3)(5,3)
{x,y}=h0?{3},{z,w}=h2?{3}
(4,3)(3,3) (6,3)(5,3)
(4,3)(3,3) (6,3)(5,3)
{x,y}=h1?{3},{z,w}=h2?{3}
27
Chapter 3
The 12n + 5 Construction
There does not exist a maximum packing of 2K5 with triples having a metamorphosis
into a 2-fold 4-cycle system of order 5. So, we begin this chapter by showing the nonexistence
for n = 5. If there were to be such a packing (and thus a packing with hinges), then we
would have to be able to use the double edges from our hinges (along with the leave) to
create two 4-cycles. Thus we would have a repeated 4-cycle. Let x be the vertex that does
not appear in this 4-cycle. Our 3 hinges can cover at most 6 edges incident with x, but
d2K5(x) = 8.
Before we can proceed to the 12n + 5 Construction, we must  rst produce a maximum
packing of 2K17 with triples having a metamorphosis into 2-fold 4-cycle system of order 17.
Example 3.1 (metamorphosis of a maximum packing of 2K17 with triples into a 2-fold
4-cycle system of order 17). We will be decomposing 2K17 with vertex set (f1;2;3;4g 
f1;2;3;4g)[f1g. We will begin with the maximum packing with hinges, since it has an
obvious correspondence to the maximum packing with triples. Our leave will be
< (4;4);1>.
28
H =
f< (1;1);(1;2);(1;4);1>; < (2;1);(2;2);(2;4);1>;
< (3;1);(3;2);(3;4);1>; < (4;1);(4;2);(4;4);1>;
< (1;3);1;(1;1);(1;2) >; < (2;3);1;(2;1);(2;2) >;
< (3;3);1;(3;1);(3;2) >; < (4;3);1;(4;1);(4;2) >;
< (1;4);(2;4);(4;4);1>; < (3;4);1;(1;4);(2;4) >;
< (1;3);(1;4);(1;1);(1;2) >; < (2;3);(2;4);(2;1);(2;2) >;
< (3;3);(3;4);(3;1);(3;2) >; < (4;3);(4;4);(4;1);(4;2) >;
< (3;4);(4;4);(1;4);(2;4) >; < (3;3);(4;3);(1;3);(2;3) >;
< (3;2);(4;2);(1;2);(2;2) >; < (3;1);(4;1);(1;1);(2;1) >;
< (1;4);(2;3);(3;2);(4;1) >; < (1;3);(2;4);(3;1);(4;2) >;
< (1;1);(2;2);(3;3);(4;4) >; < (1;2);(2;1);(3;4);(4;3) >;
< (1;3);(2;3);(3;3);(4;3) >; < (1;2);(2;2);(3;2);(4;2) >;
< (1;1);(2;1);(3;1);(4;1) >; < (3;2);(4;1);(1;4);(2;3) >;
< (3;1);(4;2);(1;3);(2;4) >; < (3;3);(4;4);(1;1);(2;2) >;
< (3;4);(4;3);(1;2);(2;1) >; < (1;4);(2;1);(3;3);(4;2) >;
< (1;2);(2;3);(3;1);(4;4) >; < (1;1);(2;4);(3;2);(4;3) >;
< (1;3);(2;2);(3;4);(4;1) >; < (3;3);(4;2);(1;4);(2;1) >;
< (3;1);(4;4);(1;2);(2;3) >; < (3;2);(4;3);(1;1);(2;4) >;
< (3;4);(4;1);(1;3);(2;2) >; < (1;4);(2;2);(3;1);(4;3) >;
< (1;1);(2;3);(3;4);(4;2) >; < (1;3);(2;1);(3;2);(4;4) >;
< (1;2);(2;4);(3;3);(4;1) >; < (3;1);(4;3);(1;4);(2;2) >;
< (3;4);(4;2);(1;1);(2;3) >; < (3;2);(4;4);(1;3);(2;1) >;
< (3;3);(4;1);(1;2);(2;4) >g
Now, we will remove the double edges from these hinges to create 4-cycles.
29
f((1;1);(1;4);(1;2);1); ((2;1);(2;4);(2;2);1);
((3;1);(3;4);(3;2);1); ((4;1);(4;4);(4;2);1);
((1;3);(1;1);1;(1;2)); ((2;3);(2;1);1;(2;2));
((3;3);(3;1);1;(3;2)); ((4;3);(4;1);1;(4;2));
((1;4);(4;4);(2;4);1); ((3;4);(1;4);1;(2;4));
((1;3);(1;1);(1;4);(1;2)); ((2;3);(2;1);(2;4);(2;2));
((3;3);(3;1);(3;4);(3;2)); ((4;3);(4;1);(4;4);(4;2));
((3;4);(1;4);(4;4);(2;4)); ((3;3);(1;3);(4;3);(2;3));
((3;2);(1;2);(4;2);(2;2)); ((3;1);(1;1);(4;1);(2;1));
((1;4);(3;2);(2;3);(4;1)); ((1;3);(3;1);(2;4);(4;2));
((1;1);(3;3);(2;2);(4;4)); ((1;2);(3;4);(2;1);(4;3));
((1;3);(3;3);(2;3);(4;3)); ((1;2);(3;2);(2;2);(4;2));
((1;1);(3;1);(2;1);(4;1)); ((3;2);(1;4);(4;1);(2;3));
((3;1);(1;3);(4;2);(2;4)); ((3;3);(1;1);(4;4);(2;2));
((3;4);(1;2);(4;3);(2;1)); ((1;4);(3;3);(2;1);(4;2));
((1;2);(3;1);(2;3);(4;4)); ((1;1);(3;2);(2;4);(4;3));
((1;3);(3;4);(2;2);(4;1)); ((3;3);(1;4);(4;2);(2;1));
((3;1);(1;2);(4;4);(2;3)); ((3;2);(1;1);(4;3);(2;4));
((3;4);(1;3);(4;1);(2;2)); ((1;4);(3;1);(2;2);(4;3));
((1;1);(3;4);(2;3);(4;2)); ((1;3);(3;2);(2;1);(4;4));
((1;2);(3;3);(2;4);(4;1)); ((3;1);(1;4);(4;3);(2;2));
((3;4);(1;1);(4;2);(2;3)); ((3;2);(1;3);(4;4);(2;1));
((3;3);(1;2);(4;1);(2;4))g
= HnD C
Finally, we will use the double edges (along with the leave) to create the last of our
4-cycles.
30
f(1;(4;3);(3;4);(4;4)); ((3;3);1;(4;4);(4;3));
((3;4);(3;3);(4;3);1); ((4;4);(3;4);1;(3;3));
((4;3);(4;4);(3;3);(3;4)); (1;(1;3);(2;4);(2;3));
(1;(1;3);(1;4);(2;3)); ((1;3);(1;4);(2;4);(2;3));
((1;3);(2;3);(1;4);(2;4)); ((1;1);(2;3);(1;2);(2;4));
((1;1);(2;3);(1;2);(2;4)); ((2;1);(1;3);(2;2);(1;4));
((2;1);(1;3);(2;2);(1;4)); ((3;1);(4;3);(3;2);(4;4));
((3;1);(4;3);(3;2);(4;4)); ((4;1);(3;3);(4;2);(3;4));
((4;1);(3;3);(4;2);(3;4)); ((1;1);(2;1);(2;2);(1;2));
((1;1);(2;1);(1;2);(2;2)); ((1;1);(1;2);(2;1);(2;2));
((3;1);(4;1);(4;2);(3;2)); ((3;1);(4;1);(3;2);(4;2));
((3;1);(3;2);(4;1);(4;2))g
 C
The 12n+5 29 Construction
We can now proceed to the 12n + 5 29 Construction, which is simply a modi cation
of the 12n + 11  23 Construction. Write 12n + 5 = 3(4n) + 5. Since 12n + 5  29,
4n 8. This is important. Let1=f11;12;13;14;15g and Q =f1;2;3;:::;4ng. Let
H(Q) = fh0;h1;:::;h2n 1g be a partition of Q into pairwise disjoint sets of size 2 (called
holes), where hi =f2i+ 1;2i+ 2g. Let (Q; ) be a commutative quasigroup of order 4n with
holes H(Q) (see [6]) and set X =1[(Q f1;2;3g). For 0 i 2n 1, let Bi = hi f1;2;3g
and Ai =1[Bi. De ne a collection of hinges as follows:
(1) Use the Example 1.7 to  nd a hinge system with our desired metamorphosis (A0;H0),
where the leave is < (1;1);(2;1) >.
31
...?A0 = B
0
leave (1,1) (2,1)
(2) For 1 i 2n 1, use Example 2.1 to  nd a hinge system with our desired metamor-
phosis (Ai;Hi), where the hole is1and the leave is < (1;2i + 1);(1;2i + 2) >.
...? B
1 B2n?1
(3,1) (4,1) (4n-1,1) (4n,1)
leave
(3) We now need to use the edges between vertices in Bi and Bj for i 6= j. De ne a
collection of hinges H? as follows. For x2hi, y2hj, i6= j, place the hinges
< (x;1);(y;1);(x y;2);(x y;3) >; < (x;2);(y;2);(x y;1);(x y;3) >;
and < (x;3);(y;3);(x y;1);(x y;2) > in H?:
...? (x,1)
(y,1)
(x?y,2)
(x?y,3)
...?
(x,2) (y,2)
(x?y,3)
(x?y,1)
32
...?
(x,3)
(y,3)
(x?y,1)
(x?y,2)
It is straightforward to see that (X =
2n 1[
i=0
Ai;H?[
2n 1[
i=0
Hi) is a hinge system of order
12n+5 with leave <11;12 >. To proceed with our metamorphosis,  rst use the prescribed
metamorphoses in (1) and (2) and place these 4-cycles in C. After removing the double edges
from our hinges in H? (from (3)), we still have the following edges remaining to use in our
metamorphosis: edges of type < (2i + 1;1);(2i + 2;1) >, 0 i 2n 1, and edges of type
< (x;k);(y;k) >, where k2f1;2;3g and x2hi, y2hj, i6= j.
Edgesleftfrom(1)and(2)
...(3,1) (4,1) (5,1) (6,1)
(4n-1,1)
(4n,1)(1,1) (2,1)
Edgesleftfrom(3)
...
...
2K2,2,...,2 on{1,2,...,4n}?{k}
fork?{1,2,3}
(3,k) (5,k) (4n-1,k)
(2,k) (4,k) (6,k) (4n,k)
(1,k)
33
(4) For k = 1, we have 2K4n on f(1;1);(2;1);:::;(4n;1)g. For 0 i n 1, on
f(4i + 1;1);(4i + 2;1);(4i + 3;1);(4i + 4;1)g, we have:
f((4i + 1;1);(4i + 2;1);(4i + 3;1);(4i + 4;1));
((4i + 1;1);(4i + 3;1);(4i + 4;1);(4i + 2;1));
((4i + 1;1);(4i + 3;1);(4i + 2;1);(4i + 4;1))g
 C
(4i+2,1)
(4i+4,1) (4i+3,1)
(4i+1,1)
? ?
(4i+2,1)
(4i+4,1) (4i+3,1)
(4i+1,1) (4i+2,1)
(4i+4,1) (4i+3,1)
(4i+1,1) (4i+2,1)
(4i+4,1) (4i+3,1)
(4i+1,1)
(5) For 0 i < i + 1 < j 2n 1, on f(2i + 1;1);(2i + 2;1);(2j + 1;1);(2j + 2;1)g, we
have:
f((2i + 1;1);(2j + 2;1);(2i + 2;1);(2j + 1;1));
((2i + 1;1);(2j + 2;1);(2i + 2;1);(2j + 1;1))g
 C.
(2i+2,1)(2i+1,1) (2j+2,1)(2j+1,1)
(2i+2,1)(2i+1,1) (2j+2,1)(2j+1,1)
0?i<i+1<j?2n?1
(6) For k = 2 and k = 3, we have 2 copies of the complete (2n)-partite graph with all
partite sets having size 2. Between each pair of partite sets fx;yg and fz;wg in these
2K2;2;:::;2, we have f(x;w;y;z);(x;w;y;z)g C.
yx wz
yx wz
{x,y}=hi?{k},{z,w}=hj?{k}, inegationslash=j, k?{2,3}
34
The result is a 2-fold 4-cycle system of order 12n + 5 with vertex set X.
Theorem 3.2. There exists a maximum packing of 2K12n+5 with triples having a metamor-
phosis into a 2-fold 4-cycle system of order 12n + 5, for all 12n + 5 17.  
35
Chapter 4
The 12n + 2 Construction
A maximum packing of 2K2 with triples has a trivial metamorphosis into a maximum
packing of 2K2 with 4-cycles; you simply have none of either! So, we begin this chapter by
producing a maximum packing of 2K14 with triples having a metamorphosis into a maximum
packing of 2K14 with 4-cycles.
Example 4.1 (metamorphosis of a maximum packing of 2K14 with triples into a maximum
packing of 2K14 with 4-cycles). We will decompose 2K14 with vertex set f0;1;:::;11g[
f11;12g. We will begin with a maximum packing with hinges, since it has an obvious
correspondence to the maximum packing with triples. Our leave will be <11;12 >.
H =
f< 0;11;11;1 >; < 2;11;1;3 >; < 4;11;3;5 >;
< 6;11;5;7 >; < 8;11;7;9 >; < 10;11;9;11 >;
< 0;12;5;7 >; < 2;12;7;9 >; < 4;12;9;11 >;
< 6;12;11;1 >; < 8;12;1;3 >; < 10;12;3;5 >;
< 0;3;11;10 >; < 1;4;0;11 >; < 2;5;1;0 >;
< 3;6;2;1 >; < 4;7;3;2 >; < 5;8;4;3 >;
< 6;9;5;4 >; < 7;10;6;5 >; < 8;11;7;6 >;
< 9;0;8;7 >; < 10;1;9;8 >; < 11;2;10;9 >;
< 0;6;2;8 >; < 1;7;3;9 >; < 2;8;4;10 >;
< 3;9;5;11 >; < 4;10;6;0 >; < 5;11;7;1 >g
Now, we will remove the double edges from these hinges to create 4-cycles.
36
f< 0;11;11;1 >; < 2;1;11;3 >; < 4;3;11;5 >;
< 6;5;11;7 >; < 8;7;11;9 >; < 10;9;11;11 >;
< 0;5;12;7 >; < 2;7;12;9 >; < 4;9;12;11 >;
< 6;11;12;1 >; < 8;1;12;3 >; < 10;3;12;5 >;
< 0;11;3;10 >; < 1;0;4;11 >; < 2;1;5;0 >;
< 3;2;6;1 >; < 4;3;7;2 >; < 5;4;8;3 >;
< 6;5;9;4 >; < 7;6;10;5 >; < 8;7;11;6 >;
< 9;8;0;7 >; < 10;9;1;8 >; < 11;10;2;9 >;
< 0;2;6;8 >; < 1;3;7;9 >; < 2;4;8;10 >;
< 3;5;9;11 >; < 4;6;10;0 >; < 5;7;11;1 >g
= HnD C
Finally, we will use the double edges (except the leave) to create the last of our 4-cycles.
f(0;11;2;12); (2;11;4;12); (4;11;6;12); (6;11;8;12);
(8;11;10;12); (10;11;0;12); (0;3;6;9); (1;4;7;10);
(2;5;8;11); (0;3;9;6); (1;4;10;7); (2;5;11;8);
(0;6;3;9); (1;7;4;10); (2;8;5;11)g
 C
Before we can begin the 12n + 2 26 Construction, we need to discuss some necessary
ingredients.
Example 4.2 (maximum packing of 2K5 with triples). We will decompose 2K5 with vertex
set f1;2;3g[f11;12g. We begin with a maximum packing with hinges, since it has an
obvious correspondence to the maximum packing with triples. Our leave will be <11;12 >.
H =f< 1;2;11;12 >;< 1;3;11;12 >;< 2;3;11;12 >g
37
1 2 3
?1 ?2
Although the above example does not have a metamorphosis into a 4-cycle system of
order 5, it will be good enough. We will also require the following lemma:
Lemma 4.3. We can always decompose 2K4n into two double 1-factors and (4n2  3n)
4-cycles.
Proof. We will prove this lemma by construction. Let V (2K4n) = f1;2;:::;4ng. Our  rst
double 1-factor will have edges of the form < i;i + 2n > for 1 i 2n.
. . .
1 2 2 n ? 1 2 n
2 n + 1 2 n + 2 4 n ? 1 4 n
Our second double 1-factor will have edges of the form < 2i 1;2(i + n) > and
< 2i;2(i + n) 1 > for 1 i n.
. . .
1 2 2 n ? 1 2 n
2 n + 1 2 n + 2 4 n ? 1 4 n
For n 0 (mod 2):
38
(1) Form two 4-cycles of the type (2i 1;2(j + n) 1;2i;2(j + n)) for 1 i n,
1 j n, i6= j.
2i ? 1 2i
2( j + n) ? 1 2( j + n)
. . .
. . .
2i ? 1 2i
2( j + n)
. . .
. . .
2( j + n) ? 1
i < j i > j
(2) The remaining edges form 2K2n = 2K4k on f1;2;:::;2ng and 2K4k on
f2n + 1;2n + 2;:::;4ng. Decompose these as in the 12n + 5 Construction.
For n 1 (mod 2):
(1) Form two 4-cycles of the type (2i 1;2(j + n) 1;2i;2(j + n)) for 2 i n,
1 j n 1, i6= j.
2i ? 1 2i
2( j + n) ? 1 2( j + n)
. . .
. . .
2i ? 1 2i
2( j + n)
. . .
. . .
2( j + n) ? 1
i < j i > j
(2) Form the following 4-cycles: (1;2;4n;4n 1);(1;4n;2;4n 1);(1;2;4n 1;4n).
1 2
4n ? 1 4n
. . .
. . .
39
(3) The remaining edges form 2K2n = 2K4k+2 onf1;2;:::;2ngminus < 1;2 > and 2K4k+2
on f2n + 1;2n + 2;:::;;4n 1;4ng minus < 4n 1;4n >. Decompose these as in the
12n + 11 Construction.  
The 12n+2 26 Construction
We can now proceed to the 12n + 2  26 Construction. Write 12n + 2 = 3(4n) + 2.
Let1= f11;12g and Q = f1;2;3;:::;4ng. Let (Q; ) be an idempotent antisymmetric
quasigroup of order 4n (that is, i i = i and i j6= j i for i6= j) and set X =
1[(Q f1;2;3g). For 1  i  4n, let Bi = hi f1;2;3g and Ai =1[Bi. De ne a
collection of hinges as follows:
(1) For 1 i 4n, use the n = 5 example (Example 4.2) to  nd a hinge system (Ai;Hi),
where
Hi =f< (i;1);(i;2);11;12 >;< (i;1);(i;3);11;12 >;< (i;2);(i;3);11;12 >g.
Now, letf((i;1);11;(i;2);12);((i;1);11;(i;3);12);((i;2);11;(i;3);12)g C and
notice that we still have 2K3 remaining on f(i;1);(i;2);(i;3)g.
(2) We now need to use edges between distinct Bi. For x < y, de ne
H? =f< (x;1);(y;1);(x y;2);(y x;2) >;< (x;2);(y;2);(x y;3);(y x;3) >;
< (x;3);(y;3);(x y;1);(y x;1) >g.
Now, let f((x;1);(x y;2);(y;1);(y x;2));((x;2);(x y;3);(y;2);(y x;3));
((x;3);(x y;1);(y;3);(y x;1))g C and notice that we still have 2K4n remaining
on f1;2;:::;4ng fkg for k = 1;2;3.
Also notice that (X = S4ni=1 Ai;H?[S4ni=1 Hi) is a hinge system of order 12n + 2 with
leave <11;12 >.
(3) For 2K4n on f1;2;:::;4ng fkg for k = 1;2;3, we will form two double 1-factors, F1
and F2, and (4n2 3n) 4-cycles, in the manner described in Lemma 4.3.
40
For each fa;bg2F1[F2, let f((a;1);(b;1);(b;2);(a;2));((a;1);(b;1);(b;3);(a;3));
((a;2);(b;2);(b;3);(a;3))g C.
(a, 1) (b, 1)
(a, 2) (b, 2)
(a, 3) (b, 3)
The result is a maximum packing with 4-cycles of 2K12n+2 with vertex set X with leave
<11;12 >.
Theorem 4.4. There exists a maximum packing of 2K12n+2 with triples having a metamor-
phosis into a maximum packing of 2K12n+2 with 4-cycles of every order 12n + 2.  
41
Chapter 5
The 12n + 8 Construction
There does not exist a maximum packing of 2K8 with triples having a metamorphosis
into a 2-fold 4-cycle system of order 8. So, we begin this chapter by showing the nonexistence
for n = 8. If there were to be such a packing (and thus a packing with hinges), then we would
have to be able to use the double edges from our hinges (along with the leave) to create  ve
4-cycles. This is equivalent to a simple graph G on 8 vertices with 10 edges having a double
cover by 4-cycles.
If we have any hope of creating these 4-cycles, G can have no vertices of degree 1. Also,
no vertex can have degree more than 4 in G, since 2K8 is 14-regular and each edge incident
with a vertex v in G accounts for either four edges incident with v in 2K8 (if it corresponds
to the double edge from a hinge in our packing) or two edges (if it corresponds to the leave).
For this reason, G has at most two vertices of degree 4, and if it has two such vertices, then
the edge between them must correspond to the leave.
Lemma 5.1. There can be no repeated 4-cycles in our double cover.
Proof. Suppose (a;b;c;d) is a repeated 4-cycle. Then none of fab;bc;cd;adg can be used in
another 4-cycle. Thus the remaining three 4-cycles must be a double cover of the remaining
six edges, i.e. a double cover of K4 with 4-cycles. Since K4 is 3-regular, the vertex set of
K4 does not meet fa;b;c;dg, since that would force the degree of some vertex to be greater
than 4 or for some edge to be used more than twice.
Now, nine of the ten edges in G are associated with hinges in our packing, so we need
36 edges from 2K4;4 to  nish them; however, 2jE(K4;4)j= 32.  
Corollary 5.2. Vertices of degree 2 in G cannot be adjacent.
42
Proof. If there were a pair of adjacent vertices of degree 2 in G, they would have to be in a
repeated 4-cycle in our double cover, a contradiction to Lemma 5.1.  
Lemma 5.3. G contains no 7-cycle
Proof. Suppose G contains a 7-cycle, C. Let e;f; and g be the remaining 3 edges. Now, C
contains no 4-cycles, so each 4-cycle must contain at least one of these edges. Without loss
of generality, assume that e and f are both covered by some 4-cycle, since e;f; and g must
each be covered by two 4-cycles and this means that there must be some 4-cycle that covers
a pair of these edges. Now, there still must be two 4-cycles that cover g, but that requires a
repeated 4-cycle, a contradiction to Lemma 5.1  
Corollary 5.4. G has no vertex of degree 0.
Proof. Suppose dG(v) = 0. Note that this means that there is no other vertex of degree 0. If
there were, neither our hinges from our packing nor the leave would cover the edges between
v and this vertex. This also means that v appears in exactly 7 of the hinges in our packing
to cover all edges incident with v in 2K8. Hence, by Lemma 5.3, we must have the following
graph as a subgraph of G:
C1
v C
2
Now, all the edges in C1 must be double covered by 4-cycles, and a 4-cycle can cover
at most a pair of its edges. Thus the remaining 3 edges in G must go from C1 to C2 to
cover the edges in C1. However, this means we must repeat C2 to double cover its edges, a
contradiction to Lemma 5.1.  
Lemma 5.5. G cannot have (4;3;3;2;2;2;2;2) as its degree sequence.
Proof. Suppose G has (4;3;3;2;2;2;2;2) as its degree sequence. Without loss of generality,
since no vertices of degree 2 can be adjacent (Corollary 5.2), G must look like this:
43
4 3 3
2 2 2 2 2
Now, redraw G to look like this:
Note that no 4-cycle can contain the dotted edges.  
Lemma 5.6. G cannot have (3;3;3;3;2;2;2;2) as its degree sequence.
Proof. Suppose G has (3;3;3;3;2;2;2;2) as its degree sequence. Call an edge pure if it is
incident with two vertices of degree 3 and call it mixed otherwise. Note that there are exactly
2 pure edges (see below).
2 2 2 2
3333
2 2 2 2
3333
By Corollary 5.2, each 4-cycle in our double cover must use an even number of mixed
edges, and thus it must also use an even number of pure edges. Hence, the pure edges must
share a vertex, v, and both of these edges must be a part of two (distinct) 4-cycles. Now,
note that the remaining edge at v must be mixed, but there is no way to cover it with a
4-cycle.  
Theorem 5.7. G does not have a double cover by 4-cycles.
Proof. G must have one of the following degree sequences, none of which has our desired
double cover by 4-cycles:
44
1. (4;4;3;3;3;3;0;0) (corollary 5.4)
2. (4;4;3;3;2;2;2;0) (corollary 5.4)
3. (4;4;2;2;2;2;2;2) (corollary 5.2)
4. (4;3;3;3;3;2;2;0) (corollary 5.4)
5. (4;3;3;2;2;2;2;2) (lemma 5.5)
6. (3;3;3;3;3;3;2;0) (corollary 5.4)
7. (3;3;3;3;2;2;2;2) (lemma 5.6)  
This leads us to the following conclusion:
Corollary 5.8. There does not exist a maximum packing of 2K8 with triples having a meta-
morphosis into a 2-fold 4-cycle system of order 8.  
Before we can begin the 12n+8 Construction, we must consider the case where 12n+8 =
32.
Example 5.9 (metamorphosis of a maximum packing of 2K32 with triples into 2-fold 4-cycle
system of order 32).
We will be decomposing 2K32 with vertex setf1;2;:::;10g f1;2;3g[f11;12g. We
will begin with the maximum packing with hinges, since it has an obvious correspondence
to the maximum packing with triples. Our leave will be <11;12 >.
45
f< (10;1);(3;1);(1;1);(2;1) >; < (10;1);(4;1);(5;1);(6;1) >;
< (10;1);(8;1);(7;1);(9;1) >; < (3;1);(4;1);(1;1);(7;1) >;
< (3;1);(8;1);(2;1);(6;1) >; < (4;1);(8;1);(6;1);(7;1) >;
< (1;1);(5;1);(7;1);(8;1) >; < (2;1);(5;1);(7;1);(8;1) >;
< (2;1);(9;1);(4;1);(6;1) >; < (1;1);(9;1);(6;1);(8;1) >;
< (5;1);(6;1);(10;1);(3;1) >; < (6;1);(7;1);(1;1);(2;1) >;
< (7;1);(9;1);(10;1);(3;1) >; < (5;1);(9;1);(3;1);(4;1) >;
< (1;1);(2;1);(10;1);(4;1) >; < (10;2);(3;2);(1;2);(2;2) >;
< (10;2);(4;2);(5;2);(6;2) >; < (10;2);(8;2);(7;2);(9;2) >;
< (3;2);(4;2);(1;2);(7;2) >; < (3;2);(8;2);(2;2);(6;2) >;
< (4;2);(8;2);(6;2);(7;2) >; < (1;2);(5;2);(7;2);(8;2) >;
< (2;2);(5;2);(7;2);(8;2) >; < (2;2);(9;2);(4;2);(6;2) >;
< (1;2);(9;2);(6;2);(8;2) >; < (5;2);(6;2);(10;2);(3;2) >;
< (6;2);(7;2);(1;2);(2;2) >; < (7;2);(9;2);(10;2);(3;2) >;
< (5;2);(9;2);(3;2);(4;2) >; < (1;2);(2;2);(10;2);(4;2) >;
< (10;3);(3;3);(1;3);(2;3) >; < (10;3);(4;3);(5;3);(6;3) >;
< (10;3);(8;3);(7;3);(9;3) >; < (3;3);(4;3);(1;3);(7;3) >;
< (3;3);(8;3);(2;3);(6;3) >; < (4;3);(8;3);(6;3);(7;3) >;
< (1;3);(5;3);(7;3);(8;3) >; < (2;3);(5;3);(7;3);(8;3) >;
< (2;3);(9;3);(4;3);(6;3) >; < (1;3);(9;3);(6;3);(8;3) >;
< (5;3);(6;3);(10;3);(3;3) >; < (6;3);(7;3);(1;3);(2;3) >;
< (7;3);(9;3);(10;3);(3;3) >; < (5;3);(9;3);(3;3);(4;3) >;
< (1;3);(2;3);(10;3);(4;3) >g
 H
46
f<11;(1;1);(1;2);(2;2) >; <11;(2;1);(1;3);(2;3) >;
<12;(1;1);(1;3);(2;3) >; <12;(2;1);(1;2);(2;2) >;
< (1;2);(1;3);11;12 >; < (1;2);(2;3);(1;1);(2;1) >;
< (2;2);(1;3);(1;1);(2;1) >; < (2;2);(2;3);11;12 >;
<11;(3;1);(3;2);(4;2) >; <11;(4;1);(3;3);(4;3) >;
<12;(3;1);(3;3);(4;3) >; <12;(4;1);(3;2);(4;2) >;
< (3;2);(3;3);11;12 >; < (3;2);(4;3);(3;1);(4;1) >;
< (4;2);(3;3);(3;1);(4;1) >; < (4;2);(4;3);11;12 >;
<11;(5;1);(5;2);(6;2) >; <11;(6;1);(5;3);(6;3) >;
<12;(5;1);(5;3);(6;3) >; <12;(6;1);(5;2);(6;2) >;
< (5;2);(5;3);11;12 >; < (5;2);(6;3);(5;1);(6;1) >;
< (6;2);(5;3);(5;1);(6;1) >; < (6;2);(6;3);11;12 >;
<11;(7;1);(7;2);(8;2) >; <11;(8;1);(7;3);(8;3) >;
<12;(7;1);(7;3);(8;3) >; <12;(8;1);(7;2);(8;2) >;
< (7;2);(7;3);11;12 >; < (7;2);(8;3);(7;1);(8;1) >;
< (8;2);(7;3);(7;1);(8;1) >; < (8;2);(8;3);11;12 >;
<11;(9;1);(9;2);(10;2) >; <11;(10;1);(9;3);(10;3) >;
<12;(9;1);(9;3);(10;3) >; <12;(10;1);(9;2);(10;2) >;
< (9;2);(9;3);11;12 >; < (9;2);(10;3);(9;1);(10;1) >;
< (10;2);(9;3);(9;1);(10;1) >; < (10;2);(10;3);11;12 >g
 H
47
f< (1;1);(3;2);(5;3);(6;3) >; < (1;1);(4;2);(6;3);(5;3) >;
< (2;1);(3;2);(6;3);(5;3) >; < (2;1);(4;2);(5;3);(6;3) >;
< (1;2);(3;1);(5;3);(6;3) >; < (1;2);(4;1);(6;3);(5;3) >;
< (2;2);(3;1);(6;3);(5;3) >; < (2;2);(4;1);(5;3);(6;3) >;
< (1;1);(5;2);(7;3);(8;3) >; < (1;1);(6;2);(8;3);(7;3) >;
< (2;1);(5;2);(8;3);(7;3) >; < (2;1);(6;2);(7;3);(8;3) >;
< (1;2);(5;1);(7;3);(8;3) >; < (1;2);(6;1);(8;3);(7;3) >;
< (2;2);(5;1);(8;3);(7;3) >; < (2;2);(6;1);(7;3);(8;3) >;
< (1;1);(7;2);(9;3);(10;3) >; < (1;1);(8;2);(10;3);(9;3) >;
< (2;1);(7;2);(10;3);(9;3) >; < (2;1);(8;2);(9;3);(10;3) >;
< (1;2);(7;1);(9;3);(10;3) >; < (1;2);(8;1);(10;3);(9;3) >;
< (2;2);(7;1);(10;3);(9;3) >; < (2;2);(8;1);(9;3);(10;3) >;
< (1;1);(9;2);(3;3);(4;3) >; < (1;1);(10;2);(4;3);(3;3) >;
< (2;1);(9;2);(4;3);(3;3) >; < (2;1);(10;2);(3;3);(4;3) >;
< (1;2);(9;1);(3;3);(4;3) >; < (1;2);(10;1);(4;3);(3;3) >;
< (2;2);(9;1);(4;3);(3;3) >; < (2;2);(10;1);(3;3);(4;3) >;
< (3;1);(5;2);(9;3);(10;3) >; < (3;1);(6;2);(10;3);(9;3) >;
< (4;1);(5;2);(10;3);(9;3) >; < (4;1);(6;2);(9;3);(10;3) >;
< (3;2);(5;1);(9;3);(10;3) >; < (3;2);(6;1);(10;3);(9;3) >;
< (4;2);(5;1);(10;3);(9;3) >; < (4;2);(6;1);(9;3);(10;3) >g
 H
48
f< (3;1);(7;2);(1;3);(2;3) >; < (3;1);(8;2);(2;3);(1;3) >;
< (4;1);(7;2);(2;3);(1;3) >; < (4;1);(8;2);(1;3);(2;3) >;
< (3;2);(7;1);(1;3);(2;3) >; < (3;2);(8;1);(2;3);(1;3) >;
< (4;2);(7;1);(2;3);(1;3) >; < (4;2);(8;1);(1;3);(2;3) >;
< (3;1);(9;2);(7;3);(8;3) >; < (3;1);(10;2);(8;3);(7;3) >;
< (4;1);(9;2);(8;3);(7;3) >; < (4;1);(10;2);(7;3);(8;3) >;
< (3;2);(9;1);(7;3);(8;3) >; < (3;2);(10;1);(8;3);(7;3) >;
< (4;2);(9;1);(8;3);(7;3) >; < (4;2);(10;1);(7;3);(8;3) >;
< (5;1);(7;2);(3;3);(4;3) >; < (5;1);(8;2);(4;3);(3;3) >;
< (6;1);(7;2);(4;3);(3;3) >; < (6;1);(8;2);(3;3);(4;3) >;
< (5;2);(7;1);(3;3);(4;3) >; < (5;2);(8;1);(4;3);(3;3) >;
< (6;2);(7;1);(4;3);(3;3) >; < (6;2);(8;1);(3;3);(4;3) >;
< (5;1);(9;2);(1;3);(2;3) >; < (5;1);(10;2);(2;3);(1;3) >;
< (6;1);(9;2);(2;3);(1;3) >; < (6;1);(10;2);(1;3);(2;3) >;
< (5;2);(9;1);(1;3);(2;3) >; < (5;2);(10;1);(2;3);(1;3) >;
< (6;2);(9;1);(2;3);(1;3) >; < (6;2);(10;1);(1;3);(2;3) >;
< (7;1);(9;2);(5;3);(6;3) >; < (7;1);(10;2);(6;3);(5;3) >;
< (8;1);(9;2);(6;3);(5;3) >; < (8;1);(10;2);(5;3);(6;3) >;
< (7;2);(9;1);(5;3);(6;3) >; < (7;2);(10;1);(6;3);(5;3) >;
< (8;2);(9;1);(6;3);(5;3) >; < (8;2);(10;1);(5;3);(6;3) >g
 H
Now, we will remove the double edges from these hinges to create 4-cycles.
49
f((10;1);(1;1);(3;1);(2;1)); ((10;1);(5;1);(4;1);(6;1));
((10;1);(7;1);(8;1);(9;1)); ((3;1);(1;1);(4;1);(7;1));
((3;1);(2;1);(8;1);(6;1)); ((4;1);(6;1);(8;1);(7;1));
((1;1);(7;1);(5;1);(8;1)); ((2;1);(7;1);(5;1);(8;1));
((2;1);(4;1);(9;1);(6;1)); ((1;1);(6;1);(9;1);(8;1));
((5;1);(10;1);(6;1);(3;1)); ((6;1);(1;1);(7;1);(2;1));
((7;1);(10;1);(9;1);(3;1)); ((5;1);(3;1);(9;1);(4;1));
((1;1);(10;1);(2;1);(4;1)); ((10;2);(1;2);(3;2);(2;2));
((10;2);(5;2);(4;2);(6;2)); ((10;2);(7;2);(8;2);(9;2));
((3;2);(1;2);(4;2);(7;2)); ((3;2);(2;2);(8;2);(6;2));
((4;2);(6;2);(8;2);(7;2)); ((1;2);(7;2);(5;2);(8;2));
((2;2);(7;2);(5;2);(8;2)); ((2;2);(4;2);(9;2);(6;2));
((1;2);(6;2);(9;2);(8;2)); ((5;2);(10;2);(6;2);(3;2));
((6;2);(1;2);(7;2);(2;2)); ((7;2);(10;2);(9;2);(3;2));
((5;2);(3;2);(9;2);(4;2)); ((1;2);(10;2);(2;2);(4;2));
((10;3);(1;3);(3;3);(2;3)); ((10;3);(5;3);(4;3);(6;3));
((10;3);(7;3);(8;3);(9;3)); ((3;3);(1;3);(4;3);(7;3));
((3;3);(2;3);(8;3);(6;3)); ((4;3);(6;3);(8;3);(7;3));
((1;3);(7;3);(5;3);(8;3)); ((2;3);(7;3);(5;3);(8;3));
((2;3);(4;3);(9;3);(6;3)); ((1;3);(6;3);(9;3);(8;3));
((5;3);(10;3);(6;3);(3;3)); ((6;3);(1;3);(7;3);(2;3));
((7;3);(10;3);(9;3);(3;3)); ((5;3);(3;3);(9;3);(4;3));
((1;3);(10;3);(2;3);(4;3))g
 HnD C
50
f(11;(1;2);(1;1);(2;2)); (11;(1;3);(2;1);(2;3));
(12;(1;3);(1;1);(2;3)); (12;(1;2);(2;1);(2;2));
((1;2);11;(1;3);12); ((1;2);(1;1);(2;3);(2;1));
((2;2);(1;1);(1;3);(2;1)); ((2;2);11;(2;3);12);
(11;(3;2);(3;1);(4;2)); (11;(3;3);(4;1);(4;3));
(12;(3;3);(3;1);(4;3)); (12;(3;2);(4;1);(4;2));
((3;2);11;(3;3);12); ((3;2);(3;1);(4;3);(4;1));
((4;2);(3;1);(3;3);(4;1)); ((4;2);11;(4;3);12);
(11;(5;2);(5;1);(6;2)); (11;(5;3);(6;1);(6;3));
(12;(5;3);(5;1);(6;3)); (12;(5;2);(6;1);(6;2));
((5;2);11;(5;3);12); ((5;2);(5;1);(6;3);(6;1));
((6;2);(5;1);(5;3);(6;1)); ((6;2);11;(6;3);12);
(11;(7;2);(7;1);(8;2)); (11;(7;3);(8;1);(8;3));
(12;(7;3);(7;1);(8;3)); (12;(7;2);(8;1);(8;2));
((7;2);11;(7;3);12); ((7;2);(7;1);(8;3);(8;1));
((8;2);(7;1);(7;3);(8;1)); ((8;2);11;(8;3);12);
(11;(9;2);(9;1);(10;2)); (11;(9;3);(10;1);(10;3));
(12;(9;3);(9;1);(10;3)); (12;(9;2);(10;1);(10;2));
((9;2);11;(9;3);12); ((9;2);(9;1);(10;3);(10;1));
((10;2);(9;1);(9;3);(10;1)); ((10;2);11;(10;3);12)g
 HnD C
51
f((1;1);(5;3);(3;2);(6;3)); ((1;1);(6;3);(4;2);(5;3));
((2;1);(6;3);(3;2);(5;3)); ((2;1);(5;3);(4;2);(6;3));
((1;2);(5;3);(3;1);(6;3)); ((1;2);(6;3);(4;1);(5;3));
((2;2);(6;3);(3;1);(5;3)); ((2;2);(5;3);(4;1);(6;3));
((1;1);(7;3);(5;2);(8;3)); ((1;1);(8;3);(6;2);(7;3));
((2;1);(8;3);(5;2);(7;3)); ((2;1);(7;3);(6;2);(8;3));
((1;2);(7;3);(5;1);(8;3)); ((1;2);(8;3);(6;1);(7;3));
((2;2);(8;3);(5;1);(7;3)); ((2;2);(7;3);(6;1);(8;3));
((1;1);(9;3);(7;2);(10;3)); ((1;1);(10;3);(8;2);(9;3));
((2;1);(10;3);(7;2);(9;3)); ((2;1);(9;3);(8;2);(10;3));
((1;2);(9;3);(7;1);(10;3)); ((1;2);(10;3);(8;1);(9;3));
((2;2);(10;3);(7;1);(9;3)); ((2;2);(9;3);(8;1);(10;3));
((1;1);(3;3);(9;2);(4;3)); ((1;1);(4;3);(10;2);(3;3));
((2;1);(4;3);(9;2);(3;3)); ((2;1);(3;3);(10;2);(4;3));
((1;2);(3;3);(9;1);(4;3)); ((1;2);(4;3);(10;1);(3;3));
((2;2);(4;3);(9;1);(3;3)); ((2;2);(3;3);(10;1);(4;3));
((3;1);(9;3);(5;2);(10;3)); ((3;1);(10;3);(6;2);(9;3));
((4;1);(10;3);(5;2);(9;3)); ((4;1);(9;3);(6;2);(10;3));
((3;2);(9;3);(5;1);(10;3)); ((3;2);(10;3);(6;1);(9;3));
((4;2);(10;3);(5;1);(9;3)); ((4;2);(9;3);(6;1);(10;3))g
 HnD C
52
f((3;1);(1;3);(7;2);(2;3)); ((3;1);(2;3);(8;2);(1;3));
((4;1);(2;3);(7;2);(1;3)); ((4;1);(1;3);(8;2);(2;3));
((3;2);(1;3);(7;1);(2;3)); ((3;2);(2;3);(8;1);(1;3));
((4;2);(2;3);(7;1);(1;3)); ((4;2);(1;3);(8;1);(2;3));
((3;1);(7;3);(9;2);(8;3)); ((3;1);(8;3);(10;2);(7;3));
((4;1);(8;3);(9;2);(7;3)); ((4;1);(7;3);(10;2);(8;3));
((3;2);(7;3);(9;1);(8;3)); ((3;2);(8;3);(10;1);(7;3));
((4;2);(8;3);(9;1);(7;3)); ((4;2);(7;3);(10;1);(8;3));
((5;1);(3;3);(7;2);(4;3)); ((5;1);(4;3);(8;2);(3;3));
((6;1);(4;3);(7;2);(3;3)); ((6;1);(3;3);(8;2);(4;3));
((5;2);(3;3);(7;1);(4;3)); ((5;2);(4;3);(8;1);(3;3));
((6;2);(4;3);(7;1);(3;3)); ((6;2);(3;3);(8;1);(4;3));
((5;1);(1;3);(9;2);(2;3)); ((5;1);(2;3);(10;2);(1;3));
((6;1);(2;3);(9;2);(1;3)); ((6;1);(1;3);(10;2);(2;3));
((5;2);(1;3);(9;1);(2;3)); ((5;2);(2;3);(10;1);(1;3));
((6;2);(2;3);(9;1);(1;3)); ((6;2);(1;3);(10;1);(2;3));
((7;1);(5;3);(9;2);(6;3)); ((7;1);(6;3);(10;2);(5;3));
((8;1);(6;3);(9;2);(5;3)); ((8;1);(5;3);(10;2);(6;3));
((7;2);(5;3);(9;1);(6;3)); ((7;2);(6;3);(10;1);(5;3));
((8;2);(6;3);(9;1);(5;3)); ((8;2);(5;3);(10;1);(6;3))g:
 HnD C
Finally, we will use the double edges (along with the leave) to create the last of our
4-cycles.
53
f((10;1);(3;1);(8;1);(4;1)); ((10;1);(3;1);(4;1);(8;1));
((10;1);(8;1);(3;1);(4;1)); ((1;1);(5;1);(2;1);(9;1));
((1;1);(5;1);(2;1);(9;1)); ((5;1);(6;1);(7;1);(9;1));
((5;1);(6;1);(7;1);(9;1)); ((10;2);(3;2);(8;2);(4;2));
((10;2);(3;2);(4;2);(8;2)); ((10;2);(8;2);(3;2);(4;2));
((1;2);(5;2);(2;2);(9;2)); ((1;2);(5;2);(2;2);(9;2));
((5;2);(6;2);(7;2);(9;2)); ((5;2);(6;2);(7;2);(9;2));
((10;3);(3;3);(8;3);(4;3)); ((10;3);(3;3);(4;3);(8;3));
((10;3);(8;3);(3;3);(4;3)); ((1;3);(5;3);(2;3);(9;3));
((1;3);(5;3);(2;3);(9;3)); ((5;3);(6;3);(7;3);(9;3));
((5;3);(6;3);(7;3);(9;3)); (11;12;(2;1);(1;1));
(11;12;(1;1);(2;1)); (11;(1;1);12;(2;1));
((1;2);(2;2);(2;3);(1;3)); ((1;2);(2;2);(1;3);(2;3));
((1;2);(1;3);(2;2);(2;3)); (11;(3;1);12;(4;1));
(11;(3;1);12;(4;1)); ((3;2);(3;3);(4;2);(4;3));
((3;2);(3;3);(4;2);(4;3)); (11;(5;1);12;(6;1));
(11;(5;1);12;(6;1)); ((5;2);(5;3);(6;2);(6;3));
((5;2);(5;3);(6;2);(6;3)); (11;(7;1);12;(8;1));
(11;(7;1);12;(8;1)); ((7;2);(7;3);(8;2);(8;3));
((7;2);(7;3);(8;2);(8;3)); (11;(9;1);12;(10;1));
(11;(9;1);12;(10;1)); ((9;2);(9;3);(10;2);(10;3));
((9;2);(9;3);(10;2);(10;3)); ((1;1);(3;2);(2;1);(4;2))g
 C
54
f((1;1);(3;2);(2;1);(4;2)) ((1;2);(3;1);(2;2);(4;1))
((1;2);(3;1);(2;2);(4;1)) ((1;1);(5;2);(2;1);(6;2))
((1;1);(5;2);(2;1);(6;2)) ((1;2);(5;1);(2;2);(6;1))
((1;2);(5;1);(2;2);(6;1)) ((1;1);(7;2);(2;1);(8;2))
((1;1);(7;2);(2;1);(8;2)) ((1;2);(7;1);(2;2);(8;1))
((1;2);(7;1);(2;2);(8;1)) ((1;1);(9;2);(2;1);(10;2))
((1;1);(9;2);(2;1);(10;2)) ((1;2);(9;1);(2;2);(10;1))
((1;2);(9;1);(2;2);(10;1)) ((3;1);(5;2);(4;1);(6;2))
((3;1);(5;2);(4;1);(6;2)) ((3;2);(5;1);(4;2);(6;1))
((3;2);(5;1);(4;2);(6;1)) ((3;1);(7;2);(4;1);(8;2))
((3;1);(7;2);(4;1);(8;2)) ((3;2);(7;1);(4;2);(8;1))
((3;2);(7;1);(4;2);(8;1)) ((3;1);(9;2);(4;1);(10;2))
((3;1);(9;2);(4;1);(10;2)) ((3;2);(9;1);(4;2);(10;1))
((3;2);(9;1);(4;2);(10;1)) ((5;1);(7;2);(6;1);(8;2))
((5;1);(7;2);(6;1);(8;2)) ((5;2);(7;1);(6;2);(8;1))
((5;2);(7;1);(6;2);(8;1)) ((5;1);(9;2);(6;1);(10;2))
((5;1);(9;2);(6;1);(10;2)) ((5;2);(9;1);(6;2);(10;1))
((5;2);(9;1);(6;2);(10;1)) ((7;1);(9;2);(8;1);(10;2))
((7;1);(9;2);(8;1);(10;2)) ((7;2);(9;1);(8;2);(10;1))
((7;2);(9;1);(8;2);(10;1))g:
 C
Now, we must consider the following examples that we will use in the 12n+ 8 Construc-
tion.
Example 5.10 (metamorphosis of a maximum packing of 2K20 with triples into 2-fold
4-cycle system of order 20).
55
We will decompose 2K20 with vertex set f1;2;3;4g f1;2;3;4;5g. We will begin with
the maximum packing with hinges, since it has an obvious correspondence to the maximum
packing with triples. Our leave will be < (4;1);(4;2) >.
f< (1;1);(1;2);(2;1);(2;2) >; < (2;1);(2;2);(3;1);(3;2) >;
< (3;1);(3;2);(1;1);(1;2) >; < (1;1);(4;1);(2;1);(3;1) >;
< (1;1);(4;2);(2;2);(3;2) >; < (1;2);(4;1);(2;2);(3;1) >;
< (1;2);(4;2);(2;1);(3;2) >; < (3;1);(4;2);(2;1);(2;2) >;
< (3;2);(4;1);(2;1);(2;2) >; < (1;3);(1;4);(1;1);(1;2) >
< (1;3);(1;5);(1;1);(1;2) >; < (1;4);(1;5);(1;1);(1;2) >;
< (2;3);(2;4);(2;1);(2;2) >; < (2;3);(2;5);(2;1);(2;2) >;
< (2;4);(2;5);(2;1);(2;2) >; < (3;3);(3;4);(3;1);(3;2) >;
< (3;3);(3;5);(3;1);(3;2) >; < (3;4);(3;5);(3;1);(3;2) >;
< (4;3);(4;4);(4;1);(4;2) >; < (4;3);(4;5);(4;1);(4;2) >;
< (4;4);(4;5);(4;1);(4;2) >; < (1;1);(3;3);(2;4);(4;4) >;
< (2;1);(4;3);(1;4);(3;4) >; < (1;1);(3;4);(2;5);(4;5) >;
< (2;1);(4;4);(1;5);(3;5) >; < (1;1);(3;5);(2;3);(4;3) >;
< (2;1);(4;5);(1;3);(3;3) >; < (1;2);(3;3);(2;5);(4;5) >;
< (2;2);(4;3);(1;5);(3;5) >; < (1;2);(3;4);(2;3);(4;3) >;
< (2;2);(4;4);(1;3);(3;3) >; < (1;2);(3;5);(2;4);(4;4) >;
< (2;2);(4;5);(1;4);(3;4) >; < (1;3);(3;1);(2;4);(4;4) >;
< (2;3);(4;1);(1;4);(3;4) >; < (1;3);(3;2);(2;5);(4;5) >;
< (2;3);(4;2);(1;5);(3;5) >; < (1;3);(3;3);(2;3);(4;3) >;
< (2;3);(4;3);(1;3);(3;3) >; < (1;3);(3;4);(2;1);(4;1) >;
< (2;3);(4;4);(1;1);(3;1) >; < (1;3);(3;5);(2;2);(4;2) >;
< (2;3);(4;5);(1;2);(3;2) >; < (1;4);(3;1);(2;5);(4;5) >;
< (2;3);(4;5);(1;2);(3;2) >; < (1;4);(3;1);(2;5);(4;5) >g
 H
56
f< (2;4);(4;1);(1;5);(3;5) >; < (1;4);(3;2);(2;3);(4;3) >;
< (2;4);(4;2);(1;3);(3;3) >; < (1;4);(3;3);(2;2);(4;2) >;
< (2;4);(4;3);(1;2);(3;2) >; < (1;4);(3;4);(2;4);(4;4) >;
< (2;4);(4;4);(1;4);(3;4) >; < (1;4);(3;5);(2;1);(4;1) >
< (2;4);(4;5);(1;1);(3;1) >; < (1;5);(3;1);(2;3);(4;3) >;
< (2;5);(4;1);(1;3);(3;3) >; < (1;5);(3;2);(2;4);(4;4) >;
< (2;5);(4;2);(1;4);(3;4) >; < (1;5);(3;3);(2;1);(4;1) >;
< (2;5);(4;3);(1;1);(3;1) >; < (1;5);(3;4);(2;2);(4;2) >;
< (2;5);(4;4);(1;2);(3;2) >; < (1;5);(3;5);(2;5);(4;5) >;
< (2;5);(4;5);(1;5);(3;5) >g:
 H
Now, we will remove the double edges from these hinges to create 4-cycles.
f((1;1);(2;1);(1;2);(2;2)); ((2;1);(3;1);(2;2);(3;2));
((3;1);(1;1);(3;2);(1;2)); ((1;1);(2;1);(4;1);(3;1));
((1;1);(2;2);(4;2);(3;2)); ((1;2);(2;2);(4;1);(3;1));
((1;2);(2;1);(4;2);(3;2)); ((3;1);(2;1);(4;2);(2;2));
((3;2);(2;1);(4;1);(2;2)); ((1;3);(1;1);(1;4);(1;2));
((1;3);(1;1);(1;5);(1;2)); ((1;4);(1;1);(1;5);(1;2));
((2;3);(2;1);(2;4);(2;2)); ((2;3);(2;1);(2;5);(2;2));
((2;4);(2;1);(2;5);(2;2)); ((3;3);(3;1);(3;4);(3;2));
((3;3);(3;1);(3;5);(3;2)); ((3;4);(3;1);(3;5);(3;2));
((4;3);(4;1);(4;4);(4;2)); ((4;3);(4;1);(4;5);(4;2));
((4;4);(4;1);(4;5);(4;2)); ((1;1);(2;4);(3;3);(4;4));
((2;1);(1;4);(4;3);(3;4)); ((1;1);(2;5);(3;4);(4;5));
((2;1);(1;5);(4;4);(3;5)); ((1;1);(2;3);(3;5);(4;3));
((2;1);(1;3);(4;5);(3;3)); ((1;2);(2;5);(3;3);(4;5));
((2;2);(1;5);(4;3);(3;5)); ((1;2);(2;3);(3;4);(4;3))g
 HnD C
57
f((2;2);(1;3);(4;4);(3;3)); ((1;2);(2;4);(3;5);(4;4));
((2;2);(1;4);(4;5);(3;4)); ((1;3);(2;4);(3;1);(4;4))
((2;3);(1;4);(4;1);(3;4)); ((1;3);(2;5);(3;2);(4;5));
((2;3);(1;5);(4;2);(3;5)); ((1;3);(2;3);(3;3);(4;3));
((2;3);(1;3);(4;3);(3;3)); ((1;3);(2;1);(3;4);(4;1));
((2;3);(1;1);(4;4);(3;1)); ((1;3);(2;2);(3;5);(4;2));
((2;3);(1;2);(4;5);(3;2)); ((1;4);(2;5);(3;1);(4;5));
((2;4);(1;5);(4;1);(3;5)); ((1;4);(2;3);(3;2);(4;3));
((2;4);(1;3);(4;2);(3;3)); ((1;4);(2;2);(3;3);(4;2));
((2;4);(1;2);(4;3);(3;2)); ((1;4);(2;4);(3;4);(4;4))
((2;4);(1;4);(4;4);(3;4)); ((1;4);(2;1);(3;5);(4;1));
((2;4);(1;1);(4;5);(3;1)); ((1;5);(2;3);(3;1);(4;3));
((2;5);(1;3);(4;1);(3;3)); ((1;5);(2;4);(3;2);(4;4));
((2;5);(1;4);(4;2);(3;4)); ((1;5);(2;1);(3;3);(4;1));
((2;5);(1;1);(4;3);(3;1)); ((1;5);(2;2);(3;4);(4;2));
((2;5);(1;2);(4;4);(3;2)); ((1;5);(2;5);(3;5);(4;5));
((2;5);(1;5);(4;5);(3;5))g:
 HnD C
Finally, we will use the double edges (along with the leave) to create the last of our
4-cycles.
58
f((3;1);(3;2);(4;1);(4;2)); ((3;1);(3;2);(4;1);(4;2));
((1;1);(4;1);(1;2);(4;2)); ((1;1);(4;1);(1;2);(4;2));
((1;1);(1;2);(3;4);(3;3)); ((2;1);(2;2);(4;4);(4;3));
((1;1);(1;2);(3;5);(3;4)); ((2;1);(2;2);(4;5);(4;4));
((1;2);(3;3);(1;1);(3;5)); ((2;2);(4;3);(2;1);(4;5));
((1;2);(3;3);(3;5);(3;4)); ((2;2);(4;3);(4;5);(4;4));
((3;3);(3;4);(1;1);(3;5)); ((4;3);(4;4);(2;1);(4;5));
((1;3);(3;2);(1;4);(3;1)); ((2;3);(4;2);(2;4);(4;1));
((1;3);(3;2);(1;5);(3;1)); ((2;3);(4;2);(2;5);(4;1));
((1;4);(3;2);(1;5);(3;1)); ((2;4);(4;2);(2;5);(4;1));
((1;3);(3;4);(1;4);(3;5)); ((2;3);(4;4);(2;4);(4;5));
((1;4);(3;3);(1;5);(3;4)); ((2;4);(4;3);(2;5);(4;4));
((1;3);(3;3);(1;5);(3;5)); ((2;3);(4;3);(2;5);(4;5));
((1;3);(1;4);(1;5);(3;4)); ((2;3);(2;4);(2;5);(4;4));
((1;3);(1;5);(1;4);(3;3)); ((2;3);(2;5);(2;4);(4;3));
((1;3);(1;5);(3;5);(1;4)); ((2;3);(2;5);(4;5);(2;4))g:
 C
Example 5.11 (metamorphosis of a maximum packing of 2K20n2K8 with triples into a
maximum packing of 2K20n2K8 with 4-cycles). We will be decomposing 2K20 with vertex set
f1;2;3;4;5g f1;2;3;4gminus 2K8 on the vertex setf1;2g f1;2;3;4g. We will begin with
the maximum packing with hinges, since it has an obvious correspondence to the maximum
packing with triples.
f< (3;1);(3;2);(3;3);(3;4) >; < (3;3);(3;4);(3;1);(3;2) >;
< (4;1);(4;2);(4;3);(4;4) >; < (4;3);(4;4);(4;1);(4;2) >;
< (5;1);(5;2);(5;3);(5;4) >; < (5;3);(5;4);(5;1);(5;2) >g
 H
59
f< (3;1);(4;1);(1;1);(2;1) >; < (3;1);(5;1);(1;1);(2;1) >;
< (4;1);(5;1);(1;1);(2;1) >; < (3;1);(4;2);(1;3);(2;3) >;
< (3;1);(5;2);(1;3);(2;3) >; < (4;1);(5;2);(1;3);(2;3) >;
< (3;1);(4;3);(1;4);(2;4) >; < (3;1);(5;3);(1;4);(2;4) >;
< (4;1);(5;3);(1;4);(2;4) >; < (3;1);(4;4);(1;2);(2;2) >;
< (3;1);(5;4);(1;2);(2;2) >; < (4;1);(5;4);(1;2);(2;2) >;
< (3;2);(4;1);(1;4);(2;4) >; < (3;2);(5;1);(1;4);(2;4) >;
< (4;2);(5;1);(1;4);(2;4) >; < (3;2);(4;2);(1;2);(2;2) >;
< (3;2);(5;2);(1;2);(2;2) >; < (4;2);(5;2);(1;2);(2;2) >;
< (3;2);(4;3);(1;1);(2;1) >; < (3;2);(5;3);(1;1);(2;1) >;
< (4;2);(5;3);(1;1);(2;1) >; < (3;2);(4;4);(1;3);(2;3) >;
< (3;2);(5;4);(1;3);(2;3) >; < (4;2);(5;4);(1;3);(2;3) >;
< (3;3);(4;1);(1;2);(2;2) >; < (3;3);(5;1);(1;2);(2;2) >;
< (4;3);(5;1);(1;2);(2;2) >; < (3;3);(4;2);(1;4);(2;4) >;
< (3;3);(5;2);(1;4);(2;4) >; < (4;3);(5;2);(1;4);(2;4) >;
< (3;3);(4;3);(1;3);(2;3) >; < (3;3);(5;3);(1;3);(2;3) >;
< (4;3);(5;3);(1;3);(2;3) >; < (3;3);(4;4);(1;1);(2;1) >;
< (3;3);(5;4);(1;1);(2;1) >; < (4;3);(5;4);(1;1);(2;1) >;
< (3;4);(4;1);(1;3);(2;3) >; < (3;4);(5;1);(1;3);(2;3) >;
< (4;4);(5;1);(1;3);(2;3) >; < (3;4);(4;2);(1;1);(2;1) >;
< (3;4);(5;2);(1;1);(2;1) >; < (4;4);(5;2);(1;1);(2;1) >;
< (3;4);(4;3);(1;2);(2;2) >; < (3;4);(5;3);(1;2);(2;2) >;
< (4;4);(5;3);(1;2);(2;2) >; < (3;4);(4;4);(1;4);(2;4) >;
< (3;4);(5;4);(1;4);(2;4) >; < (4;4);(5;4);(1;4);(2;4) >g:
 H
Now, we will remove the double edges from these hinges to create 4-cycles.
60
f((3;1);(3;3);(3;2);(3;4)); ((3;3);(3;1);(3;4);(3;2));
((4;1);(4;3);(4;2);(4;4)); ((4;3);(4;1);(4;4);(4;2));
((5;1);(5;3);(5;2);(5;4)); ((5;3);(5;1);(5;4);(5;2));
((3;1);(1;1);(4;1);(2;1)); ((3;1);(1;1);(5;1);(2;1));
((4;1);(1;1);(5;1);(2;1)); ((3;1);(1;3);(4;2);(2;3));
((3;1);(1;3);(5;2);(2;3)); ((4;1);(1;3);(5;2);(2;3));
((3;1);(1;4);(4;3);(2;4)); ((3;1);(1;4);(5;3);(2;4));
((4;1);(1;4);(5;3);(2;4)); ((3;1);(1;2);(4;4);(2;2));
((3;1);(1;2);(5;4);(2;2)); ((4;1);(1;2);(5;4);(2;2));
((3;2);(1;4);(4;1);(2;4)); ((3;2);(1;4);(5;1);(2;4));
((4;2);(1;4);(5;1);(2;4)); ((3;2);(1;2);(4;2);(2;2));
((3;2);(1;2);(5;2);(2;2)); ((4;2);(1;2);(5;2);(2;2));
((3;2);(1;1);(4;3);(2;1)); ((3;2);(1;1);(5;3);(2;1));
((4;2);(1;1);(5;3);(2;1)); ((3;2);(1;3);(4;4);(2;3));
((3;2);(1;3);(5;4);(2;3)); ((4;2);(1;3);(5;4);(2;3));
((3;3);(1;2);(4;1);(2;2)); ((3;3);(1;2);(5;1);(2;2));
((4;3);(1;2);(5;1);(2;2)); ((3;3);(1;4);(4;2);(2;4));
((3;3);(1;4);(5;2);(2;4)); ((4;3);(1;4);(5;2);(2;4));
((3;3);(1;3);(4;3);(2;3)); ((3;3);(1;3);(5;3);(2;3));
((4;3);(1;3);(5;3);(2;3)); ((3;3);(1;1);(4;4);(2;1));
((3;3);(1;1);(5;4);(2;1)); ((4;3);(1;1);(5;4);(2;1));
((3;4);(1;3);(4;1);(2;3)); ((3;4);(1;3);(5;1);(2;3));
((4;4);(1;3);(5;1);(2;3)); ((3;4);(1;1);(4;2);(2;1));
((3;4);(1;1);(5;2);(2;1)); ((4;4);(1;1);(5;2);(2;1));
((3;4);(1;2);(4;3);(2;2)); ((3;4);(1;2);(5;3);(2;2));
((4;4);(1;2);(5;3);(2;2)); ((3;4);(1;4);(4;4);(2;4));
((3;4);(1;4);(5;4);(2;4)); ((4;4);(1;4);(5;4);(2;4))g
= HnD C
61
Finally, we will use the double edges to create the last of our 4-cycles.
f((3;1);(3;2);(4;1);(4;2)); ((3;1);(3;2);(4;2);(4;1));
((3;1);(4;1);(3;2);(4;2)); ((4;3);(4;4);(5;3);(5;4));
((4;3);(4;4);(5;4);(5;3)); ((4;3);(5;3);(4;4);(5;4));
((3;3);(3;4);(5;1);(5;2)); ((3;3);(3;4);(5;2);(5;1));
((3;3);(5;1);(3;4);(5;2)); ((4;1);(5;1);(4;2);(5;2));
((4;1);(5;1);(4;2);(5;2)); ((3;3);(4;3);(3;4);(4;4));
((3;3);(4;3);(3;4);(4;4)); ((3;1);(5;1);(3;2);(5;2));
((3;1);(5;1);(3;2);(5;2)); ((3;3);(5;3);(3;4);(5;4));
((3;3);(5;3);(3;4);(5;4)); ((3;1);(4;3);(3;2);(4;4));
((3;1);(4;3);(3;2);(4;4)); ((4;1);(5;3);(4;2);(5;4));
((4;1);(5;3);(4;2);(5;4)); ((3;3);(4;1);(3;4);(4;2));
((3;3);(4;1);(3;4);(4;2)); ((4;3);(5;1);(4;4);(5;2));
((4;3);(5;1);(4;4);(5;2)); ((3;1);(5;3);(3;2);(5;4));
((3;1);(5;3);(3;2);(5;4))g:
 C
The 12n+8 44 Construction
With the above examples in hand, we can proceed to the 12n + 8  44 Construc-
tion. Write 12n + 8 = 3(4n) + 8. Since 12n + 8  44, 4n  12. This is important.
Let 1= f11;12;13;14;15;16;17;18g and Q = f1;2;3;:::;4ng. Let H(Q) =
fh0;h1;:::;hn 1g be a partition of Q into pairwise disjoint sets of size 4 (called holes),
where hi = f4i + 1;4i + 2;4i + 3;4i + 4g. Let (Q; ) be a commutative quasigroup of or-
der 4n with holes H(Q) (see [6]) and set X =1[(Q f1;2;3g). For 0  i  n 1, let
Bi = hi f1;2;3g and Ai =1[Bi. De ne a collection of hinges as follows:
(1) Use the n = 20 example (Example 5.10) to  nd a hinge system with our desired
metamorphosis (A0;H0), where the leave is <11;12 >.
62
...?
?1 ?2
A0 = B
0leave
(1,1) (2,1) (3,1) (4,1)
(2) For 1  i n 1, use the n = 20 with a hole of size 8 Construction (Example 5.11)
to  nd a hinge system with our desired metamorphosis (Ai;Hi), where the hole is1.
...? B
1
(5,1) (6,1) (7,1) (8,1)
Bn?1
(4n+1,1)
(4n+2,1)(4n+3,1)
(4n+4,1)
(3) We now need to use the edges between vertices in Bi and Bj for i 6= j. De ne a
collection of hinges H? as follows. For x2hi, y2hj, i6= j, place the hinges
< (x;1);(y;1);(x y;2);(x y;3) >; < (x;2);(y;2);(x y;1);(x y;3) >;
and < (x;3);(y;3);(x y;1);(x y;2) >g in H?
...? (x,1)
(y,1)
(x?y,2)
(x?y,3)
...?
(x,2) (y,2)
(x?y,3)
(x?y,1)
63
...?
(x,3)
(y,3)
(x?y,1)
(x?y,2)
It is straightforward to see that (X =
n[
i=1
Ai;H?[
n[
i=1
Hi) is a hinge system of order
12n+8 with leave <11;12 >. To proceed with our metamorphosis,  rst use the prescribed
metamorphoses in (1) and (2) and places the 4-cycles in C. After removing the double edges
from our hinges in H? (from (3)), we still have edges of the type < (x;k);(y;k) >, where
k2f1;2;3g and x2hi, y2hj, i6= j remaining to use in our metamorphosis.
(4) For k2f1;2;3g, we have 2 copies of the complete n-partite graph with all partite sets
having size 4. Between each pair of partite sets fx1;x2;x3;x4g and fy1;y2;y3;y4g in
these 2K4;4;:::;4, we have:
f(x1;y2;x2;y1); (x1;y2;x2;y1); (x1;y4;x2;y3); (x1;y4;x2;y3);
(x3;y4;x4;y3); (x3;y4;x4;y3); (x3;y2;x4;y1); (x3;y2;x4;y1)g
 C.
The result is a 2-fold 4-cycle system of order 12n + 8 with vertex set X.
Theorem 5.12. There exists a maximum packing of 2K12n+8 with triples having a meta-
morphosis into a 2-fold 4-cycle system of order 12n + 8 20.  
64
Chapter 6
Summary
Combining all of the results in Chapters 1, 2, 3, 4, and 5, we have a proof of Theorem
1.8.
Theorem (1.8). The spectrum for maximum packings of 2Kn with triples having a meta-
morphosis into a maximum packing of 2Kn with 4-cycles is the set of all n  2;5;8; or
11 (mod 12) 8, except n = 5 and n = 8.  
2Kn T
H
65
H\D D
?
H\D D?
?
maximum packing of 2Kn with 4-cycles
66
Bibliography
[1] S.E. McClanahan, The metamorphosis of 2-fold Triple Systems into Maximum Packings
of 2Kn with 4-cycles, Ph.D. thesis, Auburn University, August 6, 2011.
[2] T.P. Kirkman, On a Problem in Combinations, Cambridge and Dublin Math. J., (1847),
191-204.
[3] C.J. Colbourn and A. Rosa, Triple Systems, Oxford University Press, (1999), 560 pages.
[4] C. Huang and A. Rosa, On the Existence of Balanced Bipartite Designs, Utilitas Math.,
4(1973), 55-75.
[5] M. Gionfriddo and C.C. Lindner, The Metamorphosis of 2-fold Triple Systems into
2-fold 4-cycle Systems, JCMCC, 46(2003), 129-139.
[6] C.C. Lindner and C.A. Rodger, Design Theory, Second Edition, CRC Press, 2009, 264
pages.
[7] D. Sotteau, Decompositions of Km;n(K m;n) into Cycles (Circuits) of Length 2k, J. Com-
binatorial Theory, Sec. B, 30(1981), 75-78.
67