5-cycle systems
by
John Asplund
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
May 4, 2014
Keywords: 5-cycle systems, enclosing, cycle, embedding, decomposition
Copyright 2014 by John Asplund
Approved by
Christopher Rodger, Chair, Don Logan Endowed Chair of Mathematics and
Associate Dean for Research in the College of Sciences and Mathematics
Dean Ho man, Professor of Mathematics
Curt Lindner, Distinguished University Professor
Peter Johnson, Professor of Mathematics
Jessica McDonald, Assistant Professor
Abstract
A k-cycle system of a multigraph G is an ordered pair (V;C) where V is the
vertex set of G and C is a set of k-cycles, the edges of which partition the edges of
G. A k-cycle system of  Kv is known as a  -fold k-cycle system of order v. A k-cycle
system (V;C) of  Kv is said to be enclosed (embedded) in a k-cycle system (V[U;P)
of ( + m)Kv+u if C P and u;m 1 (m = 0 and u 1). We settle the enclosing
problem for  -fold 5-cycle systems when u = 1 or 2. We settle the embedding problem
for  -fold 5-cycle systems except possibly in two cases. Other analogues of this are
considered and consequently settled.
ii
Acknowledgments
First and foremost, I would like to thank Dr. Rodger for his endless patience
with me. This would not have been possible without you. Thank you for putting up
with me and mentoring me for my next stage in life.
My family have also played a pivotal role in getting me to where I am now.
Thanks to the support and encouragement of Mom, Dad, Erin, and Nancy, I made
to where I am today.
Though my friends are scattered throughout the country, they have been a con-
stant source of relief and help. These people have been able to keep me sane in times
of great strife. Thank you Bill, Braxton, Carrie, Dave K., David C., Desi, Diane,
Evie, James, Jason, Joe, Kat, Kevin, Kristina, Mel, Missy, Richard, Steve, Susan,
and Tim.
Thanks are in order for the secretaries for their help throughout my time at
Auburn University: Gwen, Carolyn, and Lori.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 De nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 Enclosings when u = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 The small cases when u = 1 . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3 Enclosings when u = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.2 Necessary tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.3 Creating closed trail systems of Gv(D) . . . . . . . . . . . . . . . . . 33
3.4 The small cases for m . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.5 The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4 5{CS of ( +m)Kv+u  Kv when u = 1 . . . . . . . . . . . . . . . . . . 67
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4.2 5{CSs of ( +m)Kv+1  Kv: The small cases . . . . . . . . . . . . . 68
4.3 5{CSs of ( +m)Kv+1  Kv: The Main Result . . . . . . . . . . . . 73
5 5-cycle systems when m = 0 . . . . . . . . . . . . . . . . . . . . . . . . . 78
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
iv
5.2 5{CSs of  Kv+u  Kv: Preliminary Results . . . . . . . . . . . . . . 78
5.3 5{CSs of  Kv+u  Kv:   5 (mod 10) and u< 3v + 1 . . . . . . . 82
5.4 5{CSs of  Kv+u  Kv: Main Result . . . . . . . . . . . . . . . . . . 88
6 Conclusion/Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
v
List of Figures
1.1 K5 (left) and 2K5 (right) . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 A 4K6 K5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 A 7-cycle (Hamilton cycle) with vertex set f0;1;:::;6g . . . . . . . . . . 3
1.4 A 5{CS of K5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.5 A Steiner triple system on 7 vertices (STS(7)) . . . . . . . . . . . . . . . 4
1.6 A 5{CS of K5 is enclosed in a 5{CS of 4K6 . . . . . . . . . . . . . . . . 5
3.1 Possible 5-cycle Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
5.1 Possible Types of 5-cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.2 Possible 5-cycle Types when   5 (mod 10) and u< 3v + 1 . . . . . . . 84
6.1 All types of 5-cycles based on the number of pure and mixed edges . . . 101
vi
List of Tables
1.1 Necessary and su cient conditions for the existence of a 5{CS(v; ) . . . 7
2.1 Di erences of edges in BnE when  is as large as possible . . . . . . . 13
2.2 Di erences of edges in BnE when  is as large as possible . . . . . . . 15
2.3 Di erences of edges in B0 and E when  is as large as possible for ?2
f2;3g and m = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.4 Di erences of edges in B0 when  is as large as possible for ?2f7;8gand
m = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.5 Di erences of edges in B when  is as large as possible for ? = 1 and m = 1 20
2.6 Di erences of edges in B when  is as large as possible for ?2f4;6g and
m = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.7 Di erences of edges in B when  is as large as possible for ? = 9 and m = 1 21
2.8 Checking  min(v;m) when v 1 or 6 (mod 10): each cell contains  min(v;m) 24
2.9 Checking  min(v;m) when v6 1 or 6 (mod 10): each cell contains  min(v;m) 25
2.10 Checking the largest values of : each cell contains max(v;m)  max(v;t(v)) 
 max(v;m t(v)) for k 0;1; and 2 (mod 3) in turn. . . . . . . . . . . . 26
2.11 Checking the smallest values of  for v 6 1 or 6 (mod 10): each cell
contains  min(v;m t(v)) + min(v;t(v))  min(v;m). . . . . . . . . . . 27
2.12 Checking the smallest values of  for v  1 or 6 (mod 10): each cell
contains  min(v;m t(v)) + min(v;t(v))  min(v;m). . . . . . . . . . . 27
3.1 Edges of di erences in B0 when  is as large as possible, v 0 (mod 10),
and m = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.2 Edges of di erences in B00 when  is as large as possible, v 0 (mod 10),
and m2f4;6;8g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
vii
3.3 Edges of di erences in BnE when  is as large as possible and v 5
(mod 10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.4 Edges of di erences in B when  is as large as possible for ? = 1 . . . . . 56
3.5 Edges of di erences in B when  is as large as possible for ?2f3;9g . . 56
3.6 Edges of di erences in B0 when  is as large as possible for ?2f4;6g . . 58
3.7 Edges of di erences in B when  is as large as possible for ? = 8 . . . . . 58
3.8 Edges of di erences in B0[E0 when  is as large as possible for ? = 2 . 61
3.9 Edges of di erences in B0nE0 when  is as large as possible for ? = 7 . 62
3.10 Checking  min(v;m) when v  0;1;5 or 6 (mod 10): each cell contains
 min(v;m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.11 Checking  min(v;m) when v 6 0;1;5 or 6 (mod 10): each cell contains
 min(v;m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.12 Checking the largest values of : each cell contains max(v;m)  max(v;t(v)) 
 max(v;m t(v)) except for some special values listed in Table 3.13. . . . 65
3.13 Checking the largest values of : each cell contains max(v;m)  max(v;t(v)) 
 max(v;m t(v)). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
3.14 Checking the smallest values of  for v6 0;1;5 or 6 (mod 10): each cell
contains  min(v;m t(v)) + min(v;t(v))  min(v;m). . . . . . . . . . . 66
3.15 Checking the smallest values of  for v  1 or 6 (mod 10): each cell
contains  min(v;m t(v)) + min(v;t(v))  min(v;m). . . . . . . . . . . 66
4.1 Edges of di erences in B0 when  is as large as possible for ?2f2;3gand
m = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.2 Edges of di erences in B0 when  is as large as possible for ?2f7;8gand
m = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.3 List of  min(v;m) values . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
4.4 Each cell contains  max(v;m)  max(v;m 1)  max(v;1) for k 0;1;2
(mod 3) in turn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
4.5 Each cell contains  min(v;m 1) + min(v;1)  min(v;m) . . . . . . . . 76
viii
5.1 Edges of di erences in B0 when  is as large as possible for ? 4 and ?6= 5 91
5.2 Edges of di erences in B0 when  is as large as possible for ? = 0 . . . . 92
5.3 Edges of di erences in B0 when  is as large as possible for ?2f1;3g . . 94
5.4 Edges of di erences in B0 when  is as large as possible for ? = 2 . . . . 94
5.5 Edges of di erences in B0 when  is as large as possible for ? = 5 . . . . 95
ix
Chapter 1
Introduction
The goal of this dissertation is to construct 5-cycle systems of a speci c family of
graphs using tools from design theory. With that said, we begin with an overview of
the terminology in Section 1.1 and history in Section 1.2 that will set the foundation
for the rest of the dissertation. At the end of Section 1.1 there will be a brief outline
of the entire dissertation.
1.1 De nitions
A graph G is an ordered pair (V(G);E(G)) where V(G) is a set of vertices and
E(G) is a set of unordered pairs of vertices called edges. A subgraph of a graph G is a
graph H such that V(H) V(G) and E(H) E(G). To simplify notation, V and E
will be used instead of V(G) and E(G) respectively when the underlying graph is not
important. Two vertices u and v are adjacent if there is an edge e =fu;vg2E(G);
e is said to join these vertices. A graph with no loops or multiple edges is called a
simple graph. On the other hand, a graph that does have loops or multiple edges is
called a multigraph. When a graph on v vertices has exactly one edge between every
pair of vertices, it is called a complete graph and is denoted Kv. The  -fold complete
graph on v vertices, denoted by  Kv, is the multigraph in which every edge of Kv is
repeated  times (see Figure 1.1). The graph ( +m)Kv+u  Kv is the graph formed
from ( + m)Kv+u on the vertex set V [U by removing the edges of the subgraph
 Kv on the vertex set V where jVj= v and jUj= u (see Figure 1.2).
Figure 1.1: K5 (left) and 2K5 (right)
It is easier to understand the structure of a graph by describing the properties
of the edges and vertices of a graph. An edge fvi;vjg is said to be incident to the
vertices vi and vj. Two edges are incident if they share a common vertex. The
1
0
4
3 2
1
5
V
U
Figure 1.2: A 4K6 K5
degree of a vertex v0 is the number of edges that are incident with v0 and is denoted
degG(v0) or deg(v0) when the underlying graph is not important. A sequence of
vertices (v0;v1;:::;vn) such that fvi;vi+1g is an edge for each 0  i n 1 in the
graph G is called a trail. A trail where v0 = vn is called a closed trail. A trail where
each vertex in the sequence is unique is called a path. Two vertices v1 and v2 in a
graph are connected if there is a path between v1 and v2. If every pair of vertices is
connected then the graph is said to be connected.
At times it will be convenient to talk about the set of integers from 0 to k 1.
This notation is given as Zk = f0;1;:::;k 1g. A graph is k-regular if the degree
of each vertex is k. A 2-regular connected graph is called a cycle. A cycle that
contains every vertex in the graph G is called a Hamilton cycle. We will encounter
situations where it will be very useful to talk about the length of a cycle. So a k-cycle
(v0;v1;:::;vk 1) is the graph on k distinct vertices v0;v1;:::;vk 1 and with edge set
E = ffvi;vi+1gj i 2 Zkg, reducing sums modulo k. The graph in Figure 1.3 is a
7-cycle and a Hamilton cycle.
A decomposition of a graph G is a set of subgraphs that partition the edges of G.
Typically, a graph is decomposed into copies of the same subgraph. This dissertation
will only focus on the decomposition of graphs into cycles. A k-cycle system, k{CS,
of a multigraph G is an ordered pair (V;C) where V is the vertex set of G and C is
a set of k-cycles, the edges of which partition the edges of G. A k-cycle system of (a
subgraph of)  Kv, k{CS(v; ), is known as a (partial)  -fold k-cycle system of order
v. The graph in Figure 1.4 is a 5{CS of K5 (i.e. a 1-fold 5-cycle system). When
k = 3 and  = 1, this structure is also called a Steiner triple system of order v, or an
STS(v). A Steiner triple system is a pair (V;B) whereV is a v-set of points andB is
a collection of 3-subsets, called blocks, such that each pair of points ofV is contained
in exactly one block. The graph in Figure 1.5 is a decomposition of K7 into 3-cycles
2
0
1
2
34
5
6
Figure 1.3: A 7-cycle (Hamilton cycle) with vertex set f0;1;:::;6g
(i.e. a Steiner triple system of order 7). Kirkman showed in [21] that there exists a
Steiner triple system of order v if and only if v 1 or 3 (mod 6).
Figure 1.4: A 5{CS of K5
A problem that has attracted much interest over the years is to  nd conditions
which guarantee that a given (partial) k{CS of  Kv (V;C) is contained in a k{CS of
( +m)Kv+u (V[U;P) with C P. If u;m = 0 then (V;C) is said to be completed
in (V [U;P) = (V;P), if m = 0 then (V;C) is said to be embedded, if u = 0 then
(V;C) is said to be immersed, and if u;m 1 then (V;C) is said to be enclosed in
(V [U;P). This dissertation will focus exclusively on enclosings and embeddings.
In particular, the focus will be placed on both the existence of enclosings of 5{CSs
and the existence of 5{CSs of ( +m)Kv+u  Kv. When  nding such cycle systems,
since the edges of  Kv have already been placed in the given k-cycles in C, it remains
to  nd a k{CS of ( + m)Kv+u  Kv. Figure 1.6 shows that a 5{CS of K5 can be
enclosed in a 5{CS of 4K6.
In order to attack these 5-cycle system problems, we need to construct 5-cycles in
a  exible manner. To this end, Skolem-type sequences are used in a creative way. In
1957, while working on the construction of Steiner triple systems, Skolem published
3
0
1
2
34
5
6
Figure 1.5: A Steiner triple system on 7 vertices (STS(7))
in [30] a construction that used sequences of integers with special properties. Later,
an equivalent formulation in terms of sequences was made with an improvement on
their construction in [1, 5, 26, 29]. Formally we de ne a Skolem sequence of order n
as a sequence of integers S = (d1;:::;d2n) such that the following two conditions are
met:
1. For every  2f1;:::;ng there exist exactly two elements dx;dy 2S such that
dx = dy =  , and
2. if dx = dy =  with x<y, then y x =  .
For example, S = (1;1;3;4;5;3;2;4;2;5) is a Skolem sequence of order 5. Note that
S can also be written as the collection of ordered pairs f(x1;y1) = (1;2);(x2;y2) =
(7;9);(x3;y3) = (3;6);(x4;y4) = (4;8);(x5;y5) = (5;10)g, the pairs being produced
by the positions in S that contain the same integer; we refer to them as Skolem pairs.
An extended Skolem sequence S = (d1;:::;d2n+1) is de ned in the same way as
a Skolem sequence, with the added condition
3. there is exactly one di2S such that di = 0 (di is commonly called the hook).
A hooked Skolem sequence is an extended Skolem sequence where d2n = 0.
The following existence results below for Skolem sequences and hooked Skolem
sequences will assist us in our construction of 5-cycles.
Theorem 1.1.1. [1] A Skolem sequence of order n exists if and only if n  0;1
(mod 4).
Theorem 1.1.2. [26] A hooked Skolem sequence of order n exists if and only if
n 2;3 (mod 4).
One way to form cycle systems is to use di erences. To explain how this is done,
 rst let the set of vertices of a graph Kv be denoted as Zv = f0;1;:::;v 1g. Let
dv(i;j) = minfji jj;v ji jjg where fi;jg is an edge in Kv. We say that the
4
K5 12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
12
3
4 5
6
Figure 1.6: A 5{CS of K5 is enclosed in a 5{CS of 4K6
edge fi;jg has di erence dv(i;j). So de ne D(v) =f1;2;:::;bv=2cg to be the set of
all di erences of edges in Kv. If v is odd then there are v edges of each di erence in
D(v). If v is even then there are v edges for each di erence except the di erence v=2.
There are v=2 edges of di erence v=2. Hence the di erence v=2 is commonly referred
to as the \half di erence."
Consider the graph K7 in Figure 1.5. The 3-cycle c = (c0;c1;c2) = (0;1;3)
contains one edge of each di erence 1, 2, and 3. We can then add 1 to each ver-
tex in c to get another 3-cycle that does not contain any of the edges in c. This
is no coincidence. The same is true if we add 2, 3, 4, 5, or 6 to c as long as the
sums are reduced modulo v = 7. So fc + j j j 2 Z7g is a set of 3-cycles that is
a decomposition of K7, i.e. f(c0 + j;c1 + j;c2 + j) jj 2Z7g is a 3{CS of K7 (see
Figure 1.5 for a visual representation). This approach works well in general for cy-
cles depending on the di erences of the edges in the cycle. Since this method is a
basic tool in the dissertation, the following notation will be useful. For any cycle
c0 = (c1;c2;:::;cx) let c0 + j = (c1 + j;c2 + j;:::;cx + j), reducing the sums modulo
v, and let C +j =fc0 +jjc02Cg.
One use of Skolem sequences is to build cyclic Steiner triple systems. A cyclic
Steiner triple system is a Steiner triple system where for each block b2B, b+ 12B.
The graph in Figure 1.5 is an example of a cycle Steiner Triple System wheref0;1;3g
would be the base block. The general construction of cyclic Steiner triple systems
using Skolem sequences is split into two cases depending on the value of v.
Remember that there exists a Steiner triple system if v 1 or 3 (mod 6). Let
5
G = K6n+1. Sof1;2;:::;3ngis the set of di erences in G that represents the edges in
G. We want to partition this set of di erences into triplesfa;b;cgsuch that a+b = c
or a+b+c 0 (mod 6n+1). This is called the  rst He ter di erence problem. Form
a Skolem or hooked Skolem sequence of order n such that (ar;br) is the rth Skolem
pair. Then T =f(r;ar +n;br +n)j1 r ng is a partition of f1;2;:::;3ng and a
solution to the  rst He ter di erence problem. Also, these di erences can be used to
form a 3{CS of K6n+1; fT +jjj2Z6n+1g.
Let G = K6n+3. Then we want to partition f1;2;:::;3n + 1gnf2n + 1g into
triples fa;b;cg such that a+b = c or a+b+c 0 (mod 6n+ 3). This is called the
second He ter di erence problem. Notice that the di erence 2n + 1 is removed so
that the number of di erences remaining is divisible by 3. Also, the di erent 2n + 1
can be used to form a 3-set by itself; (0;2n + 1;4n + 2). Form an extended Skolem
sequence of order n such that (ar;br) is the rth Skolem pair and the hook is in the nth
position. Then T =f(r;ar +n;br +n)j1 r ngis a solution to the second He ter
di erence problem. SofT +jjj2Z6n+3g[f(0;2n+ 1;4n+ 1) +jjj2Z2n+1gis a
3{CS of K6n+3.
The remainder of Chapter 1 will discuss the history of the decomposition prob-
lems related to the material presented in Chapters 2{5. In Chapter 2, the necessary
conditions for the 5{CS enclosing problem will be established. Chapter 2 will also
show that the necessary conditions for the 5{CS enclosing problem are su cient
when u = 1. Chapter 3 will show that the necessary conditions for the 5{CS en-
closing problem are su cient when u = 2. Chapters 4 and 5 will tackle the more
general problem by establishing that the necessary conditions for the existence of a
5{CS of ( +m)Kv+u  Kv are su cient when u = 1 and m = 0 respectively except
possibly in two cases in the latter case. Finally, Chapter 6 will look at open questions
related to the work in Chapters 2-5 and possible future work related to  nding 5{CS
of ( +m)Kv+u Kv and enclosings of 5{CSs. All other de nitions not mentioned can
be found in either \Design Theory" by Lindner and Rodger [22] or in \Introduction
to graph theory" by West [31].
1.2 History
In 1981, Alspach conjectured that there exists a decomposition of Kv into t cycles
of speci ed lengths m1;m2;:::;mt as long as v is odd, 3  mi v, and m1 + m2 +
   +mt = v(v 1)=2 (see [2]). He also conjectured that there exists a decomposition
of Kv into a 1-regular graph, say I, and t cycles of speci ed lengths m1;m2;:::;mt
as long as v is even, 3  mi  v, and m1 + m2 +   + mt = v(v 1)=2 v=2.
These conjectures are commonly referred to as the Alspach Conjecture. We shall
focus our discussion on the former conjecture. Lately, there have been many papers
published that aimed to solve this conjecture; most notably by Bryant and Horsley
in the following result.
Theorem 1.2.1. [9] There is an integer N such that for each integer v N, there
exists a decomposition of Kv into t cycles of lengths m1;m2;:::;mt if and only if
6
1. v is odd,
2. 3 m1;m2;:::;mt v, and
3. m1 +m2 +   +mt = v(v 1)=2.
For the remainder of this dissertation we will focus on decompositions with cycles
of the same length. Ho man, Lindner, and Rodger were able to show that a Kv can
be decomposed into m-cycles when m v 3m and v is odd. By using this result
and other methods, Alspach and Gavlas [3] were able to decompose Kv I or Kv into
m-cycles for 3 m v when m and v are both even or both odd respectively. Later,
 Sajna [28] was able to decompose Kv I or Kv into m-cycles for 3  m v when
m is odd and v even or m is even and v is odd respectively. This shows that there
exists an m{CS of Kv for all values of m and v that satisfy the obvious necessary
and su cient conditions.
The next direction for this problem is to establish there exists a k{CS of  Kv for
 > 1. For a k{CS of  Kv, de ne v to be (k; )-admissible if the degree of each vertex
is even, the number of edges in  Kv is divisible by k, and v = 1 or v k. Through
the e orts of Bermond, Hanani, Huang, Rosa, and Sotteau in [7, 8, 17, 18, 27], it
was shown that for 3  k 14 and k = 16, there exists a k{CS of  Kv if and only
if v is ( ;k)-admissible. It was not until recently that Bryant, Horsley, Maenhaut,
and Smith in [10] were able to show there exists a k{CS of  Kv if and only if v is
( ;k)-admissible. Of particular interest to this dissertation is the case when k = 5.
Theorem 1.2.2. [10] A 5{CS of  Kv exists if and only if
(a)  v(v 1) 0 (mod 5),
(b)  (v 1) 0 (mod 2), and
(c) either v = 1 or v 5.
Since the immersing problem is exactly the same as showing there exists a k{CS of
 Kv, this problem has already been completely solved by Bryant, Horsley, Maenhaut,
and Smith in [10]. Since this paper focuses solely on building 5{CS, it will cause no
confusion to refer to a ( ;5)-admissible integer as simply being  -admissible. Also
 is said to be v-admissible if the conditions in Theorem 1.2.2 are satis ed by  Kv.
Some consequences of Theorem 1.2.2 are summarized in Table 1.1.
 Possible values of v
0 (mod 10) v62f2;3;4g
1;3;7;9 (mod 10) v 1;5 (mod 10)
2;4;6;8 (mod 10) v 0;1 (mod 5)
5 (mod 10) v 1 (mod 2);v6= 3
Table 1.1: Necessary and su cient conditions for the existence of a 5{CS(v; )
Doyen and Wilson were able to prove in 1973 the following result for embeddings
7
on Steiner triple systems (i.e. embedding a 3{CS in another 3{CS); commonly called
the Doyen-Wilson theorem.
Theorem 1.2.3. [16] A Steiner triple system of order u can be embedded in a Steiner
triple system of order v>u if and only if v 2u+ 1 and v 1 or 3 (mod 6).
Rodger and Bryant were able to extend the Doyen-Wilson theorem to hold for all odd
k-cycle lengths when k 9.
Theorem 1.2.4. [12, 13] A k{CS of Kv can be embedded in a k{CS of Kv+u if and
only if the following conditions hold:
 If k = 5 then v +u 1 or 5 (mod 10) and u v=2 + 1.
 If k = 7 then v +u 1 or 7 (mod 14) and u v=3 + 1.
 If k = 9 then v +u 1 or 9 (mod 18) and u v=4 + 1.
Later, Bryant, Rodger, and Spicer were able to extend these results to embedding
k-cycle systems when k 14 in [14].
Some work has been seen in generalizing these results to establish the existence
of a k{CS of Kv+u Kv. Mendelsohn and Rosa were able to show the necessary
conditions were su cient for the existence of a 3-cycle system of Kv+u Kv. Bryant,
Ho man and Rodger were able to establish the necessary and su cient conditions for
the existence of a 5-cycle system of Kv+u Kv.
Theorem 1.2.5. [23] There exists a 3-cycle system of Kv+u Kv if and only if
u v + 1 and
1. v +u;v 1 or 3 (mod 6), or
2. v +u v 5 (mod 6).
Theorem 1.2.6. [11] There exists a 5-cycle system of Kv+u Kv if and only if
u v=2 + 1 and
1. v +u;v 1 or 5 (mod 10), or
2. v +u v 3 (mod 10), or
3. v +u;v 7 or 9 (mod 10).
Again in [14], Bryant, Rodger, and Spicer were able to show there exists a k-cycle
system of Kv+u Kv when k = 4;6;7;8;10;12; and 14.
Many papers have investigated the  -fold 3{CS enclosing problem (see for ex-
ample [15, 19, 20, 25]). The di culty of this problem is clear since as yet there is no
complete solution to this problem. Newman and Rodger [24] completely solved the
problem of enclosing a  -fold 4{CS of order v into a ( +m)-fold 4{CS of order v+u
for all u;m 1. Typically, problems involving odd cycles have many more di culties
than related even cycle problems, and that is certainly the case here. The techniques
used in the subsequent chapters for the 5{CS enclosing problem are very di erent to
those used in [24].
8
Chapter 2
Enclosings when u = 1
2.1 Introduction
In this chapter, we study the  -fold 5{CS enclosing problem, settling the fol-
lowing conjecture in the case where u = 1 (see Theorem 2.3.2). The joint work in
this chapter has been published in [4]. As in Lemma 2.1.2, all 5 conditions in this
conjecture are necessary, i.e. the di culty when studying Conjecture 2.1.1 lies in
proving su ciency.
Conjecture 2.1.1. Let u 1. Every 5{CS of  Kv can be enclosed in a 5{CS of
( +m)Kv+u if and only if
(a) ( +m)u+m(v 1) 0 (mod 2),
(b)  u2 ( +m) +m v2 +vu( +m) 0 (mod 5),
(c) if u = 1, then m(v 1) 3( +m), and
(d) if u = 2, then m v2  2( +m) (v 1)( +m)=2 0.
(e) if u 3, then dvu( + m)=4e+ 2  mv(v 1)=2 + ( + m)u(u 1)=2 where
 = 0 or 1 if vu( +m) 0 or 2 (mod 4) respectively.
Lemma 2.1.2 establishes the necessary conditions in Conjecture 2.1.1.
Lemma 2.1.2. Suppose there exists a 5{CS of  Kv. If this can be enclosed in a
5{CS of ( +m)Kv+u, then:
(a) ( +m)u+m(v 1) 0 (mod 2),
(b)  u2 ( +m) +m v2 +vu( +m) 0 (mod 5),
(c) if u = 1, then m(v 1) 3( +m),
(d) if u = 2, then m v2  2( +m) (v 1)( +m)=2 0, and
(e) if u 3, then dvu( + m)=4e+ 2  mv(v 1)=2 + ( + m)u(u 1)=2 where
 = 0 or 1 if vu( +m) 0 or 2 (mod 4) respectively.
Proof. Suppose there exists a 5{CS (V;C1) of  Kv that can be enclosed in a 5{CS
(V [U;C2) of ( +m)Kv+u. It is clear that each vertex in each of ( +m)Kv+u and
 Kv has even degree in order for the 5{CSs of ( + m)Kv+u and  Kv to exist. So
( + m)(v + u 1) and  (v 1) are both even, proving (a). Since there exists a
5{CS of ( + m)Kv+u and a 5{CS of  Kv, the number of edges of each graph must
be divisible by 5. Thus ( + m)u(u 1)=2 + ( + m)v(v 1)=2 + ( + m)uv and
 v(v 1)=2 are divisible by 5. So
( +m)
 u
2
 
+ ( +m)
 v
2
 
+ ( +m)uv  
 v
2
 
= ( +m)
 u
2
 
+m
 v
2
 
+vu( +m)
 0 (mod 5):
9
Therefore condition (b) holds.
Suppose u = 1 and let U = f1g. There must be  + m edges connecting each
vertex in V to1, so the degree of1is ( +m)v. Each 5-cycle in C2 containing1uses
two of these edges, so the number of 5-cycles containing 1 is v( + m)=2. Because
mKv has m v2 edges, and since any 5-cycle containing 1 uses 3 edges from this
graph, it follows that the number of such 5-cycles in C2 must be at most mv(v 1)=6,
so v( +m)=2 mv(v 1)=6. Thus condition (c) holds.
Suppose u = 2 and let U =f11;12g. An edge is pure if it joins two vertices in
V or if it joins two vertices in U; otherwise it is said to be mixed. Each of the  +m
edges joining11 to12 must be in a 5-cycle in C2 that contains exactly 2 pure edges
in mKv and exactly 2 mixed edges. There are 2v( +m) 2( +m) = 2(v 1)( +m)
remaining mixed edges. Furthermore, the other 5-cycles in C2 which contain at least
one vertex in U must exhaust these remaining edges. Each 5-cycle in C2 can use at
most 4 of the mixed edges, and so each 5-cycle of this type uses at least one edge in
mKv. Thus it must be that
0 m
 v
2
 
 2( +m) 2(v 1)( +m)4 = m
 v
2
 
 2( +m) 12(v 1)( +m):
Therefore condition (d) holds.
Suppose u 3. Every 5-cycle in C2 that contains a mixed edge must contain at
least one pure edge, and at most 4 mixed edges. So there are at leastdvu( +m)=4e
5-cycles in C2 containing mixed edges. Furthermore, if vu( +m) 2 (mod 4) then
at least one such 5-cycle must contain 2 mixed and 3 pure edges. (Note that since
each 5-cycle contains 0, 2, or 4 mixed edges it follows that vu( +m) is even.) Thus,
there must be at leastdvu( +m)=4e+2 pure edges in 5-cycles containing at least one
vertex in U where  2f0;1g. Sodvu( +m)=4e+2  mv(v 1)=2+( +m)u(u 1)=2
and condition (e) holds.
Notice that if u = 1 and  = 0 then the enclosing problem is equivalent to the
existence of a 5{CS of mKv+1, and Conditions Lemma 2.1.2(a-c) imply Conditions
Theorem 1.2.2(a-c) with  replaced by m and v by v + 1. So this case is already
solved.
Section 2.2 addresses the cases when v = 10k + ? for ?2 Z10 and m is small.
We take the results from Section 2.2 and use them to prove our main result of this
chapter, Theorem 2.3.2, in Section 2.3.
2.2 The small cases when u = 1
We now create the tools necessary to solve the problem of enclosing a  -fold
5{CS of order v into a ( + m)-fold 5{CS of order v + u for all m > 0 and u = 1.
For the remainder of this chapter, let u = 1 with U = f1g being the vertex set of
V(Kv+u)nV(Kv).
10
Recall that for any cycle c = (c1;:::;cx) with entries in Zv, we de ne c + j =
(c1 +j;:::;cx +j), reducing the sums modulo v, and de ne C +j =fc+jjc2Cg.
For any D f1;:::;bv=2cg; let Gv(D) be the graph with vertex set Zv and edge set
ffi;jgjdv(i;j)2Dg. Let D(v) =f1;:::;bv=2cg.
Since we consider multigraphs in this dissertation, we need to introduce the
multiset D? de ned to contain ? copies of each element in D; then Gv(D(v)?) contains
? copies of each edge with di erence in D(v). When v and ? are both even, it is useful
to let D(v) ? be formed from D(v)? by removing ?=2 copies of the di erence v=2. This
is very important notation because then the edge set of ?Kv = Gv(D(v) ?) can usefully
be de ned as
E(?Kv) =ff0;dg+jjd2D(v) ?;j2Zvg
since each occurrence of v=22D(v) ? gives rise to two copies of the edgefx;x+v=2g
in E(Gv(D(v)?)) (namely when j = x and when d = x + v=2). Another multiset
convention we use is that for any positive integer t, tD is the multiset containing t
copies of each element of D.
The following is a critical result, being used hand in hand with the observation
that there exists a 5{CS of Gv(fv=5g) and of Gv(f2v=5g) when v 0 (mod 5), since
each component is a 5-cycle. So throughout this dissertation, when v 0 (mod 5) let
(Zv; 1), (Zv; 2) and (Zv; 3) be a 5{CS of Gv(fv=5g);Gv(f2v=5g), and Gv(f1;2;3g)
respectively.
Lemma 2.2.1. If 5 divides v, then there exists a 5{CS of Gv(f1;2;3g).
Proof. (Zv;f(0;1;4;5;2)+5j;(3;4;6;8;5)+5j;(2;3;6;7;4)+5jjj2Zv=5g) is a 5{CS
of Gv(f1;2;3g) since for each d 2f1;2;3g, the edges of di erences d in the cycles
(0;1;4;5;2), (3;4;6;8;5), and (2;3;6;7;4) are of the form fx;x+dg where each x is
distinct modulo 5.
The next lemma points to our main approach to proving Theorem 2.3.2, showing
that the number of 5-cycles to be de ned on the originalv vertices during the enclosing
is divisible by v. The parameter  will play a critical role throughout the rest of the
dissertation.
Lemma 2.2.2. Suppose there exists a 5{CS (V[U;B) of ( +m)Kv+1  Kv where
U = f1g. Then the number of edges in 5-cycles avoiding the vertex 1 is  v =
(m(v 1)=2 3( +m)=2)v.
Proof. There are mv(v 1)=2 edges in mKv, 3v( +m)=2 of which occur in 5-cycles
in B, that contain the vertex 1. So
mv(v 1)2  3v( +m)2 =
 m(v 1)
2  
3( +m)
2
 
v =  v (2.1)
is the number of edges in 5-cycles in B with all vertices in V.
11
An integer x is said to be  (v;m)-admissible if there exists an integer  for which
v is  -admissible and for which Lemma 2.1.2(c) is satis ed by m;v and  such that
 = m(v 1)=2 3( + m)=2 = x. (During the enclosing process, we usually  x v
and m to consider all possible values of  . Each such value of  has an associated
value of  , and the set of all such  values are the  (v;m)-admissible integers.) The
next lemma will be used when constructing the 5-cycles in the enclosing that contain
the vertex 1; each such 5-cycle uses exactly 3 edges joining vertices in mKv.
Lemma 2.2.3. Suppose that the Conditions Lemma 2.1.2(a-c) are satis ed and that
there exists a 5{CS of  Kv. Then if v is odd, or if v is even and m is even, 3 divides
m(v 1)=2  . If v is even and m is odd then 3 divides m(v 1)=2   3=2.
Proof. If v is odd then by Lemma 2.1.2(a),  +m is even. If v is even then  is even
by Theorem 1.2.2(b) (since there exists a 5{CS of  Kv). Thus if v is odd or if v is
even and m is even, then 3 divides m(v 1)=2  . Now suppose v is even and m
is odd. Then m(v 1)=2   3=2 = 3( + m 1)=2 is clearly divisible by 3, and
otherwise m(v 1)=2  = 3( +m)=2 is divisible by 3.
The following result will be useful to refer to throughout the subsequent con-
structions. Note that it implies that all  (v;m)-admissible integers are non-negative.
Lemma 2.2.4. Suppose u = 1. Condition (c) Lemma 2.1.2 is equivalent to   0.
Proof. Clearly m(v 1) 3( +m) if and only if m(v 1)=2 3( +m)=2 0.
The following lemma is constantly used in subsequent constructions.
Lemma 2.2.5. If 0  d1  d2  d3  v=2, d1 < d3, and d2 < v=2 then c =
(d1;0;d3;d2 +d3;1i) is a 5-cycle in Kv+u with vertex set Zv[f10;11;:::;1u 1g.
Proof. The 5 vertices in c are clearly distinct.
For all positive integers v and m, let  max(v;m) and  min(v;m) be the largest
and smallest integers for  respectively satisfying Conditions (a-c) in Lemma 2.1.2
with u = 1.
It is probably no surprise that many larger enclosings can be found by combining
two smaller enclosings. This observation is the basis of the inductive proof of Theo-
rem 2.3.2. So now we focus on  nding enclosings that cannot be found using this ap-
proach. This involves the cases where m = 1, and where  2f max(v;m); min(v;m)g
for some small values of m.
Lemma 2.2.6. Let v = 10k and assume conditions (a-c) of Lemma 2.1.2 are satis ed.
If m = 1, or if m2f2;3gand  =  max(v;m), then any 5{CS of  Kv can be enclosed
in a 5{CS of ( +m)Kv+1.
12
Proof. Let (V = Z10k;A) be a 5{CS of  Kv and let U =f1g. Since (V;A) exists,  
is even by Lemma 2.1.2(a).
First, note that there aremv(v 1)=2 edges inmKv, 3v( +m)=2 of which must be
placed in 5-cycles that contain the vertex1. So we de ne  v as in Equation (2.1) of
Lemma 2.2.2 to be the number of edges we must place in 5-cycles contained completely
in mKv. Since this is divisible by v, our plan is to create 5-cycles in mKv using all
the edges of  specially chosen di erences. Note that   0 by Lemma 2.2.4.
Suppose m = 1. Then mKv = Kv contains edges of each di erence in D(v) =
f1;:::;5kg. To explicitly describe the 5-cycles containing all the edges of  di erences
(yet to be chosen), we make use of Skolem and hooked Skolem sequences.
Let P0 =f(x0i;y0i)j1 i k 1;y0i >x0igbe the collection of pairs from a Skolem
sequence or hooked Skolem sequence of order k 1 from the setf1;2;:::;2k 2gor
f1;2;:::;2k 1gif k 1 0 or 1 (mod 4) or k 1 2 or 3 (mod 4) respectively (see
Theorems 1.1.1 and 1.1.2). So each element inf1;2;:::;2k 2gorf1;2;:::;2k 1gn
f2k 2goccurs in exactly one element of P0. If k 1 0 or 1 (mod 4) then name the
Skolem pairs so that (x0k 1;y0k 1) contains 2k 3. If k 1 2 or 3 (mod 4) then name
the hooked Skolem pairs so that x0k 1 = 1. Now de ne P =f(xi = x0i+3;yi = y0i+3)j
(x0i;y0i) 2 P0g when using Skolem sequences and P = f(xi = x0i + 2;yi = y0i + 2) j
(x0i;y0i) 2 P0g when using hooked Skolem sequences; notice that 2k 2fxk 1;yk 1g
and xk 1 = 3 respectively. Then form the set of 5-cycles
B =
n
(0;xi;xi yi;3k;5k + 1 +i) +jj1 i 
j 
5
k
;j2Zv;(xi;yi)2P
o
[E 
where computations are done modulo v, and where
E =
8
>>>
>>>
<
>>>
>>>
:
? if   0 (mod 5);
 2 if   1 (mod 5);
 1[ 2 if   2 (mod 5);
 3 if   3 (mod 5);
 2[ 3 if   4 (mod 5):
(2.2)
Table 2.1 lists the di erences of edges that occur in cycles in BnE when  is as big
as possible (from Lemma 2.2.2 we see this happens when  is as small as possible and
not 0, namely when  = 2, and so when  = 5k 5).
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 4;5;:::;2k+1 3;4;:::;2k 1;2k+1
fxi yi;3kg 3k+1;3k+2;:::;4k 1 3k+1;3k+2;:::;4k 1
f3k;5k+1+ig 2k+2;2k+3;:::;3k 2k+2;2k+3;:::;3k
f5k+1+i;0g 4k;4k+1;:::;5k 2 4k;4k+1;:::;5k 2
Table 2.1: Di erences of edges in BnE when  is as large as possible
Notice that when  is as large as possible:
13
(i) if k 1 0 or 1 (mod 4), then the 5-cycle in BnE de ned when i =b =5c=
k 1 contains edges of di erences 2k = v=5 (since 2k 2 fxk 1;yk 1g) and
4k = 2v=5, and
(ii) if k 1 2 or 3 (mod 4), then the 5-cycle in BnE de ned when i =b =5c=
k 1 contains edges of di erences 3 (since xk 1 = 3) and 4k = 2v=5.
Fortunately, when  is as large as possible,   0 (mod 5) since  = 2, so E = ?.
If  is smaller, then (i)-(ii) from above shows that the edges of di erences 1;2;3;v=5
and 2v=5 do not occur in 5-cycles in BnE (since then i<k 1), hence such edges
are available to form the 5-cycles in E .
There are (m(v 1)=2  )v = (m(v 1)=2   3=2)v+3v=2 edges remaining in
mKv which do not occur in a 5-cycle in B. These remaining edges can be partitioned
by the corresponding remaining di erences as follows. Using Lemma 2.2.3 we know
that m(v 1)=2   3=2 0 (mod 3), so we can form a partition P0 of the remaining
di erences into one setfd;v=2gof size 2 with d6= v=2, and (m(v 1)=2   3=2)=3
sets of size 3. (Note that there are 3v=2 edges with di erences in fd;v=2g). De ne
C =f(d1;0;d3;d2 +d3;1) +jjfd1;d2;d3g2P0 with d1  d2  d3;j2Zvg (2.3)
[
n 
d;0;v2;v2 +d;1
 
+jj
nv
2;d
o
2P0;j2Zv=2
o
:
Since m = 1 the di erences are distinct, so by Lemma 2.2.5 the elements of C are
indeed 5-cycles. Then (Z10k[f1g;A[B[C) is a 5{CS(v + 1; + 1) that encloses
the given 5{CS (Zv;A).
Now suppose m = 2 and  =  max(v;m). By Lemma 2.1.2(c),   m(v 1)=3 
m = m(v 4)=3. Because  must also be even,  max(v;2) = 2b(v 4)=3c. So since
m = 2 and  =  max(v;2), it follows that
 = v 4 3b(v 4)=3c= 0;1; or 2 if k 1;2; or 0 (mod 3) respectively.
So de ne B = ?;  1 or  1 [  2 if k  1;2; or 0 (mod 3) respectively, and let
 = ?;fv=5g, orfv=5;2v=5gwhen  = 0;1; or 2 respectively. Using Lemma 2.2.3 we
know that jD(v) 2n j 0 (mod 3), so we can form a partition P0 of the remaining
di erences of D(v) 2n into sets of size 3. Then de ne
C =f(d1;0;d3;d2 +d3;1) +jjfd1;d2;d3g2P0 with d1  d2  d3;j2Zvg: (2.4)
By Lemma 2.2.5 the elements of C are indeed 5-cycles. Therefore, (Z10k[f1g;A[
B[C) is a 5{CS(v + 1; + 2) with  =  max(v;2).
Finally, suppose m = 3 and  =  max(v;3) = v 4. It follows that  = 0 for all
k. So we de ne B = ?. Using Lemma 2.2.3 we know that 3(v 1)=2   3=2 0
(mod 3), so we can form a partition P0 of D(v) 2[D(v) into one set fd;v=2g of size
2 with d 6= v=2, and (3(v 1)=2   3=2)=3 sets (not multisets; this is easy to
do) of size 3. De ne C as in Equation (2.3). By Lemma 2.2.5 the elements of C are
indeed 5-cycles. Therefore, (Z10k[f1g;A[B[C) is a 5{CS(v + 1; + 3) with
 =  max(v;3) that encloses the given 5{CS (Zv;A).
14
The proof of the next three results follows that of Lemma 2.2.6 closely but are
su ciently di erent and important that they are described in detail.
Lemma 2.2.7. Let v = 10k + 5 and assume conditions (a-c) of Lemma 2.1.2 are
satis ed. If m = 1 or if m2f2;3g and  =  max(v;m), then any 5{CS of  Kv can
be enclosed in a 5{CS of ( +m)Kv+1.
Proof. Let (V = Z10k+5;A) be a 5{CS of  Kv and let U =f1g. Since (V;A) exists,
 +m is even by Lemma 2.1.2(a).
By Lemma 2.2.2 the number of edges we must place in 5-cycles contained com-
pletely in mKv is  v as de ned by Equation (2.1). Since this is divisible by v, our plan
is again to create 5-cycles in mKv using all the edges of  specially chosen di erences.
By Lemma 2.2.4,   0.
Suppose m = 1. Then mKv = Kv contains edges of each di erence in D(v) =
f1;:::;5k+2g. As in Lemma 2.2.6, we will use Skolem and hooked Skolem sequences
to explicitly describe the 5-cycles containing all the edges of  di erences (yet to be
chosen).
We begin with the case where k 3. Let P0 =f(x0i;y0i)j1 i k 2;y0i >x0ig
be the collection of pairs from a Skolem sequence or hooked Skolem sequence of order
k 2 from the set f1;2;:::;2k 4g or f1;2;:::;2k 3g if k 2  0 or 1 (mod 4)
or k 2 2 or 3 (mod 4) respectively (this exists by Theorems 1.1.1 and 1.1.2 since
k 3). So each element inf1;2;:::;2k 4gorf1;2;:::;2k 3gnf2k 4goccurs in
exactly one element of P0. Now de ne P =f(xi = x0i + 3;yi = y0i + 3)j(x0i;y0i)2P0g.
Then with
ci =
8
><
>:
(0;xi;xi yi;4k + 2;7k + 3 i) if 1 i b =5c 1;
(0;2k;6k + 1;8k + 3;5k + 2) if i =b =5c and k 2 0;1 (mod 4), and
(0;2k 1;6k;8k + 2;5k + 1) if i =b =5c and k 2 2;3 (mod 4),
we form the set of 5-cycles B = fci +jj1 i b =5c;j2Zvg[E where the
computations are done modulo v, k 3, and where E is de ned as in Equation (2.2).
Table 2.2 lists the di erences of edges that occur in cycles in BnE if  is as big
as possible (from Lemma 2.2.2 this happens when  is as small as possible, namely
when  = 1 and so  = 5k 1). Fortunately, the edges of di erences 1;2;3;v=5 and
Edges Di erenceswhenk 2 0;1 (mod 4) Di erenceswhenk 2 2;3 (mod 4)
f0;xig;fxi;xi yig 4;5;:::;2k 1 4;5;:::;2k 2;2k
fxi yi;4k+2g 4k+3;4k+4;:::;5k 4k+3;4k+4;:::;5k
f4k+2;7k+3 ig 2k+3;2k+4;:::;3k 2k+3;2k+4;:::;3k
f7k+3 i;0g 3k+3;3k+4;:::;4k 3k+3;3k+4;:::;4k
ci ifi=b =5c 2k;2k+2;3k+1;4k+1;5k+2 2k 1;2k+2;3k+1;4k+1;5k+1
Table 2.2: Di erences of edges in BnE when  is as large as possible
2v=5 do not occur in 5-cycles in BnE , hence are available to form the 5-cycles in
15
E .
By Lemma 2.2.3 we know that m(v 1)=2   0 (mod 3), and so the edges
not occurring in a 5-cycle in B can be partitioned by the m(v 1)=2  remaining
di erences. Form a partition P0 of these di erences into ((v 1)=2  )=3 sets of
size 3. De ne C as in Equation (2.4). Notice that by Lemma 2.2.5 the edges in
ffd1;0g;f0;d3g;fd3;d3 + d2gg induce a 3-path in Kv. So the elements of C are all
5-cycles. Then, (Z10k+5[f1g;A[B[C) is a 5{CS(v + 1; + 1) for v = 10k + 5.
It is left to provide the required sets of 5-cycles when m = 1 and k = 0;1 and 2.
Notice that if k = 0, then   0, so this case need not be considered. Suppose k = 1,
so v = 15. Then by Lemma 2.1.2(c) and since  +m =  + 1 is even, it follows that
 2f1;3g. So ( ; )2f(1;4);(3;1)g. De ne
C =
(
f(d1;0;d3;d2 +d3;1) +jj(d1;d2;d3) = (4;5;7);j2Zvg if  = 1, and
f(d1;0;d3;d2 +d3;1) +jj(d1;d2;d3)2f(1;2;3);(4;5;7)g;j2Zvg if  = 3,
and B = E . Then (Z15 [f1g;A[B[C) is a 5{CS(16; + 1). If k = 2, so
v = 25, then   7; so  = 1;3;5; or 7 and  = 9;6;3; or 0 respectively. De ne
C=
8>
><
>>:
f(d1;0;d3;d2 +d3;1)+jj(d1;d2;d3) =(5;7;9);j2Zvg if =1,
f(d1;0;d3;d2 +d3;1)+jj(d1;d2;d3)2f(1;2;3);(5;7;9)g;j2Zvg if =3,
f(d1;0;d3;d2 +d3;1)+jj(d1;d2;d3)2f(4;5;6);(7;8;9);(10;11;12)g;j2Zvg if =5,and
f(d1;0;d3;d2 +d3;1)+jj(d1;d2;d3)2f(x;x+1;x+2)jx2f1;4;7;10gg;j2Zvg if =7;
and
B =
(
f(0;4;10;22;8) +jjj2Z25g[E if  2f1;3g,
E if  2f5;7g.
Then (Z25[f1g;A[B[C) is a 5{CS(26; + 1).
Now suppose m = 2 and  =  max(v;m). By Lemma 2.1.2(c),   m(v 4)=3,
and by Lemma 2.1.2(a),  is even. So  max(v;2) = 2b(v 4)=3c, and
 = v 4 3b(v 4)=3c= 0;1; or 2 if k 2;0; or 1 (mod 3) respectively.
So de ne B = ?;  1, or  1 [ 2 if k  2; 0; or 1 (mod 3) respectively, and let
 = ?;fv=5g, or fv=5;2v=5g when  = 0;5; or 10 respectively. Using Lemma 2.2.3
we know that jD(v)2n j 0 (mod 3), we can form a partition P0 of D(v)2n into
sets of size 3. Then de ne C as in Equation (2.4), again noting that the elements of C
are 5-cycles by Lemma 2.2.5. Therefore, (Zv[f1g;A[B[C) is a 5{CS(v+1; +2)
with  =  max(v;2).
Finally, suppose m = 3 and  =  max(v;3) = v 4. It follows that  = 0 for all
k. So we de ne B = ?. Since 3(v 1)=2   0 (mod 3) by Lemma 2.2.3, form a
partition P0 of D(v)3 into (3(v 1)=2  )=3 sets of size 3 (again it is easy to ensure
that no element of P0 contains a di erence three times). De ne C as in Equation (2.4).
Then elements of C are all 5-cycles by Lemma 2.2.5. Then, (Zv[f1g;A[B[C)
is a 5{CS(v + 1; + 3) with  =  max(v;3).
16
Lemma 2.2.8. Let v 2 or 3 (mod 5) and assume conditions (a-c) of Lemma 2.1.2
are satis ed. If m = 5 or if m2f10;15g and  =  max(v;m), then any 5{CS of  Kv
can be enclosed in a 5{CS of ( +m)Kv+1.
Proof. Let (V = Zv;A) be a 5{CS of  Kv where v = 10k+? for some ?2f2;3;7;8g
and let U =f1g. By Theorem 1.2.2(c), v 7. In each case it is shown that (V;A)
can be enclosed in a 5{CS (V [f1g;A[B[C) of ( +m)Kv+1 by de ning B and
C.
First, note that by Lemma 2.2.2, the number of edges we must place in 5-cycles
contained completely in mKv is  v as de ned in Equation (2.1). Since this is divisible
by v, our plan is to create 5-cycles in mKv using the edges of  specially chosen
di erences. Notice that   0 by Lemma 2.2.4.
Suppose m = 5. Then   0 or 5 (mod 10) if v 2;8 or 3;7 (mod 10) respec-
tively. Note that when  is as small as possible  is as large as possible: that is,
when  = 10,  = 25k 20 for ? = 2 and  = 25k 5 for ? = 8; and when  = 5,
 = 25k 10 for ? = 3 and  = 25k for ? = 7. The edges of 5Kv are considered to be
E(Gv(D(v) 4))[E(Gv(D(v))) when v is even and E(Gv(D(v)5)) when v is odd. For
v even, care must be taken with edges of di erence v=2 in E(Gv(D(v))) since there
are only v=2 of them. As in Lemma 2.2.6, we will use Skolem and hooked Skolem
sequences to explicitly describe the 5-cycles containing all the edges of  di erences
(yet to be chosen). To form the set of cycles B, we consider the cases ?2f2;3g and
?2f7;8g in turn.
Suppose ? 2 f2;3g. Let P0 = f(x0i;y0i) j 1  i  k 1;y0i > x0ig be the
collection of pairs from a Skolem sequence or hooked Skolem sequence of order
k 1 for the set f1;2;:::;2k 2g or f1;2;:::;2k 1g, if k 1  0 or 1 (mod 4)
or k 1  2 or 3 (mod 4) respectively. So each element in f1;2;:::;2k 2g
or f1;2;:::;2k 1gnf2k 2g occurs in exactly one element of P0. Now de ne
P =f(xi = x0i + 2;yi = y0i + 2)j(x0i;y0i)2P0g. Then we form the set of 5-cycles
B0 =fci 1 = (0;xi;xi yi;3k + 1;5k +?+i)j1 i k 1;(xi;yi)2Pg
where computations are done modulo v. We also need to de ne E when  is as large
as possible. Let
E0 =
8
>>><
>>>
:
(0;2k + 1;2k;2k 1;5k) if k 1 0;1 (mod 4) and ? = 2,
(0;2k;5k + 1;5k 1;5k) if k 1 2;3 (mod 4) and ? = 2,S
3
i=1f(0;2;2k + 3;1;5k + 1)g if k 1 0;1 (mod 4) and ? = 3, andS
3
i=1f(0;1;2k + 1;7k + 2;5k)g if k 1 2;3 (mod 4) and ? = 3.
Notice that when ? = 3, E0 is a multiset that contains 3 copies of the 5-cycle.
Table 2.3 lists the di erences of the edges that occur in cycles in B0 when  is as
large as possible and in E0 . Let B00 = 5B0 be the multiset consisting of 5 copies of
each 5-cycle in B0. When  is as large as possible,jB00[E0 j= 5k 4 or 5k 2 if ? = 2
or 3 respectively, sojB00[E0 j=  =5. If  is not as big as possible then   25k 25
17
in all cases (since as  increases by 10,  decreases by 15) so jB00j = 5k 5   =5.
Therefore we de ne
B =
(S
j2Zv(B
00 +j)[fc+jjc2E0
 ;j2Zvg if  is as large as possible, and
fci +jji2Z =5;ci (mod k 1) 2B00;j2Zvg otherwise.
Edges Di erenceswhenk 1 0;1 (mod4) Di erenceswhenk 1 2;3 (mod4)
f0;xig;fxi;xi yig 3;4;:::;2k 3;4;:::;2k 1;2k+1
fxi yi;3k+1g 3k+2;3k+3;:::;4k 3k+2;3k+3;:::;4k
f3k+1;5k+?+ig 2k+?;2k+1+?;:::;3k 2+? 2k+?;2k+1+?;:::;3k 2+?
f5k+?+i;0g 5k 1;5k 2;:::;4k+1 5k 1;5k 2;:::;4k+1
E if =25k 20;?=2 1;1;2k+1;3k+1;5k 1;2;2k;3k+1;5k
E0 if =25k 10;?=3 2;2;2;2k+1;2k+1;2k+1;2k+2;2k+2; 1;1;1;2k;2k;2k;2k+2;2k+2;2k+2;2k+2;5k;5k;5k;5k+1;5k+1;5k+1 5k;5k;5k;5k+1;5k+1;5k+1
Table 2.3: Di erences of edges inB0 andE when is as large as possible for?2f2;3g
and m = 5
Now suppose ?2f7;8g. Let P =f(xi;yi)j1 i k;yi >xig be the collection
of pairs from a Skolem sequence or hooked Skolem sequence of order k for the set
f1;2;:::;2kg or f1;2;:::;2k + 1g, if k  0 or 1 (mod 4) or k  2 or 3 (mod 4)
respectively. So each element in f1;2;:::;2kg or f1;2;:::;2k + 1gnf2kg occurs in
exactly one element of P. Form the set of 5-cycles B0 = fci 1 = (0;xi;xi yi;3k +
2;5k + 4 + i) j 1  i  k;(xi;yi) 2 Pg where computations are done modulo v.
Let B00 = 5B0 be the multiset consisting of 5 copies of each 5-cycle in B0. De ne
B =fci +jji2Z =5;ci (mod k) 2B00;j2Zvg. Table 2.4 lists the di erences of edges
that occur in cycles in B0.
Edges Di erenceswhenk 0;1 (mod 4) Di erenceswhenk 2;3 (mod 4)
f0;xig;fxi;xi yig 1;2;:::;2k 1;2;:::;2k 1;2k+1
fxi yi;3k+2g 3k+3;3k+4;:::;4k+2 3k+3;3k+4;:::;4k+2
f3k+2;5k+4+ig 2k+3;2k+4;:::;3k+2 2k+3;2k+4;:::;3k+2
f5k+4+i;0g 5k 5+?;5k 6+?;:::;4k 4+? 5k 5+?;5k 6+?;:::;4k 4+?
Table 2.4: Di erences of edges in B0 when  is as large as possible for ?2f7;8gand
m = 5
To de ne C, suppose ?2f2;3;7;8g. Notice that by Lemma 2.2.3 we know that
3 divides (5(v 1)=2  ) when v is odd and 3 divides (5(v 1)=2   3=2) when v is
even. Form a partition P0 of the di erences not used to form B into (5(v 1)=2  )=3
sets of size 3 when v is odd, and when v is even into (5(v 1)=2   3=2)=3 sets of size
3 (not multisets) and one setfd;v=2gof size 2 with d6= v=2, when v is even; it is easy
to ensure that d1 <d3 and d2 <v=2. Form the set C of 5-cycles as in Equation (2.4)
when v is odd and as in Equation (2.3) when v is even. By Lemma 2.2.5 the elements
of C are 5-cycles. Thus (Zv[f1g;A[B[C) is a 5{CS(v + 1; + 5) with v 2 or
18
3 (mod 5).
Now suppose m = 10 and  =  max(v;m). By Table 1.1 since v + 1 is ( + m)-
admissible  + m 0 (mod 10) and so   0 (mod 10). It follows that  = 5(v 
1) 3( + m)=2  0 (mod 5). It also follows that for any  xed values of v and
m, v-admissible values of  di er by multiples of 10, so  (v;m)-admissible integers
di er by multiples of 15 (by Lemma 2.2.2). So when  =  max(v;m) it follows from
Lemma 2.2.4 that  2f0;5;10g. (There is no need to be speci c for this proof,
but the reader may be interested to know that when m = 10 and  =  max(v;10):
 = 0;5; or 10 if k  2;0; or 1 (mod 3) respectively when v  2 or 8 (mod 10) ;
 = 0;5; or 10 if k  1;2; or 0 (mod 3) respectively when v  3 (mod 10); and
 = 0;5;10 if k 0;1; or 2 (mod 3) respectively when v 7 (mod 10).) De ne
B =
8
><
>:
? if  = 0,
f(0;1;4;2;6) +jjj2Zvg if  = 5, and
f(0;1;4;2;6) +jjj2Zvg[f(0;1;4;2;6) +jjj2Zvg if  = 10,
and let  = ?;f1;2;3;4;6g, or f1;1;2;2;3;3;4;4;6;6g when  = 0;5; or 10 respec-
tively. By Lemma 2.2.3,jD(v)10n j 0 (mod 3) when v is odd andj(D(v) 10)n j 0
(mod 3) when v is even. Form a partition P0 of D(v)10n when v is odd and D(v) 10n 
when v is even into (m(v 1)=2  )=3 sets of size 3. Then de ne the 5-cycles in
C as in Equation (2.4) (it is easy to ensure that d1 < d3 and d2 < v=2, and so by
Lemma 2.2.5 the elements of C are all 5-cycles). Thus (Zv[f1g;A[B[C) is a
5{CS(v + 1; + 10) for v 2 or 3 (mod 5) and  =  max(v;m).
Let m = 15 and  =  max(v;15) = 5(v 4). It follows that  = 0 for all k. So
de ne B = ?. Using Lemma 2.2.3 we know that 15(v 1)=2   0 (mod 3) when v
is odd and 15(v 1)=2   3=2 0 (mod 3) when v is even, so we can form a partition
P0 of D(v)15 into (15(v 1)=2  )=3 sets of size 3 when v is odd and when v is even
form a partition P0 of D(v) 14[D(v) into (15(v 1)=2   3=2)=3 sets of size 3 and
one set fd;v=2g of size 2 with d6= v=2. De ne C as in Equation (2.4) when v is odd
and as in Equation (2.3) when v is even (it is easy to ensure d1 <d3 and d2 <v=2 and
so by Lemma 2.2.5 the elements of C are all 5-cycles). Therefore (Zv[f1g;A[B[C)
is a 5{CS(v + 1; + 15) for v 2 or 3 (mod 5) with  =  max(v;15).
Lemma 2.2.9. Let v 1 or 4 (mod 5) and suppose conditions (a-c) of Lemma 2.1.2
are satis ed. If m = 1, if m2f2;3g and  =  max(v;m), if m = 3,  = 2, and v 6
(mod 10), or if m2f2;3;:::;9g,  + m = 10 and v 1 (mod 10), then any 5{CS
of  Kv can be enclosed in a 5{CS of ( +m)Kv+1.
Proof. Let (V = Zv;A) be a 5{CS of  Kv where v = 10k + ?  6 for some ? 2
f1;4;6;9gand let U =f1g. First, note that by Lemma 2.2.2, the number of edges we
must place in 5-cycles contained completely in mKv is  v as de ned in Equation (2.1)
Since this is divisible by v and   0 by Lemma 2.2.4, our plan is to create 5-cycles
in mKv using the edges of  specially chosen di erences.
Suppose m = 1. Then mKv = Kv contains edges of each di erence in D(v).
19
As in Lemma 2.2.6, we will use Skolem and hooked Skolem sequences to explicitly
describe the 5-cycles containing all the edges of  di erences (yet to be chosen).
For ? 2 f4;6;9g, let P = f(xi;yi) j 1  i  k;yi > xig be the collection
of pairs from a Skolem sequence or hooked Skolem sequence of order k for the set
f1;2;:::;2kg or f1;2;:::;2k + 1g, if k  0 or 1 (mod 4) or k  2 or 3 (mod 4)
respectively. So each element in f1;2;:::;2kg or f1;2;:::;2k + 1gnf2kg occurs in
exactly one element of P. For ? = 1, let P = f(xi;yi) j 1  i k 1;yi > xig be
the collection of pairs from a Skolem sequence or hooked Skolem sequence of order
k 1 for the set f1;2;:::;2k 2g or f1;2;:::;2k 1g, if k 1  0 or 1 (mod 4)
or k 1  2 or 3 (mod 4) respectively. So each element in f1;2;:::;2k 2g or
f1;2;:::;2k 1gnf2k 2goccurs in exactly one element of P for ? = 1. Then with
ci =
8
><
>:
(0;xi;xi yi;3k;5k + 1 +i) if ? = 1,
(0;xi;xi yi;3k 1 +?=2;5k +?=2 +i) if ?2f4;6g, and
(0;xi;xi yi;3k + 3;5k + 5 +i) if ? = 9,
form the set of 5-cycles B = fci + j j 1  i b =5c;j 2 Zv;(xi;yi) 2Pg where
computations are done modulo v. Tables 2.5, 2.6, and 2.7 list the di erences of the
edges that occur in cycles in B when  is as big as possible (that is, when  is as
small as possible).
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 1;2;:::;2k 2 1;2;:::;2k 3;2k 1
fxi yi;3kg 3k+1;3k+2;:::;4k 1 3k+1;3k+2;:::;4k 1
f3k;5k+1+ig 2k+2;2k+3;:::;3k 2k+2;2k+3;:::;3k
f5k+1+i;0g 5k 1;5k 2;:::;4k+1 5k 1;5k 2;:::;4k+1
Table 2.5: Di erences of edges in B when  is as large as possible for ? = 1 and
m = 1
Edges Di erenceswhenk 0;1 (mod4) Di erenceswhenk 2;3 (mod4)
f0;xig;fxi;xi yig 1;2;:::;2k 1;2;:::;2k 1;2k+1
fxi yi;3k 1+?=2g 3k+?=2;3k+1+?=2;:::;4k 1+?=2 3k+?=2;3k+1+?=2;:::;4k 1+?=2
f3k 1+?=2;5k+?=2+ig 2k+2;2k+3;:::;3k+1 2k+2;2k+3;:::;3k+1
f5k+?=2+i;0g 5k 1+?=2;5k 2+?=2;:::;4k+?=2 5k 1+?=2;5k 2+?=2;:::;4k+?=2
Table 2.6: Di erences of edges in B when  is as large as possible for ?2f4;6g and
m = 1
Notice that by Lemma 2.2.3 we know that 3 divides ((v 1)=2  ) when v is
odd and 3 divides ((v 1)=2   3=2) when v is even. Form a partition P0 of the
di erences not used to form B into ((v 1)=2  )=3 sets (not multisets) of size 3
when v is odd and when v is even ((v 1)=2   3=2)=3 sets (not multisets) of size
3 and one set fd;v=2g of size 2 with d6= v=2. We form the set C of 5-cycles as in
20
Edges Di erenceswhenk 0;1 (mod 4) Di erenceswhenk 2;3 (mod 4)
f0;xig;fxi;xi yig 1;2;:::;2k 1;2;:::;2k 1;2k+1
fxi yi;3k+3g 3k+4;3k+5;:::;4k+3 3k+4;3k+5;:::;4k+3
f3k+3;5k+5 +ig 2k+3;2k+4;:::;3k+2 2k+3;2k+4;:::;3k+2
f5k+5 +i;0g 5k+3;5k+2;:::;4k+4 5k+3;5k+2;:::;4k+4
Table 2.7: Di erences of edges in B when  is as large as possible for ? = 9 and
m = 1
Equation (2.4) when v is odd and as in Equation (2.3) when v is even. Notice that
by Lemma 2.2.5 the elements of C are all 5-cycles. Thus (Zv[f1g;A[B[C) is a
5{CS(v + 1; + 1) for v 1 or 4 (mod 5).
Now suppose m = 2 and  =  max(v;2). By Table 1.1: if v 1 (mod 5) then
since v + 1 is ( + m)-admissible  + m 0 (mod 10) and so   8 (mod 10); and
if v  4 (mod 5) then v is  -admissible so   0 (mod 10) (and so  + m  2
(mod 10)). It follows that  = v 1 3( + 2)=2 0 (mod 5). It also follows that
for any  xed values of v and m, v-admissible values of  di er by multiples of 10,
so  (v;m)-admissible integers di er by multiples of 15 (by Lemma 2.2.2). So when
 =  max(v;m) it follows from Lemma 2.2.4 that  2f0;5;10g. (There is no need to
be more speci c for this proof, but the reader may be interested to know that when
m = 2 and  =  max(v;2):  = 0;5; or 10 if k 0;2; or 1 (mod 3) respectively when
v  1 or 4 (mod 10); and  = 0;5;10 if k  1;0; or 2 (mod 3) respectively when
v 6 or 9 (mod 10).) De ne
B =
8
><
>:
? if  = 0,
f(0;1;4;2;6) +jjj2Zvg if  = 5, and
f(0;1;4;2;6) +jjj2Zvg[f(0;1;4;2;6) +jjj2Zvg if  = 10,
and let  = ?;f1;2;3;4;6g, or f1;1;2;2;3;3;4;4;6;6g when  = 0;5; or 10 respec-
tively. Using Lemma 2.2.3 we know that 3 divides v 1  = jD(v)2 n j if v is
odd and jD(v) 2 n j if v is even. Form a partition P0 of D(v)2 n when v is odd
and D(v) 2 n if v is even into (m(v 1)=2  )=3 sets of size 3. Form the set C
of 5-cycles as in Equation (2.4). By Lemma 2.2.5 the elements of C are all 5-cycles.
Thus (Zv[f1g;A[B[C) is a 5{CS(v + 1; + 2) with v 1 or 4 (mod 5) and
 =  max(v;2).
Now suppose m = 3 and  =  max(v;3) = (v 4). It follows that  = 0 for all
k. So de ne B = ?. Using Lemma 2.2.3 we know that 3(v 1)=2   0 (mod 3)
when v is odd and 3(v 1)=2   3=2 0 (mod 3) when v is even, so we can form
a partition P0 of the di erences in D(v)3 into (3(v 1)=2  )=3 sets of size 3 when v
is odd and when v is even form a partition P0 of the di erences in D(v) 2[D(v) into
(3(v 1)=2   3=2)=3 sets of size 3 and one set fd;v=2g of size 2 with d6= v=2; it
is easy to ensure that each element of P0 does not contain the same element 3 times.
De ne C as in Equation (2.4) when v is odd and as in Equation (2.3) when v is even.
21
By Lemma 2.2.5 the elements of C are cycles. Therefore (Zv[f1g;A[B[C) is a
5{CS(v + 1; + 3) for v 1 or 4 (mod 5),  =  max(v;3).
Let v  1 (mod 10), m 2f2;3;:::;9g, and  = 10 m. Let P = f(xi;yi) j
1  i k 1;yi > xig be the collection of pairs from a Skolem sequence or hooked
Skolem sequence of order k 1 for the set f1;2;:::;2k 2g or f1;2;:::;2k 1g, if
k 1  0 or 1 (mod 4) or k 1  2 or 3 (mod 4) respectively. So each element in
f1;2;:::;2k 2gorf1;2;:::;2k 1gnf2k 2goccurs in exactly one element of P.
Form the set of 5-cycles
B0 =fci 1 = (0;xi;xi yi;3k;5k + 1 +i)j1 i k 1;(xi;yi)2Pg
where computations are done modulo v. Let B00 = mB0. Then B00 contains m(k 1)
5-cycles, but we needb =5c= mk 3 of them. Hence for m 4 we need m 3 more
5-cycles. Let c01 = (0;2k + 1;6k + 1;k + 1;5k + 1) and c02 = (0;2k;4k;6k;8k). Also
let E contain 1, 2, 3, 3, 4, and 4 copies of c01 if m = 4; 5, 6, 7 8, and 9 respectively
and contain 1, 1, and 2 copies of c02 if m = 7; 8, and 9 respectively. Notice that the
cycles in E contain at most m copies of edges of di erences not used in cycles in B00,
namely the di erences 2k;2k+ 1;4k and 5k (B00 is also missing the di erences 2k 1
or 2k 2). Then
B =  ci +jj1 i b =5c;ci (mod k 1) 2B00;j2Zv [fc0i +jjj2Zv;c0i2E g:
Notice that regardless of our choice of m,  +m = 10. So (m(v 1)=2  ) = 15
which is divisible by 3. So we use the elements not used to form B to form a partition
P0 of the di erences in D(v)m into  ve sets of size 3. De ne C as in Equation (2.4)
(it is easy to ensure d1 < d3 and d2 < v=2 so that by Lemma 2.2.5 the elements of
C are all 5-cycles). Therefore (Zv[f1g;A[B[C) is a 5{CS(v + 1; + m) with
m2f2;3;4;5;6;7;8;9g and  +m = 10.
Let v  6 (mod 10), m = 3 and  =  min(v;3) = 2. Let P = f(xi;yi) j 1  
i k;yi > xig be the collection of pairs from a Skolem sequence or hooked Skolem
sequence of order k for the set f1;2;:::;2kg or f1;2;:::;2k + 1g, if k  0 or 1
(mod 4) or k  2 or 3 (mod 4) respectively. So each element in f1;2;:::;2kg or
f1;2;:::;2k+ 1gnf2kgoccurs in exactly one element of P. Then we form the set of
5-cycles
B0 =f(0;xi;xi yi;3k + 2;5k + 3 +i)j1 i k;(xi;yi)2Pg
where computations are done modulo v. Let B00 = mB0. Notice that b =5c = 3k =
jB00j: So de ne B =fc+jjc2B00;j2Zvg:
By Lemma 2.2.3 we know that m(v 1)=2   3=2  0 (mod 3). So we use
the elements not used to form B to form a partition P0 into one set fd;v=2g of size
2 with d6= v=2 and two sets of size 3. De ne C as in Equation (2.4) (it is easy to
ensure that d1 < d3 and d2 < v=2 so that by Lemma 2.2.5 the elements of C are
all 5-cycles). Therefore (Zv[f1g;A[B[C) is a 5{CS(v + 1; + m) with v 6
(mod 10), m = 3, and  = 2.
22
Corollary 2.2.10. Assume conditions (a-c) of Lemma 2.1.2 are satis ed. Let v 2
or 3 (mod 5). If
 m = 5, or
 m2f10;15g and  =  max(v;m),
then any 5{CS of  Kv can be enclosed in a 5{CS of ( +m)Kv+1.
Let v6 2 or 3 (mod 5). If
 m = 1,
 m2f2;3g and  =  max(v;m),
 m = 3,  =  min(v;3) = 2, and v 6 (mod 10), or
 m2f2;3;:::;9g,  +m = 10 and v 1 (mod 10),
then any 5{CS of  Kv can be enclosed in a 5{CS of ( +m)Kv+1.
Proof. By Lemmas 2.2.6, 2.2.7, 2.2.8, and 2.2.9 the results follow.
2.3 The Main Theorem
The following lemma will be used to guarantee an important point made in the
main result.
Lemma 2.3.1. If m 0 (mod 3) then  max(v;m) = m(v 4)=3.
Proof. Recall that  max(v;m) is the largest integer satisfying conditions (a-c) of
Lemma 2.1.2. By Lemma 2.1.2(c),  max(v;m)  bm(v 4)=3c = m(v 4)=3. So
the result follows since clearly
m(v 1)=3 +m(v 1) = 4m(v 1)=3
is even and
mv(v 1)=2 +mv(v 1)=3 = 5mv(v 1)=6
is divisible by 5.
In the following theorem, Corollary 2.2.10 is used as a basis for an inductive
argument to prove our main result, settling the enclosing problem when u = 1.
Theorem 2.3.2. Let m 1. A 5{CS of  Kv can be enclosed in a 5{CS of ( +
m)Kv+1 if and only if
(a)  +mv 0 (mod 2),
(b) m v2 +v( +m) 0 (mod 5), and
(c) m(v 1) 3( +m).
23
Proof. The necessity follows from Lemma 2.1.2, so we now prove the su ciency.
Since we are assuming that a 5{CS of  Kv exists, it follows by Theorem 1.2.2 that
v 62f2;3;4g, and by condition (c) v 6= 1, so we know v  5. If  = 0 then by
condition (a) mv 0 (mod 2) and by condition (b) mv(v + 1)  0 (mod 5), so by
Theorem 1.2.2 there exists a 5{CS of mKv+1. So we can assume that   1.
Let (V;A) be a 5{CS of  Kv. If v  2 or 3 (mod 5) then from Table 1.1 it
follows that m 0 (mod 5); in any other case m can potentially take on any positive
integral value. To re ect this situation, let t(v) = 1 if v 6 2 or 3 (mod 5) and let
t(v) = 5 if v 2 or 3 (mod 5). We often simply write t instead of t(v) when the value
of v is clear. Let   max(v;m) be the second largest integer for  satisfying conditions
(a-c), and let  +min(v;m) be the second smallest integer for  satisfying conditions
(a-c).
We will  rst establish that for any given v, the di erence between consecutive
values of  that satisfy conditions (a) and (b) is a constant; call this di erence  di (v).
Secondly, for all m  1 and v  5, we settle the enclosing problem for both the
smallest and the largest values of  satisfying conditions (a-c). Finally for all m 1
and v  5 it will be shown that for all  satisfying conditions (a-c) there exist
non-negative integers  1,  2 with  1 +  2 =  and positive integers m1; m2 with
m1 + m2 = m such that for 1  i  2 there exists a 5{CS of  iKv that can be
enclosed in a 5{CS of ( i +mi)Kv+1. So the union of these two enclosings completes
the proof.
We turn to the  rst step of the proof. For any given v and m, the di erence
between consecutive values of  that satisfy conditions (a) and (b) is a constant;
namely  di (v;m). Also by conditions (a) and (b),  di (v;m)  0 (mod 2) and
v di (v;m)  0 (mod 5). Therefore v di (v;m)  0 (mod 10). Since  di (v;m)
must be the smallest such value, if v  0;5 (mod 10) then  di (v;m) = 2, and if
v6 0;5 (mod 10) then  di (v;m) = 10. Notice that if m1 6= m2 then  di (v;m1) =
 di (v;m2). Therefore we de ne  di (v) =  di (v;m).
We now turn to the second step in the proof; establishing the existence of the
enclosing for the smallest and largest values of  given v and m. Beginning with the
smallest value, Tables 2.8 and 2.9 list the values of  min(v;m), and were formed using
Theorem 1.2.2 and Corollary 2.2.10. Three cases are considered in turn based on the
value of v.
m
v (mod 10) 0 1 2 3 4 5 6 7 8 9
1 0 9 8 7 6 5 4 3 2 1
6 0 4 8 2 6 0 4 8 2 6
Table 2.8: Checking  min(v;m) when v  1 or 6 (mod 10): each cell contains
 min(v;m)
24
v (mod 10)
m 0 2 3 4 5 7 8 9
odd 0 0 5 0 1 5 0 5
even 0 0 0 0 0 0 0 0
Table 2.9: Checking  min(v;m) when v 6 1 or 6 (mod 10): each cell contains
 min(v;m)
Let v  1 (mod 10). The proof is by induction on m. By Corollary 2.2.10 if
m 10 then any 5{CS of  min(v;m)Kv can be enclosed in a 5{CS of ( min(v;m) +
m)Kv+1. Suppose that for some m  11 and for all s with 1  s < m, every
5{CS of  min(v;s)Kv can be enclosed in a 5{CS of ( min(v;s) + s)Kv+1. Then,
since m 11, by induction each 5{CS (V;B1) of  min(v;m 10)Kv can be enclosed
in a 5{CS (V [f1g;B01) of ( min(v;m 10) + (m 10))Kv+1 and there exists a
5{CS (V [f1g;B2) of 10Kv+1 (by Theorem 1.2.2). Since Table 2.8 shows that
 min(v;m 10) =  min(v;m), (V;B1) is a 5{CS of  min(v;m 10)Kv =  min(v;m)Kv
which is enclosed in a 5{CS (V [f1g;B01[B2) of
(( min(v;m 10) + (m 10)) + 10)Kv+1 = ( min(v;m) +m)Kv+1:
Let v  6 (mod 10). The proof is by induction on m. We  rst settle the
base cases where m  5. By Corollary 2.2.10, if m 2 f1;3g then any 5{CS of
 min(v;m)Kv can be enclosed in a 5{CS of ( min(v;m) + m)Kv+1. Hence a 5{CS of
 min(v;1)Kv can be enclosed in a 5{CS of ( min(v;1)+1)Kv+1. Since Table 2.8 shows
 min(v;1) +  min(v;1) = 4 + 4 =  min(v;2), the union of two copies of this enclosing
proves that any 5{CS of  min(v;2)Kv can be enclosed in a 5{CS of (( min(v;1) +
1) + ( min(v;1) + 1))Kv+1 = ( min(v;2) + 2)Kv+1. Similarly, since Table 2.8 shows
that  min(v;1) +  min(v;3) = 4 + 2 =  min(v;4), any 5{CS of  min(v;4)Kv can be
enclosed in a 5{CS of (( min(v;1) + 1) + ( min(v;3) + 3))Kv+1 = ( min(v;4) + 4)Kv+1.
If m = 5, by Table 2.8  min(v;5) = 0, so this enclosing has been established. Now
suppose that for some m 6 and for all s with 1 s<m, every 5{CS of  min(v;s)Kv
can be enclosed in a 5{CS of ( min(v;s) + s)Kv+1. Then, since m 6, by induction
each 5{CS (V;B1) of  min(v;m 5)Kv can be enclosed in a 5{CS (V [f1g;B01) of
( min(v;m 5)+(m 5))Kv+1, and by Theorem 1.2.2 there exists a 5{CS(V[f1g;B2)
of 5Kv+1. Since Table 2.8 shows that  min(v;m 5) =  min(v;m), (V;B1) is a 5{CS
of  min(v;m 5)Kv =  min(v;m)Kv which is enclosed in a 5{CS (V[f1g;B01[B2)
of
(( min(v;m 5) + (m 5)) + 5)Kv+1 = ( min(v;m) +m)Kv+1:
Let v 6 1 or 6 (mod 10). The proof is by induction on m. If m = t, m is
even (including the important small value m = 2t), or v is even, then by Table 2.9
 min(v;m) = 0, so this enclosing has been established. It is left to show we can form
an enclosing of a 5{CS of ( min(v;m) + m)Kv+1 if v is odd and m is odd. Suppose
that for some m 3t (recall that m 0 (mod t)) and for all s with 1  s < m,
25
every 5{CS of  min(v;s)Kv can be enclosed in a 5{CS of ( min(v;s) +s)Kv+1. Then,
since m  2t, each 5{CS (V;B1) of  min(v;m 2t)Kv can be enclosed in a 5{CS
(V;B01) of ( min(v;m 2t)+(m 2t))Kv+1 and by Theorem 1.2.2 there exists a 5{CS
(V [f1g;B2) of 2tKv+1. Since Table 2.9 shows that  min(v;m t) =  min(v;m),
(V;B1) is a 5{CS of ( min(v;m 2t))Kv =  min(v;m)Kv which is enclosed in a 5{CS
(V [f1g;B01[B2) of
(( min(v;m 2t) + (m 2t)) + 2t)Kv+1 = ( min(v;m) +m)Kv+1:
Now turning to the largest value of  , we will enclose a 5{CS of  max(v;m)Kv in a
5{CS of ( max(v;m)+m)Kv+1. The proof is by induction on the non-negative integer
i, where m = 3ti+ and  2ft;2t;3tg. In view of Corollary 2.2.10, the largest value
of  (i.e.  =  max(v;m)) when i = 0 has already been solved. Suppose that for some
i 1 and for allswith 0 s<iwe can enclose a 5{CS of max(v;3st+ )Kv in a 5{CS
of ( max(v;3st+ )+(3st+ ))Kv+1. By Lemma 2.3.1 it follows that max(v;m) = m(v 
4)=3 when m 0 (mod 3), and so  max(v;3(i 1)t + 3t) +  max(v; ) =  max(v;m).
In particular every 5{CS of  max(v;3t(i 1) + 3t)Kv can be enclosed in a 5{CS of
(( max(v;3t(i 1)+3t)+(3t(i 1)+3t))Kv+1 and every 5{CS of  max(v; )Kv can be
enclosed in a 5{CS of ( max(v; ) + )Kv+1. The union of the previous two enclosings
proves that any 5{CS of ( max(v;3t(i 1) + 3t) + max(v; ))Kv =  max(v;m)Kv can
be enclosed in a 5{CS of
(( max(v;3t(i 1)+3t)+(3t(i 1)+3t))+( max(v; )+ ))Kv+1 = ( max(v;m)+m)Kv+1:
Since the cases wherem = tand wherev 5;m tand 2f min(v;m); max(v;m)g
have now been settled, it is left to show that for m  t, v  5, and for all  
satisfying conditions (a-c) with  min(v;m) <  <  max(v;m) there is an enclos-
ing of a 5{CS of  Kv in a 5{CS of ( + m)Kv+1. The proof is by induction
on m. Tables 2.10, 2.11 and 2.12 were formed using Theorem 1.2.2 and Corol-
lary 2.2.10, and will help in establishing each value of  that the enclosing can be
produced using two smaller enclosings. Suppose for some m 2t and for all s with
v (mod 10)
m 0 1 2 3 4 5 6 7 8 9
2t(v) 0;0;2 0;0;10 10;0;0 0;0;10 0;0;10 2;0;0 10;0;0 0;10;0 10;0;0 10;0;0
3t(v) 2;0;2 0;10;10 10;10;0 10;0;10 0;10;10 2;2;0 10;0;10 0;10;10 10;10;0 10;0;10
Table 2.10: Checking the largest values of  : each cell contains  max(v;m)  
 max(v;t(v))  max(v;m t(v)) for k 0;1; and 2 (mod 3) in turn.
1  s < m, every 5{CS of  Kv can be enclosed in a 5{CS of ( + s)Kv+1. From
Table 2.10,  max(v;m)  max(v;t)  max(v;m t) 2f0; di (v)g, so   max(v;m)  
 max(v;t) + max(v;m t). Similarly from Tables 2.11 and 2.12,
 min(v;m t) + min(v;t)  min(v;m)2f0; di (v)g;
26
v (mod 10)
m 0 2 3 4 5 7 8 9
odd 0 0 0 0 0 0 0 0
even 0 0 10 0 2 10 0 10
Table 2.11: Checking the smallest values of  for v 6 1 or 6 (mod 10): each cell
contains  min(v;m t(v)) + min(v;t(v))  min(v;m).
m
v (mod 10) 0 1 2 3 4 5 6 7 8 9
1 10 0 10 10 10 10 10 10 10 10
6 10 0 0 10 0 10 0 0 10 0
Table 2.12: Checking the smallest values of  for v  1 or 6 (mod 10): each cell
contains  min(v;m t(v)) + min(v;t(v))  min(v;m).
so  min(v;m t) + min(v;t)  +min(v;m). Thus
 min(v;m t) + min(v;t)    max(v;m t) + max(v;t):
Since for each v we know that  di (v) =  di (v;m) is constant for all m, it follows
that there exist non-negative integers  1; 2 for which  min(v;t)   1   max(v;t),
 min(v;m t)   2   max(v;m t), and  1 +  2 =  , and there exist positive
integers m1; m2 with m1 + m2 = m such that for 1  i  2 there exists a 5{CS
of  iKv that can be enclosed in a 5{CS of ( i + mi)Kv+1. So the union of these
enclosings proves that any 5{CS of ( 1 + 2)Kv =  Kv can be enclosed in a 5{CS of
(( 1 +m1) + ( 2 +m2))Kv+1 = ( +m)Kv+1.
27
Chapter 3
Enclosings when u = 2
3.1 Introduction
In this chapter, we settle Conjecture 2.1.1 in the case when u = 2; see Theo-
rem 3.5.2. Yet again, new approaches are used in the proofs that follow.
In Theorem 3.5.2 we establish that the necessary conditions in Lemma 2.1.2 are
su cient when u = 2. Section 3.2 establishes two results: the  rst is very useful
when  nding 5-cycles in mKv; the second shows that a judicious choice of parameters
suggests a good approach to placing the edges incident with the 2 new vertices into
5-cycles. In Section 3.3, some general conditions on a set D are given that guarantee
the multigraph induced by all the edges of di erences in D have a closed trail in which
each set of 3 consecutive edges induces a path. We will need to introduce Hamilton
cycles in Section 3.3 in order to construct the closed trail. The proof of Theorem 3.5.2
builds on the results in Section 3.4 when small values of m are considered, and uses
the results from Sections 3.2 and 3.3 to build the 5-cycles with mixed edges.
Recall that an integer v is said to be  -admissible and  is said to be v-admissible
if the conditions in Theorem 1.2.2 are satis ed by  Kv.
The following decomposition result will be used to simplify some of the proofs
given in Section 3.2.
Theorem 3.1.1. [6] Let 1  d1;d2 < v=2. There exist 2 edge-disjoint Hamilton
cycles in each component of Gv(fd1;d2g).
3.2 Necessary tools
We now create some tools necessary to solve the problem of enclosing a  -fold
5{CS of order v in a ( +m)-fold 5{CS of order v+u for all m> 0 and u = 2. For the
remainder of this chapter, let U =f11;12g be the vertex set of V(Kv+u)nV(Kv).
Also for each pair of positive integers v and m, let  max(v;m) and  min(v;m) be the
largest and smallest values of  respectively for which conditions (a),(b) and (d) in
Lemma 2.1.2 are satis ed when u = 2.
One such tool that we use is the concatenation of trails T1;T2;:::;Tx in Gv(D)
which we denote byQxi=1 Ti = T1T2   Tx. (For example we use the trail H1H2 formed
by concatenating two Hamilton cycles in a component of Gv(fd1;d2g) obtained using
Theorem 3.1.1.)
Again, Lemma 2.2.1 is a critical result, being used hand in hand with the obser-
vation that there exists a 5{CS of Gv(fv=5g) and of Gv(f2v=5g) when v 0 (mod 5),
since each component is a 5-cycle. So when v 0 (mod 5) let (Zv; 1), (Zv; 2) and
(Zv; 3) be a 5{CS of Gv(fv=5g);Gv(f2v=5g), and Gv(f1;2;3g) respectively.
28
For the remainder of this chapter, let
a =  +m;
b = 2vd (v + 3)( +m)5 ;
c = (3v 1)( +m) vd5 ; and
d =
8
>>>
<
>>>
:
 +m if  +m<m(v 1)=2 and v6 4 (mod 5);
 +m  if  +m<m(v 1)=2, v 4 (mod 5), and
 +m 3 (mod 5) with  2Z5, and
m(v 1)=2 if  +m m(v 1)=2.
The following lemma guarantees that these parameters are all non-negative integers,
and are specially chosen to satisfy even more properties, the reasons for requiring
Conditions (i-v) being described after the lemma.
Lemma 3.2.1. Let u = 2 and v  5. Suppose there exists a 5{CS of  Kv. Let  
and m be positive integers that satisfy the conditions of Lemma 2.1.2. Then
(i) a; b; c, and d are all non-negative integers,
(ii) b is even,
(iii) 2a+ 3b+c = vd,
(iv) 2a+ 2b+ 4c = 2v( +m),
(v) mv(v 1)=2 vd 0,
(vi) c a 2b 0,
(vii) v divides c a 2b,
(viii) m(v 1)=2 d is an integer, and
(ix) m(v 1)=2 d 0 (mod 5) when v6 0 (mod 5).
Proof. First consider Condition (i). Clearly a is a non-negative integer since  and m
are non-negative integers. To show that b and c are non-negative integers, consider
three cases that depend on the three values of d in turn.
Suppose  +m<m(v 1)=2 and v6 4 (mod 5). So d =  +m, b = ( +m)(v 
3)=5 and c = ( +m)(2v 1)=5. It is clear that b and c are non-negative integers when
v 3 (mod 5). If v 0 or 1 (mod 5) then by Lemma 2.1.2(b) it follows  + m 0
(mod 5) and so b and c are non-negative integers. If v  2 (mod 5) then m  0
(mod 5) by Lemma 2.1.2(b) and   0 (mod 5) by Theorem 1.2.2, so  + m  0
(mod 5) and thus b and c are non-negative integers.
Suppose  + m < m(v 1)=2 and v  4 (mod 5). Then d =  + m  and
29
 +m 3 (mod 5) where  2Z5. Since
2vd (v + 3)( +m) (v 3)( +m) 2v 
 3 (v 3) 2v 
  (v 9)
 0 (mod 5)
and
(3v 1)( +m) vd (2v 1)( +m) +v 
  (7v 3)
 0 (mod 5);
it follows that b and c respectively are non-negative integers.
Finally suppose that  +m m(v 1)=2, so d = m(v 1)=2, b = (mv(v 1) (v+
3)( +m))=5 and c = ((3v 1)( +m) mv(v 1)=2)=5 = (2 (3v 1) m(v2 7v+
2))=10. If v 0 or 1 (mod 5) then by Lemma 2.1.2(b),  + m 0 (mod 5), and so
mv(v 1) (v+3)( +m) mv(v 1) (mod 5) and (3v 1)( +m) mv(v 1)=2 
mv(v 1)=2 (mod 5); thus b and c are integers. If v  2 (mod 10) then m  0
(mod 10) by Lemma 2.1.2(a-b) and   0 (mod 5) by Theorem 1.2.2, and so b and
c are integers. If v 7 (mod 10) then m 0 (mod 5) by Lemma 2.1.2(a) and so b
and c are integers. If v 3 (mod 5) then   0 (mod 5) by Theorem 1.2.2 and we
write v = 5k + 3. Then
mv(v 1) (v + 3)( +m) m(5k + 3)(5k + 2) (5k + 6)( +m)
  6 
 0 (mod 5)
and
2 (3v 1) m(v2 7v + 3) 2 (15k + 8) m((5k + 3)2 7(5k + 3) + 2)
 2 (15k + 8) m(25k2 + 30k + 9 35k 21 + 2)
  m(25k2 5k 10)
  5m(5k2 k 2)
 0 (mod 10);
since 5k2 k 2 is always even, and so b and c are integers. Let v = 5k + 4. Then
  0 (mod 5), so b and c are integers since the numerator of b satis es
mv(v 1) (v + 3)( +m) m(5k + 4)(5k + 3) (5k + 7)( +m)
 5m  (5k + 7)
 0 (mod 5)
30
and the numerator in the de nition of c satis es
2 (3v 1) m(v2 7v + 2) 2 (15k + 11) m((5k + 4)2 7(5k + 4) + 2)
  m((25k2 + 40k + 16 35k 28 + 2)
  m(25k2 + 5k 10)
  5m(5k2 +k 2)
 0 (mod 10)
since 5k2 +k 2 is always even.
Lastly, we will show d is a non-negative integer. By Lemma 2.1.2(a) and using
Theorem 1.2.2(a) applied to the postulated existence of 5{CSs of  Kv and 5{CSs of
( + m)Kv+u, m(v 1) is even so d is an integer. It is clear that d 0 if  + m <
m(v 1)=2 and v6 4 (mod 5) or  +m m(v 1)=2 so we let  +m<m(v 1)=2
and v 4 (mod 5) so that d =  + m  where  + m 3 (mod 5) with  2Z5.
Since v  4 (mod 5) and by Lemma 1.2.2(b),   0 (mod 5) and so   5 which
ensures d =  +m   0 since  2Z5.
It is left to show that Conditions (ii-ix) hold. If  + m < m(v 1)=2, v 6 
4 (mod 5), and v is odd then d =  + m and so b = ( + m)(v 3)=5 is even.
If  + m < m(v 1)=2, v 6 4 (mod 5), and v is even then  + m is even by
Lemma 2.1.2(a) and so b is even. Let  +m<m(v 1)=2 and v 4 (mod 5) so that
d =  +m  and b = (m(v 3) 3 +v(  2 ))=5. If v 4 (mod 10) then   0
(mod 10) and m 0 (mod 2) by Theorem 1.2.2, so b is even. If v 9 (mod 10) then
b = (2v(m +    ) (v + 3)( + m))=5 is even. Let  + m m(v 1)=2 so that
d = m(v 1)=2 and b = (mv(v 1) (v + 3)( +m))=5. If v is odd, it is easy to see
b is even. If v is even, then  + m is also even by Lemma 1.2.2(a) and so b is even.
Thus Condition (ii) holds.
By using our de nitions of a, b, c, and d, it follows that
2a+ 3b+c = 2( +m) + 3
 2vd (v + 3)( +m)
5
 
+ (3v 1)( +m) vd5
= 10 + 10m+ 6vd 3v  3vm 9  9m+ 3v + 3vm   m vd5
= vd;
2a+ 2b+ 4c = 2( +m) + 2
 2vd (v + 3)( +m)
5
 
+ 4
 (3v 1)( +m) vd
5
 
= 10( +m) + 4vd (2v + 6)( +m) + (12v 4)( +m) 4vd5
= 2v( +m);
31
mv(v 1)=2 vd 0, and
c a 2b = (3v 1)( +m) vd5  ( +m) 2
 2vd (v + 3)( +m)
5
 
= (3v 1)( +m) vd 5  5m 4vd+ (2v + 6)( +m)5
= v( +m d):
Thus Conditions (iii-viii) hold.
It remains to prove m(v 1)=2 d 0 (mod 5) when v6 0 (mod 5). Suppose
d =  +m so that m(v 1)=2 d = (m(v 3) 2 )=2. Since   0 (mod 5) when
v 3 (mod 5), it follows that m(v 3) 2  0 (mod 5) and so m(v 1)=2 d 0
(mod 5). If v  1 (mod 5) then  + m  0 (mod 5) and so m(v 1)=2 d  0
(mod 5). If v 0 or 2 (mod 5) then   m 0 (mod 5), and so m(v 1)=2 d 0
(mod 5).
Now suppose d =  + m  . Then v  4 (mod 5) and  + m 3 (mod 5)
and consequently   0 (mod 5), so let v = 5k + 4. Since m(v 1)=2  d =
(m(v 1) 2d)=2, it follows that
m(v 1) 2d 5km+m+ 2  2 
 ( +m) + 2  3 
 3 + 2  3 
 5  3 
 0 (mod 5):
Finally suppose d = m(v 1)=2. Then clearly, m(v 1)=2 d = 0 and so
Condition (ix) is satis ed.
At this point it is worth motivating the de nitions of a, b, c, and d. There are
4 di erent types of 5-cycles in ( +m)Kv+u (V [U;E) based on the number of pure
and mixed edges in the 5-cycle and the location of the pure edges:
 2 pure edges in V, 2 mixed edges, and 1 pure edge in U,
 3 pure edges in V and 2 mixed edges,
 1 pure edge in V and 4 mixed edges, and
 5 pure edges in V.
These 5-cycles will be called Type A, Type B, Type C, and pure cycles respectively.
(See Figure 3.1 for a visual representation of these 5-cycles.)
When constructing a 5-cycle system of ( +m)Kv+u  Kv we endeavor to use a 5-
cycles of TypeA, b of typeB, and c of TypeC. Lemma 3.2.1(i) shows this choice of a,
b and c is well-de ned. Lemma 3.2.1(iv) shows that this choice would use all 2v( +m)
mixed edges. Lemma 3.2.1(ii) ensures that we can separate the 5-cycles of Type B
so that b=2 of these 5-cycles are incident to11 and b=2 of these 5-cycles are incident
32
A
?1 ?2
B
?1 ?2
C
?1 ?2
D
?1 ?2
Figure 3.1: Possible 5-cycle Types
with12. The de nition of a ensures that all  +m edges joining the two vertices in U
would be in 5-cycles. By Lemma 3.2.1(iii) the choice of d ensures that the number of
remaining edges joining vertices in V is mv(v 1)=2 (2a+3b+c) = v(m(v 1)=2 d),
which is a non-negative (by (v)) integral multiple of v or v=2 if v is odd or even
respectively. This ensures that using di erence methods is a viable technique to
employ. In view of Lemma 3.2.1(iii) de ne  by
mv(v 1)
2  (2a+ 3b+c) =
 m(v 1)
2  d
 
v =  v: (3.1)
So  is a measure of the number of di erences we should use when de ning the pure
5-cycles. Notice that if  is not an integer then v is even and  is divisible by v=2, so
the \half" di erence v=2 will play a vital role. Note that   0 by Lemma 3.2.1(v).
3.3 Creating closed trail systems of Gv(D)
The following lemmas construct closed trail systems in components of Gv(D) for
various choices of D, each closed trail satisfying the important property that each set
of 3 consecutive edges induces a path. These closed trails are eventually connected
into a single closed trail in the proof of Lemma 3.3.7. Close attention is paid to
the  rst and last edges in each closed trail so that when two such closed trails are
concatenated the property that each set of 3 consecutive edges induces a path is
maintained. In Lemma 3.3.8 this single closed trail is used to construct 5-cycles of
Types A, B, and C.
Within Lemmas 3.3.1-3.3.5, at times it will cause no confusion to simply use Ti
instead of Ti(D) to denote a trail of Gv(D) that begins on vertex i.
Lemma 3.3.1. Let D be a multiset consisting of x copies of each of v=2 and v=3.
For some integer w there exists a closed trail system fTi = Ti(xfv=2;v=3g)ji2Zwg
of Gv(D) such that for each i2Zw:
33
(i) each set of 3 consecutive edges in Ti induces a path,
(ii) the  rst edge of Ti is an edge of di erence v=3, the second edge of Ti is an edge
of di erence v=2, the last edge of Ti is an edge of di erence v=2, and the second
to last edge of Ti is an edge of di erence v=3, and
(iii) Ti starts on vertex i.
Proof. Since 2 and 3 divide v, let v = 6w. Each component of Gv(fv=3;v=2g) is
isomorphic to G6(f2;3g). Since T = (0;4;1;5;2;0;3;1;4;2;5;3;0) is a closed trail
that satis es conditions (i-ii) for Gv(f2;3g), each closed trail T can be concatenated
with another copy of T such that Conditions (i-ii) are satis ed. For each i2Zw, form
Ti from Qxi=1T by multiplying each vertex in the trail by w and adding i. Clearly
conditions (i-iii) are satis ed by each closed trail Ti. So fTi = Ti(xfv=2;v=3g)ji2
Zwg is a closed trail system of Gv(D) that satis es Conditions (i-iii).
Lemma 3.3.2. Let D be a multiset consisting of x copies of each of v=3 and v=6.
For some integer w there exists a closed trail system fTi = Ti(xfv=3;v=6g)ji2Zwg
of Gv(D) such that for each i2Zw:
(i) each set of 3 consecutive edges in Ti induces a path,
(ii) Ti can be chosen so that the  rst two edges are either fi;i + v=6g and fi +
v=6;i+v=6 v=3g or fi;i v=6g and fi v=6;i v=6 +v=3g,
(iii) the last two edges of Ti are both edges of di erence v=3, and
(iv) Ti starts on vertex i.
Proof. Each component of Gv(fv=3;v=6g) is isomorphic to G6(f1;2g), so the re-
quired closed trail Ti can be formed from the concatenation of x copies of T =
(0;1;5;4;3;2;0;5;3;1;2;4;0) or of T0 = (0;5;1;2;3;4;0;1;3;5;4;2;0) (the choice is
determined by the choice for the  rst and second edges of Ti in Condition (ii)) by
multiplying each vertex in the trail by v=6 and adding i. Clearly Conditions (i-iv)
are satis ed in each closed trail Ti. SofTi = Ti(xfv=3;v=6g)ji2Zvgis the required
closed trail system.
Lemma 3.3.3. Let D be a multiset consisting of x copies of v=3 and x elements in
xf1;2;:::;bv=2cgnxfv=2;v=3;v=6g. Let j;k2Z2. For some integer w there exists
a closed trail system fTi = Ti(D;v=3)ji2Zwg of Gv(D) such that for each i2Zw:
(i) each set of 3 consecutive edges in Ti induces a path,
(ii) the  rst edge in Ti is an edge of a di erence in Dnxfv=3g,
(iii) the last two edges of Ti are edges of di erence v=3,
(iv) Ti starts on vertex i, and
(v) if d2D and gcd(fd;v=3;vg) = 1 then E(Gv(fd;v=3g)) E(T0(D;v=3)).
Proof. LetD be a multiset consisting ofxcopies ofv=3 andxelements inxf1;2;:::;bv=2cgn
xfv=2;v=3;v=6g. For each d 2 Dnxfv=3g, name the components of Gv(fd;v=3g)
with G(d;i) for 0  i < gcd(fd;v=3;vg), where G(d;i) contains vertex i. It is
34
convenient to de ne G(d;i) to be empty if i gcd(fd;v=3;vg). Notice that this im-
plies that if gcd(fd;v=3;vg) = 1 then G(d;0) = Gv(fd;v=3g), so E(Gv(fd;v=3g))  
E(T0(D;v=3)) as required by (v). We now form the required trail system as follows
in which for each i2Zv Ti(D;v=3) is a closed trail of Sd2Dnxfv=3gG(d;i).
By Theorem 3.1.1, for each i2 Zv and each d2Dnxfv=3g, if G(d;i) is not
empty then it can be decomposed into 2 Hamilton cycles H1(d;i) and H2(d;i). It
is clear that in Gv(fv=3;dg) there are no 2-cycles, and since v=6 62 D the only 3-
cycles are components of Gv(fv=3g). Clearly the edges between the vertices in c(i) =
fi;i+v=3;i+ 2v=3gcannot all appear in the same Hamilton cycle, so we can assume
that:
(1) H1(d;i) and H2(d;i) contain one and two edges in c(i) respectively, and
(2) vertex i is incident with edges of di erence d and v=3 in each of H1(d;i) and
H2(d;i).
Let H1(d;i) start on the edge of di erence d, so by (2) it will end on the edge of
di erence v=3. Let H2(d;i) start on the edge of di erence d, so by (1) it must end on
two consecutive edges of di erence v=3. Then clearly H1(d;i)H2(d;i) has no set of 3
consecutive edges that induces a cycle. Since v=6 62D and each H2(d;i) ends with
two edges of di erence v=3, Ti(D;v=3) = Qd2Dnxfv=3gH1(d;i)H2(d;i) also satis es
condition (i), so Ti(D;v=3) is a closed trail ofSd2Dnxfv=3gG(d;i) satisfying Conditions
(i-iv). Therefore fTi(D;v=3)ji2Zvg is the required closed trail system.
Lemma 3.3.4. Let D be a multiset consisting of x copies of v=2 and x elements in
xf1;2;:::;bv=2cgnxfv=2;v=3;v=4g. For some integer w there exists a closed trail
system fTi = Ti(D;v=2)ji2Zwg of Gv(D) such that for each i2Zw:
(i) each set of 3 consecutive edges in Ti induces a path,
(ii) Ti can be chosen so that if the  rst and second edges of Ti are ei;1 and ei;2
respectively then (ei;1;ei;2) is any of the 8 elements of f(fi;i + ( 1)jyg;fi +
( 1)jy;i + ( 1)jy + ( 1)kzg) j fy;zg = fv=2;dg;j;k 2 Z2g for some d 2
Dn(xfv=2g),
(iii) exactly one of the last two edges of Ti has di erence v=2,
(iv) Ti starts on vertex i,
(v) if d2D and gcd(fd;v=2;vg) = 1 then E(Gv(fd;v=2g)) E(T0(D;v=2)),
(vi) for any d i 2Dnxfv=2g if Ti contains an edge of di erence d i then the last
edge of Ti has di erence d i or v=2, and
(vii) if 22Dn(xfv=2g) then E(Gv(f2;v=2g)) 
E(Sgcd(fv;v=2;2g)i=0 Ti(D;v=2)).
Proof. Let D be a multiset consisting of x copies of v=2 and x elements in xf1;2;:::;
bv=2cgnxfv=2;v=3;v=4g. For each d 2 Dn(xfv=2g), let v0(d) = v=gcd(v;v=2;d)
and d0 = d=gcd(v;v=2;d). Then each component of Gv(fv=2;dg) is isomorphic to
Gv0(d)(fv0(d)=2;d0g). Note that one of v0(d)=2 and d0 must be odd. If at least one of
35
them is even then a closed trail of Gv0(d)(fv0(d)=2;d0g) can be formed by alternately
adding v0(d)=2 and d0 (or choosing to always add  d0 instead of d0) to reach the
next vertex in the closed trail, starting with an edge of either di erence. If both
v0(d)=2 and d0 are odd then a closed trail of Gv0(d)(fv0(d)=2;d0g) can be formed by
alternately adding v0(d)=2 and d0 (or  d0) for the  rst half of the closed trail, again
starting with an edge of either di erence, then alternately adding v0(d)=2 and  d0
(or d0 respectively) for the second half of the closed trail. In this manner we can
 nd closed trails in each component of Gv(fv=2;dg) which provide the required four
choices for the  rst two edges, always ensuring that exactly one of the last two edges
has di erence v=2 which satis es Conditions (ii) and (iii).
We now concatenate these closed trails to form the required closed trail system.
Care must be taken to ensure that no 2-cycle and no 3-cycle is formed when two
such trails (constructed using two elements of Dnxfv=2g) are concatenated. This is
always ensured by carefully choosing the  rst two edges of the second path, based on
the last two edges of the  rst path. If the  rst path ends on an edge of di erence v=2
then since d6= v=4 the choice of starting the second path with an edge of di erence d
or  d allows 3-cycles to be avoided. If the  rst path ends on an edge of di erence d0
then starting the second path on an edge of di erence v=2 and appropriately choosing
the second edge to have di erence d or  d again avoids any 3-cycle.
For each d 2 Dnxfv=2g in turn, if i < gcd(fv=2;d;vg) then let Pi(fv=2;dg)
be the closed trail starting on vertex i in Gv(fv=2;dg) chosen as just described (so
this is dependent on the order in which the trails are concatenated), and if i  
gcd(fv=2;d;vg) then let Pi(fv=2;dg) = ?. Notice that if gcd(fd;v;v=2g) = 1, then
Gv(fv=2;dg) is a single component, so E(Gv(fd;v=2g)  E(T0(D;v=2)) as required
by (v). Also, notice that if 2 2Dn(xfv=2g) then gcd(f2;v=2;vg) = 1 or 2 and so
Pi(fv=2;2g) = ? for i> 0 or i> 1 respectively as required by (vii). So for each i2
Zv, Ti(D;v=2) = Qd2Dn(xfv=2g) Pi(fv=2;dg) where the last trail in this concatenation
is Pi(fv=2;d ig) (so (vi) is satis ed) is a closed trail satisfying Conditions (i-vii) and
so fTi(D;v=2) ji2Zvg is a closed trail system of Gv(D) satisfying all of the listed
conditions.
In subsequent proofs, edges of di erence v=5 can be problematic when trying to
avoid 3-cycles in the concatenation process, so the following lemma uses a di erence
d at the end of the trail to separate edges of di erence v=5 from edges in the next
trail. To this end, d cannot be 2v=5.
Lemma 3.3.5. Let D be a multiset in which each element is v=5 except for a single
element
d 62 fv=2;v=3;v=5;2v=5g. For some integer w there exists a closed trail system
fTi = Ti(D;v=5)ji2Zwg of Gv(D) such that for each i2Zw:
(i) each set of 3 consecutive edges in Ti induces a path,
(ii) the  rst and second edges of Ti can be chosen to be either fi;i + v=5g and
fi+v=5;i+ 2v=5g, or fi;i v=5g and fi v=5;i 2v=5g,
36
(iii) Ti ends on an edge of di erence d ,
(iv) Ti starts on vertex i, and
(v) if gcd(fd ;v;v=5g) = 1 then E(Gv(fd ;v=5g)) E(T0(D;v=5)).
Proof. Let D be a multiset in which each element is v=5 except for a single element
d 62fv=2;v=3;v=5;2v=5g and let k = (v=5)=gcd(fv=5;d g). Let Pi(fv=5g) be the
Euler tour of one component of Gv((jDj 1)fv=5g) chosen to meet condition (ii) (use
edges of di erence ( 1)jv=5 for the appropriate j2Z2 to form the trail). The edges
of di erence d can be used to connect these tours as follows. If i < gcd(fv=5;d g)
then let
Ti =
 Y
j2Zk
Pjd +i(fv=5g)(jd +i;(j + 1)d +i)
!
(kd +i;(k+1)d +i;:::; d +i;i):
Then fTi = Ti(D;v=5) ji2Zwg is a closed trail system of Gv(D) satisfying Condi-
tions (i-iv). Notice that this implies that if gcd(fd ;v=5g) = 1 then Gv(fd ;v=5g) is
a single component, so E(Gv(fd;v=5g)) E(T0(D;v=5))).
Each of the Lemmas 3.3.1 to 3.3.5 extend the de nition of Ti so that if i2ZvnZw
then de ne Ti = ?. This will be important because trails formed using di erent
lemmas will be concatenated, but the value of w may change from lemma to lemma.
Lemma 3.3.6. Let the following trails be of those described in Lemmas 3.3.1, 3.3.2,
3.3.3, 3.3.4, and 3.3.5 with multisets D1;D2;D3;D4 and D5 respectively are the mul-
tisets satisfying the respective conditions. Then in each case by judicious use of the
choice available (as described in each lemma) for the  rst two edges of the second com-
ponent trail, the following trails can be constructed so that each set of 3 consecutive
edges induces a path:
(i) Ti((jD1j=2)fv=2;v=3g)Ti((jD2j=2)fv=3;v=6g),
(ii) Ti((jD1j=2)fv=2;v=3g)Ti(D3;v=3),
(iii) Ti((jD1j=2)fv=2;v=3g)Ti(D4;v=2),
(iv) Ti((jD1j=2)fv=2;v=3g)Ti(D5;v=5),
(v) Ti((jD1j=2)fv=2;v=3g)Ti(fdg) where d62fv=2;v=3;v=6g,
(vi) Ti((jD2j=2)fv=3;v=6g)Ti(D3;v=3),
(vii) Ti((jD2j=2)fv=3;v=6g)Ti(D4;v=2),
(viii) Ti((jD2j=2)fv=3;v=6g)Ti(D5;v=5),
(ix) Ti((jD2j=2)fv=3;v=6g)Ti(fdg) where d62fv=3;v=2g,
(x) Ti(D3;v=3)Ti(D4;v=2),
(xi) Ti(D3;v=3)Ti(D5;v=5),
(xii) Ti(D3;v=3)Ti(fdg) where d62fv=2;v=3;v=6g,
(xiii) Ti(D4;v=2)Ti(D5;v=5),
(xiv) Ti(D4;v=2)Ti(fdg) where d62fv=2;v=3g,
(xv) Ti(D5;v=5)Ti(fdg) where d62fv=2;v=3g, and
37
(xvi) Ti(fdg)Ti(fd0g) where d;d062fv=2;v=3;v=5g.
Proof. Let D1;D2;D3;D4; and D5 be multisets de ned as D is in Lemmas 3.3.1, 3.3.2,
3.3.3, 3.3.4, and 3.3.5 respectively. Then by Lemmas 3.3.1, 3.3.2, 3.3.3, 3.3.4, and
3.3.5, let Ti(jD1j=2fv=2;v=3g), Ti(jD2j=2fv=3;v=6g), Ti(D3;v=3), Ti(D4;v=2), and
Ti(D5;v=5) respectively be the closed trails that satisfy the conditions outlined in
the lemmas. Since in each case d;d062fv=3;v=2g, each set of 3 consecutive edges in
Ti(fdg) and in Ti(fd0g) induces a path. Therefore, in each case it remains to show
that the choices permitted in how the trails start and end described in the lemmas
can be exploited to ensure that as two trails are concatenated, the last 1 or 2 of the
edges from the  rst closed trail followed by the  rst 2 or 1 edges from the second
closed trail respectively induce a path.
Suppose the  rst of the two closed trails to be concatenated isTi(jD1j=2fv=2;v=3g).
By Lemma 3.3.1, the last edge of Ti(jD1j=2fv=2;v=3g) is an edge of di erence v=2, and
the second to last edge of Ti(jD1j=2fv=2;v=3g) is an edge of di erence v=3. Choose
the  rst and second edge of the following trail as indicated:
 choose the  rst and second edge of Ti(jD2j=2fv=3;v=6g) to be fi;i + v=6g and
fi+v=6;i+v=6 v=3gorfi;i v=6gandfi v=6;i v=6 +v=3grespectively;
 choose the  rst and second edge of Ti(D3;v=3) to befi;i v=3gandfi v=3;i 
v=3 d3gorfi;i+v=3gandfi+v=3;i+v=3 +d3grespectively where d3 2D3;
 choose the  rst and second edge of Ti(D4;v=2) to befi;i+v=2gandfi+v=2;i+
v=2 +d4gorfi;i+v=2gandfi+v=2;i+v=2 d4grespectively where d4 2D4;
 choose the  rst and second edge of Ti(D5;v=5) to befi;i+v=5gandfi+v=5;i+
2v=5g or fi;i v=5g and fi v=5;i 2v=5g respectively; and
 choose the  rst and second edge of Ti(fdg) to be fi;i + dg and fi + d;i + 2dg
respectively (since d62fv=2;v=3;v=6g).
This settles Conditions (i-v).
Now suppose the  rst closed trail is Ti(jD2j=2fv=3;v=6g). By Lemma 3.3.2, the
last two edges of Ti(jD2j=2fv=3;v=6g) are edges of di erence v=3. The  rst vertex in
the second trail is of course vertex i, so it remains to ensure that the second and third
vertices in the second trail avoid v=3 and 2v=3 respectively, or avoid 2v=3 and v=3
respectively, depending on whether the last edge of Ti(jD1j=2fv=2;v=3g) isfi;i+v=3g
or fi;i + 2v=3g. In Case (vi) this is accomplished by Lemma 3.3.3(ii), noting that
v=662D3. This settles conditions (vi-ix).
If Ti(D3;v=3) is the  rst closed trail then since the last two edges of Ti(D3;v=3)
are both edges of di erence v=3, it follows that each set of 3 consecutive edges
in Ti(D3;v=3)Ti(D4;v=2), Ti(D3;v=3)Ti(D5;v=5), and Ti(D3;v=3)Ti(fdg) (since d62
fv=2;v=3;v=6g) induces a path.
Suppose the  rst closed trail is Ti(D4;v=2). Since one of the last two edges of
the closed trail Ti(D4;v=2) is an edge of di erence v=2 and we have a choice of the
 rst two edges of the closed trail Ti(D5;v=5), it follows that each set of 3 consecutive
edges in the closed trail Ti(D4;v=2)Ti(D5;v=5) induces a path. Since d62fv=2;v=3g,
38
we can choose the  rst two edges of T(fdg) to be either fi;i + dg and fi;i + 2dg or
fi;i dgandfi;i 2dgrespectively so that each set of 3 consecutive edges in either
Ti(D4;v=2)Ti((d)) induces a path.
If the  rst closed trail isTi(D5;v=5) then sinced 2D5 withd 62fv=2;v=3;v=5;2v=5g
and d62fv=2;v=3;v=5g, it follows that we can choose the  rst two edges of Ti(fdg)
so that each set of 3 consecutive edges in the closed trail Ti(D5;v=5)Ti(fdg) induces
a path.
Finally, we can choose the  rst two edges of Ti(fd0g) so that each set of 3 con-
secutive edges in the closed trail Ti(fdg)Ti(fd0g) induces a path.
For the multiset D, let mD(d) be the number of times d appears in D. Note that
we use the convention Gv(fdg) = Gv(d) to simplify notation.
Lemma 3.3.7. Let v> 3 be a positive integer. Let D be a multiset, each element of
which is in f1;2;:::;bv=2cg. Suppose that the following conditions hold:
(A) D contains a di erence
d1 =
(
1 if v6 5 (mod 10) and
4 otherwise
and another di erence d2 62fv=2;v=3;v=4;v=5;2v=5g (possibly d2 is a second
copy of d1 in D);
(B) 2mD(v=2) + mD(v=4) +   jDnfd1gj, where  = 1 if fv=2;v=3;v=6g Dn
fd1;d2g, mD(v=2) mD(v=3), and mD(v=6) >mD(v=2) min(fmD(v=2);mD(v=3)g)
and  = 0 otherwise; and
(C) 2mD(v=3) +mD(2v=5) jDnfd1gj.
Then there exists a closed trail of Gv(D) in which each set of 3 consecutive edges
induces a path.
Proof. To form the required trail decomposition  of Gv(D), we begin by partitioning
D into the 7 sets D1;D2;:::;D6, and fd1g de ned as follows. D will actually be
partitioned in two di erent ways, depending on the occurrences of the di erences
v=2, v=3, and v=6 in D (as will be seen, this actually relates to whether or not a
single copy of the di erence v=3 needs to be set aside so as to be paired with a single
copy of the di erence v=6).
First, suppose that
fv=2;v=3;v=6g Dnfd1;d2g, mD(v=2) mD(v=3), (3.2)
and mD(v=6) >mD(v=2) min(fmD(v=2);mD(v=3)g);
so in Condition (B),  = 1. Let D1 consist of z1 = mD(v=3) 1 copies of both v=2 and
v=3. Let D4 contain mD(v=2) z1 copies of v=2 and mD(v=2) z1 di erences in  4 =
Dn(D1[(mD(v=2) z1)fv=2g[mD(v=4)fv=4g[fd1;v=3g); choose these di erences
39
in D4 so that d2 2D4 only if D1 [D4 contains all of the di erences in D besides
mD(v=4)fv=4g[fd1;v=3g(i.e. only ifjD1j+jD4j=jDn(mD(v=4)fv=4g[fd1;v=3g)j).
Notice that D1[D4 contains z1 +mD(v=2) z1 = mD(v=2) copies of v=2 (that is, all
the copies of v=2 in D). Also notice that by Condition (B) it follows that
j 4j=jDnfd1gj mD(v=2) z1 mD(v=4) 1
 2mD(v=2) +mD(v=4) + 1 mD(v=2) z1 mD(v=4) 1
= mD(v=2) z1  0
so  4 is large enough for D4 to be well de ned. Let D2 contain 1 copy of both v=6 and
v=3. Notice that D1[D2 contains all copies of v=3 in D. If 2v=52Dn(D1[D2[D4)
then let D5 contain mDn(D1[D2[D4)(v=5) copies of v=5 and a single copy of d2 and
otherwise, let D5 = ?. Let D6 contain all of the remaining di erences in D (i.e.
D6 = Dn(D1[D2[D4[D5[fd1g)).
Second, suppose that
fv=2;v=3;v=6g6 Dnfd1;d2g, mD(v=2) <mD(v=3), (3.3)
or mD(v=6) mD(v=2) min(fmD(v=2);mD(v=3)g);
so, in Condition (B),  = 0. Let D1 consist of z2 = min(fmD(v=3);mD(v=2)g) copies
of both v=2 and v=3. Let D4 contain mD(v=2) z2 copies of v=2 and mD(v=2) z2
di erences in  4 = Dn(D1 [(mD(v=2) z2)fv=2g[mD(v=4)fv=4g); choose these
di erences in D4 so that D4 contains as many copies of v=6 as possible and so that
d2 2D4 only if D1[D4 contains all of the di erences in Dn(fd1g[mD(v=4)fv=4g)
(i.e. only ifjD1j+jD4j=jDn(fd1g[mD(v=4)fv=4g)j). As before, D1[D4 contains
z2 +mD(v=2) z2 = mD(v=2) copies of v=2. Also notice that by Condition (B)
j 4j=jDnfd1gj mD(v=2) z2 mD(v=4)
 2mD(v=2) +mD(v=4) mD(v=2) z2 mD(v=4)
= mD(v=2) z2  0;
so  4 is large enough for D4 to be well de ned. It will be important to note that
under the set of assumptions in (3.3) one of the two following statements (i.e. (3.4)
or (3.5)) must hold.
Case A: Suppose fv=2;v=3;v=6g D and either mD(v=2) <mD(v=3) or
mD(v=6) mD(v=2) min(fmD(v=2);mD(v=3)g).
Notice that in this case if mD(v=2) < mD(v=3) then v=3 2 Dn(D1 [D4), and if
mD(v=6)  mD(v=2) min(fmD(v=2);mD(v=3)g) then since D4 contains as many
copies of v=6 as possible, v=662Dn(D1[D4). Therefore the following is true:
if v=62Dn(D1[D4) then v=32Dn(D1[D4). (3.4)
Case B: Suppose fv=2;v=3;v=6g6 D. Then clearly the following is true:
if v=62D then fv=2;v=3g6 D. (3.5)
40
Returning to the partitioning of D under the set of assumptions in (3.3), D2 contains
min(fmD(v=3) z2;mDnD4(v=6)g) copies of both v=6 and v=3. Note that by (3.4) in
Case A,
if v=62Dn(D1[D4) then D2 6= ?. (3.6)
Let D3 contain y = mD(v=3) z2 min(fmD(v=3) z2;mDnD4(v=6)g) copies of v=3
together with y di erences in
 3 = Dn(mD(v=2)fv=2g[mD(v=3)fv=3g[mD(v=6)fv=6g[mD(2v=5)f2v=5g[fd1g);
choose these di erences in D3 so that d2 2D3 only if D1[D2[D3[D4 contains every
di erence in D except d1 (i.e. only if jD1j+jD2j+jD3j+jD4j=jDnfd1gj). Notice
that since D3 6= ? implies mD(v=2) <mD(v=3) and mD(v=3) z2 >mdnD4(v=6), and
by Condition (C)
j 3j=
(
jDnfd1gj mD(v=2) mD(v=3) mD(v=6) mD(2v=5) if D3 6= ?,
0 if D3 = ?
 
(
2mD(v=3) +mD(2v=5) mD(v=2) mD(v=3) mD(v=6) mD(2v=5) if D3 6= ?,
0 if D3 = ?
=
(
mD(v=3) mD(v=2) mD(v=6) if D3 6= ?,
0 if D3 = ?
=
(
mD(v=3) min(fmD(v=2);mD(v=3)g) min(fmD(v=3) z2;mD(v=6) if D3 6= ?,
0 if D3 = ?
=
(
mD(v=3) z2 min(fmD(v=3) z2;mD(v=6) if D3 6= ?,
0 if D3 = ?.
Then there are enough di erences in Dn(D1[D2[D4) to construct D3 as described.
If 2v=52Dn(D1[D2[D3[D4) then let D5 contain mDn(D1[D2[D3[D4)(v=5) copies
of v=5 and a single copy of d2 and otherwise, let D5 = ?. Let D6 contain all of the
remaining di erences, i.e. D6 = Dn(D1[D2[D3[D4[D5[fd1g).
Form a sequence S = (s1;s2;:::;sjD6j+5), the elements of which form a partition
of Dnfd1gwhere sk = Dk for k2f1;2;3;4;5gand s6;s7;:::;sjD6j+5 are sets of size 1
such that SjD6j+5k=6 sk = D6; by choosing the copies of the di erence 2 in D6 to appear
as early as possible in this ordering, we can assume that
if sjD6j+5 =f2g then D6 =jD6jf2g. (3.7)
Then by Lemmas 3.3.1-3.3.5 for eachi2Zv we can form closed trailsTi(jD1j=2fv=2;v=3g),
Ti(jD2j=2fv=3;v=6g), Ti(D3;v=3), Ti(D4;v=2), and Ti(D5;v=5) respectively in each of
which every set of 3 consecutive edges induces a path. Let d3, d4, and d5 be v=3, v=2,
and v=5 respectively. So for each i2Zv de ne the concatenation
 i = Ti(jD1j=2fv=2;v=3g)Ti(jD2j=2fv=3;v=6g)
 5Y
k=3
Ti(Dk;dk)
!0
@
jD6j+5Y
k=6
Ti(sk)
1
A
41
of these trails. Note that some of the trails being concatenated may be empty.
It will be important later to note that
if d2Dnfd1g and gcd(fd;vg) = 1 then E(Gv(fdg))  0; (3.8)
this follows by applying Condition (v) of Lemmas 3.3.3{3.3.5 to Ti(Dk;dk) for k = 3,
4, and 5 respectively, and the fact that Gv(fdg) is connected (so Ti(fdg) is empty if
i> 0).
To check that  i has the property that no 3 consecutive edges form a cycle of
length 2 or 3 consider the following. So the concatenation of each pair of trails making
up  i must be considered, but this is done in Lemma 3.3.6 in all cases except one. The
one case not considered is when Ti(jD1j=2fv=2;v=3g) is followed in the concatenation
by Ti(fv=6g) (this could happen if Dk = ? for 2 k 5). If this case did occur, a 3-
cycle would necessarily be formed by 3 consecutive edges (such as (i+v=3;i;i+v=6;i+
v=3)), but by (3.4), (3.5), and (3.8) the concatenation Ti(jD1j=2fv=2;v=3g)Ti(fv=6g)
will never occur in  i. If we are in Case B then this is clearly so since by (3.5)
one of the trails Ti(jD1j=2fv=2;v=2g) or Ti(fv=6g) is empty. Otherwise, since the
value of w is the same in Lemmas 3.3.1 and 3.3.2, if Ti(jD1j=2fv=2;v=3g) 6= ?
then Ti(jD2j=2fv=3;v=6g) 6= ?. If v=6 2 Dn(D1 [D4) then since D2 6= ? (by
(3.6) if we are in Case A), the concatenation of Ti(jD1j=2fv=2;v=3g)Ti(fv=6g) will
never occur in  i under the set of assumptions in (3.2), nor under the set of assump-
tions in (3.3). In other words, in such a situation  i is formed by  rst concatenat-
ing Ti(jD1j=2fv=2;v=3g) with Ti(jD2j=2fv=3;v=6g) which is then concatenated with
Ti(fv=6g). Thus by Lemma 3.3.6 every set of 3 consecutive edges in  i induces a path.
It is critical to note that for all i 1 we still have the choice available (as de-
scribed in Lemmas 3.3.4 to 3.3.5) for the  rst two edges in the  rst component trail
in  i.
The  nal step in forming the required closed trail  is to use the edges in Gv(fd1g)
to connect the trails  i, i2Zv. In so doing, we need to make sure that we maintain
the property that every set of 3 consecutive edges induces a path. To do so, it turns
out that even more  exibility is required since  0 and  d1 may need to be replaced
with the reverse trails   0 and   d1 respectively de ned formally in the next paragraph.
This choice is only needed for  0 and  d1 and not for  i if i62f0;d1g;  0 is unusual
because from (3.8), if there is an edge of di erence d where gcd(fd;vg) = 1, then this
edge could only be found in  0.
Let fi;i + fi;1g and fi + fi;1;i + fi;1 + fi;2g be the  rst and second edges of  i
respectively. Similarly, letfi fi; 1 fi; 2;i fi; 1gandfi fi; 1;igbe the second-
to-last and last edges of  i respectively. Let   i denote the closed trail formed by
mapping each edge fj1;j2g in the closed trail  i to the edge f j1 + 2i; j2 + 2ig in
  i . (Informally,   i is formed from  i by also starting at vertex i but proceeding in
the reverse direction around the v vertices.) So clearly   i has the property that each
set of 3 consecutive edges induces a path.
42
We are now ready to de ne the required trail  .
If d1 2f f0; 1; (f0; 1 +f0; 2)g, or (Pi)
if both  f0; 1 = 2d1 and  d1 = ? (Pii)
(i.e. d1 is one of the second-to-last or third-to-last vertices in  0 or 2d1 is the second-
to-last vertex in  0), then begin  with   0 =   0 and
otherwise begin  with   0 =  0. (Piii)
Extend the trail by concatenating   0 with the trail (0;d1) of length 1. Recursively
de ne   i for all i 1 as follows:
For all i+d1  0 if  i+d1 6= ?,   i 6= ?, and d1 +fi+d1;1 2f0; f i; 1g (Piv)
where fi;i f i; 1g is the last edge of   i
then   i+d1 =   i+d1 and otherwise   i+d1 =  i+d1. For all i2Zv let fi;i + f i;1g, fi +
f i;1;i+f i;1 +f i;2g, fi f i; 1 f i; 2;i f i; 1g, and fi f i; 1;ig be the  rst, second,
second-to-last, and last edges of   i . Then de ne
 =
v 1Y
i=0
   
id1(id1;(i+ 1)d1)
 :
It now requires some e ort to show that all sets of 3 consecutive edges that contain
the edge fid1;(i+ 1)d1g induce a path, especially in the case when i = 0.
If  0(0;d1) contains 3 consecutive edges that form a cycle then d1 must be one of
the last 3 vertices of  0, so d1 2f0; f0; 1; (f0; 1 + f0; 2)g, in which case   0 =   0
since clearly d1 6= 0. To see that not both  0(0;d1) and   0 (0;d1) can contain 3
consecutive edges that form a cycle, suppose that d1 2f f0; 1; (f0; 1 + f0; 2)g\
ff0; 1;f0; 1+f0; 2g. First note that if d1 = (f0; 1+f0; 2) then d1 6= f0; 1+f0; 2 and
so d1 = f0; 1. Similarly if d1 = f0; 1 then d1 = f0; 1 +f0; 2. Therefore in either case
the last two edges of  0 and   0 have di erence d1 and 2d1. But by Lemmas 3.3.1{3.3.5
the last two edges of  0 are either both edges of the same di erence or one of the two
edges has di erence v=5 or v=2. Since d1 62fv=2;v=5g, it is impossible for both  0
and   0 to have the vertex d1 in the last three vertices of either trail. So   0 has been
chosen such that every 3 consecutive edges in   0 (0;d1) induces a path.
To see that   0 (0;d1)  d1 when   d1 6= ? does not contain 3 consecutive edges
that form a cycle, consider the following. First note that since gcd(fd1;vg) = 1
by Condition (A), if d = d1 and d 2 Dnfd1g (that is, d is another copy of d1 in
D) then E(Gv(fdg))  E(  0 ), so neither the second vertex of  d1 nor the second
vertex of   d1 equal 0 (the last vertex of   0 ). Second, note that if Td1(D4;v=2) is the
 rst trail in   d1 then by Lemma 3.3.4(ii), the preamble in Lemma 3.3.6, and since
mD(v=2)fv=2g D1[D4, we can choose the  rst edge of Td1(D4;v=2) to not have
di erence v=2. Therefore if d1 +fd1;1 = f 0; 1 (so   d1 =   d1) then d1 fd1;1 6= f 0; 1
43
(for otherwise subtracting this equality from the  rst implies that 2fd1;1 = 0, so
fd1;1 = v=2), so the second vertex of   d1 is not  f 0; 1.
To  nish showing that each set of 3 consecutive edges in   0 (0;d1)  d1 induces a
path, we must show that the third vertex of  d1, namely d1 + f d1;1 + f d1;2, is not
v. Since d1 > 0, we need only show that the third vertex of   d1, namely d1 +
f d1;1 + f d1;2, is not 0. This requires considering several cases, most of which rely on
the claim that gcd(ff d1;1;f d1;2;vg) 6= 1 (later we will show that if f d1;1 6= f d1;2 then
gcd(ff d1;1;f d1;2g)6= 1). If f d1;1 = f d1;2 (that is, the  rst two edges of   d1 have the same
di erence) then d1 +f d1;1 +f d1;2 6= 0, for otherwise d1 = v f d1;1 f d1;2 = v 2f d1;1
contradicting the facts that gcd(fv;f d1g) 6= 1 (by (3.8)), and gcd(fv;d1g) = 1 (by
Condition (A)). So we can assume that the  rst two edges of   d1 have di erent
di erences: f d1;1 6= f d1;2. Then the  rst trail in   d1 cannot be Td1(fdg) for any
di erence d2D6 and cannot be Td1(D5;v=5) by Lemma 3.3.5 (since in such cases
the  rst two edges have the same di erences). So the  rst trail in   d1 must be
one of the following: Td1(jD1j=2fv=2;v=3g), Td1(jD2j=2fv=3;v=6g), Td1(D3;v=3), or
Td1(D4;v=2). If the  rst trail of   d1 is Td1(D3;v=3) or Td1(D4;v=2) then d1 + f d1;1 +
f d1;2 6= 0 since otherwise gcd(fv;d1g) = 1 would mean that gcd(ff d1;1;f d1;2g) = 1
which contradicts Lemma 3.3.3(v) or 3.3.4(v) respectively (since then gcd(fd;d0;vg) =
1 where d0 = v=3 or v=2 respectively and ff d1;1;f d1;2g = fd;d0g, so these edges
would be in   0 instead of   d1). If the  rst trail of   d1 is Td1(jD1j=2fv=2;v=3g) or
Td1(jD2j=2fv=3;v=6g) then d1 + f d1;1 + f d1;2 6= 0 since otherwise by the descriptions
of the  rst two edges of the trails constructed in Lemmas 3.3.1 and 3.3.2, d1 = v=6
so by Condition (A), v = 6, and so these edges would be in   0 instead of in   d1. So
each set of 3 consecutive edges in   0 (0;d1)  d1 induce a path.
It remains to show that in each of the following trails T every set of 3 consecutive
edges induces a path:
 if i 0 then T = (i;i+d1;i+ 2d1;i+ 3d1);
 if i> 0,   i+d1 6= ?, and   i 6= ? then T =   i (i;i+d1)  i+d1;
 if i 0,   i+d1 = ?, and   i+2d1 6= ? then T = (i;i + d1;i + 2d1)  i+2d1 (by (A)
and the de nition of  this only happens if d1 = 4, so v 5 (mod 10)); and
 if i 0,   i 6= ?,   i+d1 = ? then T =   i (i;i+d1;i+ 2d1).
We consider each of the four remaining cases in turn.
First since v> 3, it is clear that (i;i+d1;i+ 2d1;i+ 3d1) is a path.
Second suppose that i> 0,   i+d1 6= ?, and   i 6= ?. We only need to consider the
sets of 3 consecutive edges in   i (i;i + d1)  i+d1 which contain fi;i + d1g. So we need
to show that: (a) i+d1 6= i f i; 1, (b) i6= i+d1 +f i+d1;1, (c) i+d1 6= i f i; 1 f i; 2,
(d) i f i; 1 6= i + d1 + f i+d1;1, and (e) i 6= i + d1 + f i+d1;1 + f i+d1;2. By (3.8)
and since i > 0, i + d1 6= i f i; 1 and i 6= i + d1 + f i+d1;1, so (a) and (b) are
true. To see (d),  rst we use the  exibility in the component trails (see preamble of
Lemma 3.3.6) to ensure that   i+d1 does not begin with an edge of di erence v=2. This
follows by condition (ii) in Lemmas 3.3.4 and 3.3.5, opting to choose the  rst edge
in Ti+d1(D4;v=2) to not have di erence v=2 if it is the  rst trail in   i+d1. Therefore
44
by (Piv), if i f i; 1 = i+d1 +fi+d1;1 then   i+d1 =   i+d1 so i f i; 1 6= i+d1 fi+d1;1
(for otherwise subtracting this equality from the  rst implies that 2fi+d1;1 = 0, so
fi+d1;1 = v=2). Similarly by (Piv), if i f i; 1 = i + d1  fi+d1;1 then   i+d1 =  i+d1
so i f i; 1 6= i + d1 + fi+d1;1. So (d) is true regardless of which of the two values,
 fi+d1;1, f i+d1;1 takes on. We now turn to showing (c) is true. If f i; 1 = f i; 2
then i + d1 6= i f i; 1  f i; 2 for otherwise d1 =  f i; 1  f i; 2 contradicting that
gcd(fv;f i; 1g)6= 1 (by (3.8)) and gcd(fv;d1g) = 1 (by (A)). So we can assume that
f i; 1 6= f i; 2. Notice that the last trail in   i cannot be Ti(fdg) for any di erence
d2D6 and cannot be Ti(jD2j=2fv=3;v=6g) or Ti(D3;v=3) by Lemmas 3.3.2 and 3.3.3
(since in such cases the last two edges have the same di erence). So the last trail
in   i must be one of the following: Ti(jD1j=2fv=2;v=3g), Ti(D4;v=2), or Ti(D5;v=5).
If the last trail of   i is Ti(D4;v=2) or Ti(D5;v=5) then i + d1 6= i f i; 1  f i; 2
since otherwise gcd(fv;d1g) = 1 would mean that gcd(fv;f i; 1;f i; 2g) = 1 which
contradicts Lemma 3.3.4(v) or Lemma 3.3.5(v) respectively since i > 0. If the last
trail of   i is Ti(jD1j=2fv=2;v=3g) then i+d1 6= i f i; 1 f i; 2 since otherwise d1 = v=6
since f i; 1 + f i; 2 2fv=2 v=3g which implies v = 6 by Condition (A) and so these
edges would be in   0 instead of   i . This shows that (c) is true. Lastly, we focus on
showing (e) is true. Since i+d1 > 0, we need only show that the third vertex of   i+d1,
namely i+d1 +f i+d1;1 +f i+d1;2, is not i. This requires considering several cases, most
of which rely on the claim that gcd(ff i+d1;1;f i+d1;2;vg) 6= 1 (later we will show that
if f i+d1;1 6= f i+d1;2 then gcd(ff i+d1;1;f i+d1;2g) 6= 1). If f i+d1;1 = f i+d1;2 (that is, the
 rst two edges of   i+d1 have the same di erence) then i+d1 +f i+d1;1 +f i+d1;2 6= i, for
otherwise i+d1 = i+v f i+d1;1 f i+d1;2 = i+v 2f i+d1;1 contradicting the facts that
gcd(fv;f i+d1g) 6= 1 (by (3.8)), and gcd(fv;d1g) = 1 (by Condition (A)). So we can
assume that the  rst two edges of   i+d1 have di erent di erences: f i+d1;1 6= f i+d1;2.
Then the  rst trail in   i+d1 cannot be Ti+d1(fdg) for any di erence d 2 D6 and
cannot be Ti+d1(D5;v=5) by Lemma 3.3.5 (since in such cases the  rst two edges
have the same di erences). So the  rst trail in   i+d1 must be one of the following:
Ti+d1(jD1j=2fv=2;v=3g), Ti+d1(jD2j=2fv=3;v=6g), Ti+d1(D3;v=3), or Ti+d1(D4;v=2). If
the  rst trail of   i+d1 is Ti+d1(D3;v=3) or Ti+d1(D4;v=2) then i+d1+f i+d1;1+f i+d1;2 6= i
since otherwise gcd(fv;d1g) = 1 would mean that gcd(ff i+d1;1;f i+d1;2g) = 1 which
contradicts Lemma 3.3.3(v) or 3.3.4(v) respectively (since then gcd(fd;d0;vg) = 1
where d0 = v=3 or v=2 respectively and ff i+d1;1;f i+d1;2g = fd;d0g, so these edges
would be in   0 instead of   i+d1). If the  rst trail of   i+d1 is Ti+d1(jD1j=2fv=2;v=3g)
or Ti+d1(jD2j=2fv=3;v=6g) then i + d1 + f i+d1;1 + f i+d1;2 6= i since otherwise by the
descriptions of the  rst two edges of the trails constructed in Lemmas 3.3.1 and 3.3.2,
d1 = v=6 so by Condition (A), v = 6, and so these edges would be in   0 instead of
in   i+d1. So each set of 3 consecutive edges in (i;i + d1)  i+d1 and   0 (0;d1)  d1 induce
a path. So every set of 3 consecutive edges in   i (i;i + d1)  i+d1 induces a path when
i> 0.
Third, suppose that i  0,   i+d1 = ?, and   i+2d1 6= ?, so v  5 (mod 10).
We will show that (i;i + d1;i + 2d1;i + 2d1 + f i+2d1;1) is a path; that is we show
that i 6= i + 2d1 + f i+2d1;1. This follows since d1 = 4 and v  5 (mod 10), so
45
gcd(f2d1;vg) = 1 and by (3.8), gcd(ffi+2d1;1;vg)6= 1 since i+ 2d1 > 0.
Finally, suppose that i 0,   i 6= ?, and   i+d1 = ?. We consider the sets of 3
consecutive edges in   i (i;i + d1;i + 2d1) that contain the edge fi;i + d1g. We will
break this into 2 cases: v 5 (mod 10) and v6 5 (mod 10).
Case 1: Suppose v  5 (mod 10); then d1 = 4. We need to show that i + 4 6=
i (f i; 1 +f i; 2), i f i; 1 6= i+ 4, and i f i; 1 6= i+ 8. By (3.8) and since v is odd,
if  f i; 1 = 4 or 8 then i = 0. By (Pi), if  f0; 1 = 4 then   0 =   0 so  f 0; 1 =  4,
and if  f0; 1 =  4 then   0 =  0 so  f 0; 1 =  4; so in any case  f 0; 1 6= 4 (since
v 6= 8). Similarly, by (Pii), if  f0; 1 = 8 then   0 =   0 so  f 0; 1 =  8, and if
 f0; 1 = 8 then   0 =  0 so  f 0; 1 = 8; so in any case  f 0; 1 6= 8 (since v6= 16).
If f i; 1 f i; 2 = 4 and the last two edges of   i have the same di erence, then since v
is odd this di erence is 2, so by (3.8), i = 0. So by (Pi), if f0; 1 f0; 2 = 4 and the
last two edges of   i have the same di erence then   0 =   0 so  f 0; 1 f 0; 2 =  4,
and if  f0; 1  f0; 2 =  4 and the last two edges of   i have the same di erence
then   0 =  0 so  f 0; 1  f 0; 2 =  4. Therefore in any case  f 0; 1  f 0; 2 6= 4. If
 f i; 1 f i; 2 = 4 and the last two edges of   i have di erent di erences then since
v is odd, the last trail in   i must be Ti(D5;v=5) (see Lemmas 3.3.1-3.3.5). But then
by the de nition of D5, since D5 6= ?, it follows that 2v=5 2D6, which contradicts
Ti(D5;v=5) being the last trail in   i (since the number of components in Gv(D5)
is at most the number of components in Gv(f2v=5g), so Ti(D5;v=5) 6= ? implies
Ti(D6)6= ?). Thus each set of 3 consecutive edges in   i (i;i+ 4;i+ 8) induces a path.
Case 2: Suppose v6 5 (mod 10); then d1 = 1. We need to show that i f i; 1 
f i; 2 6= i + 1, i f i; 1 6= i + 1, and i f i; 1 6= i + 2; this is done by contradiction.
First suppose that either f i; 1 = 1 or f i; 1 = 2 and v is odd; so by (3.8) it follows
that i = 0. Then by (Pi), if  f0; 1 = 1 then   0 =   0 so  f 0; 1 =  1, and by
(Piii) if  f0; 1 =  1 then   0 =  0 so  f 0; 1 =  1; so in any case  f 0; 1 6= 1.
Similarly, by (Pii), if  f0; 1 = 2 then   0 =   0 so  f 0; 1 =  2, and by (Piii) if
 f0; 1 = 2 then   0 =  0 so  f 0; 1 = 2; so in any case  f 0; 1 6= 2. Next suppose
that f i; 1 f i; 2 = 1 and the last two edges of   i have the same di erence. Then v is
odd and this di erence is (v 1)=2, so by (3.8) and since gcd(f(v 1)=2;vg) = 1, i = 0.
By (Pi), if  f0; 1 f0; 2 = 1 and the last two edges of   i have the same di erence
then   0 =   0 so  f 0; 1 f 0; 2 =  1, and by (Piii) if  f0; 1 f0; 2 =  1 and the
last two edges of   i have the same di erence then   0 =  0 so f 0; 1 f 0; 2 = 1; so
in any case f 0; 1 f 0; 2 6= 1. Now suppose that f i; 1 f i; 2 = 1 and the last two
edges of   i have a di erent di erence. Then by Lemmas 3.3.1-3.3.5, the last trail in   i
must be one of the following trails: Ti(jD1j=2fv=2;v=3g), Ti(D4;v=2), or Ti(D5;v=5).
As described at the end of Case 1, Ti(D5;v=5) cannot be the last trail in   i (since
2v=562D6, so D5 = ?, eliminating this case). If Ti(jD1j=2fv=2;v=3g) is the last trail
in   i then f i; 1 f i; 2 6= 1 since otherwiseff i; 1;f i; 2g=fv=2; v=3gand so v = 6
and i = 0, and thus by (Pi) we reach a contradiction (since if f0; 1 f0; 2 = 1 or 1
then   0 =   0 or  0 respectively and so in any case f 0; 1 f 0; 2 = 1). If Ti(D4;v=2)
is the last trail in   i , then  f i; 1 f i; 2 6= 1 since otherwise ff i; 1;f i; 2g=fv=2;dg
where d2D4nmD4(v=2)fv=2gand gcd(fd;v=2;vg) = 1, so i = 0 and thus by (Pi) we
46
reach a contradiction (since if f 0; 1 f 0; 2 = 1 or 1 then   0 =   0 or  0 respectively
and so  f 0; 1 f 0; 2 = 1).
Finally suppose that  f i; 1 = 2 and v is even. The last trail in   i must be one
of the following: Ti(jD1j=2fv=2;v=3g), Ti(jD2j=2fv=3;v=6g), Ti(D3;v=3), Ti(D4;v=2),
Ti(D5;v=5), and Ti(f2g). If Ti(jD2j=2fv=3;v=6g) or Ti(D3;v=3) is the last trail in   i
then by Lemma 3.3.2 or 3.3.3 respectively, the last edge has di erence  v=3 = 2,
so v = 6 and i = 0, and thus by (Pii) and (Piii) we reach a contradiction (since if
 f0; 1 = 2 or  2 then   0 =   0 or  0 respectively and so in any case  f 0; 1 6= 2).
Similarly, if Ti(jD1j=2fv=2;v=3g) is the last trail in   i then by Lemma 3.3.1(ii),
 f i; 1 =  v=2 = 2 and so v = 4, thus v=3 62 D, a contradiction. As previously
described, if Ti(D5;v=5) is the last trail in   i then 2v=562D6, so D5 = ?, eliminating
this case.
We now address when the last trail of   i is Ti(f2g) or Ti(D4;v=2), still assuming
that  f i; 1 = 2 (so the last edge of   i has di erence 2) and v is even. By (3.8)
and Lemma 3.3.4(vii) respectively, it follows that i 1 (to apply Lemma 3.3.4(vii)
note that gcd(fv;v=2;2g) 2f1;2g). Suppose Ti(D4;v=2) is the last trail in   i and
i = 0. Then by (Pii), if  f0; 1 = 2 then   0 =   0 so  f 0; 1 =  2 and by (Piii)
if  f0; 1 =  2 then   0 =  0 so  f 0; 1 =  2; in any case  f 0; 1 6= 2. If Ti(f2g) is
the last trail in   i then i = 1 (to see that i6= 0, note that if i = 0 then since v is
even   i+1 would contain edges of di erence 2 thus contradicting   i+1 = ?). So in all
cases we can now assume that i = 1. If T1(f2g) is the last trail in   1 , then by (3.7),
mD6(2)f2g= D6. Furthermore, if T1(D4;v=2) is the last trail in   1 and if D4 contains
a di erence d not inf2;v=2gthen by Lemma 3.3.4(vi) we avoid this case (since then
the last edge in   i would have di erence d 6= 2); so in this situation we can assume
that D4 = mD4(2)f2g[mD4(v=2)fv=2g.
We are now in the case where   j = ? for all j 2, and the last trail in   1 is either
T1(D4;v=2) or T1(f2g) where D4 = mD4(2)f2g[mD4(v=2)fv=2g or D6 = mD6(2)f2g
respectively. This situation is most easily handled by altering the de nitions of   1
and   v 1 by changing  1 and  v 1. The only change is that one of three things is done
to avoid the new   1 ending on an edge of di erence 2:
T1(D4;v=2)
jD6j+5Y
k=6
T1(f2g) or T1(D4;v=2) or
jD6j+5Y
k=6
T1(f2g) is detached (3.9)
from the end of   1 and moved to   v 1 as described below.
This will ensure that   1 (as newly de ned) does not end on an edge of di erence 2.
We will then need to check that the altering of these trails does not create a 2-cycle
or 3-cycle.
For any trail T, let  (T) be formed by mapping each vertex ?2Zv in the trail
to the vertex ?+v 2 modulo v. Let
T0v 1(D4;v=2) =
(
 (T1(D4;v=2)) if D4 = mD4(2)f2g[mD4(v=2)fv=2g, and
? otherwise.
47
So if T0v 1(D4;v=2)6= ? then it is a closed trail that begins on vertex v 1 and satis es
E(T0v 1(D4;v=2)) = E(T1(D4;v=2)). To see this equality, notice that E(T1(D4;v=2))
is the set of edges in the component of Gv((jD4j=2)f2;v=2g) that contains vertex 1;
since v is even, this component also contains the vertex v 1. Similarly de ne
T0v 1(f2g) =
(
 (T1(f2g)) if 22D6, and
? otherwise:
Clearly E(T1(f2g)) = E(T0v 1(f2g)) and T0v 1(f2g) starts and ends on vertex v 1.
Since  preserves the distance between consecutive vertices in the trails T1(D4;v=2)
and T1(f2g), each set of 3 consecutive edges in T0v 1(D4;v=2) and T0v 1(f2g) induces
a path. Also,  preserves the  exibility to choose the  rst 2 edges in T0v 1(D4;v=2)
as stated in Lemma 3.3.4(ii) and the  exibility in our choice of the direction of the
 rst edge in T0v 1(f2g). Let
T01(D4;v=2) =
(
? if D4 = mD4(2)f2g[mD4(v=2)fv=2g, and
T1(D4;v=2) otherwise.
We are now ready to rede ne
 1 = T1(jD1j=2fv=2;v=3g)T1(jD2j=2fv=3;v=6g)T1(D3;v=3)T01(D4;v=2)
(notice that T1(D5;v=5) = ? since 2v=562D6) and rede ne
 v 1 = T0v 1(D4;v=2)
0
@
jD6j+5Y
k=6
T0v 1(f2g)
1
A:
As described in (Piv), for all j 0 if  j+1 6= ?,   j 6= ?, and 1+fj+1;1 2f0;f j; 1gthen
we still de ne   j+1 =   j+1 and otherwise   j+1 =  j+1. Since  j = ? for 2 j v 2
and since v> 3 in this lemma,   j =  j for 2 j v 1. Then since d1 = 1 we de ne
 =
v 1Y
j=0
   
j (j;j + 1)
 =   
0 (0;1) 
 
1 (1;2;3;:::;v 1) 
 
v 1(v 1;0):
In view of (3.9) it remains to check there are no 2-cycles or 3-cycles in the following
trails:   1 (1;2;3) and (v 3;v 2;v 1)  v 1(v 1;0). Note that since v > 3 in
this lemma we know that there are at least 2 edges of di erence 1 between   1 and
  v 1. Since   1 cannot end on an edge of di erence 2 by (3.9) (recall that   1 does not
contain the trail T1(f2g) and if T01(D4;v=2)6= ? then D4 contains a di erence not in
f2;v=2gand by Lemma 3.3.4(vi) we let it be d so that we avoid this case), following
the original argument shows that each set of 3 consecutive edges in   1 (1;2;3) induces
a path. By exercising the choice we have in the  rst two edges of T0v 1(D4;v=2)
and T0v 1(f2g), we can ensure that the  rst edge of   v 1 is fv 1;1g (an edge of
48
di erence 2); so f v 1;1 = 2. It remains to show that: v 3 6= v 1 + f v 1;1 = 1;
v 2 6= v 1 + f v 1;1 = 1 (this situation cannot occur since v > 3); v 2 6=
v 1+f v 1;1 +f v 1;2 = 1+f v 1;2; v 1 f v 1; 1 6= 0; and v 1 f v 1; 2 f v 1; 1 6= 0.
If v = 4 then by Lemma 3.3.4(v) all edges would occur in   0 , contradicting i = 1 so
the  rst situation cannot occur. By Lemma 3.3.4 and the de nition of T0v 1(f2g) the
second edge of   v 1 is either f1;3g or f1;v=2 + 1g. Then v 2 6= 1 + f v 1;2 since
otherwise v = 5 (a contradiction since v6 5 (mod 10)) or v = 6 (a contradiction since
2 = v=3 and v=362D4) respectively, so the third situation is resolved. Since each edge
in   v 1 is an edge of di erence 2 or v=2, it follows that v 1 f v 1; 1 6= 0. Also since
each edge in   v 1 is an of di erence 2 or v=2, it follows that v 1 f v 1; 2 f v 1; 1 6= 0
since otherwise v = 6, and so 2 = v=3, a contradiction since v=362D4.
Thus  is a closed trail such that every set of 3 consecutive edges in  induce a
path.
The following lemma makes use of the trail decomposition formed in Lemma 3.3.7
to construct a 5{CS using a speci ed number of 5-cycles of Type A, B and C. This
lemma will be useful in using up all of the di erences needed to cover the 2v( +m)
mixed edges in ( + m)Kv+u  Kv. De ne degH(v) to be the degree of v in the
multigraph of H.
For any two vertex disjoint graphsG1 andG2, de neG1_tG2 to be the multigraph
with vertex set V(G1_tG2) = V(G1)[V(G2) and edge set E(G1_tG2) = E(G1)[
E(G2)[(tffg1;g2gjg1 2V(G1);g2 2V(G2)g).
Lemma 3.3.8. Suppose that:
(a) the conditions of Lemma 3.2.1 are satis ed;
(b) there exist sets  1 and  2 of di erences such that  1[ 2  D(v)m;
(c) D =  1 satis es the conditions of Lemma 3.3.7;
(d) j 1j+j 2j= d; and
(e) j 2j= (c a 2b)=v.
Then there exists a 5{CS of Gv( 1[ 2)_ +m ( +m)K2.
Remark. Since v divides c a 2b by Lemma 3.2.1(vii) and c a 2b  0 by
Lemma 3.2.1(vi), j 2j is a non-negative integer.
Proof. De ne the required 5{CS (Zv[f11;12g;B) by following the approach de-
scribed at the end of Section 3.2. The set B will contain a, b, and c 5-cycles of Types
A, B, and C respectively, thereby exhausting all the pure edges (as will be seen).
By Lemma 3.2.1(iii), 2a + 3b + c = vd = v(j 1j+j 2j). Edges in Gv( 1) will
be used to construct b paths of length 3 and 2b paths of length 1 (that are combined
into b=2 trails of length 10), and a paths of length 2 and a paths of length 1 (that are
combined into a paths of length 3); the remaining c a 2b ( 0 by Lemma 3.2.1(vi))
paths of length 1 will occur in Gv( 2). SojE(Gv( 1))j= 3b+ 2a+ 2b+a = 3a+ 5b
and since Gv( 1) is regular, each vertex has degree 2(3a + 5b)=v. Notice that by
49
Lemma 3.2.1(iii,vii), 3a+ 5b is divisible by v.
By Lemma 3.3.7 there exists a closed trail T = (g0;g1;:::;gvj 1j = g0) of Gv( 1)
such that every 3 consecutive edges induces a path. The number of edges in T is
vd (c a 2b) = 5b+ 3a (by Lemma 3.2.1(iii)). Arbitrarily break T into a set T1
of b=2 trails that are each of length 10 and a set T2 of a paths that are each of length
3. Form a set of 5-cycles B1 = St2T1[T2 B1(t) where
B1(gi;gi+1;gi+2;gi+3) =f(gi;gi+1;gi+2;12;11);(gi+2;gi+3;12;gi+1;11)g and
B1(gi;gi+1;:::;gi+10) =f(gi;gi+1;12;gi+2;11);(gi+1;gi+2;gi+3;gi+4;11);
(gi+4;gi+5;11;gi+3;12);(gi+5;gi+6;11;gi+7;12);
(gi+6;gi+7;gi+8;gi+9;12);(gi+9;gi+10;12;gi+8;11)g
Notice that for each t2T1[T2, each vertex gj in t is joined to11 and12 in the
5-cycles in B1(t) with degt(gj)=2 edges except for the  rst vertex and the last vertex
in t if they are di erent, in which case they have odd degree in t, so the  rst is joined
with one less edge to 12 than to 11 and the last is joined with one less edge to 11
than to12. Therefore, since T is a closed trail, each vertex z in Zv is joined to each
of 11 and 12 with degT(z)=2 = (3a+ 5b)=v edges in 5-cycles in B1.
Finally, let B2 = f(z;z + d;12;z + d + 1;11) jd2  2;z 2 Zvg. Then each
vertex in Zv is joined to each of11 and12 with 2(c a 2b)=v edges in 5-cycles in
B2.
Then each vertex z2Zv is joined to each of11 and12 with degT(z)=2 = (3a+
5b)=v edges in cycles in B1 and 2(c a 2b)=v edges in cycles in B2. So each z2Zv is
joined to each of11 and12 with ((3a+5b)+2(c a 2b))=v = (a+b+2c)=v =  +m
by Lemma 3.2.1(iv). Also, each of the a trails of length 3 give rise to a edges joining
11 to 12 in B1, so 11 and 12 are joined by a =  + m edges as required. Since
clearly the edges of Gv( 1[ 2) occur in T, (Zv[f11;12g;B = B1[B2) is a 5{CS
of Gv( 1[ 2)_ +m ( +m)K2.
3.4 The small cases for m
As in Chapter 2, we embark on a proof of the main result by beginning with a
section which will prove the necessary conditions in Lemma 2.1.2 are su cient when
u = 2 and m is small.
Lemma 3.4.1. Let v = 10k,  and m be integers satisfying Conditions (a,b,d) of
Lemma 2.1.2 with u = 2. If m = 2 or if m2f4;6;8g and  + m = 10, then any
5{CS of  Kv can be enclosed in a 5{CS of ( +m)Kv+2.
Proof. Let (V = Z10k;A) be a 5{CS of  Kv and let V [U be the vertex set of
( + m)Kv+2 where U =f11;12g. Since (V;A) exists, it follows by Theorem 1.2.2
that  is even. So m is even by Lemma 2.1.2(a). By Lemma 2.1.2(b)  + m  0
(mod 5). Therefore  + m  0 (mod 10). Thinking of v=2 as half a di erence,
we begin by choosing  v edges to place in cycles containing only pure edges (edges
50
between vertices only in V or only in U) using  di erences in Dv(m) where  is
de ned in Equation (3.1).
Suppose m = 2. Then mKv = 2Kv contains edges of each di erence in D(v)2.
Care must be taken with edges of di erence v=2 in E(Kv) since there are only v=2
of them. To explicitly describe the 5-cycles containing all the edges of  di erences
(yet to be chosen), we make use of Skolem and hooked Skolem sequences.
By Theorem 1.1.1 or 1.1.2, let P0 =f(x0i;y0i)j1 i k 2;y0i >x0ig correspond
to a Skolem sequence or hooked Skolem sequence of order k 2 with k 2 0 or 1
(mod 4) or k 2 2 or 3 (mod 4) respectively; so each element inf1;2;:::;2k 4g
orf1;2;:::;2k 3gnf2k 4grespectively occurs in exactly one element of P0. Now
de ne P = f(xi = x0i + 3;yi = y0i + 3) j (x0i;y0i) 2P0g if k 2  0 or 1 (mod 4) and
P =f(xi = x0i + 4;yi = y0i + 4)j(x0i;y0i)2P0g if k 2 2 or 3 (mod 4). Then with
ci =
8>
>>><
>>>>
:
(0;xi;xi yi;3k;5k + 1 +i) if 1 i k 2,
(0;3;4k + 2;k + 2;5k + 1) if i = k 1,
(0;2k + 1;4k + 1;2k;5k) if i = k and k 2 0;1 (mod 4), and
(0;4;2k + 4;5k + 4;5k) if i = k and k 2 2;3 (mod 4),
(3.10)
(where arithmetic is done modulov) form the set of 5-cyclesB0 =  c00i 1 = cij1 i k 
where computations are done modulo v. Table 3.1 lists the di erences of edges that
occur in cycles in B0 when  is as big as possible (from Lemma 3.2.1 this happens when
 is as small as possible and not 0, namely when  = 8, and so when  = 10k 11;
notice that when  = 0 a 5{CS of  Kv can be enclosed in a 5{CS of ( + m)Kv+u
by Theorem 1.2.2).
Edges Di erenceswhenk 2 0;1 (mod 4) Di erenceswhenk 2 2;3 (mod 4)
f0;xig;fxi;xi yig 4;5;:::;2k 1 5;6;:::;2k 1;2k+1
fxi yi;3kg 3k+1;3k+2;:::;4k 2 3k+1;3k+2;:::;4k 2
f3k;5k+1+ig 2k+2;2k+3;:::;3k 1 2k+2;2k+3;:::;3k 1
f5k+1+i;0g 5k 2;5k 3;:::;4k+1 5k 2;5k 3;:::;4k+1
ci ifi=k 1 3;3k;4k 1;4k 1;5k 1 3;3k;4k 1;4k 1;5k 1
ci ifi=k 2k;2k+1;2k+1;3k;5k 4;4;2k;3k;5k
Table 3.1: Edges of di erences in B0 when  is as large as possible, v 0 (mod 10),
and m = 2
Let B00 = 2B0 be the multiset consisting of 2 copies of each 5-cycle in B0. De ne
B =fc00i +jji2Z =5;c00i (mod k) 2B00;j2Zvg[E where
E =
8
>>>
>>>
<
>>>
>>>
:
? if   0 (mod 5),
 2 if   1 (mod 5),
 1[ 2 if   2 (mod 5),
 3 if   3 (mod 5), and
 2[ 3 if   4 (mod 5).
51
Notice that ck 1 and ck will only be chosen once in B. Fortunately when  is as
small as possible ( is as large as possible)  = 10k 11 and so E = ? allowing the
di erences 3 and 2k to be used in ck 1.
Let D be the set of di erences not used in forming B. Notice that because m = 2,
f1;2g D. Partition D into two sets  1 and  2 so thatf1;2g  1 and  1 contains
as many di erences that are not v=2, v=3, v=4, v=5, or 2v=5 as possible. When  is
as small as possible, it is clear that Conditions (A-D) of Lemma 3.3.7 are satis ed.
Whenever  increases  will decrease by 10 and so 10 more di erences (at most 2
copies of the same di erence) will be contained in D. So there are still enough di erent
di erences for any  value ensuring  1 still satis es Conditions (A-D) of Lemma 3.3.7.
So by Lemma 3.3.8, there exists a 5{CS (V [U;C) of Gv(D)_ +m ( +m)K2 which
covers each mixed edge 2v( + m) times. Thus, (Z10k[f11;12g;A[B[C) is a
5{CS(v + 2; + 2) that encloses the given 5{CS (Zv;A).
Let m 2f4;6;8g and  + m = 10, so d =  + m. Now let P0 = f(xi;yi) j
1  i k 2;yi > xig be the collection of pairs from a Skolem sequence or hooked
Skolem sequence of order k 2 for the set f1;2;:::;2k 4g or f1;2;:::;2k 3g if
k 2  0 or 1 (mod 4) or k 2  2 or 3 (mod 4) respectively. So each element
in f1;2;:::;2k 4g or f1;2;:::;2k 3gnf2k 4g occurs in exactly one element of
P0. Now de ne P = f(xi = x0i + 3;yi = y0i + 3) j (x0i;y0i) 2P0g when using Skolem
sequences and P = f(xi = x0i + 4;yi = y0i + 4) j (x0i;y0i) 2 P0g. Then with ci as
de ned in Equation (3.10) form the set of 5-cycles B0 = 2fci + j jj 2Zv;1  i 
k 2;(xi;yi) 2Pg[fci + jjj2Zv;i2fk 1;kgg where computations are done
modulo v. Let P =f(xi;yi)j1 i k 1;yi >xigbe the collection of pairs from a
Skolem sequence or hooked Skolem sequence of order k 1 for the setf1;2;:::;2k 2g
orf1;2;:::;2k 3gif k 1 0 or 1 (mod 4) or k 1 2 or 3 (mod 4) respectively.
Then with
c0i =
(
(0;xi;xi yi;3k;5k +i) if 1 i k 1, and
(0;2k;6k;9k;5k) if i = k,
for the set of 5-cycles B00 = 2fc0i +jjj2Zv;1 i k 1;(xi;yi)2P g[fc0k +jj
j2Zvg. Table 3.2 lists the di erences of edges that occur in cycles in B00 when  is
as large as possible (from Lemma 3.2.1 this happens when  is as small as possible
and not 0; notice that when  = 0 a 5{CS of  Kv can be enclosed in a 5{CS of
( +m)Kv+u by Theorem 1.2.2).
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 1;2;:::;2k 2 1;2;:::;2k 3;2k 1
fxi yi;3kg 3k+1;3k+2;:::;4k 1 3k+1;3k+2;:::;4k 1
f3k;5k+ig 2k+1;2k+3;:::;3k 1 2k+1;2k+3;:::;3k 1
f5k+i;0g 5k 1;5k 2;:::;4k+1 5k 1;5k 2;:::;4k+1
c0i ifi=k 2k;3k;4k;4k;5k 2k;3k;4k;4k;5k
Table 3.2: Edges of di erences in B00 when  is as large as possible, v 0 (mod 10),
and m2f4;6;8g
52
Then ((m 2)=2)B00[B0 contains (m 2)(2k 1)=2 + 2k 2 = mk m=2 1 5-
cycles, but we needb =5c= mk 3 of them. Hence we need 1 or 2 more 5-cycles when
m = 6 or m = 8 respectively. Let c001 = (0;2k;4k;6k;3k) and c002 = (0;2k;4k;7k;3k).
Then let E0 contain 1 copy of c01 if m = 6 or 8 and 1 copy of c02 if m = 8. Notice that
the 5-cycles in E0 contain at most m copies of di erences of edges not used in cycles
in ((m 2)=2)B00[B0. Form the set of 5-cycles
B = ((m 2)=2)B00[B0[fc00i +jjj2Zv;c00i 2E0 g[E :
Since  +m = 10, it follows that (m(v 1)=2  ) =  +m = 10. Let D be the
set of di erences not used in B. Notice that D contains at least (m 2)=2 copies of
f2k;3k;2k 1gwhenk 1 0 or 1 (mod 4) or (m 2)=2 copies off2k;3k;2k 2gwhen
k 1  2 or 3 (mod 4). Partition D into two sets  1 and  2 so that f1;2g  1
and  1 contains as many di erences that are not v=2, v=3, v=4, v=5, or 2v=5 as
possible. In this way, Conditions (A-D) of Lemma 3.3.7 are satis ed for  1. So by
Lemma 3.3.8 there exists a 5{CS (V [U;C) of Gv(D)_ +m ( +m)K2 ensuring we
cover each mixed edge 2v( + m) times. Therefore (Zv[f11;12g;A[B[C) is a
5{CS(v + 2; +m) with m2f4;6;8g and  +m = 10.
The proof of the next several results follows that of Lemma 3.4.1 closely, but are
su ciently di erent and important that they are described in detail.
Lemma 3.4.2. Let v = 10k + 5 and assume Conditions (a,b,d) of Lemma 2.1.2 are
satis ed. If m = 1 or if m2f2;3;4g and  + m = 5, then any 5{CS of  Kv can be
enclosed in a 5{CS of ( +m)Kv+2.
Proof. Let (V = Z10k+5;A) be a 5{CS of  Kv and let U = f11;12g. Since (V;A)
exists,  + m 0 (mod 5) by Lemma 2.1.2(a). As in Lemma 3.2.1, the number of
edges we must place in 5-cycles contained completely in mKv is
 m(v 1)
2  
2a+ 3b+c
v
 
v =  v:
Since this is divisible by v, our plan is again to create 5-cycles in mKv using all the
edges of  specially chosen di erences. By Lemma 3.3.8,  is a non-negative integer.
Suppose m = 1. Then mKv = Kv contains edges of each di erence in D(v) =
f1;:::;5k+2g. As in Lemma 3.4.1, we will use Skolem and hooked Skolem sequences
to explicitly describe the 5-cycles containing all the edges of  di erences (yet to be
chosen).
We begin with the case where k 2. Let P0 =f(x0i;y0i)j0 i k 3;y0i >x0ig
be the collection of pairs from a Skolem sequence or hooked Skolem sequence of order
k 2 from the set f1;2;:::;2k 4g or f1;2;:::;2k 3g if k 2  0 or 1 (mod 4)
or k 2 2 or 3 (mod 4) respectively (this exists by Theorems 1.1.1 and 1.1.2 since
k 3). So each element inf1;2;:::;2k 4gorf1;2;:::;2k 3gnf2k 4goccurs in
exactly one element of P0. Now de ne P =f(xi = x0i + 4;yi = y0i + 4)j(x0i;y0i)2P0g
53
when using Skolem sequences and P =f(xi = x0i+5;yi = y0i+5)j(x0i;y0i)2P0gwhen
using hooked Skolem sequences. Notice that because  = 5k+2 d = 5k+2 ( +m)
or 0 (depending on d) and  + m 0 (mod 5), it follows that   0 or 2 (mod 5).
Then with
ci =
(
(0;xi;xi yi;4k + 2;7k + 3 i) if 0 i k 3 and;
(0;3k + 1;6k + 3;k + 2;5k + 3) if i = k 2, (3.11)
we form the set of 5-cycles B =  ci (mod k 1) +jji2Zb =5c;j2Zv [E where the
computations are done modulo v, k 2, and where E = ? or  1[ 2 if   0 or
2 (mod 5) respectively. Table 3.3 lists the di erences of edges that occur in cycles in
BnE if  is as big as possible (from Lemma 3.2.1 this happens when  is as small as
possible, namely when  = 4 and so  = 5k 3). By design, the edges of di erences
Edges Di erenceswhenk 2 0;1 (mod 4) Di erenceswhenk 2 2;3 (mod 4)
f0;xig;fxi;xi yig 5;6;:::;2k 6;7;:::;2k;2k+2
fxi yi;4k+2g 4k+3;4k+4;:::;5k 4k+3;4k+4;:::;5k
f4k+2;7k+3 ig 3k;3k 1;:::;2k+3 3k;3k 1;:::;2k+3
f7k+3 i;0g 3k+3;3k+4;:::;4k 3k+3;3k+4;:::;4k
ci ifi=k 2 3k+1;3k+2;4k+1;5k+1;5k+2 3k+1;3k+2;4k+1;5k+1;5k+2
Table 3.3: Edges of di erences in BnE when  is as large as possible and v 5
(mod 10)
v=5 and 2v=5 do not occur in 5-cycles in BnE , hence are available to form the
5-cycles in E . If k = 0 then v = 5 and it is easy to see how to form the set 5-cycles
B using di erences so that B contains only 5-cycles with 5 pure edges. If k = 1 then
  6  , so form the set of 5-cycles B =b =5cf(0;1;3;6;10) +jjj2Zvg[E .
Let D be the set of di erences not used in B. From Table 3.3 we knowf2;4g2D.
Partition D into two sets  1 and  2 so that f2;4g  1 and  1 contains as many
di erences that are not v=3, v=5, or 2v=5 as possible. If  is as small as possible ( 
is as large as possible) then  = 5k 3 and so 1;3 2D. From this, it is clear that
Conditions (A-D) in Lemma 3.3.7 are satis ed for  1 when  is as small as possible.
As  increases ( decreases) the di erences used to construct ci are added to D. In
this manner, it is clear that  1 still satis es Conditions (A-D) of Lemma 3.3.7. So
by Lemma 3.3.8 we can form a 5{CS (V [U;C) of Gv(D)_ +m ( +m)K2 ensuring
we cover each mixed edge 2v( +m) times. Thus, (Z10k+5[f11;12g;A[B[C) is
a 5{CS(v + 2; + 1) that encloses the given 5{CS (Zv;A).
Let m2f2;3;4g and  +m = 5. Now let P0 =f(xi;yi)j0 i k 3;yi >xig
be the collection of pairs from a Skolem sequence or hooked Skolem sequence of
order k 2 for the set f1;2;:::;2k 4g or f1;2;:::;2k 3g if k 2  0 or 1
(mod 4) or k 2 2 or 3 (mod 4) respectively. So each element inf1;2;:::;2k 4g
or f1;2;:::;2k 3gnf2k 4g occurs in exactly one element of P0. Now de ne
P = f(xi = x0i + 4;yi = y0i + 4) j (x0i;y0i) 2 P0g when using Skolem sequences and
54
P =f(xi = x0i + 5;yi = y0i + 5)j(x0i;y0i)2P0g when using hooked Skolem sequences.
De ne the set B0 =fciji2Zk 1;(xi;yi)2Pg. Then we form the set of 5-cycles
B =fci (mod k 1) ji2Zb =5cg[(m 1)Gv( 1)[(m 1)Gv( 2)[(m 1)Gv( 3)
where computations are done modulo v and ci is de ned in Equation (3.11). Since
b =5c= mk 1 and B0 contains mk m 5-cycles, the extra m 1 copies of Gv( 1)[
Gv( 2)[Gv( 3) ensures we have enough pure 5-cycles in B.
Notice that regardless of our choice of m,  +m = 5 and so (m(v 1)=2  ) =
 +m = 5. Let D be the set of di erences not used in B. There are m copies of both
4 and 2k+ 2 in D when k 2 0 or 1 (mod 10) and m copies of both 4 and 5 when
k 2 2 or 3 (mod 4). Partition D into two sets  1 and  2 so thatf4;2k+2g  1
or f4;5g  1 if v  0 or 1 (mod 4) or v  2 or 3 (mod 4) respectively and  1
contains as many di erences that are not v=3, v=5, or 2v=5 as possible. It is clear
that  1 satis es Conditions (A-D) of Lemma 3.3.7. By Lemma 3.3.8 there exists
a 5{CS (V [U;C) of Gv(D)_ +m ( + m)K2 ensuring we cover each mixed edge
2v( +m) times. Therefore (Zv[f11;12g;A[B[C) is a 5{CS(v+ 2; +m) with
m2f2;3;4g and  +m = 5.
Lemma 3.4.3. Let v 1;3; or 9 (mod 10) and assume Conditions (a,b,d) of
Lemma 2.1.2 are satis ed. If m = 1 or if m 2f2;3;4g,  + m = 5 and v  1
(mod 10), then any 5{CS of  Kv can be enclosed in a 5{CS of ( +m)Kv+2.
Proof. Let (V = Z10k+?;A) be a 5{CS of  Kv and let U = f11;12g. Since (V;A)
exists,  + m 0 (mod 5) when v 1 (mod 10) and   0 (mod 5) when v 3;9
(mod 10) by Lemma 2.1.2(a-b). As in Lemma 3.2.1, the number of edges we must
place in 5-cycles contained completely in mKv is
 m(v 1)
2  
2a+ 3b+c
v
 
v =  v:
Since this is divisible by v, our plan is again to create 5-cycles in mKv using all the
edges of  specially chosen di erences.
Suppose m = 1. Then mKv = Kv contains edges of each di erence in D(v) =
f1;:::;5k + (? 1)=2g. As in Lemma 3.4.1, we will use Skolem and hooked Skolem
sequences to explicitly describe the 5-cycles containing all the edges of  di erences
(yet to be chosen).
If ? = 1, let P0 = f(x0i;y0i) j 1  i k 2;y0i > x0ig be the collection of pairs
from a Skolem sequence or hooked Skolem sequence of order k 2 from the set
f1;2;:::;2k 4g or f1;2;:::;2k 3g if k 2  0 or 1 (mod 4) or k 2  2 or 3
(mod 4) respectively. If ? 2f3;9g let P0 = f(x0i;y0i) j 1  i  k 1;y0i > x0ig be
the collection of pairs from a Skolem sequence or hooked Skolem sequence of order
k 1 from the set f1;2;:::;2k 2g or f1;2;:::;2k 1g if k 1  0 or 1 (mod 4)
or k 1  2 or 3 (mod 4) respectively. So each element in f1;2;:::;2k 4g or
f1;2;:::;2k 3gnf2k 4g occurs in exactly one element of P0 when ? = 1. When
?2f3;9g, each element inf1;2;:::;2k 2gorf1;2;:::;2k 1gnf2k 4goccurs in
55
exactly one element of P0. Now de ne P =f(xi = x0i + 4;yi = y0i + 4)j(x0i;y0i)2P0g
if ? = 1 or P =f(xi = x0i + 3;yi = y0i + 3)j(x0i;y0i)2P0g if ?2f3;9g. Then with
ci =
8>
>>>>
>>>
>><
>>>
>>>>
>>>
:
(0;xi;xi yi;3k;5k + 2 +i) if 1 i b =5c 1 and ? = 1,
(0;xi;xi yi;3k + 1 +?=3;8k 2 +? i) if 1 i b =5c 1 and ?2f3;9g,
(0;2k + 1;6k;10k;5k + 1) if i =b =5c, k 2 0;1 (mod 4) and ? = 1,
(0;4;2k + 4;2;5k + 1) if i =b =5c, k 2 2;3 (mod 4), and ? = 1,
(0;xi;xi yi;3k + 1 +?=3;8k 2 +? i) if i =b =5c and ? = 3,
(0;2k + 2;5k + 4;k;4k + 3) if i =b =5c, k 1 0;1 (mod 4), and ? = 9, and
(0;2k + 1;5k + 4;2k + 2;6k + 5) if i =b =5c, k 1 2;3 (mod 4), and ? = 9,
(3.12)
we form the set of 5-cycles B = fci +jj1 i b =5c;j2Zv;(xi;yi)2Pg where
the computations are done modulo v. By Lemma 3.2.1,   0 (mod 5). Tables 3.4
and 3.5 list the di erences of edges that occur in cycles in B if  is as big as possible
(from Lemma 3.2.1 this happens when  is as small as possible, namely when  = 4
if ? = 1 and  = 5 if ?2f3;9g and so  = 5k 5 if ?2f1;3g and  = 5k if ? = 9).
Edges Di erenceswhenk 2 0;1 (mod 4) Di erenceswhenk 2 2;3 (mod 4)
f0;xig;fxi;xi yig 5;6;:::;2k 5;6;:::;2k 1;2k+1
fxi yi;3kg 3k+1;3k+2;:::;4k 2 3k+1;3k+2;:::;4k 2
f3k;5k+2+ig 2k+3;2k+4;:::;3k 2k+3;2k+4;:::;3k
f5k+2+i;0g 5k 2;5k 3;:::;4k+1 5k 2;5k 3;:::;4k+1
ci ifi=b =5c 2k+1;4k 1;4k;5k 1;5k 4;2k;2k+2;5k 1;5k
Table 3.4: Edges of di erences in B when  is as large as possible for ? = 1
Edges Di erenceswhenk 1 0;1 (mod4) Di erenceswhenk 1 2;3 (mod4)
f0;xig;fxi;xi yig 4;5;:::;2k+1 4;5;:::;2k;2k+2fx
i yi;3k+?=3g 3k+1+?=3;3k+2+?=3;:::;4k 1+?=3 3k+1+?=3;3k+2+?=3;:::;4k 1+?=3f3k+?=3;8k 2+? ig 5k 2+2?=3;5k 3+2?=3;:::;4k+2?=3 5k 2+2?=3;5k 3+2?=3;:::;4k+2?=3
f8k 2+? i;0g 2k+3;2k+4;:::;3k+1 2k+3;2k+4;:::;3k+1c
i ifi=b =5cand?=9 2k+2;3k+2;3k+3;4k+3;4k+4 2k+1;3k+2;3k+3;4k+3;4k+4
Table 3.5: Edges of di erences in B when  is as large as possible for ?2f3;9g
Let D be the set of di erences not used in B. When  is as large as possible ( 
is as small as possible)
D =
8
>>>
>>>
<
>>>
>>>
:
f1;2;3;2k + 2g if v 1 (mod 10) and k 0;1 (mod 4),
f1;2;3;4k 1;4kg if v 1 (mod 10) and k 2;3 (mod 4),
f1;2;3;2k + 2;5k + 1g if v 3 (mod 10),
f1;2;3;4k + 5g if v 9 (mod 10) and k 0;1 (mod 4), and
f1;2;3;4k + 5g if v 9 (mod 10) and k 2;3 (mod 4).
56
Partition D into two sets  1 and  2 so thatf1;2;3g  1 and  1 contains as many
di erences that are notv=3 as possible. It is clear that  1 satis es Conditions (A-D) of
Lemma 3.3.7 when  is as small as possible. As  increases, the di erences represented
in ci are added to D. So it is clear that Conditions (A-D) of Lemma 3.3.7 will still
be satis ed and so by Lemma 3.3.8 there exists a 5{CS (V [U;C) of Gv(D)_ +m
( + m)K2 ensuring we cover each mixed edge 2v( + m) times. Thus, (Z10k+?[
f11;12g;A[B[C) is a 5{CS(v + 2; + 1) that encloses the given 5{CS (Zv;A).
Let v 1 (mod 10), m2f2;3;4g, and  +m = 5. As before, let P0 =f(x0i;y0i)j
1  i k 2;y0i > x0ig be the collection of pairs from a Skolem sequence or hooked
Skolem sequence of order k 2 from the set f1;2;:::;2k 4g or f1;2;:::;2k 3g
if k 2  0 or 1 (mod 4) or k 2  2 or 3 (mod 4) respectively. Now de ne P =
f(xi = x0i + 4;yi = y0i + 4)j(x0i;y0i)2P0g. Then with ci de ned as in Equation (3.12)
we form the set of 5-cycles B0 =fcij1 i k 1;(xi;yi)2Pgwhere computations
are done modulo v. Let B00 = mB0. Then B00 contains m(k 1) 5-cycles, but we need
b =5c= mk 1 of them. Hence we need m 1 more 5-cycles. Let
c0i =
8
>>>
<
>>>
:
(0;1;2k + 3;2k + 5;3) if i = 1 and k 0;1 (mod 4),
(0;1;5;7;3) if i = 2 and k 0;1 (mod 4),
(0;1;4k;4k + 2;3) if i = 1 and k 2;3 (mod 4), and
(0;1;4k + 1;4k + 3;3) if i = 2 and k 2;3 (mod 4).
Then let E0 contain 1, 1, or 2 copies of c01 if m = 2; 3, or 4 respectively, and contain
1 or 2 copies of c02 if m = 3 or 4 respectively. Notice that the cycles in E0 contain at
most m copies of edges not used in cycles in B00. Then we form the set of 5-cycles
B =fci +jj1 i b =5c;ci (mod k 1) 2B00;j2Zvg[fc0i +jjj2Zv;c0i2E0 g:
Notice that regardless of our choice of m,  + m = 5. So (m(v 1)=2  ) =
 +m = 5. Let D be the set of di erences not used in B. It is clear that D contains
1;2; and 3. Partition D into two sets  1 and  2 so that f1;2;3g  1 and  1
contains as many di erences that are not v=3 as possible. Since v=5;2v=5;v=262D,
it is clear that  1 satis es Conditions (A-D) of Lemma 3.3.7 and by Lemma 3.3.8
there exists a 5{CS (V[U;C) of Gv(D)_ +m( +m)K2 ensuring we cover each mixed
edge 2v( +m) times. Therefore (Zv[f11;12g;A[B[C) is a 5{CS(v+ 2; +m)
with m2f2;3;4g and  +m = 5.
Lemma 3.4.4. Let v 4;6; or 8 (mod 10) and assume Conditions (a,b,d) of
Lemma 2.1.2 are satis ed. If m = 2 or if m 2f4;6;8g,  + m = 10, and v  6
(mod 10), then any 5{CS of  Kv can be enclosed in a 5{CS of ( +m)Kv+2.
Proof. Let (V = Zv;A) be a 5{CS of  Kv and let U =f11;12g. Since (V;A) exists,
  m 0 (mod 2) by Lemma 2.1.2(a). As in Lemma 3.2.1, the number of edges we
must place in 5-cycles contained completely in mKv is
 m(v 1)
2  
2a+ 3b+c
v
 
v =  v:
57
Since this is divisible by v, our plan is again to create 5-cycles in mKv using all the
edges of  specially chosen di erences.
Suppose m = 2. Then mKv = 2Kv contains edges of each di erence in D(v)2.
Care must be taken with edges of di erence v=2 in E(Kv) since there are only v=2
of them. As in Lemma 3.4.1, we will use Skolem and hooked Skolem sequences to
explicitly describe the 5-cycles containing all the edges of  di erences (yet to be
chosen).
Let P0 =f(x0i;y0i)j1 i k 1;y0i >x0igbe the collection of pairs from a Skolem
sequence or hooked Skolem sequence of order k 1 from the setf1;2;:::;2k 2gor
f1;2;:::;2k 1g if k 1  0 or 1 (mod 4) or k 1  2 or 3 (mod 4) respectively.
So each element inf1;2;:::;2k 2gorf1;2;:::;2k 1gnf2k 2goccurs in exactly
one element of P0. Now de ne P =f(xi = x0i + 3;yi = y0i + 3)j(x0i;y0i)2P0g. Then
with
ci =
(
(0;xi;xi yi;3k + 1;5k + 3 +i) if 1 i k 1 and ?2f4;6g,
(0;xi;xi yi;3k + 3;5k + 5 +i) if 1 i k 1 and ? = 8, (3.13)
we form the set of 5-cycles B0 =  c00i 1 = cij1 i k 1 where the computations
are done modulo v. Notice that based on our choice of d,   0 (mod 5). Tables 3.6
and 3.7 list the di erences of edges that occur in cycles in B if  is as big as possible
(from Lemma 3.2.1 this happens when  is as small as possible, namely when  = 10
if v 4;8 (mod 10) or  = 8 if v 6 (mod 10)).
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 4;5;:::;2k+1 4;5;:::;2k;2k+2
fxi yi;3k+1g 3k+2;3k+3;:::;4k 3k+2;3k+3;:::;4k
f3k+1;5k+3+ig 2k+3;2k+4;:::;3k+1 2k+3;2k+4;:::;3k+1
f5k+3+i;0g 5k 4+?;5k 5+?;:::;4k 2+? 5k 4+?;5k 5+?;:::;4k 2+?
Table 3.6: Edges of di erences in B0 when  is as large as possible for ?2f4;6g
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 4;5;:::;2k+1 4;5;:::;2k;2k+2
fxi yi;3k+3g 3k+4;3k+5;:::;4k+2 3k+4;3k+5;:::;4k+2
f3k+3;5k+5+ig 2k+3;2k+4;:::;3k 2k+3;2k+4;:::;3k
f5k+5+i;0g 5k+2;5k+1;:::;4k+4 5k+2;5k+1;:::;4k+4
Table 3.7: Edges of di erences in B when  is as large as possible for ? = 8
Let c00k 1 = (0;1;3;4k + 4;4k + 1) or (0;1;3;3k + 4;3k + 1) if ? 2 f4;6g or
? = 8 respectively. Now let B00 = 2B0[fc00k 1g so that we form the set of 5-cycles
B =fc00i +jji2Z =5;c00i (mod k) 2B00;j2Zvg.
Let D be the set of di erences not used in B. It is clear that 1;2; and 3 are
contained in D. Partition D into two sets  1 and  2 so that f1;2;3g  1 and  1
58
contains as many di erences that are not v=2, v=3, or v=4 as possible. Since v=5 and
2v=5 are not in D, it follows that  1 satis es Conditions (A-D) of Lemma 3.3.7 and
so by Lemma 3.3.8 there exists a 5{CS (V[U;C) of Gv(D)_ +m( +m)K2 ensuring
we cover each mixed edge 2v( +m) times. Thus, (Z10k+?[f11;12g;A[B[C) is
a 5{CS(v + 2; + 2) that encloses the given 5{CS (Zv;A).
Let ? = 6, m2f4;6;8g, and  +m = 10. Let P0 =f(x0i;y0i)j1 i k 1;y0i >
x0ig be the collection of pairs from a Skolem sequence or hooked Skolem sequence of
order k 1 from the set f1;2;:::;2k 2g or f1;2;:::;2k 1g if k 1  0 or 1
(mod 4) or k 1 2 or 3 (mod 4) respectively. Now de ne P =f(xi = x0i + 3;yi =
y0i + 3) j (x0i;y0i) 2P0g. Then with ci de ned as in Equation (3.13) we form the set
of 5-cycles B0 = fc00i 1 = ci j 1  i k 1;(xi;yi) 2Pg where computations are
done modulo v. Let B00 = mB0 be the multiset consisting of m copies each 5-cycle
in B0. Then B00 contains m(k 1) 5-cycles, but we need b =5c = mk 2 + m=2
of them. Hence we need 3m=2 2 more 5-cycles. Let c01 = (0;1;3;4k + 5;4k + 2),
c02 = (0;2k + 1;4k + 2;8k + 5;4k + 3), c03 = (0;2k + 2;4k + 4;8k + 6;4k + 3), c04 =
(0;2k+1;6k+3;k;5k+3), and c05 = (0;1;3;4k+4;4k+1). If k 1 0 or 1 (mod 4),
then let E0 :
 contain 1, 2, or 2 copies of c01 if m = 4; 6, or 8 respectively,
 contain 1, 2, or 4 copies of c03 if m = 4, 6, or 8 respectively,
 and contain m=2 copies of c05.
If k 1 2 or 3 (mod 4), then let E0 :
 contain 0, 1, or 2 copies of c01 if m = 4; 6, or 8 respectively,
 contain 1, 2, or 2 copies of c02 if m = 4, 6, or 8 respectively,
 contain 1, 1, and 2 copies of c04 if m = 4, 6, or 8 respectively,
 and contain m=2 copies of c05.
Notice that the cycles in E0 contain at most m copies of edges not used in cycles in
B00. Then
B =fci +jj1 i k 1;ci (mod k) 2B00;j2Zvg[fc0i +jjj2Zv;c0i2E0 g:
Notice that regardless of our choice of m,  + m = 10. So (m(v 1)=2  ) =
 +m = 10. Let D be the set of di erences not used in B. By examining the remaining
di erences we know di erences 1 and 2 appear in D and when k 0 or 1 (mod 4), D
has at most 1 copy of v=2. Partition D into two sets  1 and  2 so that f1;2g  1
and  1 contains as many di erences that are not v=2, v=3, or v=4 as possible. Based
on the remaining di erences it is clear that v=3, v=6, 2v=5; and v=5 are not in D.
Thus  1 satis es Conditions (A-D) of Lemma 3.3.7 and by Lemma 3.3.8 there exists
a 5{CS (V [U;C) of Gv(D)_ +m ( + m)K2 ensuring we cover each mixed edge
2v( +m) times. Therefore (Zv[f11;12g;A[B[C) is a 5{CS(v+ 2; +m) with
m2f4;6;8g and  +m = 10.
59
Lemma 3.4.5. Let v 2 (mod 10) and assume Conditions (a,b,d) of Lemma 2.1.2
are satis ed. If m = 10 then any 5{CS of  Kv can be enclosed in a 5{CS of ( +
m)Kv+2.
Proof. Let (V = Z10k+2;A) be a 5{CS of  Kv and let U = f11;12g. Since (V;A)
exists,   m 0 (mod 10). As in Lemma 3.2.1, the number of edges we must place
in 5-cycles contained completely in mKv is
 m(v 1)
2  
2a+ 3b+c
v
 
v =  v:
Since this is divisible by v, our plan is again to create 5-cycles in mKv using all the
edges of  specially chosen di erences.
Suppose m = 10. Then mKv = Kv contains edges of each di erence in D(v)10.
Care must be taken with edges of di erence v=2 in E(Kv) since there are only v=2
of them. As in Lemma 3.4.1, we will use Skolem and hooked Skolem sequences to
explicitly describe the 5-cycles containing all the edges of  di erences (yet to be
chosen).
Let P0 =f(x0i;y0i)j1 i k 2;y0i >x0igbe the collection of pairs from a Skolem
sequence or hooked Skolem sequence of order k 2 from the setf1;2;:::;2k 4gor
f1;2;:::;2k 3g if k 2  0 or 1 (mod 4) or k 2  2 or 3 (mod 4) respectively.
So each element inf1;2;:::;2k 4gorf1;2;:::;2k 3gnf2k 4goccurs in exactly
one element of P0. Now de ne P =f(xi = x0i + 4;yi = y0i + 4)j(x0i;y0i)2P0g. Then
with
c0i =
8
>>>
<
>>>
:
(0;4;4k + 3;8k + 3;4k + 2) if i = 1,
(0;1;3;3k + 3;3k) if i = 2,
(0;1;3;2k + 4;2k + 1) if i = 3 and k 2 0;1 (mod 4), and
(0;1;3;2k + 3;2k) if i = 3 and k 2 2;3 (mod 4),
we form the set of 5-cycles
B0 =fci 1 = (0;xi;xi yi;3k;5k + 1 +i)j1 i k 2;(xi;yi)2Pg
and let E0 contain 10 copies of c01, 5 copies of c02, and 2 copies of c03. Notice that based
on our choice of d,   0 (mod 5). Table 3.8 lists the di erences of edges that occur
in cycles in B0[E0 if  is as big as possible (from Lemma 3.2.1 this happens when
 is as small as possible and not 0, namely when  = 10 and so  = 50k 15). Then
form the set of 5-cycles B =fci +jji2Zz1;ci (mod k 1) 2B0;j2Zvg[z2fc+jjj2
Zv;c2E0 g where z1 = min(f =5;m(k 2)g), z2 = max(f0; =5 m(k 2)g), and
the computations are done modulo v.
Let D be the set of di erences not used in B. There are 3 copies of 1, 2, and 3 in
D when  is as small as possible ( is as large as possible. Partition D into two sets
 1 and  2 so that f1;2;3g  1 and  1 contains as many di erences that are not
60
Edges Di erenceswhenk 2 0;1 (mod 4) Di erenceswhenk 2 2;3 (mod 4)
f0;xig;fxi;xi yig 5;6;:::;2k 5;6;:::;2k 1;2k+1
fxi yi;3kg 3k+1;3k+2;:::;4k 2 3k+1;3k+2;:::;4k 2
f3k;5k+1+ig 2k+2;2k+3;:::;3k 1 2k+2;2k+3;:::;3k 1
f5k+1+i;0g 5k;5k 1;:::;4k+3 5k;5k 1;:::;4k+3
c0i ifi=1 4;4k 1;4k;4k+1;4k+2 4;4k 1;4k;4k+1;4k+2
c0i ifi=2 1;2;3;3k;3k 1;2;3;3k;3k
c0i ifi=3 1;2;3;2k+1;2k+1 1;2;3;2k;2k
Table 3.8: Edges of di erences in B0[E0 when  is as large as possible for ? = 2
v=2, v=3, or v=4 as possible. As  increases, the di erences from ci are added to D.
In this manner, it is clear that  1 will satisfy Conditions (A-D) of Lemma 3.3.7 and
by Lemma 3.3.8 there exists a 5{CS (V [U;C) of Gv(D)_ +m ( + m)K2 ensuring
we cover each mixed edge 2v( +m) times. Thus, (Z10k+2[f11;12g;A[B[C) is
a 5{CS(v + 2; + 10) that encloses the given 5{CS (Zv;A).
Lemma 3.4.6. Let v 7 (mod 10) and assume Conditions (a,b,d) of Lemma 2.1.2
are satis ed. If m = 5, then any 5{CS of  Kv can be enclosed in a 5{CS of ( +
m)Kv+2.
Proof. Let (V = Z10k+7;A) be a 5{CS of  Kv and let U = f11;12g. Since (V;A)
exists,  +m 0 (mod 5). As in Lemma 3.2.1 in Lemma 3.4.1, the number of edges
we must place in 5-cycles contained completely in mKv is
 m(v 1)
2  
2a+ 3b+c
v
 
v =  v:
Since this is divisible by v, our plan is again to create 5-cycles in mKv using all the
edges of  specially chosen di erences.
Suppose m = 5. Then mKv = Kv contains edges of each di erence in D(v)5.
As in Lemma 3.4.1, we will use Skolem and hooked Skolem sequences to explicitly
describe the 5-cycles containing all the edges of  di erences (yet to be chosen).
Let P0 =f(x0i;y0i)j1 i k 1;y0i >x0igbe the collection of pairs from a Skolem
sequence or hooked Skolem sequence of order k 1 from the setf1;2;:::;2k 2gor
f1;2;:::;2k 1g if k 1  0 or 1 (mod 4) or k 1  2 or 3 (mod 4) respectively.
So each element inf1;2;:::;2k 2gorf1;2;:::;2k 1gnf2k 2goccurs in exactly
one element of P0. Now de ne P =f(xi = x0i + 3;yi = y0i + 3)j(x0i;y0i)2P0g. Then
with
c0i =
8
>>>
>>>
<
>>>
>>>
:
(0;2k + 2;5k + 4;k + 1;5k + 3) if i = 1 and k 1 0;1 (mod 4),
(0;1;2;5;3) if i = 2 and k 1 0;1 (mod 4),
(0;1;2k + 2;7k + 5;4k + 3) if i = 1 and k 1 2;3 (mod 4),
(0;2;4;1;4k + 3) if i = 2 and k 1 2;3 (mod 4), and
(0;2;2k + 3;2k + 1;5k + 3) if i = 3 and k 1 2;3 (mod 4),
61
we form the set of 5-cycles
B0 =fci 1 = (0;xi;xi yi;3k + 2;5k + 4 +i)j1 i k 1;(xi;yi)2Pg
and let E0 contain 5 copies of c01 and 1 copy of c02 if k 1  0 or 1 (mod 4), and
contain 4 copies of c01, 1 copy of c02 and 1 copy of c03 if k 1 2 or 3 (mod 4). Notice
that based on our choice of d,   0 (mod 5). Table 3.9 lists the di erences of edges
that occur in cycles in B0[E0 if  is as big as possible (from Lemma 3.2.1 this
happens when  is as small as possible, namely when  = 5 and so  = 25k + 5).
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 4;5;:::;2k+1 4;5;:::;2k;2k+2
fxi yi;3k+2g 3k+3;3k+4;:::;4k+1 3k+3;3k+4;:::;4k+1
f3k+2;5k+4+ig 2k+3;2k+4;:::;3k+1 2k+3;2k+4;:::;3k+1
f5k+4+i;0g 5k+2;5k+1;:::;4k+4 5k+2;5k+1;:::;4k+4
c0i ifi=1 2k+2;3k+2;4k+2;4k+3;5k+3 1;2k+1;3k+2;4k+3;5k+3
c0i ifi=2 1;1;2;3;3 2;2;3;4k+2;4k+3
c0i ifi=3  2;2;2k+1;3k+2;5k+3
Table 3.9: Edges of di erences in B0nE0 when  is as large as possible for ? = 7
De ne B = fci + j ji2Zz1;ci (mod k 1) 2B0;j 2Zvg[z2fc + j jj 2Zv;c2
E0 g where z1 = min(f =5;m(k 1)g), z2 = max(f0; =5 m(k 1)g), and the
computations are done modulo v.
Let D be the set of di erences not used in B. By examining Table 3.9, we see
that 1, 2, and 3 are contained in D and v=2;v=4;v=5;2v=5;v=6 62D. Partition D
into two sets  1 and  2 so that f1;2;3g  1 and  1 contains as many di erences
that are not v=3 as possible. Thus  1 satis es Conditions (A-D) of Lemma 3.3.7
and by Lemma 3.3.8 there exists a 5{CS of Gv(D)_ +m ( + m)K2 ensuring we
cover each mixed edge 2v( + m) times. Thus, (Z10k+7[f11;12g;A[B[C) is a
5{CS(v + 2; + 5) that encloses the given 5{CS (Zv;A).
We can gather results from Lemmas 3.4.1, 3.4.2, 3.4.3, 3.4.4, 3.4.5, and 3.4.6 into
the following corollary.
Corollary 3.4.7. Let v, m and  be integers satisfying Conditions (a,b,d) of
Lemma 2.1.2 with u = 2. Let v = 10k +? with ?2f0;1;:::;9g. If:
 m = 1 when ?2f1;3;5;9g, m = 2 when ?2f0;4;6;8g, m = 5 when ? = 7,
m = 10 when ? = 2,
 m2f2;3;4g,  +m = 5, and ?2f1;5g, or
 m2f4;6;8g,  +m = 10, and ?2f0;6g,
then any 5{CS of  Kv can be enclosed in a 5{CS of ( +m)Kv+2.
62
3.5 The Main Result
The following lemma will be useful in constructing the enclosings when  is as
large as possible.
Lemma 3.5.1. Let v and m be integers satisfying conditions (a), (b), and (d) of
Lemma 2.1.2 with u = 2 and let  =  max(v;m). Then a 5{CS of  Kv can be
enclosed in a 5{CS of ( +m)Kv+2.
Proof. Let (V = Zv;A) be a 5{CS of  Kv. Since  =  max(v;m),  +m m(v 1)=2
and so it follows that d = m(v 1)=2 by de nition and so  = 0 from Equation (3.1).
Partition D = D(v)m into two sets  1 and  2 so that f1;2g  1,  1 contains as
many di erences that are not v=2, v=3, v=4, v=5, or 2v=5 as possible. So Conditions
(A-D) of Lemma 3.3.7 hold for  1 and therefore by Lemma 3.3.8 a 5{CS of  Kv can
be enclosed in a 5{CS of ( +m)Kv+2 when  =  max(v;m).
In the following theorem, Corollary 3.4.7 is used as a basis for an inductive
argument to settle the enclosing problem when u = 2.
Theorem 3.5.2. A 5{CS of  Kv can be enclosed in a 5{CS of ( +m)Kv+2 if and
only if
(a) ( +m)u+m(v 1) 0 (mod 2),
(b)  u2 ( +m) +m v2 +vu( +m) 0 (mod 5),
(c) m v2  2( +m) 12(v 1)( +m) 0.
Proof. The necessity follows from Lemma 2.1.2(a,b,d), so we now prove the su ciency.
Since we are assuming that a 5{CS of  Kv exists, it follows by Theorem 1.2.2 that
v62f2;3;4g, and by Condition (c) that v6= 1, so we know that v 5. If  = 0 then
by Condition (a) m(v + 1)  0 (mod 2) and by Condition (b) m(v + 2)(v + 1)  0
(mod 5), so by Theorem 1.2.2 there exists a 5{CS of mKv+2 as required; so we can
assume that   1.
Let (V;A) be a 5{CS of  Kv. If v  0;4;6; or 8 (mod 10), v  2 (mod 10),
or v  7 (mod 10), then it follows from Theorem 1.2.2 that   m  0 (mod 2),
  m 0 (mod 10), and  +m 0 (mod 5) respectively; in any other case m can
potentially take on any positive integral value. To re ect this situation, let
t(v) =
8>
>><
>>>:
1 if v 1;3;5;9 (mod 10),
2 if v 0;4;6;8 (mod 10),
5 if v 7 (mod 10), and
10 if v 2 (mod 10).
We often simply write t instead of t(v) when the value of v is clear. For any choice
of v and m, let   max(v;m) be the second largest integer satisfying Conditions (a-c),
and let  +min(v;m) be the second smallest integer satisfying Conditions (a-c).
63
We will  rst establish that for any given v, the di erence between consecutive
values of  that satisfy Conditions (a) and (b) is a constant; call this di erence
 di (v). Secondly, for all m 1 and v 5, we settle the enclosing problem for both
the smallest and the largest values of  satisfying Conditions (a-c). Finally for all
m  1 and v  5 it will be shown that for all  satisfying Conditions (a-c) there
exist non-negative integers  1,  2 with  1 + 2 =  and positive integers m1;m2 with
m1 + m2 = m such that for 1  i  2 there exists a 5{CS of  iKv that can be
enclosed in a 5{CS of ( i +mi)Kv+2. So the union of these two enclosings completes
the proof.
We turn to the  rst step of the proof. For any given v and m, the di erence be-
tween consecutive values of  that satisfy Conditions (a) and (b) is a constant; namely
 di (v;m). Also by Conditions (a) and (b) and by Theorem 1.2.2,   0 (mod 2)
if v is even and (2v + 1) di (v;m)  0 (mod 5). Therefore  di (v;m)  0 (mod 5)
and  di (v;m) 0 (mod 10) if v is odd and even respectively. Since  di (v;m) must
be the smallest such value, if v is odd then  di (v;m) = 5, and if v is even then
 di (v;m) = 10. Notice that if m1 6= m2 then  di (v;m1) =  di (v;m2). Therefore
we de ne  di (v) =  di (v;m).
We now turn to the second step in the proof; establishing the existence of the
enclosing for the smallest and largest values of  given v and m. Beginning with the
smallest value, Tables 3.10 and 3.11 list the values of  min(v;m), and were formed
using Theorem 1.2.2 and Corollary 3.4.7. Three cases are considered in turn based
on the value of v.
m (mod 5t)
v (mod 10) 0 t 2t 3t 4t
0,6 0 8 6 4 2
1,5 0 4 3 2 1
Table 3.10: Checking  min(v;m) when v 0;1;5 or 6 (mod 10): each cell contains
 min(v;m)
v (mod 10)
m 2 3 4 7 8 9
any m 0 0 0 0 0 0
Table 3.11: Checking  min(v;m) when v6 0;1;5 or 6 (mod 10): each cell contains
 min(v;m)
Let v  0;1;5; or 6 (mod 10). The proof is by induction on m. By Corol-
lary 3.4.7 if m 4t then any 5{CS of  min(v;m)Kv can be enclosed in a 5{CS of
( min(v;m) + m)Kv+2. If m = 5t, by Table 2.8  min(v;10) = 0, so this enclosing has
been established by Theorem 1.2.2 and we say there exists a 5{CS (V[f11;12g;B2)
64
of (5t)Kv+2. Suppose that for some m 6t and for all s with 1 s<m, every 5{CS
of  min(v;s)Kv can be enclosed in a 5{CS of ( min(v;s)+s)Kv+2. Then, since m 6t,
by induction each 5{CS (V;B1) of  min(v;m 5t)Kv can be enclosed in a 5{CS
(V[f11;12g;B01) of ( min(v;m 5t) + (m 5t))Kv+2. Since Table 3.10 shows that
 min(v;m 5t) =  min(v;m), (V;B1) is a 5{CS of  min(v;m 5t)Kv =  min(v;m)Kv
which is enclosed in a 5{CS (V [f11;12g;B01[B2) of
(( min(v;m 5t) + (m 5t)) + 5t)Kv+2 = ( min(v;m) +m)Kv+2:
Now let v6 0;1;5 or 6 (mod 10). By Table 3.11  min(v;m) = 0, and so clearly
a 5{CS of  min(v;m) can be enclosed in a 5{CS of ( +m)Kv+u.
Now turning to the largest value of  , by Lemma 3.5.1 a 5{CS of  max(v;m)Kv
can be enclosed in a 5{CS of ( max(v;m) +m)Kv+2.
Since the cases where m = t (by Corollary 2.2.10) and where v 5; m t and
 2f min(v;m); max(v;m)g have now been settled, it is left to show that for m t,
v  5, and for all  satisfying Conditions (a-c) with  min(v;m) <  <  max(v;m)
there is an enclosing of a 5{CS of  Kv in a 5{CS of ( + m)Kv+2. The proof is by
induction on m. Tables 3.12, 3.13, 3.14 and 3.15 were formed using Theorem 1.2.2
and Corollary 3.4.7, and will help in establishing for each value of  that the enclosing
can be produced using two smaller enclosings.
v (mod 10)
m 0 1 2 3 4 5 6 7 8 9
2t(v) 0 5 0 5 10 0 0 0 10 5
3t(v) 0 0 0 0 10 0 10 0 0 5
Table 3.12: Checking the largest values of  : each cell contains  max(v;m)  
 max(v;t(v))  max(v;m t(v)) except for some special values listed in Table 3.13.
v
m 10 12 13 15 16 17 18 22 23 27 28 32 33
2t(v) 0 10 5 0 10 5 10 0 5 0 10 0 5
3t(v) 10 10 5 5 0 0 10 10 5 5 10 10 5
Table 3.13: Checking the largest values of  : each cell contains  max(v;m)  
 max(v;t(v))  max(v;m t(v)).
Suppose for some integer m  2t and for all s with 1  s < m, every 5{CS
of  Kv can be enclosed in a 5{CS of ( + s)Kv+2. From Tables 3.12 and 3.13,
 max(v;m)  max(v;t)  max(v;m t)2f0; di (v)g, so   max(v;m)  max(v;t) +
 max(v;m t). Similarly from Tables 3.14 and 3.15,  min(v;m t) +  min(v;t) 
 min(v;m)2f0; di (v)g, so  min(v;m t)+ min(v;t)  +min(v;m). Thus  min(v;m 
t) +  min(v;t)     max(v;m t) +  max(v;t). Since for each v we know that
65
v (mod 10)
m 2 3 4 7 8 9
any m 0 0 0 0 0 0
Table 3.14: Checking the smallest values of  for v6 0;1;5 or 6 (mod 10): each cell
contains  min(v;m t(v)) + min(v;t(v))  min(v;m).
m
v (mod 10) 0 t 2t 3t 4t
0,6 10 0 10 10 10
1,5 5 0 5 5 5
Table 3.15: Checking the smallest values of  for v  1 or 6 (mod 10): each cell
contains  min(v;m t(v)) + min(v;t(v))  min(v;m).
 di (v) =  di (v;m) is constant for all m, it follows that there exist non-negative inte-
gers  1; 2 for which  min(v;t)  1   max(v;t),  min(v;m t)  2   max(v;m t),
and  1 +  2 =  , and there exist positive integers m1; m2 with m1 + m2 = m such
that for 1  i 2 there exists a 5{CS of  iKv that can be enclosed in a 5{CS of
( i+mi)Kv+2. So the union of these enclosings proves that any 5{CS of ( 1+ 2)Kv =
 Kv can be enclosed in a 5{CS of (( 1 +m1) + ( 2 +m2))Kv+2 = ( +m)Kv+2.
66
Chapter 4
5{CS of ( +m)Kv+u  Kv when u = 1
4.1 Introduction
We now turn to generalizing the enclosing results from Chapter 2. As said in
Chapter 1, showing that a 5{CS of  Kv can be enclosed in a 5{CS of ( + m)Kv+u
is the same as showing there exists a 5{CS of ( +m)Kv+u  Kv in the case where
there exists a 5{CS of  Kv. We will solve some of the cases where the existence
condition of a 5{CS of  Kv is removed.
Not surprisingly, necessary conditions for the existence of a 5{CSof ( +m)Kv+u 
 Kv in Lemma 4.1.1 are very similar to the necessary conditions in Lemma 2.1.2.
Lemma 2.1.2 will help establish some of the conditions in the following lemma since
in Lemma 2.1.2 the existence of a 5{CS of  Kv was not used to establish every
condition in this theorem.
Lemma 4.1.1. Let v, m, and u be positive integers. If there exists a 5{CS of ( +
m)Kv+u  Kv, then:
(a) ( +m)u+m(v 1) 0 (mod 2),
(b) (v +u 1)( +m) 0 (mod 2),
(c)  u2 ( +m) +m v2 +vu( +m) 0 (mod 5),
(d) if u = 1, then m(v 1) 3( +m),
(e) if u = 2, then m v2  2( +m) (v 1)( +m)=2 0, and
(f) if u 3, then dvu( + m)=4e+ 2  mv(v 1)=2 + ( + m)u(u 1)=2 where
 = 0 or 1 if vu( +m) 0 or 2 (mod 4) respectively.
Proof. Suppose there exists a 5{CS of ( + m)Kv+u  Kv (V [U;C) where V
contains the vertices in  Kv, U contains the other vertices in ( + m)Kv+u and
C contains the 5-cycles that partition the edge set of ( + m)Kv+u  Kv. Then
each vertex in V and U must have even degree. So ( + m)(u 1) + ( + m)v =
( +m)(v+u 1) and m(v 1) + ( +m)u are both even, proving (a) and (b). Since
the number of edges in ( + m)Kv+u  Kv must be divisible by 5, it follows that
( +m) u2 +m v2 +vu( +m) 0 (mod 5), thus Condition (c) is established.
Using the work in the proof of Lemma 2.1.2, Conditions (d-f) can be estab-
lished since the existence of a 5{CS of  Kv is not used to prove Conditions (c-e) of
Lemma 2.1.2. For completeness, we will reconstruct the argument here.
Suppose u = 1. Then every 5-cycle that contains at least one mixed edge must
have three pure edges joining vertices in V and two mixed edges that join a vertex in
V to a vertex in U. Since there are mv(v 1)=2 pure edges, mv(v 1)=6 v( +m)=2,
thus establishing Condition (d).
67
Suppose u = 2. Then there must be  + m 5-cycles that contain a pure edge
joining vertices in U. These 5-cycles also contain 2( +m) pure edges joining vertices
in V and 2( + m) mixed edges joining a vertex in V to a vertex in U. Since each
remaining 5-cycle must contain at least one pure edge and at most four mixed edges
and there are 2(v 1)( +m) remaining mixed edges, it follows that
0 m
 v
2
 
 2( +m) 2(v 1)( +m)4 = m
 v
2
 
 2( +m) (v 1)( +m)2 :
Therefore Condition (e) holds.
Suppose u 3. Then every 5-cycle must contain at least one pure edge and at
most 4 mixed edges. So there are at leastdvu( +m)=4e5-cycles that contain mixed
edges. But if vu( + m)  2 (mod 4), one of these 5-cycles must contain 2 mixed
edges and 3 pure edges. So dvu( +m)=4e+ 2  mv(v 1)=2 + ( +m)u(u 1)=2
which shows Condition (f) holds.
For the remainder of the chapter let U = f1g. Sections 4.2 and 4.3 will show
that the conditions in Lemma 4.1.1 are su cient for u = 1. In particular, Section 4.2
addresses the cases when m is small. These cases will then be used to construct the
main result of this chapter, Theorem 4.3.1, in Section 4.3.
4.2 5{CSs of ( +m)Kv+1  Kv: The small cases
Notice that when v6 2;3;7; or 8 (mod 10) the  -admissible values and corre-
sponding m values that satisfy Theorem 2.3.2 are identical to the  -admissible values
and corresponding m values that satisfy Lemma 4.1.1(a-d). So all we need to do is
prove that a 5{CS of ( +m)Kv+1  Kv exists when v 2;3;7 or 8 (mod 10).
We will use almost the same de nitions used in Chapter 2. The only change is
that we will de ne D(v) ? di erently. When v and ? are both even, it is useful to
let D(v) ? be formed from D(v)? by removing ?=2 copies of the di erence v=2. Let
D(v) ? = D(v)? if v and ? are not both even.
Lemma 2.2.2 shows that the number of 5-cycles de ned on the vertices in V is
divisible by v and also points to the main approach used in Lemma 4.2.3. So the
parameter
 = m(v 1) 3( +m)2 ;
restated here, will play a critical role throughout the remainder of the paper.
The next lemma will be used when constructing the 5-cycles that contain the
vertex 1; each such 5-cycle uses exactly 3 pure edges in V.
Lemma 4.2.1. Suppose that Conditions (a-d) in Lemma 4.1.1 are satis ed. Then if
v is odd, or if both v and m are even, 3 divides m(v 1)=2  . If v is even and m
is odd then 3 divides m(v 1)=2   3=2.
68
Proof. If v is odd then by Lemma 4.1.1(a),  + m is even. If v is even then  is
even by Lemma 4.1.1(b). So if v is odd or if both v and m are even, then 3 divides
m(v 1)=2  . Now suppose v is even and m is odd. It is clear that 3 divides
m(v 1)=2   3=2 = 3( + m 1)=2 and otherwise 3 divides m(v 1)=2  =
3( +m)=2.
The next lemma will be most useful in Theorem 4.3.1 but also plays a role in
Lemma 4.2.3.
Lemma 4.2.2. If Conditions (a-d) of Lemma 4.1.1 hold for any given v and m when
u = 1 then the di erence between consecutive  values that are admissible is 10 if
v6 0 (mod 5) and is 2 if v 0 (mod 5). Furthermore, this di erence is constant
for all values of m.
Proof. For any given v and m, the di erence between consecutive values of  that
satisfy Conditions (a-d) of Lemma 4.1.1, say  di (v;m), is a constant. Also by
Lemma 4.1.1(a-c),  di (v;m) 0 (mod 2) and v di (v;m) 0 (mod 5), so
v di (v;m)  0 (mod 10) if v 6 0 (mod 5) and v di (v;m)  0 (mod 2) if v  0
(mod 5). Since  di (v;m) must be the smallest such value,  di (v;m) = 10 or
 di (v;m) = 2 if v 6 0 (mod 5) or v  0 (mod 5) respectively. Notice that if
m1 6= m2 then  di (v;m1) =  di (v;m2).
For all positive integers v and m, let  max(v;m) and  min(v;m) be the largest
and smallest integers for  respectively satisfying Conditions (a-d) in Lemma 4.1.1
with u = 1. As was done in Chapters 2 and 3 we will use smaller 5{CSs to build
larger 5{CSs. This is the basis behind the induction proof given in Theorem 4.3.1.
So we now focus on  nding a 5{CS of ( + m)Kv+1  Kv where m = 1 and where
 2f min(v;m); max(v;m)g for small values of m.
Lemma 4.2.3. Let v 2;3;7 or 8 (mod 10) and assume Conditions (a-d) of
Lemma 4.1.1 are satis ed. Suppose at least one of the following statements is satis ed:
 m = 1 or both m2f2;3g and  =  max(v;m);
 m = 2,  = 2, and v 2 (mod 10);
 ( ;m)2f(2;4);(1;7)g and v 3 (mod 10); or
 ( ;m)2f(2;4);(6;2);(4;3)g and v 8 (mod 10).
Then there exists a 5{CS of ( +m)Kv+1  Kv.
Proof. Let V = Zv be the vertex set of  Kv where v = 10k+? for some ?2f2;3;7;8g
and let U =f1g. If  = 0 then there exists a 5{CS of mKv by Theorem 1.2.2. Notice
that by Lemma 4.1.1(d), v 7 when   1. By de ning the sets of 5-cycles B and C,
our aim is to show that there exists a 5{CS (V[f1g;B[C) of ( +m)Kv+1  Kv.
First, note that by Lemma 2.2.2, the number of edges,  v, from 5-cycles with
5 pure edges is given in Equation (3.1). Since this is divisible by v, our plan is to
69
create these 5-cycles using the edges of  specially chosen di erences. Notice that by
Lemma 4.1.1(d),   0 and by Lemma 4.1.1(c),
m
 v
2
 
+v( +m) ?( +m) + 3?(v 1)m 0 (mod 5)
so that  3( +m) 4m(v 1) (mod 5). Then
m(v 1) 3( +m) m(v 1) + 4m(v 1)
 5m(v 1)
 0 (mod 5);
and so   0 (mod 5). This shows we can distribute the  specially chosen di erences
into 5-cycles.
Supposem = 1 or (m; ;?)2f(2;2;2);(4;2;3);(7;1;3);(2;6;8);(3;4;8)g. Skolem
and hooked Skolem sequences will be used to explicitly describe the 5-cycles contain-
ing all the edges of  di erences (yet to be chosen). To form the set of 5-cycles B we
consider the cases ?2f2;3g and ?2f7;8g in turn.
Suppose ?2f2;3g. Let P0 =f(x0i;y0i)j1 i k 1;y0i >x0ig be the collection
of pairs from a Skolem sequence or hooked Skolem sequence of order k 1 for the
set f1;2;:::;2k 2g or f1;2;:::;2k 1g, if k 1  0 or 1 (mod 4) or k 1  
2 or 3 (mod 4) respectively (see Theorems 1.1.1 and 1.1.2). So each element in
f1;2;:::;2k 2gorf1;2;:::;2k 1gnf2k 2goccurs in exactly one element of P0.
Now de ne P =f(xi = x0i + 2;yi = y0i + 2)j(x0i;y0i)2P0g. Then let
B0 =fci 1 = (0;xi;xi yi;3k;5k + 1 +i)j1 i k 1;(xi;yi)2Pg (4.1)
where computations are done modulo v. So for m = 1, form the set of 5-cycles
B = fci + j ji2Z =5;ci 2B0;j2Zvg. Table 4.1 lists the di erences of the edges
that occur in 5-cycles in B0 when  is as large as possible. When  is as large as
possible ( is as small as possible),  =5 = k 2 or k 1 and jB0j= k 2 or k 1 if
? = 2 or 3 respectively. So B0 is large enough to construct the required 5-cycles.
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 3;4;:::;2k 3;4;:::;2k 1;2k+1
fxi yi;3kg 3k+1;3k+2;:::;4k 1 3k+1;3k+2;:::;4k 1
f3k;5k+1+ig 2k+2;2k+3;:::;3k 2k+2;2k+3;:::;3k
f5k+1+i;0g 5k 2+?;5k 3+?;:::;4k+? 5k 2+?;5k 3+?;:::;4k+?
Table 4.1: Edges of di erences in B0 when  is as large as possible for ?2f2;3gand
m = 1
If m = 2,  = 2, and ? = 2, then  =5 = 2k 1 and j2B0j= 2k 2, so we must
create one more 5-cycle constructed from di erences. To account for this, form the
70
set of 5-cycles B = 2fci +jji2Zk 1;j2Zv;ci2B0g[fc+jjj2Zvg where B0 is
de ned in Equation (4.1) and
c =
(
(0;1;2;2k + 3;2k + 1) if k 1 0;1 (mod 4), and
(0;2k;2k + 1;1;2) if k 1 2;3 (mod 4).
The 5-cycle c uses the di erences 1, 1, 2, 2k + 1, and 2k + 1 if k 1 0;1 (mod 4)
and the di erences 1, 1, 2, 2k, and 2k if k 1 2;3 (mod 4).
If m = 4,  = 2, and ? = 3, then  =5 = 4k 1 when  is as large as possible and
j4B0j = 4k 4, so we must construct three more 5-cycles using di erences. In this
case form the set of 5-cycles B = 4fci +jji2Zk 1;j2Zv;ci2B0g[fc0i (mod 2) +jj
j2Zv;i2Z3g where B0 is de ned in Equation (4.1) and
c0i =
8
><
>:
(0;1;2k + 2;4k + 3;4k + 1) if i = 0 and k 1 0;1 (mod 4),
(0;1;4k + 2;2k + 2;2) if i = 0 and k 1 2;3 (mod 4), and
(0;1;2;4k + 3;4k + 1) if i = 1.
(4.2)
Notice that c00 is repeated twice in B so that if k 1 0 or 1 (mod 4) thenfc0i (mod 2) j
i2Z3guses three copies of the di erence 2 and four copies of each of the di erences
1;2k+ 1; and 4k+ 1 and if k 1 2 or 3 (mod 4) thenfc0i (mod 2) ji2Z3guses three
copies of the di erence 2 and four copies of each of the di erences 1;2k; and 4k + 1.
If m = 7,  = 1, and ? = 3, then  =5 = 7k 1, and j7B0j = 7k 7, and so we
need to construct six more 5-cycles using di erences. Under these conditions form
the set of 5-cycles
B = 7fci +jji2Zk 1;j2Zv;ci2B0g[3fc00 +jjj2Zvg[fc01 +jjj2Zvg
[2fc02 +jjj2Zvg
where B0 is de ned in Equation (4.1), c00 and c01 are de ned as in Equation (4.2), and
c02 = (0;1;4k + 3;8k + 4;4k + 2). The 5-cycles c00;c01 and c02 in B use four copies of
di erence 2, six copies of the di erence 4k+ 2, seven copies of both of the di erences
1 and 4k + 1, and six copies of the di erence 2k + 1 or the di erence 2k if k 0 or 1
(mod 4) or k 2; or 3 (mod 4) respectively.
Now suppose ?2f7;8g. Let P =f(xi;yi)j1 i k;yi >xig be the collection
of pairs from a Skolem sequence or hooked Skolem sequence of order k for the set
f1;2;:::;2kg or f1;2;:::;2k + 1g, if k  0 or 1 (mod 4) or k  2 or 3 (mod 4)
respectively. De ne
B0 =fci 1 = (0;xi;xi yi;3k + 2;5k + 4 +i)j1 i k;(xi;yi)2Pg (4.3)
where computations are done modulov. So form = 1 or (m; ;?)2f(2;6;8);(3;4;8)g,
form the set of 5-cycles B =fci +jji2Z =5;ci (mod k) 2B0;j2Zvg. Table 4.2 lists
the di erences of edges that occur in cycles in B when  is as large as possible ( 
71
is as small as possible). Since jmB0j = mk and  =5 is k, k 2, 2k 1, or 3k when
(m; ;?) is (1;1;7), (1;8;8), (2;6;8), or (3;4;8) respectively (note that if m = 1 and
 is as large as possible then  = 1 or 8 when ? = 7 or 8 respectively), there are
enough 5-cycles in mB0 to construct all of the 5-cycles with only pure edges when
m = 1 or (m; ;?)2f(2;6;8);(3;4;8)g.
Edges Di erenceswhenk 0;1 (mod 4) Di erenceswhenk 2;3 (mod 4)
f0;xig;fxi;xi yig 1;2;:::;2k 1;2;:::;2k 1;2k+1
fxi yi;3k+2g 3k+3;3k+4;:::;4k+2 3k+3;3k+4;:::;4k+2
f3k+2;5k+4+ig 2k+3;2k+4;:::;3k+2 2k+3;2k+4;:::;3k+2
f5k+4+i;0g 5k 5+?;5k 6+?;:::;4k 4+? 5k 5+?;5k 6+?;:::;4k 4+?
Table 4.2: Edges of di erences in B0 when  is as large as possible for ?2f7;8gand
m = 1
Now suppose m = 4,  = 2, and ? = 8. Since  =5 = 4k + 1 and j4B0j= 4k, we
need to construct one more 5-cycle constructed from di erences. So we form the set
of 5-cycles
B = 4fci+jji2Zk;j2Zv;ci2B0g[f(0;4k+3;2k+1;6k+4;8k+6)+jjj2Zvg
where B0 is de ned in Equation (4.3) and computations are done modulo v and
(0;4k + 3;2k + 1;6k + 4;8k + 6) uses the di erences 2k + 2, 2k + 2, 2k + 2, 4k + 3,
and 4k + 3.
To de ne C, let ?2f2;3;7;8g. By Lemma 4.2.1, if v is odd or both v and m
are even then 3 divides (m(v 1)=2  ) and if v is even and m is odd then 3 divides
(m(v 1)=2   3=2). So form a partition P0 from the di erences not used to form
B into (m(v 1)=2  )=3 sets of size 3 (not multisets) when either v is odd or both
v and m are even, and form a partition P0 from the di erences not used to form B
into (m(v 1)=2   3=2)=3 sets of size 3 (not multisets) and one setfd;v=2gof size
2 with d6= v=2 when v is even and m is odd; this ensures that d1 <d3 and d2 <v=2
when de ning C. Form the set C of 5-cycles
C =f(d1;0;d3;d2 +d3;1) +jjfd1;d2;d3g2P0 with d1  d2  d3;j2Zvg: (4.4)
when either v is odd or both v and m are even and the set
C =f(d1;0;d3;d2 +d3;1) +jjfd1;d2;d3g2P0 with d1  d2  d3;j2Zvg (4.5)
[
n 
d;0;v2;v2 +d;1
 
+jj
nv
2;d
o
2P0;j2Zv=2
o
:
when v is even and m is odd. By Lemma 2.2.5 the elements of C are 5-cycles. Thus
(Zv[f1g;B[C) is a 5{CS of ( +m)Kv+1  Kv with v 2;3;7; or 8 (mod 10)
and m = 1 or with ( ;m;?)2f(2;2;2);(2;4;3);(1;7;3);(2;6;8);(3;4;8);(4;2;8)g.
Now suppose m = 2,  =  max(v;m), and v  2;3;7; or 8 (mod 10). For
72
any  xed values of v and m, v-admissible values of  di er by multiples of 10
by Lemma 4.2.2, and so the corresponding  values di er by multiples of 15 (by
Lemma 2.2.2). So when  =  max(v;m) it follows that  2f0;5;10g. Then let
B =
8
><
>:
? if  = 0,
f(0;1;4;2;6) +jjj2Zvg if  = 5, and
f(0;1;4;2;6) +jjj2Zvg[f(0;1;4;2;6) +jjj2Zvg if  = 10,
and let  = ?;f1;2;3;4;6g, or f1;1;2;2;3;3;4;4;6;6g when  = 0;5; or 10 respec-
tively. By Lemma 4.2.1, jD(v) 2n j 0 (mod 3). Notice that if v = 7, m = 2, and
 max(7;2) = 2, then  = 0 and if v = 8 and m = 2 then there are no 5{CSs of
( +m)Kv+1  Kv by Lemma 4.1.1(a-d), so these 5-cycles in B can be constructed.
Form C using Equation (4.4) when either v is odd or both v and m are even and
using Equation (4.5) when v is even and m is odd. Thus (Zv[f1g;B[C) is a 5{CS
of ( + 2)Kv+1  Kv for v 2;3;7; or 8 (mod 10) and  =  max(v;m).
Let m = 3 and  =  max(v;3) = v 4. It follows that  = 0 for all k, so de ne
B = ?. Form C using Equation (4.4) when either v is odd or both v and m are even
and using Equation (4.5) when v is even and m is odd. Therefore (Zv[f1g;B[C)
is a 5{CS of ( +3)Kv+1  Kv for v 2;3;7; or 8 (mod 10) with  =  max(v;3).
4.3 5{CSs of ( +m)Kv+1  Kv: The Main Result
In the following theorem, Lemma 4.2.3 is used as a basis for an inductive argu-
ment to settle the problem of constructing a 5{CS of ( +m)Kv+1  Kv.
Theorem 4.3.1. Let m 1. There exists a 5{CS of ( +m)Kv+1  Kv if and only
if
(a) ( +m) +m(v 1) 0 (mod 2),
(b) v( +m) 0 (mod 2),
(c) m v2 +v( +m) 0 (mod 5), and
(d) m(v 1) 3( +m).
Proof. The necessity follows from Lemma 4.1.1, so we now prove the su ciency. If
 = 0 then by Condition (a), mv 0 (mod 2) and by Condition (c), mv(v+1)=2 0
(mod 5); so by Theorem 1.2.2 there exists a 5{CS of mKv+1. So we can also assume
that   1. By Condition (d), v62f1;2;3;4g, so we know v 5. By Theorem 2.3.2,
there exists a 5{CS of ( + m)Kv+1  Kv for v6 2;3;7 or 8 (mod 10), and so we
can assume v 2;3;7 or 8 (mod 10). Let V be the vertex set of  Kv and U contains
the other vertex in ( +m)Kv+1.
First we establish that for any given v, the di erence between consecutive val-
ues of  that satisfy Conditions (a-c) is a constant; call this di erence  di (v).
Lemma 4.2.2 establishes this claim and shows that  di (v) = 10. Secondly, for all
m 1 and v 5, we will show there exists a 5{CS of ( + m)Kv+1  Kv for both
73
the smallest and the largest values of  satisfying all of the conditions. Finally for
all m 1 and v 5 it will be shown that for all  satisfying Conditions (a-d) there
exist non-negative integers  1,  2 with  1 + 2 =  and positive integers m1;m2 with
m1 + m2 = m such that there exists a 5{CS of ( i + mi)Kv+1  iKv for i2f1;2g.
So the union of these two 5{CSs completes the proof.
We turn to the second step in the proof; establishing the existence of the 5{CS
for the smallest and largest values of  given v and m. Beginning with the smallest
value, Table 4.3 lists the values of  min(v;m) using Theorem 1.2.2 and Lemma 4.2.3.
Four cases are considered in turn based on the value of v.
m (mod 10)
v (mod 10) 1 2 3 4 5 6 7 8 9 0
2 6 2 8 4 0 6 2 8 4 0
3 3 6 9 2 5 8 1 4 7 0
7 1 2 3 4 5 6 7 8 9 0
8 8 6 4 2 0 8 6 4 2 0
Table 4.3: List of  min(v;m) values
Suppose v 2 (mod 10). By Lemma 4.2.3 a 5{CS of ( min(v;m) + m)Kv+1 
( min(v;m))Kv exists for m = 1 and 2. From examining Table 4.3 it is easy to see
that the union of
( min(v;m0) +m0)Kv+1 ( min(v;m0))Kv
and
( min(v;m00) +m00)Kv+1 ( min(v;m00))Kv
where m0 = 1 and m00 = 2 or m0 = 2 and m00 = 2 form a 5{CS of ( min(v;m) +
m)Kv+1 ( min(v;m))Kv where m = 3 or 4 respectively. Now we can form a 5{CS
of ( min(v;5) + 5)Kv+1 ( min(v;5))Kv by taking the union of
( min(v;2) + 2)Kv+1 ( min(v;2))Kv
and
( min(v;3) + 3)Kv+1 ( min(v;3))Kv:
We will revisit what happens when m > 5 after we examine when v 8 (mod 10)
and m 5
Suppose v 8 (mod 10). It follows by Lemma 4.2.3 that a 5{CS of ( min(v;m)+
m)Kv+1 ( min(v;m))Kv exists for m 4. By Theorem 2.3.2, there exists a 5{CS
of ( min(v;5) + 5)Kv+1 ( min(v;5))Kv:
Now suppose v 2 or 8 (mod 10). By using the 5{CS of 5Kv+1 (which exists by
Theorem 1.2.2) we will generate the remaining 5{CS for m 6 when  =  min(v;m)
using induction on m. Suppose for some m  6 and for all s with 1  s < m
74
there exists a 5{CS of (( min(v;s) +s)Kv+1 ( min(v;s))Kv. Since  min(v;m 5) =
 min(v;m) and there exists a 5{CS (V [U;B1) of
( min(v;m 5) +m 5)Kv+1 ( min(v;m 5))Kv
and a 5{CS (V[U;B2) of 5Kv+1, it follows that there exists a 5{CS (V[U;B1[B2)
of
(( min(v;m 5) +m 5) + 5)Kv+1 ( min(v;m 5))Kv
= ( min(v;m) +m)Kv+1 ( min(v;m))Kv:
Suppose v 3 (mod 10). By Lemma 4.2.3 we have established the existence of
a 5{CS of ( min(v;m) + m)Kv+1  ( min(v;m))Kv for m2f1;4;7g. So by taking
the union of particular copies of these 5{CSs, we can form a 5{CS of ( min(v;m) +
m)Kv+1  ( min(v;m))Kv for m 2 f2;5;8g. Since we can build a 5{CS of ( +
m)Kv+1   Kv in the case when  is as small as possible and m2f1;2;4;5;7;8g,
we can form a 5{CS of ( min(v;m) + m)Kv+1 ( min(v;m))Kv for m2f3;6;9;10g
by taking the union of 2 particular copies of the 5{CSs we have already constructed.
We will revisit what happens when m> 10 after we examine when m 10 and v 7
(mod 10).
Suppose v  7 (mod 10). Then by Lemma 4.2.3, we can form a 5{CS of
( min(v;1) + 1)Kv+1  ( min(v;1))Kv. So by using successive copies of this 5{CS,
we can form a 5{CS of ( min(v;m) +m)Kv+1 ( min(v;m))Kv for 2 m 10.
Now suppose v 3 or 7 (mod 10). We will use induction on m in this proof to
construct the remaining 5{CSs. As before, we use the existence of a 5{CS of 10Kv+1
(which exists by Theorem 1.2.2) and a 5{CS of ( min(v;m)+m)Kv+1 ( min(v;m))Kv
for m 10 to assist constructing larger 5{CSs. So suppose for some m 11 and for
all s with 1  s < m, there exists a 5{CS of ( min(v;s) + s)Kv+1 ( min(v;s))Kv.
Since there exists a 5{CS (V [U;B1) of
( min(v;m 10) +m 10)Kv+1 ( min(v;m 10))Kv;
since there exists a 5{CS (V[U;B2) of 10Kv+1, and since min(v;m 10) =  min(v;m),
it follows that there exists a 5{CS (V [U;B1 [B2) of ( min(v;m) + m)Kv+1  
( min(v;m))Kv:
We now turn to constructing a 5{CS of ( +m)Kv+1  Kv where  =  max(v;m).
The proof of this case will be done using induction on m. By Lemma 4.2.3, there
exists a 5{CS of ( max(v;m)+m)Kv+1 ( max(v;m))Kv for m2f1;2;3g. Suppose for
some m 4 and for all s with 1 s<m there exists a 5{CS of ( max(v;s)+s)Kv+1 
( max(v;s))Kv. Since  max(v;m 3)+ max(v;3) =  max(v;m) by Lemma 2.3.1, there
exists a 5{CS of
( max(v;m 3) +m 3)Kv+1 ( max(v;m 3))Kv
and there exists a 5{CS of ( max(v;3) + 3)Kv+1 ( max(v;3))Kv, it follows that the
union of these two 5{CSs form a 5{CS of ( max(v;m) +m)Kv+1 ( max(v;m))Kv:
75
v (mod 10)
m 2 3 7 8
2 0,10,0 10,0,0 0,0,10 0,10,0
3 10,10,0 10,0,10 0,10,10 10,10,0
Table 4.4: Each cell contains  max(v;m)  max(v;m 1)  max(v;1) for k 0;1;2
(mod 3) in turn
m (mod 10)
v (mod 10) 1 2 3 4 5 6 7 8 9 0
2 0 10 0 10 10 0 10 0 10 10
3 0 0 0 10 0 0 10 0 0 10
7 0 0 0 0 0 0 0 0 0 10
8 0 10 10 10 10 0 10 10 10 10
Table 4.5: Each cell contains  min(v;m 1) + min(v;1)  min(v;m)
Now we turn to the third and  nal step of the proof using induction on m.
We have settled the cases where m = 1 and where v  5, m > 1, and  2
f min(v;m); max(v;m)g. It is left to show that there exists a 5{CS of ( +m)Kv+1 
 Kv for m > 1 and  min(v;m) <  <  max(v;m). Tables 4.4 and 4.5 were formed
using Theorem 1.2.2 and Lemma 4.2.3. These tables will be instrumental in showing
that we can use smaller 5{CSs to form larger 5{CSs. Suppose for some m > 1 and
for all s with 1 s<m there exists a 5{CS of ( +s)Kv+1  Kv. By Lemma 4.2.2,
 di (v) is constant for all m. From Table 4.4,
 max(v;m)  max(v;m 1)  max(v;1)2f0; di (v)g
and so
 max(v;m)  di (v;m) =  max(v;m) 10  max(v;m 1) + max(v;1):
Similarly, Table 4.5 shows that
 min(v;m 1) + min(v;1)  min(v;m)2f0; di (v)g;
and so
 min(v;m 1) + min(v;1)  min(v;m) + di (v;m) =  min(v;m) + 10:
This shows that
 min(v;m 1) + min(v;1)    max(v;m 1) + max(v;1):
It follows that there exist non-negative integers  1; 2 for which  min(v;1)   1  
 max(v;1),  min(v;m 1)   2   max(v;m 1), and  1 +  2 =  , and there exist
76
positive integers m1;m2 with m1 + m2 = m such that for 1  i 2 there exists a
5{CS of ( i +mi)Kv+1  iKv. So the union of these two 5{CSs proves there exists
a 5{CS of ( +m)Kv+1  Kv.
77
Chapter 5
5-cycle systems when m = 0
5.1 Introduction
In this chapter we will settle the existence of a 5{CS of  Kv+u  Kv except
possibly in the case where   5 (mod 10), v  v + u 1 (mod 2), and  < v  
(u 1)=3 and in the case where   2;4;6; or 8 (mod 10), u  0 (mod 5), and
v=2 + 1  u < v=2 + 5. To do this, the methods in this chapter draw heavily from
the proof methods established in Chapters 2-4. Notice that if a 5{CS of a  Kv exists
then this is an embedding type problem rather than an enclosing.
In the literature there is one result which settles the existence problem in the
special case where  = 1 and m = 0 [11]:
Theorem 5.1.1. [11] There exists a 5-cycle system of Kv+u Kv if and only if
(a) u v=2 + 1, and
(b) v + u v 3 (mod 10), or v + u;v 1 or 5 (mod 10), or v + u;v 7 or 9
(mod 10).
Section 5.2 will  nd some necessary conditions for the existence of a 5{CS of
 Kv+u  Kv and establish some conditions that will be useful in subsequent lemmas.
In Section 5.3, we will establish that these necessary conditions are su cient in the
case where   5 (mod 10) and u < 3v + 1. Section 5.4 will show the necessary
conditions in Lemma 5.2.1 are su cient except possibly in two cases.
5.2 5{CSs of  Kv+u  Kv: Preliminary Results
The following necessary conditions are remarkably similar to those given in The-
orem 5.1.1. Notice that Theorem 5.1.1 proves the conditions in Lemma 5.2.1(a) are
necessary and su cient by taking  copies of a 5{CS of Kv+u Kv. Recall that an
edge is said to be pure if it joins two vertices in V or if it joins two vertices in U;
otherwise it is said to be mixed.
Lemma 5.2.1. If there exists a 5{CS of  Kv+u  Kv then u v=2 + 1 and
(a) If   1;3;7; or 9 (mod 10) then
v +u;v 1 or 5 (mod 10);
v +u;v 7 or 9 (mod 10); and
v +u v 3 (mod 10).
(b) If   0 (mod 10) then any v 1 and v +u 5.
78
(c) If   2;4;6; or 8 (mod 10) then
v +u;v 0;1 (mod 5);
v +u;v 2;4 (mod 5); and
v +u v 3 (mod 5).
(d) If   5 (mod 10) then v and v +u are both odd.
Proof. Let V[U be the vertex set of  Kv+u  Kv where V is the vertex set of  Kv
and U contains the other vertices in  Kv+u. If there exists a 5{CS of  Kv+u  Kv
then there is a limit to the number of mixed edges since every 5-cycle must contain
at least one pure edge. Hence 4 u(u 1)=2  vu shows u v=2 + 1.
As said before, Condition (a) holds by Theorem 5.1.1. If   0 (mod 10) then
it is clear that v +u 5 and so Condition (b) holds. Suppose that   5 (mod 10).
Since the degree of each vertex must be even, v 1 and v + u 1 must be even
and so v and v + u must be odd showing that Condition (d) is satis ed. Suppose
that   2;4;6; or 8 (mod 10). Since the number of edges must be divisible by 5,
 vu+ u(u 1)=2 0 (mod 5) and so vu+u(u 1)=2 u(2v+u 1) 0 (mod 5)
since  6 0 (mod 5). Condition (c) follows from this observation.
There are three types of 5-cycles in  Kv+u  Kv based on the number of mixed
and pure edges in each 5-cycle:
 one pure edge and four mixed edges,
 three pure edges and two mixed edges, and
  ve pure edges.
De ne these 5-cycles as Type A, B, and C respectively. A visual representation of
these 5-cycles is given in Figure 5.1. For the remainder of the chapter, let a, b, and
V
U
A
V
U
B
V
U
C
Figure 5.1: Possible Types of 5-cycle
c be the number of 5-cycles of Type A, B, and C respectively. We may need to set
aside certain di erences in order for the conditions of Lemma 5.2.2 to hold when
u 0 (mod 5); let  be the number of di erences we choose to set aside. If   5
(mod 10) then  is always chosen to be 0. Furthermore, if u6 0 (mod 5) then  is
always chosen to be 0 and if u 0 (mod 5) then  is always chosen to be in Z5. Since
there are  vu mixed edges and   u2 pure edges in  Kv+u  Kv, we can set up the
79
following system of equations:
4a+ 2b =  vu (5.1)
a+ 3b+ 5c =  
 u
2
 
  u:
For the rest of the chapter, de ne
a =
(
0 if u 3v + 1 and
u( (3v u+ 1) + 2 )=10 if u< 3v + 1,
b =
(
 uv=2 if u 3v + 1 and
u( (2u v 2) 4 )=10 if u< 3v + 1, (5.2)
c =
(
u( (u 1 3v) 2 )=10 if u 3v + 1 and
0 if u< 3v + 1.
The values for a, b, and c in (5.2) satisfy the system of equations in (5.1).
The following lemma will be useful for showing that the values for a, b, and c in
(5.2) satisfy conditions that will become vital in Theorem 5.4.1.
Lemma 5.2.2. Suppose that  , v; and u satisfy Conditions (a-d) of Lemma 5.2.1
and  6 1;3;7; or 9 (mod 10). If u 3v + 1 then
(i) b and c are non-negative integers,
(ii) u divides c,
(iii) vu divides b if  is even or vu=2 divides b if  is odd,
(iv)  (u 1)=2  (u 3v 1)=2 0 (mod 3) when  is even, and
(v)  (u 1)=2  (u 3v 1)=2 3v=2 0 (mod 3) when  is odd,
If v=2 + 1 u< 3v + 1 then
(vi) a and b are non-negative integers if   0 (mod 5),
(vii) a and b are non-negative integers if   2;4;6; or 8 (mod 10) and u 6 0
(mod 5) or if   2;4;6; or 8 (mod 10), u 0 (mod 5), and u v=2 + 5,
(viii) u divides a, and
(ix) u divides b if  is even.
Proof. Suppose that u  3v + 1. Then it is clear that b and c are non-negative.
Conditions (i) and (ii) will be discussed in turn based on  .
If   0 (mod 10) then c is an integer and u divides c. If   5 (mod 10) then
v is odd, u is even, and  = 0 so that u 1 3v is even and consequently c is an
integer and u divides c. Now suppose that   2;4;6; or 8 (mod 10). Then it is
clear that b =  uv=2 is an integer. Suppose u6 0 (mod 5) so that  = 0. We will
now show that (u 1 3v)  0 (mod 5) to show c is an integer and u divides c
80
when   2;4;6; or 8 (mod 10). Let u = 5k1 +e1 and v = 5k2 +e2 satisfy Condition
(c) of Lemma 5.2.1 where e1 2Z5 nf0g (i.e. u6 0 (mod 5)) and e2 2Z5. So the
possible values of e1 and e2 are (e1;e2) 2f(1;0);(4;1);(2;2);(3;4)g. It follows that
(u 1 3v) e1 1 3e2  0 (mod 5) which shows c is an integer and consequently
u divides c. Now let e1 = 0 so that u 0 (mod 5). Then
 (u 1 3v) 2     3 e2 2 (mod 10)
and so for some  2Z5,   3 e2 2  0 (mod 10). This shows that c is an integer
and u divides c and so Conditions (i) and (ii) are satis ed.
Since b =  uv=2, it is clear that vu divides b or vu=2 divides b when  is even or
odd respectively, so Condition (iii) holds. Since  (u 1)=2  (u 3v 1)=2 = 3v =2
is divisible by 3 when  is even and  (u 1)=2  (u 3v 1)=2 3v=2 = 3v(  1)=2
is divisible by 3 when  is odd, it follows that Conditions (iv) and (v) are satis ed.
Finally, suppose that u < 3v + 1. Then a 0 and b 0 (since u v=2 + 5 if
  2;4;6; or 8 (mod 10) in this lemma). To show a is divisible by u we go through
the same argument as before. If   0 (mod 10) then a is an integer and u divides
a. If   5 (mod 10) then v is odd, u is even, and  = 0 so that 3v u + 1 is even
and consequently a is an integer and u divides a. Now suppose that   2;4;6; or
8 (mod 10). Also, suppose u 6 0 (mod 5) so that  = 0. We will now show that
(u 1 3v) 0 (mod 5) to show a is an integer and u divides a when   2;4;6; or
8 (mod 10). Let u = 5k1 +e1 and v = 5k2 +e2 satisfy Condition (c) of Lemma 5.2.1
where e1 2 Z5 nf0g (i.e. u 6 0 (mod 5)) and e2 2 Z5. So the possible values of
e1 and e2 are (e1;e2) 2f(1;0);(4;1);(2;2);(3;4)g. It follows that (3v u + 1)  
3e2 e1 + 1 0 (mod 5) which shows a is an integer and consequently u divides a.
Now let e1 = 0 so that u 0 (mod 5). Then
 (3v u+ 1) 2   + 3 e2 2 (mod 10)
and so for some  2Z5,  + 3 e2 2  0 (mod 10). This shows that a is an integer
and u divides a and so Conditions (i) and (ii) are satis ed. Conditions (vi), (vii),
and (ix) will be discussed in turn based on  .
When   0 (mod 10), Conditions (vi) and (ix) are satis ed. If   5 (mod 10)
then u is even and so it is clear that Condition (vi) is satis ed. Let   2;4;6; or
8 (mod 10). Suppose that u 6 0 (mod 5) so that  = 0. Let u = 5k1 + e1 and
v = 5k2 + e2 satisfy Condition (c) of Lemma 5.2.1 where e1 2Z5nf0g and e2 2Z5,
so the possible values of e1 and e2 are (e1;e2)2f(1;0);(4;1);(2;2);(3;4)g. Hence it
is clear that  (2u v 2)   (2e1  e2  2)  0 (mod 10), so b is an integer and
u divides b. Now suppose that u 0 (mod 5). If  is even then there exists some
 2 Z5 such that  (2u v 2) 8   ( e2  2) 8  0 (mod 10). Therefore
Conditions (vii) and (ix) are satis ed.
When u 0 (mod 5) and  6 5 (mod 10), we need to use edges of 0; 1, 2, 3, or
4 di erences (depending on the value of  ) to construct some 5-cycles with only pure
81
edges so that the total number of remaining di erences needed to build 5-cycles with
only pure edges is divisible by 5. To this end, remember back to Section 2.2 where
Gu(D(u) ?) is de ned as containing ? copies of each edge with di erence in D(u) .
We will use the observation that there exists a 5{CS of Gu(fu=5g) and a 5{CS of
Gu(f2u=5g) when u  0 (mod 5), since each component is a 5-cycle. Recall that
Gu(f1;2;3g) is a 5{CS by Lemma 2.2.1 when u 0 (mod 5). Let (Zu; 1), (Zu; 2)
and (Zu; 3) be a 5{CS of Gu(fu=5g);Gu(f2u=5g), and Gu(f1;2;3g) respectively
when u 0 (mod 5).
5.3 5{CSs of  Kv+u  Kv:   5 (mod 10) and u< 3v + 1
In this section we will focus solely on the case where   5 (mod 10) and u <
3v+ 1. This case needs to be dealt with separately from the other values of  . Recall
that if   5 (mod 10) then u is even, v is odd, and  = 0. Partition the vertices
of U into two sets U0 and U1 of equal size such that U = U0[U1 = Zu=2 Z2 and
(x;i)2Ui for x2Zu=2 and i2f0;1g. Let G =  Kv+u  Kv. Then for each i2Z2,
G[Ui] =  Ku=2  = Gu=2(D(u=2)  ). For each i2Z2 the edge joining (j0;i) to (j1;i) is
said to have di erence du=2(j0;j1) and the di erence is said to be Type i. So for each
i2Z2 the edges joining vertices in Ui are partitioned by the set D0i = D(u=2)  of
Type i di erences. Each edge joining a vertex (j0;0) in U0 to a vertex (j1;1) is U1 is
said to have Type 2 di erence j1 j0 (mod u=2) (so the edges of Type 2 di erence
j form a 1-factor in G[U]); so the edges joining a vertex in U0 to a vertex in U1
are partitioned by the set D02 =  Zu=2 of Type 2 di erences. For each i 2 Z2 let
Si =fu=4gif u 0 (mod 4) and Si = ? if u 2 (mod 4). Let S2 = 2fu=4gif u 0
(mod 4) and S2 = ? if u 2 (mod 4). For each i2Z3 let Di = D0inSi.
Since we are currently focused on the case when u < 3v + 1, by de nition of c
there are no Type C 5-cycles. Additionally, we can split the classi cation of Type A
and Type B 5-cycles further based on how many edges these 5-cycles have in E(U0),
E(U1), or between vertices in U0 and U1:
 1 pure edge, and it either joins vertices in U0 or it joins vertices in U1,
 1 pure edge, that joins a vertex in U0 to a vertex in U1,
 3 pure edges, each joining a vertex in U0 to a vertex in U1,
 3 pure edges, one edge between vertices in U0, one edge between vertices in U1,
and one edge joining a vertex in U0 to a vertex in U1, and
 3 pure edges, one edge of di erence in Si for some i2Z2 joining two vertices
in Ui and two edges of di erence in S2 joining a vertex in U0 to a vertex in U1.
De ne these 5-cycles as Type A1, A2, B1, B2, and B3 respectively; so fA1;A2g is a
partition of A and fB1;B2;B3g is a partition of B. A visual representation of these
5-cycles are given in Figure 5.2. For the remainder of the paper, let a1, a2, b1, b2,
and b3 represent the number of 5-cycles of TypeA1, A2, B1, B2, andB3 respectively.
Recall that since   5 (mod 10),  is chosen to be 0. Notice that a1 + a2 = a and
82
b1 + b2 + b3 = b. The following system of equations de nes the relationship between
these values and the number of edges in  Ku:
a1 +a2 = a;
b1 +b2 +b3 = b; (5.3)
a2 + 3b1 +b2 + 2b3 =  (u=2)2; and
a1 + 2b2 +b3 =  u(u 2)=4:
It is easy to check that one solution of this system is:
a1 =
(
 u(3v u+ 1)=10 if 7u 6(2 +v) > 0 and
 u(u 2)=4 b3 if 7u 6(2 +v) 0,
a2 =
(
0 if 7u 6(2 +v) > 0 and
 u( 7u+ 6(2 +v))=20 +b3 if 7u 6(2 +v) 0,
b1 =
(
 u(2v +u+ 4)=40 b3=2 if 7u 6(2 +v) > 0 and
 u(2u v 2)=10 b3 if 7u 6(2 +v) 0, (5.4)
b2 =
(
 u(7u 6(2 +v))=40 b3=2 if 7u 6(2 +v) > 0 and
0 if 7u 6(2 +v) 0,
b3 =
(
u=2 if u 0 (mod 4) and
0 otherwise.
For the rest of the chapter, let a1, a2, b1, b2, and b3 be de ned as in (5.4). The following
lemma will be useful for showing that a1, a2, b1, and b2 satisfy the conditions that
will become vital in Lemma 5.3.2.
Lemma 5.3.1. Suppose that   5 (mod 10), that v=2 + 1  u < 3v + 1, and that
 , v, and u satisfy Condition (d) of Lemma 5.2.1. Then
(A) a1, a2, b1, and b2 are non-negative,
(B) a1, a2, b1, and b2 are each divisible by u=2, and
(C) a1=(u=2) and a2=(u=2) are even.
Proof. Let  = 10k3+5 so then by Lemma 5.2.1(d) we can let u = 2k1 and v = 2k2+1.
First suppose that 7u 6(2 + v) > 0. Then a1  0 since u < 3v + 1 in this
lemma, and a2 = 0. By de nition,
b1 =  u(2v +u+ 4)40  b32
  u(2v +u+ 4)40  u4
= u( (2v +u+ 4) 10)40
 0:
83
V
A1
U0 U1
V
A2
U0 U1
V
B1
U0 U1
V
B2
U0 U1
V
B3
U0 U1
Figure 5.2: Possible 5-cycle Types when   5 (mod 10) and u< 3v + 1
Since u> (6v + 12)=7 and (6v + 13)=7 is never an even integer, u (6v + 14)=7 and
so
b2  u( (7u 6(2 +v)) 10)40
 (6v + 14)( ((6v + 14) 6(2 +v)) 10)280
= (3v + 7)(  5)70  0:
So Condition (A) is satis ed. By de nition
a1 = 2k1(3k2 k1 + 2)(2k3 + 1);
b1 = k1(2k2 +k1 + 3)(2k3 + 1)=2 b3=2;and
b2 = k1(7k1 6k2 9)(2k3 + 1)=2 b3=2:
Clearly, a1 and a2 = 0 are divisible by u=2 = k1 and both a1=(u=2) = 2a1=u = a1=k1
and 2a2=u are even, so (C) is satis ed. If b3 = 0 then k1 is odd and so 2b1=u = b1=k1
and 2b2=u = b2=k1 are integers. If b3 = u=2 then u 0 (mod 4) and k1 is even, so
2b1=u =  (2v +u+ 4)20  b3u
= (2k2 +k1 + 3)(2k3 + 1)2  12
= 1 +k2 + 3k3 + 2k2k3 + k1(2k3 + 1)2
84
and
2b2=u =  (7u 6(2 +v))20  b3u
= (7k1 6k2 9)(2k3 + 1)2  12
= k1(14k3 + 7)2  9k3 3k2(2k3 + 1) 5:
So it follows that 2b1=u and 2b2=u are integers and thus Condition (B) is satis ed.
Finally, suppose that 7u 6(2 + v)  0. It is clear by (5.4) that a1, a2, and b2
are non-negative. Since v is odd, u6= v=2 + 1 and thus u v=2 + 3=2. So if b3 = u=2
then
b1   (v + 3)(2(v=2 + 3=2) v 2)20  v + 34 = (  5)(v + 3)20  0;
and so b1 is non-negative. If b3 = 0 then it is clear that b1 is non-negative. Since
u=2 = k1 and since
a1 = 5k1(k1 1)(2k3 + 1) b3;
a2 = k1(6k2 7k1 + 9)(2k3 + 1) +b3; and
b1 = k1(4k1 2k2 3)(2k3 + 1) b3;
it follows that Condition (B) is satis ed. If b3 = u=2 then k1 is even and so 2a1=u =
a1=k1 = 5(k1 1)(2k3 + 1) 1 is even. If b3 = 0 then k1 is odd and so 2a1=u = 5(k1 
1)(2k3+1) is even. If b3 = u=2 then k1 is even and so 2a2=u = (6k2+9 7k1)(2k3+1)+1
is even. If b3 = 0 then k1 is odd and so 2a2=u = (6k2 + 9 7k1)(2k3 + 1) is even; thus
Condition (C) is satis ed.
The next lemma will cover one of the cases of Theorem 5.4.1 using a slightly
di erent technique to construct the 5-cycles from the methods used so far.
Lemma 5.3.2. Suppose that   5 (mod 10), that v=2 + 1  u < 3v + 1, and
that  , v, and u satisfy Condition (d) of Lemma 5.2.1. Then there exists a 5{CS of
 Kv+u  Kv.
Proof. By Lemma 5.2.1(d), u is even and v is odd. Let U = Zu=2 Z2 be the vertex
set of  Ku. Let V = f1i j i 2 Zvg be the vertex set of  Kv. For i 2 Z2 let
Di = D(u=2)  nSi where Si =fu=4g if u 0 (mod 4) and Si = ? if u 2 (mod 4).
Let D2 = Zu=2 nS2 where S2 = 2fu=4g if u  0 (mod 4) and S2 = ? if u  2
(mod 4). Note that D0[S0, D1[S1, and D2[S2 represent all of the Type 0, 1 and
2 di erences respectively in  Ku and will be used to keep track of the edges in  Ku
as we decompose the edges of  Kv+u  Kv into 5-cycles.
The aim of this construction will be to form a1, a2, b1, b2, and b3 5-cycles of Type
A1,A2,B1,B2, andB3 respectively which will be shown to decompose  Kv+u  Kv
85
into 5-cycles. By Lemma 5.3.1, a1, a2, b1, b2, and b3 are non-negative integers. For
each i2Z2a1=u+2a2=u de ne
si =
8
><
>:
(i;i+ 1) if i is even and i<r,
(i+ 1;i) if i is odd and i<r, and
(2i;2i+ 1) if i r,
where calculations are done modulo v and
r = min(fv(  1)=2;2a1=u+ 2a2=ug):
The ordered pairs si will be used in the formation of the Type A 5-cycles to denote
the subscripts of the vertices in V.
Before forming the 5-cycles, the sets of di erences D0, D1, and D2 need to
be partitioned appropriately. By Lemma 5.3.1, 2b1=u, 2b2=u and 2a2=u are inte-
gers, and by the system of equations in (5.3), 3(2b1=u) + (2b2=u) + (2a2=u) = jD2j.
Form a partition T0[T [T000 of D2 where T0 contains 2b1=u sets of size 3 such
that at least one di erence in each set in T0 is unique (for example, if u 6 then
the ith set of T0 could be f3i;3i + 1;3i + 2g reducing the sum modulo u=2, and if
u = 4 then each element in T0 is f0;1;1g or f0;0;1g), T = fg0;g1;:::;g2b2=u 1g
contains 2b2=u di erences, and T000 = fq0;q1;:::;q2a2=u 1g contains the remaining
2a2=u di erences. By the system of equations in (5.3), jDij = 2b2=u + a1=u for
i 2 Z2 and by Lemma 5.3.1, 2a1=u is an even integer. Form a partition F1 [F2
of the set D0 so that F1 = ff0;f1;:::;f2b2=u 1g contains 2b2=u di erences and F2 =
fq2a2=u;q2a2=u+1;:::;q2a2=u+a1=u 1gcontains a1=u di erences. Form a partition H1[H2
of the set D1 such that H1 = fh0;h1;:::;h2b2=u 1g contains 2b2=u di erences and
H2 =fq2a2=u+a1=u;q2a2=u+a1=u+1;:::;q2a2=u+2a1=u 1gcontains a1=u di erences. The dif-
ferences in F2[H2 will be used to build the TypeA1 5-cycles. Let T00 =fffi;gi;higj
fi2F1;gi2T ;hi2H1;i2Z2b2=ug. Later, the di erences in T0 will be used to build
the TypeB1 5-cycles, the di erences in T00 will be used to build the TypeB2 5-cycles,
and the di erences in T000 will be used to build the Type A2 5-cycles.
Let t = 0 if qj2F2 and let t = 1 if qj2H2. De ne a mapping
f : sj!
(
f(0;j);(qj;j + 1)g if 0 j 2a2=u 1 and
f(0;t);(qj;t)g if 2a2=u j 2a2=u+ 2a1=u 1.
The function f will be used to indicate the pure edge in the Type A 5-cycle that is
to be joined with the corresponding vertices in V.
For each x2Zv let n(x) be the number of ordered pairs infsiji2Z2a1=u+2a2=ug
that contain x. Partition T into v sets T0;T1;:::;Tv 1 such that jTxj=   n(x) for
x < v 1 and jTv 1j =   n(v 1) 2b3=u. We will now show:   n(x)  0 for
x2Zv 1,   n(v 1) 2b3=u 0, and Pv 1j=0jTjj=jTj.
First we aim to show Pv 1j=0jTjj = jTj. The other 2 inequalities are established
86
in the next paragraph. Notice that Pv 1j=0 n(j) = 2(2a1=u+ 2a2=u). Then
v 1X
j=0
jTjj= 2b3=u+
v 1X
j=0
(  n(j))
=  v 2b3=u 2(2a1=u+ 2a2=u)
=
8<
:
 v 2b3=u 2
  (3v u+1)
5 +
2(0)
u
 
if 7u 6(2 +v) > 0 and
 v 2b2=u 2
  (u 2)
2  
2b3
u +
 ( 7u+6(2+v))
10 +
2b3
u
 
if 7u 6(2 +v) 0,
=  (2u v 2)=5 2b3=u
= (2=u)( u(2u v 2)=10 b3)
=
8<
:
2
u
  u((7u 6v 12)+(2v+u+4))
40  b3
 
(use this expression if 7u 6(2 +v) > 0)
2
u
  u(2u v 2)
10  b3 + 0
 
(use this expression if if 7u 6(2 +v) 0),
= 2b1=u+ 2b2=u =jTj:
If r = 2a1=u + 2a2=u and r < v(  1)=2, then clearly at most   1 ordered
pairs in fsiji2Z2a1=u+2a2=ug contain x for all x2Zv. Suppose that r = v(  1)=2
so that r 2a1=u+ 2a2=u. If we can show that v=2 > 2a1=u+ 2a2=u r then for all
x2Zv,   1 ordered pairs in fsiji2Zrg contain x and at most 1 ordered pair in
fsiji2Z2a1=u+2a2=unZrgcontain x (to be precise, n(x) =  if x 2(2a1=u+2a2=u r)
and n(x) =   1 otherwise); so it follows that for all x2Zv 1,   n(x)  0 and
that   n(v 1) 2b3=u 0. So if r = v(  1)=2 then
v
2 +r 
2a1
u  
2a2
u =
 v
2  
2a1
u  
2a2
u
=
8
<
:
 v
2  
  (3v u+1)
5 +
2(0)
u
 
if 7u 6(2 +v) > 0 and
 v
2  
  (u 2)
2  
2b3
u +
 (6(v+2) 7u)
10 +
2b3
u
 
if 7u 6(2 +v) 0,
=
(5 v 2 (3v u+1)
10 if 7u 6(2 +v) > 0 and5 v 5 (u 2)  (6v+12 7u)
10 if 7u 6(2 +v) 0,
=  (2u v 2)10 > 0;
The last inequality follows from u v=2 + 1 and v is odd. Therefore   n(x) 0
if x2Zv 1 and   n(v 1) 2b3=u 0.
We will now de ne the sets of 5-cycles that will form our desired 5{CS. Let
p1 = p3 = 1 and p2 = 0 if Ti T0 and p1 = 0 and p2 = p3 = 1 if Ti T00. Then de ne
C0 =f(1t1;(0;i);(qi;i+ 1);1t2;(qi + 1;i+ 1)) + (j;j)ji2Z2a2=u;
f(si) =f(0;i);(qi;i+ 1)g;j2Zu=2;si = (t1;t2)g;
C1 =f(1t1;(0;t);(qi;t);1t2;(0;t+ 1)) + (j;j)ji2ZrnZ2a2=u;
f(si) =f(0;t);(qi;t)g;j2Zu=2;si = (t1;t2)g;
C2 =f(12i;(0;t);(qi;t);12i+1;(0;t+ 1)) + (j;j)ji2Z2a1=u+2a2=u r;
87
f(si) =f(0;t);(qi;t)g;j2Zu=2g;
C3 =f(1i;(0;0);(d1;p1);(d1 d2;p2);(d1 d2 +d3;p3)) + (j;j)ji2Zv;
fd1;d2;d3g2Ti;j2Zu=2; if Ti T0 then d2 is uniqueg; and
C4 =
8
><
>:
? if b3 = 0, and
f((0;0);(v=4;0);(0;1);1v 1;(v=4;1)) + (j;j)jj2Zu=2g if b3 = u=2,
[f((0;1);(v=4;1);(0;0);1v 1;(v=4;0)) + (j;j)jj2Zu=2g
where calculations are done modulo v (including indicies). So C0 is a set of TypeA2
5-cycles, C1[C2 is a set of Type A1 5-cycles, C3 is a set of Type B1 and B2 5-cycles
(depending on whether Ti T0 or Ti T00 respectively), and C4 is a set of Type B3
5-cycles. Since every di erence in D0 [S0, D1 [S1, and D2 [S2 represents every
edge in  Ku, it follows that C0 [C1 [C2 [C3 [C4 decomposes the edges in  Ku
into 5-cycles. It remains to show that each vertex in U is joined to each vertex in V
 times.
By Lemma 5.3.1, 2a2=u is even, so by de nition of si, each 1i in V is adjacent
to every vertex in U the same number of times, though 1i may not be joined to
every vertex in U the same number of times as 1j. To be precise, 1i is adjacent
to every vertex in U n(i) times. Since jTij =   n(i) for i < v 1 and jTv 1j =
  n(v 1) 2b3=u, it follows that in C3,1i is adjacent to each vertex in U   n(i)
times for i<v 1 and 1v 1 is adjacent to each vertex in U   n(i) 2b3=u times.
Thus (V[U;C0[C1[C2[C3[C4) is a 5{CS of  Kv+u  Kv when   5 (mod 10)
and u< 3v + 1.
5.4 5{CSs of  Kv+u  Kv: Main Result
The following theorem uses the same proof method used in Chapters 2-4 to prove
the su cient conditions for a 5{CS of  Kv+u  Kv except possibly in the case where
  5 (mod 10) and  < v  (u 1)=3 or in the case where   2;4;6; or 8
(mod 10), u 0 (mod 5), and v=2 + 1 u<v=2 + 5. When  is odd, u is even, and
 <v (u 1)=3, the proof falls apart since each set of di erences used to construct
the 5-cycles of Type B or C will join each vertex in U to a vertex in V twice unless
the di erence u=2 is used. As long as the number of copies of di erence u=2 is more
than the number of vertices in V (i.e.   v) the method described in Theorem 5.4.1
is viable. If   2;4;6; or 8 (mod 10), u 0 (mod 5), and v=2 + 1  u < v=2 + 5,
then b is possibly negative since  can be at most 4.
Theorem 5.4.1. There exists a 5{CS of  Kv+u  Kv if and only if u v=2 + 1
and
(a) If   1;3;7; or 9 (mod 10) then
v +u;v 1 or 5 (mod 10);
v +u;v 7 or 9 (mod 10); and
88
v +u v 3 (mod 10),
(b) If   0 (mod 10) then any v 1 and v +u 5,
(c) If   2;4;6; or 8 (mod 10) then
v +u;v 0;1 (mod 5);
v +u;v 2;4 (mod 5); and
v +u v 3 (mod 5),
(d) If   5 (mod 10) then v and v +u are both odd,
except possibly in the case where   5 (mod 10), v  v + u  1 (mod 2), and
 < v (u 1)=3 or in the case where   2;4;6; or 8 (mod 10), u 0 (mod 5),
and v=2 + 1 u<v=2 + 5.
Proof. Let V = f10;11;:::;1v 1g, u = 10k + ?  2 for ? 2 Z10, and Zu is the
vertex set of U. The necessary conditions were proved in Lemma 5.2.1 so it remains
to show su ciency. As said before, Condition (a) is satis ed by Theorem 5.1.1. So
we can assume   2. Notice that if v = 0 then  Kv+u  Kv =  Ku in which case
a 5{CS exists by Theorem 1.2.2 and if v = 1 then  Kv+u  Kv =  Ku+1 in which
case a 5{CS exists again by Theorem 1.2.2; so v 2
First suppose u 3v + 1. By Lemma 5.2.2, b and c are non-negative integers, u
divides c, and vu divides b. We will need 3b pure edges to construct the 5-cycles of
Type B and so the number of pure edges that will be used to construct the Type C
5-cycles is
 u(u 1)=2 3b = ( (u 3v 1)=2)u = (5c=u+ )u =  u
where  represents the number of di erences used to construct the 5-cycles of Type
C. It is clear that when  = 0,   0 (mod 5). Recall that the value of  indicates
the number of di erences that will be used to construct  i for i2f1;2;3g (possibly
more than one of the 5{CSs of  i will be constructed) so that the    remaining
di erences needed to construct TypeC 5-cycles is divisible by 5. Our plan is to create
TypeC 5-cycles using  specially chosen di erences. Skolem sequences will be used as
in the proof to Lemma 4.2.3 to construct the TypeC 5-cycles. First, we will construct
the set B of Type C 5-cycles by addressing the values of ? in turn. Afterwards, we
will construct the set C of Type B 5-cycles.
Suppose ? 4 and ?6= 5. Then let P = f(xi;yi) j 1  i k;yi > xig be the
collection of pairs from a Skolem sequence or hooked Skolem sequence of order k for
the setf1;2;:::;2kgorf1;2;:::;2k+1gif k 0 or 1 (mod 4) or k 2 or 3 (mod 4)
respectively (see Theorems 1.1.1 and 1.1.2). If ? = 8 then let t1 = 1 and let t1 = 0
otherwise. If ? = 9 then let t2 = 1 and let t2 = 0 otherwise. Then let
B0 =fci 1 = (0;xi;xi yi;3k 1+b?=2c t1;5k+i+b?=2c+t2)j1 i k;(xi;yi)2Pg:
Table 5.1 lists the di erences of edges that occur in 5-cycles in B0 when  is as large as
possible ( is as small as possible). Notice that  jB0j=  k  =5 =  (10k+? 3v 
89
1)=10 if ? 7 or v 3. If v = 2 and ?2f8;9g, then  =10 and  =5 more 5-cycles using
di erences need to be constructed when ? = 8 and 9 respectively. To account for this,
form the set of 5-cycles B =fci+jjj2Zu;i2Zt01;ci (mod k) 2B0g[t02fc+jjj2Zug
where t01 = min(f k; =5g), t02 = max(f0; =5  kg) and
c =
8
><
>:
(0;2k + 1;4k + 2;6k + 4;4k + 3) if ? = 8 and k 0;1 (mod 4),
(0;2k;6k + 3;4k + 3;6k + 5) if ? = 8 and k 2;3 (mod 4), and
(0;2k + 2;4k + 4;6k + 6;3k + 3) if ? = 9.
Since v 2 and
 
5   k =
 (u 3v 1)
10   k =  
 
k + ?10
 
  k  (3v + 1)10 =  (? 3v 1)10 ;
it follows that the number of times c is copied is no more than  =10 or  =5 times if
? = 8 or 9 respectively. So we have enough copies of the di erences to construct c.
Suppose ? = 0. Let P0 = f(x0i;y0i) j 0  i k 3;y0i > x0ig be the collection of
pairs from a Skolem sequence or hooked Skolem sequence of order k 2 for the set
f1;2;:::;2k 4g or f1;2;:::;2k 3g if k 2  0 or 1 (mod 4) or k 2  2 or 3
(mod 4) respectively. If k 2 0 or 1 (mod 4) then name the Skolem pairs so that
(x0k 3;y0k 3) contains 2k 3. De ne P = f(xi = x0i + 3;yi = y0i + 3) j (x0i;y0i) 2P0g;
notice that 2k 2fxk 3;yk 3g when k 2  0 or 1 (mod 4). To simplify how the
5-cycles are de ned, let z =b =2c. Then with c0i = (0;xi;xi yi;3k;5k + 2 +i) and
ci =
8
>>>>
>>>
>>>>
>>>
>><
>>>>
>>>
>>>>
>>>
>>:
(0;1;3;2k + 4;2k + 1) if i = 1 and k 2 0;1 (mod 4),
(0;1;3;3k + 3;3k) if i = 1 and k 2 2;3 (mod 4),
(0;3k;6k;k + 1;5k + 1) if i = 2 and k 2 0;1 (mod 4),
(0;2k;6k 1;k;5k 1) if i = 2 and k 2 2;3 (mod 4),
(0;1;2;5;3) if 3 i z + 1,
(0;2;2k + 3;7k + 2;5k + 1) if z + 2 i 2z and k 2 0;1 (mod 4),
(0;2;2k + 2;7k + 1;5k + 1) if z + 2 i 2z and k 2 2;3 (mod 4),
(0;3k;6k;2k + 1;6k + 1) if 2z + 1 i 3z 1, and
c0i 3z (mod k 2) if 3z i  (k 2) + 3z 1,
form the set of 5-cycles B0 = fc00i 1 = ci j 1  i  (k 2) + 3z 1;(xi;yi) 2Pg.
Notice that when  = 2, ci will never be any of the following 5-cycles: (0;1;2;5;3),
(0;2;2k+3;7k+2;5k+1), (0;2;2k+2;7k+1;5k+1), or (0;3k;6k;2k+1;6k+1). So
when k> 2 we form the set of 5-cycles B =fc00i +jji2Zb =5c;j2Zu;c00i 2B0g[E 
90
Edges
Di erences
when
k 
0;1
(mo
d4)
Di erences
when
k 
2;3
(mo
d4)
f0;
xig
;fx
i;x
i 
yig
1;2
;::
:;2
k
1;2
;::
:;2
k 
1;2
k+
1
fxi
 y
i;3
k 
1+
b?=
2c 
t1g
if?
6=8
3k
+b
?=2
c;3
k+
1+
b?=
2c;
:::
;4k
 1
+b
?=2
c
3k
+b
?=2
c;3
k+
1+
b?=
2c;
:::
;4k
 1
+b
?=2
c
fxi
 y
i;3
k 
1+
b?=
2c 
t1g
if?
=8
3k
+3
;3k
+4
;::
:;4
k+
2
3k
+3
;3k
+4
;::
:;4
k+
2
f3k
 1
+b
?=2
c 
t1;5
k+
i+
b?=
2c+
t2g
if?
 7
2k
+2
;2k
+3
;::
:;3
k+
1
2k
+2
;2k
+3
;::
:;3
k+
1
f3k
 1
+b
?=2
c 
t1;5
k+
i+
b?=
2c+
t2g
if?
 8
2k
+3
;2k
+4
;::
:;3
k+
2
2k
+3
;2k
+4
;::
:;3
k+
2
f5k
+i
+b
?=2
c+
t2;0
gif
?6=
9
5k
 1
+d
?=2
e;5
k 
2+
d?=
2e;
:::
;4k
+d
?=2
e
5k
 1
+d
?=2
e;5
k 
2+
d?=
2e;
:::
;4k
+d
?=2
e
f5k
+i
+b
?=2
c+
t2;0
gif
?=
9
5k
+3
;5k
+2
;::
:;4
k+
4
5k
+3
;5k
+2
;::
:;4
k+
4
Table
5.1:
Edges
of
di erences
in
B0
when
 
is
as
large
as
possible
for
? 
4and
?6=
5
91
where computations are done modulo u and where
E =
8
>>>
>>>
<
>>>
>>>
:
? if   0 (mod 5),
 2 if   1 (mod 5),
 1[ 2 if   2 (mod 5),
 3 if   3 (mod 5), and
 2[ 3 if   4 (mod 5).
Since
   
5 =
 (10k 3v 1) 2 
10 =  k 
 (3v + 1) + 2 
10   k 
 7 
10
 
;
sincejB0j= 2k 2 if  = 2, and sincejB0j=  (k 2)+3z  (k 2)+3( =2 1=2) =
 k  =2 3=2 if  > 2, it follows that b(   )=5c jB0j, so we have enough 5-
cycles to construct the Type C 5-cycles. Notice that if v = 2 and  < 8, then  6 2
(mod 5). Fortunately, when  is as large as possible then either  6 2 (mod 5) or
d =5e<jB0j, so in either case 2k is an available di erence and thus the appropriate
di erences are available to be used in E . Table 5.2 lists the di erences of edges that
occur in cycles in B0 when  is as large as possible. Note that j2Zk 2 in the table
and that the  rst 4 rows of this table represent ci when i 3z. If u = 10 (k = 1), then
since b =5c = b3 =10c  =2 when v = 2 and d =5e = 0 when v = 3 (v 3 since
u 3v + 1 in this case), it follows that B =b =5cf(0;1;2;5;3) +jjj2Zug[E is
a set of 5-cycles with the required number of 5-cycles. If u = 20, let
S =f(0;1;4;5;2);(3;4;6;8;5);(2;3;6;7;4);(0;u=5;2u=5;3u=5;4u=5); (5.5)
(0;2u=5;4u=5;u=5;3u=5)g;
f = min(f  1;b =5cg), and g = max(f0;b =5c (  1)g). So then form the set
Edges Di erenceswhenk 2 0;1 (mod4) Di erenceswhenk 2 2;3 (mod4)
f0;xjg;fxj;xj yjg 4;5;:::;2k( copies) 4;5;:::;2k 1;2k+1 ( copies)fx
j yj;3kg 3k+1;3k+2;:::;4k 2 ( copies) 3k+1;3k+2;:::;4k 2 ( copies)f3k;5k+j+2g 2k+2;2k+3;:::;3k 1 ( copies) 2k+2;2k+3;:::;3k 1 ( copies)
f5k+j+2;0g 5k 2;5k 3;:::;4k+1 ( copies) 5k 2;5k 3;:::;4k+1 ( copies)c
i ifi=1 1;2;3;2k+1;2k+1 1;2;3;3k;3kc
i ifi=2 3k;3k;4k;5k 1;5k 1 2k;4k 1;4k 1;5k 1;5k 1c
i if3 i z+1 1;1;2;3;3 (z 1 copies) 1;1;2;3;3 (z 1 copies)c
i ifz+2 i 2z 2;2k+1;2k+1;5k 1;5k 1 (z 1 copies) 2;2k;2k;5k 1;5k 1 (z 1 copies)c
i if2z+1 i 3z 1 3k;3k;4k 1;4k 1;4k(z 1 copies) 3k;3k;4k 1;4k 1;4k(z 1 copies)
Table 5.2: Edges of di erences in B0 when  is as large as possible for ? = 0
of 5-cycles B = ffs + j j j 2 Zu;s 2 Sg[gf(0;5;11;6;13) + j j j 2 Zug[E .
Notice that S represents the set of 5-cycles formed from the di erences 1, 2, 3, u=5,
and 2u=5 and f represents the number of copies of the 5-cycles constructed from the
92
di erences 1, 2, 3, u=5, and 2u=5. Also notice that B uses at least one less than the
number of copies of the di erences 1, 2, 3, u=5, and 2u=5 in D(u)  . Since v 2 and
b =5c (  1) =
  (u 3v 1)
10
 
 (  1)  + 1  (3v + 1)10 ;
it follows that there are enough copies of the di erences to construct the cycle
(0;5;11;6;13) g times. So there are enough copies of each di erence to construct
the 5-cycles in B.
Suppose ?2f1;3g. Let P =f(xi;yi)j1 i k 1;yi >xig be the collection
of pairs from a Skolem sequence or hooked Skolem sequence of order k 1 for the set
f1;2;:::;2k 2g or f1;2;:::;2k 1g if k 1  0 or 1 (mod 4) or k 1  2 or 3
(mod 4) respectively. Then let
B0 =fci 1 = (0;xi;xi yi;3k;5k +i)j1 i k 1;(xi;yi)2Pg:
Table 5.3 lists the di erences of edges that occur in cycles in B0 when  is as large as
possible ( is as small as possible). Since  =5 < k and jB0j=  (k 1) we need to
form  more 5-cycles constructed from di erences. (Note that by Conditions (b-d),  
must be even since u is odd.) To account for these 5-cycles, form the set of 5-cycles
B =fci+jji2Zx1;j2Zu;ci (mod k 1) 2B0g[x2fc00+jjj2Zug[x3fc01+jjj2Zug
where x1 = min(f =5; (k 1)g), x2 = max(f0;b( =5   (k 1))=2cg), x3 =
max(f0;d( =5  (k 1))=2eg), and where
c0i =
8
>>>
<
>>>
:
(0;2k + 1;4k;2k;6k); if ? = 1, k 0;1 (mod 4), and i2Z2,
(0;2k 2;4k 2;2k;4k) if ? = 1, k 2;3 (mod 4), and i = 0,
(0;2k + 1;4k + 2;8k + 2;4k + 1) if ? = 1, k 2;3 (mod 4), and i = 1, and
(0;2k;4k + 1;1;4k + 2) if ? = 3 and i2Z2.
Since v 2 and
 
5  
 (k 1)
2 =
 (u 3v 1)
10  
 (k 1)
2
=  k +  (? 3v 1)10   k2   2
=  k2 +  (? 3v 6)10
<  k2
there are enough copies of the di erences to construct c0i.
Suppose ? = 2. Let P0 = f(x0i;y0i) j 0  i k 2;y0i > x0ig be the collection
of pairs from a Skolem sequence of order k 1 for the set f1;2;:::;2k 2g or
93
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 1;2;:::;2k 2 1;2;:::;2k 3;2k 1
fxi yi;3kg 3k+1;3k+2;:::;4k 1 3k+1;3k+2;:::;4k 1
f3k;5k+ig 2k+2;2k+3;:::;3k 2k+2;2k+3;:::;3k
f5k+i;0g 5k 1+d?=2e;5k 2+d?=2e;:::; 5k 1+d?=2e;5k 2+d?=2e;:::;4k+1+d?=2e 4k+1+d?=2e
Table 5.3: Edges of di erences in B0 when  is as large as possible for ?2f1;3g
f1;2;:::;2k 1g if k 1  0 or 1 (mod 4) or k 1  2 or 3 (mod 4) respectively.
De ne P =f(xi = x0i + 1;yi = y0i + 1)j(x0i;y0i)2P0g. Then with
ci =
8>
>><
>>>:
(0;xi;xi yi;3k + 1;5k +i+ 1) if 0 i k 2,
(0;1;3k + 1;7k + 2;4k + 1) if i = k 1 and k 1 0;1 (mod 4),
(0;1;2k + 1;5k + 1;2k) if i = k and k 1 0;1 (mod 4), and
(0;1;2k;6k + 1;3k + 1) if i = k 1 and k 1 2;3 (mod 4),
let
B0 =fc0i = ci (mod k 1) ji2Z (k 1);(xi;yi)2Pg
[fc0i = ck 1 j (k 1) i  (k 1) +b =2c 1g
[fc0i = ckj (k 1) +b =2c i  (k 1) + 2b =2c 1g
if k 1 0 or 1 (mod 4) or let
B0 =fc0i = ci (mod k 1) ji2Z (k 1);(xi;yi)2Pg[fc0i = ck 1 j (k 1) i  kg
if k 1  2 or 3 (mod 4). Since jB0j   (k 1) + 2b =2c 1   k 2 and
 =5  k  =2 we have enough 5-cycles to form the appropriate number of 5-cycles
of Type C. Table 5.4 lists the di erences of edges that occur in 5-cycles in B0 when
 is as large as possible ( is as small as possible). Then form the set of 5-cycles
B =fc0i + jji2Z =5;j2Zu;c0i2B0g. If k = 0 then it is impossible to construct a
5-cycle.
Edges Di erenceswhenk 1 0;1 (mod 4) Di erenceswhenk 1 2;3 (mod 4)
f0;xig;fxi;xi yig 2;3;:::;2k 1 2;3;:::;2k 2;2k
fxi yi;3k+1g 3k+2;3k+3;:::;4k 3k+2;3k+3;:::;4k
f3k+1;5k+i+1g 2k+1;2k+2;:::;3k 1 2k+1;2k+2;:::;3k 1
f5k+i+1;0g 5k;5k 1;:::;4k+2 5k;5k 1;:::;4k+2
ci ifi=k 1;3k;3k+1;4k+1;4k+1 1;2k 1;3k;3k+1;4k+1
ci ifi=k+1 1;2k;2k;3k;3k+1  
Table 5.4: Edges of di erences in B0 when  is as large as possible for ? = 2
Suppose ? = 5. Let P0 = f(x0i;y0i) j 1  i k 2;y0i > x0ig be the collection of
94
pairs from a Skolem sequence or hooked Skolem sequence of order k 2 for the set
f1;2;:::;2k 4g or f1;2;:::;2k 3g if k 2  0 or 1 (mod 4) or k 2  2 or 3
(mod 4) respectively. De ne P =f(xi = x0i + 3;yi = y0i + 3)j(x0i;y0i)2P0g. Let
ci =
(
(0;xi;xi yi;4k + 2;7k + 3 i) if 1 i k 2, and
(0;3k + 1;6k + 3;k + 2;5k + 3) if i = k 1,
and B0 = fc0i 1 = ci j 1  i  k 1;(xi;yi) 2 Pg. Table 5.4 lists the di erences
of edges that occur in cycles in B0 when  is as large as possible ( is as small as
possible). Let S be de ned as in Equation (5.5) and let w1 = min(f (k 1);b =5cg)
and w2 = max(f0;b =5c  (k 1)g). So when k 1 we form the set of 5-cycles
B =fc0i +jji2Zw1;j2Zu;c0i (mod k) 2B0g[w2fs+jjs2S;j2Zug[E 
where computations are done modulo u. Table 5.5 lists di erences of edges in the
5-cycles in B0 when  is as large as possible. Since b =5c< k and since
j 
5
k
  (k 1) =
  (u 3v 1)
10
 
  (k 1)
=
 
 k + 4 3v10
 
+   k
=  +
 4 3v
10
 
   1;
it follows that E can be constructed and there are enough copies of the di erences
1, 2, 3, u=5, and 2u=5 in S. If k = 0 then let ci = (0;1;2;3;4) for 0 i b =2c 1
and let ci = (0;2;4;1;3) for i >b =2c 1, and so we can form the set of 5-cycles
B =fci +jji2Zb =5c;j2Zv.
Edges Di erenceswhenk 2 0;1 (mod 4) Di erenceswhenk 2 2;3 (mod 4)
f0;xig;fxi;xi yig 4;5;:::;2k 1 4;5;:::;2k 2;2k
fxi yi;4k+2g 4k+3;4k+4;:::;5k 4k+3;4k+4;:::;5k
f4k+2;7k+3 ig 3k;3k 1;:::;2k+3 3k;3k 1;:::;2k+3
f7k+3 i;0g 3k+3;3k+4;:::;4k 3k+3;3k+4;:::;4k
ci ifi=k 1 3k+1;3k+2;4k+1;5k+1;5k+2 3k+1;3k+2;4k+1;5k+1;5k+2
Table 5.5: Edges of di erences in B0 when  is as large as possible for ? = 5
We will now construct the TypeB 5-cycles. By Lemma 5.2.2,  (u 1)=2  or
 (u 1)=2   3v=2 is divisible by 3 when  is even or odd respectively. We can form
a partition P0 of the remaining di erences after constructing B into  (u 1)=2  
sets of size 3 when  is even or into v sets fd;u=2g of size 2 with d 6= u=2 and
 (u 1)=2   3v=2 sets of size 3 when  is odd; make sure that at most one copy of
v=2 is in each set of size 3 and that there is at least one unique element in each set of
size 3. These last conditions are easily met based on the di erences used to construct
95
each B. Since   v in this lemma, there are enough copies of the di erence u=2 to
construct the desired partition. Letfd1;i;d2;i;d3;igandfdi;u=2gdenote the ith set in
P0 and let p be the number of sets of size 3 in P0. If  is even then de ne
C =f(d1;i;0;d3;i;d2;i +d3;i;1i (mod v)) +jjfd1;i;d2;i;d3;ig2P0 (5.6)
with d1;i d2;i d3;i;j2Zug (5.7)
and de ne
C =f(d1;i;0;d3;i;d2;i +d3;i;1i (mod v)) +jjfd1;i;d2;i;d3;ig2P0
with d1;i d2;i d3;i;j2Zug
[f(di;0;u=2;u=2 +di;1i+p (mod v)) +jjfu=2;dig2P0;j2Zu=2g
when  is odd. This ensures the elements of C are indeed 5-cycles. Then (Zu[
f10;:::;1v 1g;B[C) is a 5{CS of  Kv+u  Kv when u 3v + 1.
Finally suppose u< 3v+1. If  is odd then by Lemmas 5.1.1 and 5.3.2 there is a
5{CS of  Kv+u  Kv. So let  be even. Remember that if   2;4;6; or 8 (mod 10)
and u 0 (mod 5) then u v=2 + 5. So u6= 5 when   2;4;6; or 8 (mod 10). By
Lemma 5.2.2, a and b are non-negative integers, u divides a, and either u divides b
when u is odd or u=2 divides b when u is even.
If u = 1 or 2, then it is impossible to form any 5-cycles. If u = 3 then v = 2;3;
or 4. Since only Type A 5-cycles can be formed when u = 3, since there are 3 
pure edges in  Kv+3  Kv, and since there are 6 or 9 mixed edges when v = 2
or 3 respectively, it follows that there are not enough mixed edges to form all of the
required 5-cycles, so a 5{CS cannot be formed under these conditions. When u = 3
and v = 4, de ne
C =f(0;10;2;11;1) +jjj2Z3g and C0 =f(0;12;2;13;1) +jjj2Z3g:
So then (Zu[f10;11;12;13g; C[ C0) is a 5{CS of  K7  K4 when u< 3v+1.
Suppose u2f4;5g. Then form a partition P0 of 3b=u of the di erences in D(u)  
into b=u sets of size 3 such that each set in P0 is f1;1;2g. Let fd1;i;d2;i;d3;ig denote
the ith set of size 3 in P0 and let D = fd1;d2;:::;dsg be the remaining di erences
after forming the partition P0. De ne C as in Equation (5.6) and
B =f(0;di;12i 2+jP0j (mod v);di + 1;12i 1+jP0j (mod v)) +jjdi2D;j2Zug:
Then (Zu[f10;11;:::;1vg;B[C) is a 5{CS of  Kv+4  Kv.
Suppose u = 6. Then form a partition P0 = P1[P2 of 3b=u of the di erences in
D(u)  into b=u sets of size 3 such that P0 does not containf2;2;2gorf3;3;3gand P2
only contains copies off1;2;3g. It is easy to ensure that P0 does not containf2;2;2g
or f3;3;3g since every di erence in D(u)  is available when P0 is constructed. Let
fd1;i;d2;i;d3;ig denote the ith set of size 3 and D = fd1;d2;:::;dsg be the remaining
96
di erences after forming the partition P0. De ne
C1 =f(0;d2;i;d1;i +d2;i;d1;i +d2;i +d3;i;1i (mod v)) +jjfd1;i;d2;i;d3;ig2P1;j2Zu;
d1;i d2;i d3;ig;
C2 =f(d1;i;0;d3;i;d2;i +d3;i;1i+jP1j (mod v)) +jjfd1;i;d2;i;d3;ig2P2;j2Zu;g; and
B =f(0;di;12i 2+jP0j (mod v);di + 1;12i 1+jP0j (mod v)) +jjdi2D;j2Zug:
Then C = C1 [C2 and so (Zu [f10;11;:::;1vg;B[C1 [C2) is a 5{CS of
 Kv+6  Kv.
Let u  7. Let P contain the  di erences used to construct E . Form a
partition P0 of 3b=u di erences from D(u)  nP into b=u sets of size 3 so that each set
contains at most one copy of v=2 and at least one unique element. We can guarantee
that such a partition can be constructed since u 7 and every di erence in D(u)  nP 
is available at this point. Notice that 3b=u is no larger than D(u)  nP . De ne
C =f(d1;i;0;d3;i;d2;i +d3;i;1i (mod v)) +jjfd1;i;d2;i;d3;ig2P0
with d1;i d2;i d3;i;j2Zug[
where calculations are performed modulo u. Thus C is a set of 5-cycles. Let D =
fd1;d2;:::;dsg be the remaining di erences in D(u)  after partitioning 3b=u +  of
the di erences into P0. Then de ne
B =f(0;di;12i 2+jP0j (mod v);di + 1;12i 1+jP0j (mod v)) +jjdi2D;j2Zug[E :
Then (Zu[f10;:::;1v 1g;B[C) is a 5{CS of  Kv+u  Kv for u< 3v + 1.
97
Chapter 6
Conclusion/Future Work
As a result of the work in this dissertation, we have come up with some notable
conjectures that will be investigated in the future.
Conjecture 6.1.1. Suppose there exists a 5{CS of  Kv. If the necessary conditions
in Lemma 2.1.2 are satis ed then a 5{CS of  Kv can be enclosed in a 5{CS of
( +m)Kv+u when u 3.
Conjecture 6.1.2. If the necessary conditions in Lemma 4.1.1 are satis ed then
there exists a 5{CS of ( +m)Kv+u  Kv for m 1 and u 2
Conjecture 6.1.3. Suppose there exists a 5{CS of  Kv F where F is a 1-factor.
Then a 5{CS of  Kv F can be enclosed in a 5{CS of ( +m)Kv+u if and only if
1. ( +m)u+m(v 1) 1 0 (mod 2),
2.  u2 ( +m) +m v2 +vu( +m) +v=2 0 (mod 5),
3. if u = 1, then (m(v 1) + 1) 3( +m),
4. if u = 2, then m v2 +v=2 2( +m) (v 1)( +m)=2 0, and
5. if u 3, then dvu( + m)=4e+ 2  v(m(v 1) + 1)=2 + ( + m)u(u 1)=2
where  = 0 or 1 if vu( +m) 0 or 2 (mod 4) respectively.
Conjecture 6.1.4. Suppose there exists a 5{CS of  Kv. Then a 5{CS of  Kv can
be enclosed in a 5{CS of ( +m)Kv+u F where F is a 1-factor if and only if
1. ( +m)u+m(v 1) 1 0 (mod 2),
2.  u2 ( +m) +m v2 +vu( +m) (v +u)=2 0 (mod 5),
3. if u = 1, then (m(v 1) 1) 3( +m),
4. if u = 2, then m v2  v=2 2( +m) (v 1)( +m)=2 0, and
5. if u 3, then dvu( + m)=4e+ 2  v(m(v 1) 1)=2 + ( + m)u(u 1)=2
where  = 0 or 1 if vu( +m) 0 or 2 (mod 4) respectively.
Conjecture 6.1.5. There exists a 5{CS of ( + m)Kv+u  Kv F where F is a
1-factor if and only if
1. ( +m)u+m(v 1) 1 0 (mod 2),
2. ( +m)(v +u 1) 1 0 (mod 2),
3.  u2 ( +m) +m v2 +vu( +m) (v +u)=2 0 (mod 5),
4. if u = 1, then (m(v 1) 1) 3( +m),
5. if u = 2, then m v2  v=2 2( +m) (v 1)( +m)=2 0, and
98
6. if u 3, then dvu( + m)=4e+ 2  v(m(v 1) 1)=2 + ( + m)u(u 1)=2
where  = 0 or 1 if vu( +m) 0 or 2 (mod 4) respectively.
Conjecture 6.1.6. If the necessary conditions in Lemma 2.1.2 are satis ed then a
t{CS of  Kv can be enclosed in a t{CS of ( +m)Kv+u when t> 5 is odd.
Since it was shown in Chapter 4 when a 5{CS of ( + m)Kv+1   Kv exists
by using the construction in Chapter 2, we believe that a proof of Conjecture 6.1.2
when u = 2 can be found by using the construction in Chapter 3. This will soon be
investigated.
The di culty in extending our work from Chapters 2 and 3 to u 3 in Con-
jecture 6.1.1 can be seen from the extra work in Section 3.3 to prove Theorem 3.5.2.
A trail with a very special property was constructed so as to form a 5{CS subgraph
that contains all of the mixed edges of a given multiset of di erences. Unless an-
other method is developed to obtain such a trail we believe that it will be extremely
di cult to use the methods presented in this dissertation to solve Conjecture 6.1.1
when u = 3 or 4. There may be some hope when u  5 since Theorem 5.4.1 can
be used to construct some 5{CSs of ( + m)Kv+u  Kv. If there exists a 5{CS
of ( + m)Kv+u ( + m)Kv+u and there exists a 5{CS of mKv, then there exists
a 5{CS of ( + m)Kv+u  Kv. By checking the conditions in Theorem 1.2.2 and
Theorem 5.4.1, if v 1 (mod 10) or  + m 0 (mod 10) then there exists a 5{CS
of ( +m)Kv+u  Kv. Also, if   1;3;7; or 9 (mod 10) and one of the following is
true if:
  +m 0 (mod 5),
  +m 2;4;6; or 8 (mod 10) and u 0 or 1 (mod 5), or
  +m 1;3;7; or 9 (mod 10) and u 6 (mod 10),
then there exists a 5{CS of ( + m)Kv+u  Kv. Once  6 0;1;3;7; or 9 (mod 10)
then the cases settled using this approach become very sparse.
There may be other hope in  nishing the u 5 case in Conjecture 6.1.1 since
the types of 5-cycles based on the number of pure edges and mixed edges are equal
for every u 5. A visual representation of these 5-cycles is given in Figure 6.1.
The same comments made for Conjecture 6.1.1 can be said for Conjecture 6.1.2
since the work on the enclosing problem in this dissertation was used to construct a
5{CS of ( +m)Kv+u  Kv.
Following the two parts of the Alspach Conjecture stated in Section 1.2, there
is a second version of the enclosing problem as described in Conjectures 6.1.3 and
6.1.4. Since having a 5{CS of  Kv F or a 5{CS of ( +m)Kv+u F are di erent
problems, there are two conjectures that involve a 1-regular graph as opposed to the
one conjecture that involves a 1-regular graph in the Alspach Conjecture. Both of
these conjectures can be extended by removing the condition that a 5{CS of  Kv F
or a 5{CS of  Kv exists respectively. Removing this condition in Conjecture 6.1.4
will give us Conjecture 6.1.5 while removing this condition from Conjecture 6.1.3 will
99
give us Conjecture 6.1.2.
Another avenue to take this research is to consider enclosing problems for t{CSs
for t> 7 and t odd, as seen in Conjecture 6.1.6. Conjectures 6.1.3-6.1.5 can also be
generalized to t{CSs. As long as a t-cycle can be constructed using di erences, the
arguments described in this dissertation could be transferred to create constructions
of other t{CSs. The hitch in this idea is that we have not explored whether a t-cycle
can be constructed using Skolem sequences. There may also be options other than
Skolem sequences to build the t-cycles using di erences. This too will be investigated
soon.
100
V
U
V
U
V
U
V
U
V
U
V
U
V
U
V
U
Figure
6.1:
All
typ
es
of
5-cycles
based
on
the
num
ber
of
pure
and
mixed
edges
101
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104
Index
( +m)Kv+u  Kv, 1
1-factor, 98
degH(v), 49
 -fold k-cycle system, 2
_t, 49
k-cycle, 2
k-cycle system, 2
k{CS(v; ), 2
adjacent, 1
admissible, 7, 12, 19, 21, 28, 68, 69, 73
Alspach Conjecture, 6, 99
blocks, 2
closed trail system, 33{37
complete, 3
component, 37, 42, 44
connected, 2, 33, 42
cycle, 2
cycle system, 1, 9, 32, 78
decomposition, 2
degree, 2, 9, 10, 49, 50, 67, 79
Doyen-Wilson theorem, 8
edge, 1
embedded, 3, 7, 8, 78
enclosed, 3, 9, 98
 rst He ter di erent problem, 6
graph, 1, 2, 4, 5, 9, 11
 -fold complete, 1
complete, 1
multigraph, 1
regular, 2
simple, 1
Hamilton cycle, 2, 28
hook, 4
immersed, 3
incident, 1, 28
multigraph, 1, 28, 49
path, 2, 16, 28, 33{39, 41{46, 48{50
regular, 2
second He ter di erence problem, 6
Skolem pairs, 4, 13, 90
Skolem sequence, 4, 13, 15, 17, 18, 20, 22,
51{55, 57{61, 70, 71, 89, 90, 93,
95
extended, 4
hooked, 4, 13, 15, 17, 18, 20, 22, 51{
55, 57{61, 70, 71, 89, 90, 93, 95
Steiner triple system, 2{5, 8
cyclic, 5
subgraph, 1, 99
trail, 2, 28, 33, 34, 36{39, 42{50, 99
closed, 2, 28, 33{36, 38, 39, 41, 42,
48{50
vertex, ii, 1{5, 9{13, 28, 33{38, 42{45, 47{
51, 67{69, 73, 78, 79, 82, 85, 86,
88, 89
105