, 103

(factorization)

(combinatorial design)

(star decomposition)

(path decomposition)(cycle decomposition)

(complete bipartite decomposition)

(antidirected cycle)

(caterpillar

factorization)(arboricity)

(linear arboricity)(star arboricity)

(one-factor)

(crown)

kk(linear

k-arboricity)

Path, Cycle and Star Decompositions of

Graphs

Student: ming-ju Lee Advisor: Dr. Chiang Lin

Department of Mathematics

National Central University

Abstract

Graph decomposition is an important subject of graph theory since many

mathematical structures are linked to it and its result can be widely applied

in other fields. The factorization is a special type of graph decomposition,

and it has close connections to combinatorial designs. There are various de-

composition problems such as clique decomposition, star decomposition, path

decomposition, cycle decomposition, complete bipartite decomposition, and so

on. Nowadays, they continue to be popular topics of research.

In this thesis, the problems of antidirected cycle decomposition and directed

path decomposition of some digraph, and that of caterpillar factorization of

some graphs are investigated. There are various close connections between

graph decomposition and the arboricity of a graph. In this thesis, we also

show that the linear arboricity and the star arboricity of some graphs.

There are seven chapters in this thesis. In Chapter 1, some basic definitions

and notations are introduced. In Chapter 2, we first establish a necessary and

sufficient condition for the isomorphic path decomposition of complete tripar-

i

tite multigraphs. We then investigate the isomorphic directed path decompo-

sition of complete tripartite multidigraphs.

In Chapter 3, we give a necessary and sufficient condition for the existence

of the antidirected cycle decompositions of complete symmetric graphs. In

Chapter 4, we give a necessary and sufficient condition for the existence of

the antidirected cycle decompositions of complete symmetric graphs minus a

one-factor.

In Chapter 5, the caterpillar factorization of crowns are studied. We first

establish a necessary and sufficient condition for the balanced caterpillar fac-

torization of crowns. Then we give a necessary and sufficient condition for the

directed caterpillar factorization of symmetric crowns.

In Chapter 6, we first consider the problem of the star arboricity of crowns.

A lower bound is given. Then we investigate the star arboricity of some special

crowns. In Chapter 7, we consider the problem of the linear k-arboricity of

complete graph Kn for some specific k.

ii

Contents

1 Introduction 1

1.1 Introduction to decompositions and factorizations of graphs . . 1

1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Overview of the thesis . . . . . . . . . . . . . . . . . . . . . . . 6

2 Path Decompositions of Complete Tripartite Multigraphs 9

2.1 Introduction and preliminaries . . . . . . . . . . . . . . . . . . . 9

2.2 Path decompositions of Kn,n,n for odd n and for even . . . . 12

2.3 Directed path decompositions of Kn,n,n for odd n . . . . . . . . 19

3 Antidirected Cycle Decompositions of Complete Symmetric

Graphs 25

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.3 Antidirected cycle decompositions of Kn . . . . . . . . . . . . . 28

4 Antidirected Cycle Decompositions of (Kn I) 43

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.2 Antidirected cycle decompositions of (Kn I) . . . . . . . . . 44

5 Caterpillar Factorization of Crowns and Directed Caterpillar

Factorization of Symmetric Crowns 61

iii


5.2 Caterpillar factorization of crown . . . . . . . . . . . . . . . . . 64

5.3 Directed caterpillar factorization of symmetric crowns . . . . . . 66

6 The Star Arboricity of Crowns 71


6.2 A lower bound of sa(Cn,2k) . . . . . . . . . . . . . . . . . . . . . 74

6.3 An upper bound of sa(Cn,2k) . . . . . . . . . . . . . . . . . . . . 80

7 The Linear k-Arboricity of Complete Graphs 83


7.2 Lemmas and result . . . . . . . . . . . . . . . . . . . . . . . . . 84

8 Bibliography 95

iv

Chapter 1

Introduction

1.1 Introduction to decompositions and fac-

torizations of graphs

Graph decomposition is an important subject in graph theory. The first pa-

pers dealing directly with decompositions of graphs appeared at the turn of the

19th century. Since that time the interest in graph decomposition has been on

increase. Many combinatorial, algebraic, and other mathematical structures

are linked to decompositions of graphs, which give their study a great theo-

retical importance. On the other hand, results on graph decompositions can

be applied in coding theory, design of experiments, computer and communi-

cation network, and other fields. Nowadays, graph decomposition ranks the

most prominent area in graph theory, even in combinatorics. There are lots

of intriguing problems and significant results. Excellent surveys of them are

provided by Beineke [10], Bermond and Sotteau [12], Bosak [15], Chung and

Graham [21] and Ushio [83].

There are various decomposition problems, such as clique decomposition [33],

star decomposition [40, 55, 71, 74, 75, 76, 81, 82, 88], path decomposition [43,

1

73], cycle decomposition [17, 44, 72] and so on.

The star decomposition of complete graphs was solved by Yamamoto et

al. [88], and that of complete multigraphs was solved by Tarsi [74]. Tarsi [75]

also gave a sufficient condition for a complete graph to be decomposed into a

given sequence of stars (not necessarily isomorphic). As for the star decom-

position of complete multipartite graphs, the solutions for complete bipartite

graphs and complete bipartite multigraphs were given by Yamamoto et al. [88]

and Shyu [71] respectively, and that for complete m-partite graphs were given

by Ushio et al. [81] (for equal size partite sets) and Tazawa [76] (for different

size partite sets).

The path decomposition of a complete multigraph has been raised and

solved completely by Tarsi. [73]

The problem of cycle decomposition of complete bipartite graphs has been

raised and solved by Sotteau [72], and that of complete tripartite graph with

equal size was solved by Cavenagh [17]. Some results for complete graphs

and complete m-partite graphs were given by Jackson [44]. The problem of

cycle decomposition of complete graphs has been completely solved recently

by Sajna [67].

Graph factorization is a special type of graph decomposition, and contin-

ues to be a popular topic of research. There are many types of factoriza-

tion problems, such as star factorization [23, 42, 57, 87, 89], path factoriza-

tion [12, 38, 59, 90], directed star factorization [84, 85, 86] and so on.

The star factorization of complete graphs was solved by Yu [89]. The path

factorization of a complete multigraph has been solved completely by Hor-

ton [38] and Bermond et al. [12]. The problem of directed star factoriza-

tion of symmetric complete bipartite digraphs has been completely settled by

Ushio [84], and some results for symmetric complete tripartite digraphs were

2

given by Ushio [85, 86].

To find the arboricity of some graphs is also a popular topic in graph theory.

There are many types of arboricity problems, such as star arboricity, linear

arboricity, linear k-arboricity.

The notion of star arboricity was introduced in [2], where the star arboric-

ity of the complete graphs was solved. The concept of linear arboricity was

introduced Harary [34], but the first results about it were published ten years

later in [3] together with various result on some other covering and packing

invariants for graphs.

1.2 Preliminaries

This section describes some basic definitions and notations in graph theory.

A graphG is an ordered pair (V (G), E(G)), where V (G) is a finite nonempty

set and E(G) is a set of unordered pairs of distinct elements of V (G). The

elements in V (G) are called vertices of G and the elements in E(G) are edges.

We call V (G) the vertex set of the graph G and E(G) the edge set. For

convenience, we use uv to denote an edge {u, v} and u, v are called the ends

of the edge {u, v}. Two edges are adjacent if they have a common vertex. In

general, if one allows more than one edge(but a finite number) to join pairs of

vertices, the resulting graph is called a multigraph. For a graph G, we use G

to denote the multigraph obtained from G by replacing each edge e of G by

edges with the same ends as e.

The degree of a vertex v in a graph G, denoted by dG(v) or simply d(v), is

the number of edges of G incident with v. A graph is regular (d-regular) if all

its vertices have the same degree (d).

A directed graph or digraph D is an ordered pair (V (D), E(D)), where V (D)

is a finite nonempty set and E(D) is a set of ordered pairs of distinct elements

3

of V (D). The elements in V (D) are called vertices of D and the elements in

E(D) are arcs. We call V (D) the vertex set of the digraph D and E(D) the

arc set. If e = u v is an arc of a digraph D, then u is the tail of e, and v

is its head. A digraph is called is called symmetric if, whenever u v is an

arc of D then v u is as well. For a digraph D, we use D to denote the

multiple digraph obtained from D by replacing each arc e of D by arcs with

the same tail and head as e.

Given any graph G, we can obtain a digraph D from G by assigning a

direction to each edge of G. We call D an orientation of G. For a digraph

G, we use G to denote the symmetric digraph obtained from G by replacing

each edge uv by two arcs u v and v u.

An (v0, vr)-walk of length r in a graph G is a sequence of vertices of the

form v0, v1, v2, , vr where vivi+1 E(G) for i = 0, 1, 2, , r1; this walk is

denoted by v0v1v2 vr and v0 is called the starting vertex and vr the ending

vertex. A trail is a walk in which all edges are distinct. A path is a trail in

which all vertices are distinct. A cycle is a trail in which all vertices are distinct

except that the starting vertex is the same as the ending vertex. Denote by

Pn a path on n vertices and Cn a cycle on n vertices. A cycle of a graph G

containing every vertex of G is called a Hamiltonian cycle of G. An Euler trail

of a graph G is a trail containing all edges of G.

A graph in which every two distinct vertices are adjacent is a complete

graph. The complete graph on n vertices is denoted by Kn. A bipartite graph

is a graph whose vertex set can be partitioned into two sets A and B such that

each edge joins a vertex of A to a vertex of B. A complete bipartite graph is a

bipartite graph in which every vertex in the first set is adjacent to every vertex

in the second set. If these two sets contain m and n vertices, respectively, the

complete bipartite graph is denoted by Km,n.

4

An r-partite graph is a graph whose vertex set can be partitioned into r

subsets so that no edge has both ends in any one subset. A complete r-partite

graph is an r-partite graph in which each vertex is joined to every vertex that

is not in the same subset.

For positive integers k n, the crown Cn,k is the graph with vertex set

{a1, a2, , an, b1, b2, , bn} and edge set {aibj : 1 i n, j i + 1, i +

2, , i+ k (mod n)}.

A graph contains no cycles is called an acyclic graph or a forest. A tree is

a connected acyclic graph. A star Sl is the complete bipartite graph K1,l. For

l 2, the vertex of degree l in Sl is the center and any vertex of degree 1 is

an endvertex.

A decomposition of a graph G is a family {Hi : i L} of subgraphs of G

such that each edge of G is contained in exactly one member of {Hi : i L}.

Let H be a graph. An H decomposition of a graph G is a decomposition of

G into subgraphs isomorphic to H.

Suppose thatG andH are graphs. AnH-factor ofG is a spanning subgraph

of G such that each component of which is isomorphic to H. (A K2-factor is

also called an one-factor.) An H-factorization of G is a set of H-factors of

G which partitions the edges of G.

The arboricity of G, denoted by a(G), is the minimum number of forests

needed to decompose the edge set of G. A linear forest is a forest in which

each component is a path. The linear arboricity of G, denoted by la(G), is the

minimum number of linear forests needed to decompose the edge set of G. A

star forest is a forest in which each component is a star. The star arboricity

of G, denoted by sa(G), is the minimum number of star forests needed to

decompose the edge set of G.

5

1.3 Overview of the thesis

There are seven chapters in this thesis.

In Chapter 1, some basic definitions and notations are introduced.

In Chapter 2, we consider the isomorphic path decomposition of complete

tripartite multigraph, and the directed path decomposition of complete sym-

metric tripartite multigraph. We prove the following results.

Theorem 2.1.1. Let n be an odd integer. Then Kn,n,n has a Pk-decomposition

if and only if 2 k 3n and |E(Kn,n,n)| 0 (mod k 1).

Theorem 2.1.2. Let be an even integer. Suppose that k is an integer such

that 2 k 3n, k 6= 3n 1, and |E(Kn,n,n)| 0 (mod k 1). Then Kn,n,nhas a Pk-decomposition.

Theorem 2.1.3. Let n 3 be an odd integer. Suppose that k is an integer

such that 2 k 3n1 and |E(Kn,n,n)| 0 (mod k1). Then Kn,n,n has

aPk-decomposition.


of the isomorphic antidirected cycle decompositions of the complete symmetric

graphs. We prove the following result.

Theorem 3.1.1. Let m and n be positive integers with 4 m n and

m be even. Then the complete symmetric digraph Kn can be decomposed into

antidirected cycles of length m if and only if n is odd and m divides the number

of arcs in Kn.


of the isomorphic antidirected cycle decompositions of the complete symmetric

graphs minus a one-factor. We prove the following result.

Theorem 4.1.1. Let m and n be positive integers with 4 m n and m

be even. Let I be an one-factor of Kn. Then the digraph (Kn I) can be

6

decomposed into antidirected cycles of length m if and only if n is even and m

divides the number of arcs in (Kn I).

Suppose that G andH are graphs. An H-factor of G is a spanning subgraph

of G, each component of which is isomorphic to H. An H-factorization of G is

a set of H-factors of G which partition the edges of G. For digraphs G and H,

H-factor of G and H-factorization of G are similarly defined. A caterpillar is

a tree of order at least three which contains a path such that each vertex not

on the path is adjacent to a vertex on the path. A connected bipartite graph

with bipartition (C,D) is balanced if |C| = |D|.

In Chapter 5, we study the problem of caterpillar factorization of crowns and

directed caterpillar factorization of symmetric crowns. We have the following

results.

Theorem 5.2.3. Let G be a balanced caterpillar of order t. Then Cn,k has a

G-factorization if and only if n 0 (mod t2) and k 0 (mod t 1).

Theorem 5.3.2. Let G be a caterpillar of order t, and let G be a digraph

obtained from G by orienting each edge of G arbitrarily.

(1) Suppose that G is balanced. Then Cn,k has a G factorization

if and only if n k, n 0 (mod t2), k 0 (mod t 1).

(2) Suppose that G is nonbalanced. Then Cn,k has a G factorization

if and only if n 0 (mod t), k 0 (mod t 1).

In Chapter 6, we study the problem of the star arboricity of crowns. We

have the following results.

Theorem 6.1.5 sa(C9s,4) = 3 where s is a positive integer.

Theorem 6.1.6 sa(Cn,2k) k+2 for positive integer k where 4 2k n1,

except when k = 2 and n = 9s for some positive integer s.

7

Theorem 6.1.7 sa(Cn,2k) k + 2 if k 2 and n = ak + b for nonnegative

integers 2a b 0 or if n = ak 1 for a 3.

In Chapter 7, we are interested in finding the linear k-arboricity of the

complete graph. We have the following result.

Theorem 7.2.11. If dn/3e 1 k dn/2e 2, then lak(Kn) = dn(n1)2(n3) e =

dn/2e+ 2.

8

Chapter 2

Path Decompositions of

Complete Tripartite

Multigraphs

2.1 Introduction and preliminaries

If H1, H2, , Hr are edge-disjoint subgraphs of a multigraph G such that

every edge of G appears in some Hi, then we say that G is decomposed into

H1, H2, , Hr. Furthermore if each Hi (i = 1, 2, , r) is isomorphic to a

multigraph H, we say that G has an H-decomposition. The above terms

are similarly defined for multidigraphs. The H-decomposition problems of

a multigraph G are widely investigated when G is a complete graph or a

complete r-partite graph and H is a path or a cycle. This paper studies the

path decomposition of the complete tripartite graph.

For a multigraph G, we use the symbol G to denote the multidigraph

obtained from G by replacing each edge e by two opposite arcs connecting the

endvertices of e. Let be a positive integer. For a multigraph H, we use the

9

symbol H to denote the multigraph obtained from H by replacing each edge

e by edges each of which has the same endvertices as e. Similarly, for a

multidigraph H, we use the symbol H to denote the multidigraph obtained

from H by replacing each arc e by arcs each of which has the same tail and

head as e. For a multigraph G, it is easy to see that (G) = (G); we use

the symbol G to denote (G).

For a positive integer k, let Pk denote a path on k vertices and letPk denote

a directed path on k vertices.

LetKn denote the complete graph on n vertices. Tarsi [73] established a cri-

teria for Pk-decompositions of Kn. Recently Meszka and Skupien [58] solved

thePk-decomposition problem of K

n. A path in a digraph is antidirected

if the two adjacent arcs of the path have opposing orientations. The isomor-

phic antidirected path decomposition problem of Kn was studied by Shyu and

Lin [69].

Let Km1,m2,,mr denote the complete r-partite graph with parts of sizes m1,

m2, ,mr, respectively. In [78] Truszczynski solved thePk-decomposition

problem of Km,n, and considered the Pk-decomposition of Km,n. The Pk-

decomposition problem of Km,n was completely solved by Parker [61]. The

condition for P4-decomposition of Km1,m2,,mr was obtained by Kumar [48].

For a simple graph G, we use E(G) to denote the edge set of G. Kotzig [46]

proved that a connected graphG has a P3-decomposition if and only if |E(G)|

0 (mod 2), and proved that a 3-regular graph G has a P4-decomposition if and

only if G has a 1-factor. Heinrich, Liu, and Yu [35] proved that a connected

4-regular graph G has a P4-decomposition if and only if |E(G)| 0 (mod 3).

Jacobson, Truszczynski, and Tuza [45] proved that every 4-regular bipartite

graph has a P5-decomposition.

In this chapter, we consider the Pk-decomposition of Kn,n,n and the

10

Pk-decomposition of K

n,n,n. For a multigraph (multidigraph, respectively)

G, we also use E(G) to denote the edge set (arc set, respectively) of G. We

will obtain the following results.

Theorem 2.1.1 Let n be an odd integer. Then Kn,n,n has a Pk-decomposition

if and only if 2 k 3n and |E(Kn,n,n)| 0 (mod k 1).

Theorem 2.1.2 Let be an even integer. Suppose that k is an integer such

that 2 k 3n, k 6= 3n 1, and |E(Kn,n,n)| 0 (mod k 1). Then Kn,n,nhas a Pk-decomposition.

Theorem 2.1.3 Let n 3 be an odd integer. Suppose that k is an integer

such that 2 k 3n1 and |E(Kn,n,n)| 0 (mod k1). Then Kn,n,n has

aPk-decomposition.

For our discussions we need the following notations and terms. Let G

be a multigraph. Suppose that W1 is a walk v0v1 vk and W2 is a walk

vkvk+1 vl in G. Then the sum of W1 and W2, denoted by W1 +W2, is a

walk v0v1 vkvk+1 vl. If W is a walk in G, then the girth of W , denoted

by g(W ), is the least number of edges between two appearances of the same

vertex along W . A trail is a walk with no repeated edges. An Euler trail

of G is a trail in G which traverses every edge of G. For multidigraphs, the

following terms are similarly defined: the sum of directed walks, the girth of a

directed walk, the directed trail, and the directed Euler trail.

In [73] Tarsi obtained the path decomposition of Kn by cutting Euler

trails. We state the result of cutting method in the following remark. This

remark and its directed version were henceforth used in many papers, e.g. [35,

58, 61, 69, 78].

11

Remark 2.1.4 Suppose that a multigraph G contains an Euler trail with girth

g, and l1, l2, , lr are positive integers g1 such that |E(G)| = l1+l2+ +

lr. Then G can be decomposed into paths of lengths l1, l2, , lr, respectively.

Letting l1 = l2 = = lr in the above lemma, we have the following.

Remark 2.1.5 Suppose that a multigraph G contains an Euler trail with girth

g, and k is a positive integer such that 2 k g and |E(G)| 0 (mod k1).

Then G has a Pk-decomposition.

The directed version of the above remark is the following:

Remark 2.1.6 Suppose that a multidigraph G contains a directed Euler trail

with girth g, and k is a positive integer g and |E(G)| 0 (mod k1). Then

G has aPk-decomposition.

2.2 Path decompositions of Kn,n,n for odd n

and for even

In this section, we investigate the problem of decomposing the complete tri-

partite multigraph Kn,n,n into the path Pk for odd n and also for even .

We begin with odd n. For a multigraph G, and subsets A,B of V (G) with

A B = , we use G(A,B) to denote the set of all edges joining vertex in A

and vertex in B. We begin with some lemmas.

Lemma 2.2.1 Let n 3 be an odd integer. Then

(1) Kn,n,n has an Euler trail with girth 3n 6,

(2) Kn,n,n has an Euler trail with girth 3n 3 if 2.

12

Proof. For = 1, 2, 3, , let (A,B,C) be the tripartition of Kn,n,n where

A = {a0, a1, , an1}, B = {b0, b1, , bn1} and C = {c0, c1, , cn1}.

An edge joining ai and bi+k (i = 0, 1, , n 1; k = 0, 1, , n 1) where

the indices are taken modulo n is said to be an edge between A and B with

label k. Equivalently, an edge joining ai and bj is said to be an edge between A

and B with label (j i)(mod n) . Similarly an edge joining bi and ci+k is said

to be an edge between B and C with label k, and an edge joining ci and ai+k is

said to be an edge between C and A with label k.

(1) Let = 1. For each i = 0, 1, 2, , n 1, let Di be the following walk in

Kn,n,n:

a0bic2ia1bi+1c2i+1a2bi+2c2i+2 an1bi+n1c2i+n1a0

where the indices are taken modulo n. Note that each Di is a Hamiltonian

cycle of Kn,n,n, and consists of all edges between A and B with label i, all

edges between B and C with label i, and all edges between C and A with

label (1 2i)(mod n). Thus Kn,n,n(A,B) is a disjoint union of D0(A,B),

D1(A,B), , Dn1(A,B), and Kn,n,n(B,C) is a disjoint union of D0(B,C),

D1(B,C), , Dn1(B,C). Also since n is odd, we have {(1 2i)(mod n) :

i = 0, 1, 2, , n1} = {0, 1, 2, , n1}; thus Kn,n,n(C,A) is a disjoint union

of D0(C,A), D1(C,A), , Dn1(C,A). Hence E(Kn,n,n) is a disjoint union

of E(D0), E(D1), , E(Dn1). Let T be the walk D0 +D1 + +Dn1. We

thus see that T is an Euler trail in Kn,n,n.

Now we evaluate g(T ). For i = 0, 1, , n 2, Di +Di+1 is the trail a0bic2ia1bi+1c2i+1 a2bi+2c2i+2 an1bi+n1 c2i+n1 a0bi+1c2i+2 a1bi+2c2i+3 an1bic2i+1 a0. We see that in Di+Di+1 there are 3n edges between two appearances

of aj (j = 0, 1, , n 1), 6n 3 edges between two appearances of bi, 3n 3

edges between two appearances of bj (j = 0, 1, 2, , i1, i+1, i+2, , n1),

6n 6 edges between two appearances of cj (j = 2i, 2i+ 1) and 3n 6 edges

13

between two appearances of cj (j = 0, 1, 2, , 2i 1, 2i+2, 2i+3, , n 1).

Thus g(Di +Di+1) = 3n 6, which implies g(T ) = 3n 6, and hence T is a

required Euler trail of Kn,n,n.

(2) Let 2. For i = 0, 1, 2, , n 1, let Di be, as in the proof of (1), the

following trail:

a0bic2ia1bi+1c2i+1a2bi+2c2i+2 an1bi+n1c2i+n1a0,

and let Ei be the following trail:

a0bi+1c2i+1a1bi+2c2i+2a2bi+3c2i+3 an1bi+nc2i+na0

where the indices are taken modulo n. Let G = Kn,n,n be a subgraph of

Kn,n,n. As in (1), E(G) is a disjoint union of E(D0), E(D1), , E(Dn1).

Note also that each Ei is a Hamiltonian cycle of Kn,n,n, and consists of all

edges between A and B with label (i + 1)(mod n), all edges between B and

C with label i, and all edges between C and A with label (2i)(mod n). By

arguments as in (1), E(G) is a disjoint union of E(E0), E(E1), , E(En1).

Let T be the following trail:

D0 +D0 + +D0 1 copies of D0

+E0 +

D1 +D1 + +D1 1 copies of D1

+E1 +

+

+

Dn2 +Dn2 + +Dn2 1 copies of Dn2

+En2 +

Dn1 +Dn1 + +Dn1 1 copies of Dn1

+En1.

Then T is an Euler trail of Kn,n,n. To determine g(T ), we show in the

following that (i) g(Di+Di) = 3n for i = 0, 1, , n1, (ii) g(Di+Ei) = 3n3

14

for i = 0, 1, , n 1, and (iii) g(Ei +Di+1) = 3n 3 for i = 0, 1, , n 2.

Note that both Di and Ei are Hamiltonian cycles in Kn,n,n.

(i) For i = 0, 1, , n 1, there are 3n edges between two appearances of the

same vertex in Di +Di. Thus g(Di +Di) = 3n.

(ii) For i = 0, 1, , n 1, we see that Di + Ei is the trail a0bic2i a1bi+1c2i+1a2bi+2c2i+2 an1 bi+n1c2i+n1 a0bi+1c2i+1 a1bi+2c2i+2 a2bi+3c2i+3 an1bic2ia0. In Di + Ei, there are 3n edges between two appearances of aj (j =

0, 1, , n1), 6n3 edges between two appearances of bi, 3n3 edges between

two appearances of bj (j = 0, 1, 2, , i1, i+1, i+2, , n1), 6n3 edges

between two appearances of c2i and 3n 3 edges between two appearances of

cj (j = 0, 1, 2, , 2i 1, 2i+ 1, 2i+ 2, , n 1). Thus g(Di +Ei) = 3n 3.

(iii) For i = 0, 1, , n 2, Ei + Di+1 is the trail a0bi+1c2i+1 a1bi+2c2i+2a2bi+3c2i+3 an1bic2i a0bi+1c2i+2 a1bi+2c2i+3 a2bi+3c2i+4 an1 bi c2i+1 a0.

There are 3n edges between two appearances of aj (j = 0, 1, , n 1), 3n

edges between two appearances of bj (j = 0, 1, , n 1), 6n 3 edges be-

tween two appearances of c2i+1 and 3n 3 edges between two appearances of

cj (j = 0, 1, 2, , 2i, 2i+ 2, 2i+ 3, , n 1). Thus g(Ei +Di+1) = 3n 3.

From (i), (ii) and (iii), we obtain g(T ) = 3n3. Thus T is a required Euler

trail of Kn,n,n.

Lemma 2.2.2 Let G be a graph of order t such that G can be decomposed into

Hamiltonian cycles. Suppose that and k are positive integers with 2 k t

and k 1|t. Then G has a Pk-decomposition.

Proof. Suppose that G is decomposed into Hamiltonian cycles H1, H2,

, Hv. Then G is decomposed into H1, H2, , Hv. Since k t, k1|t,

and Hi (1 i v) has an Euler trail with girth t, each Hi has a Pk-

decomposition. Thus G has a Pk-decomposition.

15

In the proof of (1) in Lemma 2.2.1, we see that if n 3 is an odd inte-

ger, then Kn,n,n can be decomposed into Hamiltonian cycles D0, D1, , Dn1.

More generally, Lasker and Auerbach [49] proved that the complete m-partite

graph Km(n) can be decomposed into Hamiltonian cycles if and only if (m1)n

is even. Thus Kn,n,n can be decomposed into Hamiltonian cycles for any posi-

tive integer n. We are ready to prove the main result of this section.

Theorem 2.1.1

Let n be an odd integer. Then Kn,n,n has a Pk-decomposition if and only if

2 k 3n and |E(Kn,n,n)| 0 (mod k 1).

Proof. The necessity is trivial. Now we prove the sufficiency.

The case n = 1 is trivial. We assume that n 3. By the assumption, k

is a positive integer with 2 k 3n and |E(Kn,n,n)| 0 (mod k 1) (i.e.

(k 1)|3n2). We distinguish two cases for = 1 and 2.

Case 1. = 1.

By Lemma 2.2.1(1), Kn,n,n has an Euler trail with girth 3n 6. Hence by

the cutting method, Kn,n,n has a Pk-decomposition if k 3n 6. So we only

need to consider 3n 5 k 3n. Since n is odd and (k 1)|3n2, we have

that k is even. So it remains to consider the following subcases: k = 3n 5,

3n 3, 3n 1.

Subcase 1.1. k = 3n 5.

From the assumption that (3n 6)|3n2, we have (n 2)|n2, which implies

(n2)|4 for 4 = n2(n+2)(n2). This implies n2 = 1 since n is odd. Thus

n = 3 and k = 4. As mentioned in the paragraph preceding the theorem, K3,3,3

can be decomposed into Hamiltonian cycles. Then by Lemma 2.2.2, K3,3,3 has

P4-decomposition. This completes Subcase 1.1.


16

From the assumption that (3n 4)|3n2, we have (3n 4)|16 for 16 =

3 3n2 (3n+ 4)(3n 4). This is impossible since n is odd.


From the assumption that (3n2)|3n2, we have (3n2)|4 since 4 = 33n2

(3n+ 2)(3n 2). Thus n = 1 since n is odd. This is a contradiction since we

assumed that n 3.

Case 2. 2.

By Lemma 2.2.1(2), Kn,n,n has an Euler trail with girth 3n 3. Hence

Kn,n,n has a Pk-decomposition if k 3n 3. So we only need to consider

k = 3n 2, 3n 1, 3n. We will show that (k 1)|3 for these k.


From the assumption that (3n3)|3n2, we have (n1)|n2, which implies

(n 1)| since gcd(n 1, n) = 1. Thus 3(n 1)|3 (i.e. (k 1)|3).


Since n is odd, it is easy to see that gcd(3n2, n) = 1, and hence gcd(3n

2, n2) = 1. Combining this with the assumption (3n 2)|3n2, we obtain

(3n 2)|3 (i.e. (k 1)|3).

Subcase 2.3. k = 3n.

It is trivial that gcd(3n 1, n) = 1, and hence gcd(3n 1, n2) = 1. Com-

bining this with the assumption (3n 1)|3n2, we obtain (3n 1)|3 (i.e.

(k 1)|3).

Now we have that Kn,n,n has order 3n and can be decomposed into Hamil-

tonian cycles, and that k 3n, (k 1)|3n. Thus by Lemma 2.2.2, Kn,n,nhas a Pk-decomposition. This completes Case 2.

Now we consider to decompose Kn,n,n into Pk for even .

Lemma 2.2.3 Let be an even integer and n 2 be an integer. Then Kn,n,nhas an Euler trail with girth 3n 3.

17

Proof. Let (A,B,C) be the tripartition of Kn,n,n where A = {a0, a1,

, an1}, B = {b0, b1, , bn1} and C = {c0, c1, , cn1}. For i = 0, 1, 2, , n

1, let Di and Ei be the walks we construct in Lemma 2.1. Let F be the walk:

D0 +D0 + +D0 2copies of D0

+E0 + E0 + + E0 2copies of E0

+

D1 +D1 + +D1 2copies of D1

+E1 + E1 + + E1 2copies of E1

+

+

+

Dn1 +Dn1 + +Dn1 2copies of Dn1

+En1 + En1 + + En1 2copies of En1

.

Then F is an Euler trail of Kn,n,n.

For i = 0, 1, , n 1, we see that Di + Ei is the trail a0bic2i a1b1+ic1+2ia2b2+ic2+2i an1 bn1+icn1+2i a0b1+ic1+2i a1b2+ic2+2i a2b3+ic3+2i an1 bic2i a0. In Di + Ei, there are 3n edges between two appearances of aj (j =

0, 1, , n 1), 3n 3 edges between two appearances of bj (j = 0, 1, , i

1, i+ 1, , n 1), 6n 3 edges between two appearances of bi, 3n 3 edges

between two appearances of cj (j = 0, 1, , 2i 1, 2i + 1, , n 1) and

6n 3 edges between two appearances of c2i. And for i = 0, 1, , n 1,

Ei+Di+1 is the trail a0b1+ic1+2i a1b2+ic2+2i a2b3+ic3+2i an1bic2i a0b1+ic2+2ia1 b2+i c3+2i a2b3+i c4+2i an1 bi c1+2ia0. There are 3n edges between two

appearances of aj (j = 0, 1, , n 1), 3n edges between two appearances

of bj (j = 0, 1, , n 1), 3n 3 edges between two appearances of cj (j =

0, 1, , 2i, 2i + 2, , n 1) and 6n 3 edges between two appearances of

c2i+1. Thus g(F ) = 3n 3.

The proof of the following result is the same as that of Case 2 in Theo-

rem 2.1.1 except that Subcase 2.2 is not needed here.

Theorem 2.1.2

18

Let be an even integer. Suppose that k is an integer such that 2 k 3n,

k 6= 3n 1, and |E(Kn,n,n)| 0 (mod k 1). Then Kn,n,n has a Pk-

decomposition.

2.3 Directed path decompositions of Kn,n,n for

odd n

In this section, we investigate the problem of decomposing the complete tri-

partite multidigraph Kn,n,n into the directed pathPk for odd n. Let us begin

with n = 1.

The result for the decomposition of Kn into directed Hamiltonian paths

is the following [15, 58]: Kn can be decomposed into directed Hamiltonian

paths if and only if neither n = 3 and is odd nor n = 5 and = 1. It follows

from the case n = 3 that K1,1,1 has aP3-decomposition if and only if is

even.

Thus we can see that K1,1,1 has aPk-decomposition if and only if either

is odd and k = 2 or is even and k = 2, 3.

For a multigraph G, the trivial relationship between Pk-decomposition of

G andPk-decomposition of G

is the following.

Remark 2.3.1 For a multigraph G, we have the following:

(1) If G has a Pk-decomposition, then G has a

Pk-decomposition.

(2) If G has aPk-decomposition, then 2G has a Pk-decomposition. 2

Suppose that G is a multidigraph. For A,B V (G) with A B = , let

G(A,B) denote the set of all arcs joining vertex in A to vertex in B.

Lemma 2.3.2 Let n 3 be an odd integer. Then Kn,n,n has a directed Euler

trail with girth 3n 4.

19

Proof. Let (A,B,C) be the tripartition of Kn,n,n where A = {a0, a1,

, an1}, B = {b0, b1, , bn1} and C = {c0, c1, , cn1}.

An arc joining ai to bi+k (i = 0, 1, , n 1, k = 0, 1, , n 1) where the

indices are taken modulo n is said to be an arc from A to B with label k. The

following terms are similarly defined: arc from B to A with label k, arc from

B to C with label k, arc from C to B with label k, arc from A to C with label

k, and arc from C to A with label k.

For i = 0, 1, 2, , n 1, let D i be the directed trail:

a0 bi c2i a1 bi+1 c2i+1 a2 bi+2 c2i+2

an1 bi+n1 c2i+n1 a0,

and letF i be the directed trail:

a0 c2i+1 bi+1 a1 c2i+2 bi+2 a2 c2i+3 bi+3

an1 c2i+n bi+n a0

where the indices are taken modulo n.

Let G = Kn,n,n be a subgraph of Kn,n,n. Note that each

Di consists of the

following arcs: all arcs in G from A to B with label i, all arcs in G from B

to C with label i, and all arcs in G from C to A with label (1 2i)(mod n).

Thus G(A,B) is a disjoint union ofD0(A,B),

D1(A,B), ,

Dn1(A,B), and

G(B,C) is a disjoint union ofD0(B,C),

D1(B,C), ,

Dn1(B,C). And since

n is odd, we have {(12i)(mod n) : i = 0, 1, 2, , n1} = {0, 1, 2, , n1};

thus G(C,A) is a disjoint union ofD0(C,A),

D1(C,A), ,

Dn1(C,A). Note

also that eachFi consists of the following arcs: all arcs in G from A to C with

label (2i+1)(mod n), all arcs in G from C to B with label (i)(mod n) and all

arcs in G from B to A with label (i)(mod n). It is obvious that {(i)(mod

n) : i = 0, 1, 2, , n 1} = {0, 1, 2, , n 1}. And since n is odd, we have

{(2i + 1)(mod n) : i = 0, 1, 2, , n 1} = {0, 1, 2, , n 1}. Thus G(C,B)

20

is a disjoint union ofF0(C,B),

F1(C,B), ,

F n1(C,B), and G(B,A) is

a disjoint union ofF0(B,A),

F1(B,A), ,

F n1(B,A), and G(A,C) is a

disjoint union ofF0(A,C),

F1(A,C), ,

F n1(A,C). Since E(G) is a disjoint

union of G(A,B), G(B,C), G(C,A), G(A,C), G(C,B) and G(B,A), from

above we have that E(G) is a disjoint union of E(D0), E(

D1), , E(

Dn1),

E(F0), E(

F1), , E(

F n1).

LetT be the directed trail:

D0 +

D0 + +

D0

copies ofD0

+F0 +

F0 + +

F0

copies ofF0

+

D1 +

D1 + +

D1

copies ofD1

+F1 +

F1 + +

F1

copies ofF1

+

+

+Dn1 +

Dn1 + +

Dn1

copies ofDn1

+F n1 +

F n1 + +

F n1

copies ofF n1

.

We see thatT is a directed Euler trail of Kn,n,n.

To evaluate g(T ), we first show that (i) g(

D i+

D i) = 3n, g(

F i+

F i) = 3n

for i = 0, 1, , n 1 (ii) g(D i +F i) = 3n 4 for i = 0, 1, , n 1 and (iii)

g(F i +

D i+1) = 3n 2 for i = 0, 1, , n 2. Note that each

Di is a directed

Hamiltonian cycle of Kn,n,n, and so is eachFi .

(i) This is trivial.

(ii) For i = 0, 1, , n 1, we see that D i+F i is the directed trail a0 bi

c2i a1 bi+1 c2i+1 an1 bi+n1 c2i+n1 a0 c2i+1

bi+1 a1 c2i+2 bi+2 an1 c2i bi a0. InD i +

F i, there

are 3n arcs between two appearances of aj (j = 0, 1, , n 1), 6n 2 arcs

between two appearances of bi, 3n2 arcs between two appearances of bj (j =

0, 1, , i 1, i+1, , n 1), 6n 4 arcs between two appearances of c2i and

21

3n4 arcs between two appearances of cj (j = 0, 1, , 2i1, 2i+1, , n1)

. Thus g(D i +

F i) = 3n 4.

(iii) For i = 0, 1, , n 2, F i +D i+1 is the directed trail a0 c2i+1

bi+1 a1 c2i+2 bi+2 an1 c2i bi a0 bi+1

c2i+2 a1 bi+2 c2i+3 an1 bi c2i+1 a0. InF i +

D i+1,

there are 3n arcs between two appearances of aj (j = 0, 1, , n 1), 3n 1

arcs between two appearances of bj (j = 0, 1, , n 1), 6n 2 arcs between

two appearances of c2i+1 and 3n 2 arcs between two appearances of cj (j =

0, 1, , 2i, 2i+ 2, , n 1). Thus g(F i +D i+1) = 3n 2.

From (i), (ii) and (iii), we obtain g(T ) = 3n 4.

Now we prove the main result of this section.

Theorem 2.1.3

Let n 3 be an odd integer. Suppose that k is a positive integer such that

2 k 3n 1 and |E(Kn,n,n)| 0 (mod k 1). Then Kn,n,n has aPk-decomposition.

Proof. Since |E(Kn,n,n)| 0 (mod k 1) (i.e. (k 1)|6n2), by Lem-

mas 2.3.2, Kn,n,n has aPk-decomposition if 2 k 3n 4. So we only need

to consider 3n 3 k 3n 1. We distinguish two cases: Case 1. k = 3n 3

or k = 3n 1, Case 2. k = 3n 2.

Case 1. k = 3n 3 or k = 3n 1.

Then k1 is odd. Combining this with (k1)|6n2, we obtain (k1)|3n2.

By Theorem 2.1.1 and Remark 2.3.1, Kn,n,n has aPk-decomposition.

Case 2. k = 3n 2.

From the assumption that (3n3)|6n2, we have (n1)|2n2, which implies

(n 1)|2 since gcd(n, n 1) = 1. Hence (k 1)|6.

For i = 0, 1, , n1, letD i,F i be the directed trails defined in Lemma 2.3.2,

22

and letW i be the directed trail:

Fi +

Fi + +

Fi

copies ofFi

+D i+1 +

D i+1 + +

D i+1

copies ofD i+1

where the indices are taken modulo n. In the proof of Lemma 2.3.2, we see

that a subgraph Kn,n,n of Kn,n,n can be decomposed into

D0,D1, ,

Dn1,

F0,F1, ,

F n1. Thus K

n,n,n can be decomposed into

W 0,

W 1, ,

W n1.

Also from (iii) in the proof of Lemma 2.3.2, we see that g(F i +

D i+1) =

3n 2 for i = 0, 1, , n 2; in fact the equality also holds for i = n 1, i.e.,

g(F n1 +

D 0) = 3n 2. Thus g(

W i) = 3n 2 for i = 0, 1, , n 1. Now we

have k 1 g(W i) for each i, (k 1)|6 and the length ofW i is 6n. Thus

we can cut eachW i from the starting vertex into

6nk1 directed paths of length

k 1. Hence Kn,n,n is decomposed into directed paths of length k 1.

The trivial necessity for thePk-decomposition of K

n,n,n is that 2 k 3n

and |E(Kn,n,n)| 0 (mod k 1). Comparing with Theorem 2.1.3, we pose

the following problem.

Problem. Suppose that n 3 is odd. Do the conditions that k = 3n and

|E(Kn,n,n)| 0 (mod k 1) (i.e. (3n 1)|6n2) imply the existence of thePk-decomposition of K

n,n,n (i.e. the existence of the directed Hamiltonian

path decomposition of Kn,n,n)?

From Theorem 2.1.1 and Remark 2.3.1, we see that if n is odd and (3n

1)|3n2, then Kn,n,n has aP 3n-decomposition. Therefore the answer to the

above problem is in the affirmative if (3n 1)|3n2.

23

Chapter 3

Antidirected Cycle

Decompositions of Complete

Symmetric Graphs

3.1 Introduction

Recall that Kn denote the complete graph of order n, Kn denote the complete

symmetric digraph of order n, which is obtained from Kn by replacing each

edge e by two opposite arcs with the same ends as e, and Cm denote the m-

cycle. If the edges of a graph G can be decomposed into subgraphs isomorphic

to a graph H, then we say that G has an H-decomposition.

Among the earliest results concerning graph decomposition is the problem

of partitioning the edges of Kn into cycles. The trivial necessary conditions for

this decomposition are m n, n is odd, and that m must divide n(n1)/2. In

[6, 67], it is shown that these necessary conditions are also sufficient. As to the

Cm-decomposition of Kn I, where n is even, and I is a one-factor of Kn, the

necessary conditions arem n, andm divides n(n2)/2. These conditions are

25

also known to be sufficient [6, 67]. In [6], the authors establish necessary and

sufficient conditions for decomposing the complete graph of even order minus a

1-factor into even cycles and the complete graph of odd order into odd cycles.

Then in [67], the author settles the remaining cases: the complete graph of even

order minus a 1-factor into odd cycles and the complete graph of odd order

into even cycles. This question can be extended to isomorphic directed cycles

decomposition of the complete symmetric digraph Kn. The problem has been

solved completely by Brian Alspach et al. [7]; they showed that the complete

symmetric graph Kn can be decomposed into directed cycles of length m if and

only if m n and m divides n(n 1) except (n,m) {(4, 4), (6, 3), (6, 6)}.

A cycle in a digraph is antidirected if the two adjacent arcs of the cycle have

opposite orientations. It is easy to see that the length of an antidirected cycle is

even. In a digraph G, an antidirected Hamiltonian cycle is an antidirected cycle

that passes every vertex of G. In [62, 65, 77], the existences of antidirected

Hamiltonian cycles in tournaments were investigated. A path in a digraph

is antidirected if two adjacent arcs of the path have opposite orientations.

The antidirected path decomposition of directed crowns has been investigated

in [69].

A natural question is to ask when Kn admits a decomposition into antidi-

rected cycles of length m. Let x V (Kn), each antidirected cycle passing

through x contributes either zero or two of the outdegree of x in Kn. Thus

outdegree of x in Kn is even; and hence n is odd. Hence the trivial necessary

conditions are that 4 m n 1, m is even, n is odd and that m|n(n 1).

The goal of this chapter is to show that the trivial necessary conditions are

also sufficient, that is, we prove the following theorem:

Theorem 3.1.1 For an even integer m 4 and an odd integer n with m < n,

the complete symmetric digraph Kn can be decomposed into antidirected cycles

26

of length m if and only if m divides the number of edges in Kn.

3.2 Definitions

Let us begin this section with a few definitions.

For a simple graph G, we use G to denote the digraph obtained from G by

replacing each edge e of G by two opposite arcs with the same ends as e. For

vertices u and v in a digraph D, we will use the notation u v to denote the

arc from u to v. The antidirected cycle on m vertices is denoted byCm. For a

graph G, we write G = H1H2 if G is the edge-disjoint union of the subgraphs

H1 and H2. If G = H1H2 Hk where H1 = H2 = = Hk = H, then

the graph G can be decomposed into subgraphs isomorphic to H and we say

that G has an H-decomposition. We shall also write H|G.

Similarly, a digraph D can be decomposed into copies of a digraph H if the

arc set of D can be partitioned into sets, each inducing a copy of H. We will

also say that D has an H-decomposition and write H|D.

Let n be a positive integer and S {1, 2, . . . , n 1}. The circulant digraphX (n;S) with set S is the digraph whose vertices are u0, u1, . . . , un1 with an

arc from ui to uj if and only if j i S. We will often write s for n s

when n is understood.

Many of our decompositions arise from the action of a rotation on a fixed

subdigraph. The next definition makes this precise.

For an integer l and an antidirected walk W : ui1 ui2 ui3 ui4

uik of a digraph with vertex set {u0, u1, , un1}. We use W + l to

denote the antidirected walk ui1+l ui2+l ui3+l ui4+l uik+l,

where the subscripts are taken modulo n.

For a subset A of Zn, the notation A will denote the set {a Zn|a A}.

27

3.3 Antidirected cycle decompositions of Kn

The following lemmas provide a useful tool for antidirected cycle decomposi-

tions.

Lemma 3.3.1 Let m be an even integer. If G has a Cm-decomposition, then

G has aCm-decomposition.

Proof. Since G has a Cm-decomposition, G has a Cm-decomposition. For

even m, Cm can be decomposed into two antidirected cycles of length m. Thus

G has aCm-decomposition .

Lemma 3.3.2 Suppose that m 4 is an even integer. Then Km+1 has aCm-decomposition.

Proof. Let the vertices of Km+1 be u0, u1, . . . , um, and the label of the arc

from ui to uj is j i.

Case 1. m = 4t, t a positive integer.

We construct an antidirected cycle C (of length 4t) consisting of arcs with

labels in the following order: 4t, 4t 1, 4t 2, 4t 3, . . . , 2t + 3, 2t + 2, 2t +

1, 2t 1, 2t 2, 2t 3, 2t 4, . . . , 2, 1 and 2t as follows. Let C be the following

antidirected cycle:

u0 u4t u1 u4t1 u2 u3t+2 ut1 u3t+1 ut

u3t1 ut+1 u3t2 ut+2 u2t+1 u2t1 u2t u0.

It follows that the set {C + i|i = 0, 1, ,m} is a partition of the arc set of

K4t+1 into antidirected cycles of length 4t.

Case 2. m = 4t+ 2, t a positive integer.

We construct an antidirected cycle C (of length 4t + 2) consisting of arcs

with labels in the following order: 2t+1, 2t+2(= 2t1 (mod 4t+3)), 2t+3(=

28

2t (mod 4t+ 3)), 2, 3, 4, 5, . . . , 2t 1, 2t, 4t+ 2(= 1 (mod 4t+ 3)), 1, 2t+

4, 2t+ 5, 2t+ 6, 2t+ 7, . . . , 4t 1, 4t, 4t+ 1 as follows. Let C be the following

antidirected cycle:

u0 u2t+1 u4t+2 u2t+2 u2t u2t+3 u2t1

u2t+4 ut+2 u3t+1 ut+1 ut ut1 u3t+3

ut2 u3t+4 ut3 u4t u1 u4t+1 u0.

It follows that the set {C + i|i = 0, 1, ,m} is a partition of the arc set of

K4t+3 into antidirected cycles of length 4t+ 2. This completes the proof.

D. Sotteau [72] showed that the complete bipartite graph Km,n can be

decomposed into cycles of length 2k if and only if m and n are even integers

not less than k and 2k divides mn.

The following Lemma is the same as Lemma 3.2 in [7], exceptCm is used

instead ofCm.

Lemma 3.3.3 Let m 4 be an even integer. If Kn has aCm-decomposition

for odd n satisfying m < n < 2m with m|n(n 1), then Kn has aCm-

decomposition for all odd n > m satisfying m|n(n 1).

Proof. Suppose that Kn can be decomposed into antidirectedm-cycles when-

everm is even, n is odd, m|n(n1), andm < n < 2m. Letm and n be positive

integers with m even and n odd such that 4 m < n and m|n(n 1). Write

n = qm + r + 1 for integers q 2 and even r with 0 r < m 1. Observe

that m|n(n 1) implies m|r(r+1) as well. Label one vertex of Kn with x and

partition the remaining vertices of Kn into q 1 sets of m vertices and one set

of m+ r vertices. Each set of m vertices together with vertex x induces a sub-

digraph isomorphic to Km+1, and the arcs between any two of these sets of m

vertices induce a subdigraph isomorphic to Km,m. The remaining set of m+ r

29

vertices together with vertex x induce a subdigraph isomorphic to Km+r+1,

and the arcs between the set of m + r vertices and any one of the sets of m

vertices induce a subdigraph isomorphic to Km,m+r. By the above mentioned

result of Sotteau and Lemma 3.3.1, we haveCm |Km,m and

Cm |Km,m+r. Since

m|(m+ r + 1)(m+ r), we haveCm |Km+r+1 by hypothesis. By Lemma 3.3.2,

Cm |Km+1 and thus

Cm |Kn.

Observe that if n(n1) is an even multiple of m, then m divides n(n1)/2.

Hence a decomposition of the complete graph Kn into cycles of length m exists

by [67]. Lemma 3.3.1 then gives a decomposition of Kn into antidirected m-

cycles. Therefore, we may assume that n(n 1) is an odd multiple of m. This

leads to the following lemma which has been proved in [7].

Lemma 3.3.4 If m, n and r are positive integers such that n = m + r + 1

with m even, n odd, and n(n 1) m (mod 2m), then r 0 (mod 4).

In the proof of the following Lemma, we employ the decomposition tech-

niques used in the proof of Lemma 3.4 in [7].

Lemma 3.3.5 Let m and n be positive integers satisfying m 0 (mod 4).

If A = {a1, a2, . . . , am/2}, where a1, a2, . . . , am/2 are positive integers satisfying

a1 < a2 < < am/2 < n/2, thenCm |

X (n;A).

Proof. Label the vertices of the circulant digraphX (n;A) with u0, u1,

. . . , un1. We have ui uj E(X (n;A)) if and only if j i A.

Suppose m = 2k with k even, we let C be the antidirected cycle of length

m starting at u0, where the labels of the m arcs of C are:

a1, a2, a3, a4, . . . , ak1, ak,ak1,ak2, . . . ,a1,ak,

30

that is, C is the cycle:

C : u0 ua1 ua1a2 ua1a2+a3 ua1a2+a3a4

ua1a2++ak1 ua1a2++ak1ak ua1a2+ak2ak

ua1a2++ak3ak ua1a2ak ua1ak uak u0.

Since the subscripts of u are taken modulo n, C can be written as

C : un ua1 un+a1a2 ua1a2+a3 un+a1a2+a3a4

ua1a2++ak1 un+a1a2++ak1ak ua1a2+ak2ak

un+a1a2++ak3ak un+a1a2ak un+a1ak

unak un.

Since a1 < a2 < < am/2 < n/2, it is not difficult to see that the alternate

vertices starting with un on C will have strictly decreasing subscripts, and the

subscripts are between n and n + a1 ak, while alternate vertices starting

with ua1 on C will have strictly increasing subscripts, and the subscripts are

between a1 and nak. Since nak < n+a1ak, the vertices in C are distinct.

Hence, it follows that the set {C + i|i = 0, 1, , n 1} is a decomposition ofX (n;A) into antidirected m-cycles.

Lemma 3.3.6 Let m be positive integers satisfying m 2 (mod 4) and n

be an odd integer. If A = {1, a2, a3, . . . , am22, (n 3)/2, (n 1)/2}, where

a2, a3, . . . , am22 are positive integers satisfying 1 < a2 < a3 < < am

22 1 a2 + a3 a4 + + ak2 >

a2 + a3 a4 + + ak2, and n12 >n+1

2 a2, the vertices in C are distinct.

Hence, it follows that the set {C + i|i = 0, 1, , n 1} is a decomposition ofX (n;A) into antidirected m-cycles.

32

In the proof of the following Lemma, we employ the decomposition tech-

niques used in Case 1 in the proof of Theorem 3.1 in [7].

Lemma 3.3.7 Let m be a positive integer satisfying m 0 (mod 4), and

n be an odd integer, 4 m < n < 2m and n(n 1) m (mod 2m).

Then there exists a subset B {1, 2, , n12} with |B| = n1m

2such that

Cm |

X (n;B).

Proof. Suppose n = m + r + 1 for some integers r with 0 < r < m 1. By

Lemma 3.3.4, we know that r 0 (mod 4). Let r = 2ea where a is odd and

e 2. Thus r(r + 1) = 2ea(2ea+ 1). Since n(n 1) is an odd multiple of m,

and m is even, we see that r(r+1) is an even multiple of m. Then m = 2dab,

where a|a, b|(2ea+ 1), and 1 d < e. Thus

n = m+ r + 1 = 2dab + 2ea+ 1 = b(2da +2ea+ 1

b).

Since m 0 (mod 4), d 2. Define the antidirected path P0,0 of length

2da as follows:

P0,0 : u0 u2 u1 u3 u2 u2d2a u2d2a+1

u2d2a+2 u2d2a u2d2a+3 u2d2a1

u2d1a+1 u2d1a+1 ul+1 ul.

We see that the labels of the arcs of P0,0 are 2, 3, 4, 5, . . . , 2d1a1, 2d1a+

1, 2d1a + 2, 2d1a + 3, 2d1a + 4, . . . , 2da, l + 2d1a and 1.

The definition of P0,0 does not make sense when 2d2a = 1, that is, d = 2

and a = 1. In this case let

P0,0 : u0 u3 u1 ul+1 ul.

Let P0,1 = P0,0 + l. Since l > 2da implies that 2d1a + 1 < l 2d1a + 1,

the vertices of P0,1 are distinct from the vertices of P0,0 except for ul, which is

33

the last vertex of P0,0 and the first vertex of P0,1. Similarly, the antidirected

paths

P0,0, P0,1 = P0,0 + l, P0,2 = P0,0 + 2l, . . . , P0,b1 = P0,0 + (b 1)l

are vertex-disjoint except that the path P0,i begins at the last vertex of P0,i1

for 1 i b1 and P0,b1 ends at u0. Thus C0 = P0,0P0,1 P0,b1 is an

antidirected m-cycle and the labels of the arcs of C0 are 2, 3, 4, 5, . . . , 2d1a

1, 2d1a + 1, 2d1a + 2, 2d1a + 3, 2d1a + 4, . . . , 2da, l + 2d1a and 1. Let

C 0 denote the antidirected m-cycle obtained by reversing the orientation of all

the arcs of C0. The family 0 of antidirected m-cycles defined by

0 = {C0 + i|0 i l 1} {C 0 + i|0 i l 1}

is a decomposition of ~X(n;B0) into antidirected m-cycles where

B0 = {1, 2, 3, . . . , 2d1a 1, 2d1a + 1, 2d1a + 2, . . . , 2da, l + 2d1a}.

Let b = r/(2d+1a) = 2ed1a/a. If b = 1, take B = B0. Suppose that

b 2. First for 1 i b 1, let Pi,0 be the path obtained by adding il to the

subscripts of every other vertex of P0,0 starting with u2, that is,

Pi,0 : u0 u2+il u1 u3+il u2 u2d2a+il

u2d2a+1 u2d2a+2+il u2d2a

u2d2a+3+il u2d2a1 u2d1a+1+il

u2d1a+1 u(i+1)l+1 ul.

Next, for each i, 1 i b 1, obtain the paths Pi,1, Pi,2, . . . , Pi,b1 by letting

multiples of l act on Pi,0 in the same way they acted on P0,0. Furthermore,

the path Pi,j begins at the last vertex of Pi,j1 for 1 j b 1 and the

last vertex of Pi,b1 is u0. Thus, for each i with 1 i b 1, we have that

Ci = Pi,0 Pi,1 Pi,2 Pi,b1 is an antidirected m-cycle, where the labels

34

of the arcs of Ci are 1+il, 2+il, 3+il, . . . , 2d1a1+il, 2d1a+1+il, 2d1a+

2 + il, . . . , 2da + il, (i+ 1)l + 2d1a. As before, let C i denote the antidirected

m-cycle obtained by reversing the orientation of all the arcs of Ci.

Let B = {1 + il, 2 + il, 3 + il, . . . , 2d1a 1 + il, 2d1a + 1 + il, 2d1a +

2 + il, . . . , 2da + il, (i + 1)l + 2d1a|0 i b 1}. Now the longest arc

in B has label bl + 2d1a in absolute value. Since 2ea < 2dab implies that

2eda/a + 1 b, it follows that

bl + 2d1a =2ed1al

a+ 2d1a =

2ed1al

a+

l

2 2

ea+ 1

2b

bl

2 2

ea+ 1

2b 2a + 1.

Therefore, C1 = Q0,0 Q0,1 Q0,b1 is an antidirected cycle of length m.

Let C 1 denote the antidirected m-cycle obtained by reversing the orientation

of all the arcs of C1. The family 1 of antidirected m-cycles defined by

1 = {C1 + i|0 i l 1} {C 1 + i|0 i l 1}

is a decomposition of the circulantX (n;B1) into antidirected m-cycles

where

B1 = { a + 2, l + 2, l + 3, l + 4, . . . , a + l, a + l + 2, a + l + 3,

38

. . . , 2a + l 1, 2a + l, 2a + l + 1}.

Note that label (l + a + 1) appears in 0 but not in 1. Similarly, label

(a + 2) appears in 1 but not in 0.

Now we must show that we can modify the preceding two families of directed

m-cycles often enough to use r arc labels. Recall that r 0 (mod 4), that

is, e 2.

As previously noted, the family 0 uses 4a labels. If r = 4a, then we may

stop. If r > 4a, then the families 0 and 1 use a total of 8a labels. If

r = 8a, then we may stop. If r > 8a, then let P2,0 be the antidirected path

obtained from P0,0 by adding 2l to the subscripts of every other vertex of P0,0

starting with u2. We obtain a family of antidirected m-cycles, denoted 2,

from P2,0 in the same way 0 is obtained from P0,0. If necessary, we obtain a

family 4 from P0,0 by adding 4l and so on.

To show that r arc labels can be used, we must calculate the maximum

number of times the preceding extensions can be carried out. Let b = r/(4a).

If b is even, the longest arc in b1 has label 2a + (b 1)l + 1 in absolute

value. Since both m and r are even multiples of a, and since m > r, we have

m 2a r. It follows that

2a + (b 1)l + 1 = 2a + ( r4a 1)l + 1

= l r + 1b

+rl

4a l + 1

=rbl

4ab r + 1

b+ 1 2ea, the vertices on P0,0 are distinct, hence P0,0 is a path.

Let P0,1 = P0,0 + l. Since l > 2ea implies that 3 2e2a + 1 + l > 2e2a,

the vertices of P0,1 are distinct from the vertices of P0,0 except for ul, which is

the last vertex of P0,0 and the first vertex of P0,1. Similarly, the antidirected

paths

P0,0, P0,1 = P0,0 + l, P0,2 = P0,0 + 2l, . . . , P0,2b1 = P0,0 + (2b 1)l


for 1 i 2b 1 and P0,2b1 ends at u0. Thus C0 = P0,0 P0,1

P0,2b1 is an antidirected m-cycle and the labels of the arcs of C0 are

1, 2, 3, 4, . . . , 2e1a,2e1a+1,2e1a+2,2e1a+3, . . . ,1, and 2e1a

l. Since l > 2ea and n 2 = 2bl 6l in this case, we have 2e1a l >

3l/2 > n22. The labels of arcs in C0 are half positive and half negative.

Hence the family 0 of antidirected m-cycles defined by

0 = {C0 + i|0 i l 1}

49

is a decomposition ofX (n 2;L) into antidirected m-cycles where

L = { 1, 2, 3, . . . , 2e1a,2e1a + 1,2e1a + 2,

2e1a + 3, . . . ,1,2e1a l}.

Let b = r/(2ea) = a/a. If b = 1, we stop. If b 3, since m > r and

a = ab, we have 2e+1ab = m > r = 2eab implies 3 b 2b 1. Renew the

antidirected path P0,0 of length 2ea as follows:

P0,0 : u0 u1 u1 u2 u2 u2e2a+1 u2e2a

u2e2a1 u2e2a+1 u2e2a2 u2e2a+2

u2e1a u2e1a ul.

Let P0,1 = P0,0 + l. Since l > 2ea implies that 2e1a + l > 2e1a, the

vertices of P0,1 are distinct from the vertices of P0,0 except for u0, which is the

first vertex of P0,0 and the last vertex of P0,1. Similarly, the antidirected paths

P0,0, P0,1 = P0,0 + l, P0,2 = P0,0 + 2l, . . . , P0,2b1 = P0,0 + (2b 1)l

are vertex-disjoint except that the path P0,i ends at the first vertex of P0,i1 for

1 i 2b 1 and P0,2b1 ends at ul. Thus C0 = P0,0P0,1 P0,2b1 is

an antidirected m-cycle and the labels of the arcs of C0 are 1, 2, 3, . . . , 2e1a

1, 2e1a+1, 2e1a+2, . . . , 2ea and 2e1a+ l. Let C 0 denote the antidirected

m-cycle obtained by reversing the orientation of all the arcs of C0.

If b > 3, then obtain the path Pi,0 by adding il to the subscripts of every

other vertex of P0,0 starting with u1, that is,

Pi,0 : u0 u1+il u1 u2+il u2 u2e2a+1

u2e2a+il u2e2a1 u2e2a+1+il u2e2a2

u2e2a+2+il u2e1a u2e1a+il ul,

50

for 1 i (b 3)/2. Next, obtain the paths Pi,1, Pi,2, . . . , Pi,2b1 by letting

multiples of l act on Pi,0 in the same way they acted on P0,0. Furthermore,

the path Pi,j ends at the first vertex of Pi,j1 for 1 j 2b 1 and the last

vertex of Pi,2b1 is ul. Thus, for each i with 1 i (b 3)/2, we have that

Ci = Pi,0Pi,1Pi,2 Pi,2b1 is an antidirected m-cycle, where the labels

of the arcs of Ci are 1+ il, 2+ il, . . . , 2e1a 1+ il, 2e1a+1+ il, 2e1a+2+

il, . . . , 2ea+ il, 2e1a+(i+1)l. Here we jump the label 2e1a+ il in Ci, which

is used in Ci1. Since l > 2ea, n 2 = 2bl and b 2b 1, the largest label

of arcs in Ci is 2e1a + ( b1

2)l < bl

2 (2b1

2)l < n2

2. As before, let C i denote

the antidirected m-cycle obtained by reversing the orientation of all the arcs

of Ci.

Lastly, define the path P b12

,0 as follows:

P b12

,0 : u0 ubl1 u1 ubl2 u2 u2e2a1

ubl2e2a ul+2e2a+1 ubl2e2a1 ul+2e2a+2

ul+2e1a ubl2e1a ul.

Then, obtain the paths P b12

,1, P b12

,2, . . . , P b12

,2b1 by letting multiples of l act

on P b12

,0 in the same way they acted on P0,0. For convenient, let the (i+1)-th

segment contains the vertices uil, uil+1, . . . , u(i+1)l1. The path P b12

,0 use the

first, second, . . ., (2e2a 2)-th vertices in the first segment, the (2e2a)-th,

. . ., (2e1a + 1)-th vertices in the (2b 1)-th segment, and the last 2e1a

vertices in the (b 1)-th segment, and ul. Since l > 2ea, the paths P b12

,0,

P b12

,1, P b12

,2, . . . , P b12

,2b1 are vertex-disjoint except for uil which is the first

vertex of P b12

,i and the last vertex of P b12

,i1 for 1 i 2b1, and P b12

,2b1

ends at u0. Thus, C b12

= P b12

,0P b12

,1P b12

,2b1 is an antidirected m-cycle,

where the labels of the arcs of C b12

are bl 1, bl 2, bl 3, . . . , bl 2e1a+

1, (b+1)l2e1a1, (b+1)l2e1a2, (b+1)l2e1a3, . . . , (b+1)l2ea

and (b 1)l 2e1a. Since n 2 = 2bl, the label set of C b12

is congruent to

51

bl 1, bl 2, bl 3, . . . , bl 2e1a + 1,n2 2e1a + l,n

2 2e1a + l

1,n2 2e1a + l 2, . . . ,n

2 2ea + l + 1 and (b 1)l 2e1a.

In the following, we show that the labels in C b12are distinct from Ci for 1

i (b3)/2. If b 2b3, then largest label of arcs in Ci for 1 i (b3)/2

is 2e1a+( b12)l, which is less than 2e1a+(b2)l = (b1)l+(2e1a l)

b 12

l+2e1a

(where b12l + 2e1a is the largest label of arcs in Ci for 1 i (b 3)/2),

and the second smallest positive label of arcs in C b12

is bl 2e1a+1, which

is equal to ( b+12)l 2e1a + 1, then greater than 2e1a + ( b1

2)l (the largest

label of arcs in Ci for 1 i (b 3)/2). Now we check that the smallest

positive label of arcs in C b12is (b1)l2e1a. Since b = 2b1, b = a/a and

l 3 2e1a + 1, we have b12l+ 2e1a=(b 1)l+ 2e1a > (b 1)l 2e1a,

and b32l + 2ea=(b 2)l + 2ea = (b 1)l l + 2ea < (b 1)l 3 2e1a

1+2ea=(b 1)l2e1a1< (b 1)l2e1a. That is, the smallest positive

label of arcs in C b12

is between the largest label and the second largest label

of arcs in C b12

for 1 i (b 3)/2. Thus the labels in C b12

are distinct from

Ci for 1 i (b 3)/2.

Let L = {1+il, 2+il, . . . , 2e1a1+il, 2e1a+1+il, . . . , 2ea+il, 2e1a+

(i+1)l|0 i (b3)/2}{bl1, bl2, bl3, . . . , bl2e1a+1,n22e1a+

52

l,n22e1a+ l1,n

22e1a+ l2, . . . ,n

22ea+ l+1, (b1)l2e1a},

hence L contains 2ea b12

+ 2e1a = 2e1ab = 2e1a = r/2 positive integers

and r/2 negative integers. Let j = {Cj + i, C j + i|0 i l 1} for

j = 0, 1, . . . , (b 3)/2; and let (b1)/2 = {C(b1)/2 + i|0 i l 1}. Thus,

the collection

{0,1, . . . ,(b3)/2,(b1)/2}

is a partition of the arc set of the circulant digraphX (n 2;L) into antidi-

rected cycles of length m.

Lemma 4.2.5 Let m,n, r be positive integers satisfying m 0 (mod 8),

r 2 (mod 4), n = m+ r, 8 m < n < 2m and n(n 2) m (mod 2m).

Then there exists a subset L {1, 2, , n42} with |{i > 0 : i L}| = |{i 2b implies that b

+12

+ l > 3b12

, the vertices

of P0,1 are distinct from the vertices of P0,0 except for ul, which is the first

vertex of P0,1 and the last vertex of P0,0. Similarly, the antidirected paths

P0,0, P0,1 = P0,0 + l, P0,2 = P0,0 + 2l, . . . , P0,2e1a1 = P0,0 + (2e1a 1)l


for 1 i 2e1a 1 and P0,2e1a1 ends at u0. Thus C0 = P0,0 P0,1

P0,2e1a1 is an antidirected m-cycle. Since l 2b + 1, e 3, we have

b l b 2b 1 = b 1 < b + 1 and l b < l < 2e2al = n22, the

55

labels of the arcs in C0 are all distinct and with half positive and half negative.

The family 0 of antidirected m-cycles defined by

0 = {C0 + i|0 i l 1}

is a decomposition of the circulantX (n 2;L) into antidirected cycles of

length m where L = {1,2, . . . ,b + 1, b, b 1, . . . , 1, b l}.

Subcase 2.2. b 3 and a = a.

Since l = 2b+ a/a 7 and a = a, we have b 3. Define the antidirected

path P1,0 by:

P1,0 : u0 ul+1 u1 ul+2 u2 ub+1 ul+b ul.

Let P1,1 = P1,0+l and P1,2 = P1,0+2l. Since l > 2b implies thatb+1+2l >

l+ b, the vertices of P1,2 are distinct from the vertices of P1,0 and the vertices

of P1,1 are distinct from the vertices of P1,0 except for u0, which is the first

vertex of P1,0 and the last vertex of P1,1. Similarly, the antidirected paths

P1,0, P1,1 = P1,0 + l, P1,2 = P1,0 + 2l, . . . , P1,2e1a1 = P1,0 + (2e1a 1)l

are vertex-disjoint except that the path P1,i ends at the first vertex of P1,i1

for 1 i 2e1a 1 and P1,2e1a1 ends at ul. Thus C1 = P1,0 P1,1

P1,2e1a1 is an antidirected m-cycle, and the labels of the arcs of C1 are

1 + l, 2 + l, . . . , 2b 1 + l and b + 2l. Let C 1 denote the antidirected m-cycle

obtained by reversing the orientation of all the arcs of C1.

If b > 3, then obtain the path Pi,0 by adding 2(i 1)l to the subscripts of

every other vertex of P1,0 starting with ul+1 for 2 i (b 1)/2, that is,

Pi,0 : u0 u1+l+2(i1)l u1 u2+l+2(i1)l u2

ub+1 ub+l+2(i1)l ul,

for 2 i (b1)/2. Next, obtain the paths Pi,1, Pi,2, . . . , Pi,2e1a1 by letting

multiples of l act on Pi,0 in the same way they acted on P1,0. Thus, for each i

56

with 1 i (b1)/2, we have that Ci = Pi,0Pi,1Pi,2 Pi,2e1a1 is an

antidirected m-cycle, where the labels of the arcs of Ci are 1+ l+2(i 1)l, 2+

l+2(i 1)l, 3+ l+2(i 1)l, . . . , 2b 1+ l+2(i 1)l and b+2il. As before,

let C i denote the antidirected m-cycle obtained by reversing the orientation of

all the arcs of Ci.

The largest label of arcs in Ci is b + (b 1)l for 1 i (b 1)/2.

Since b 3, we have b = r2b

= 2e2a+1b

2e3a = 2e3a, this implies that

b + (b 1)l b + 2e3al l = n24

+ b l < n24. Let L = {1 + l +

2(i 1)l, 2+ l+2(i 1)l, 3+ l+2(i 1)l, . . . , 2b1+ l+2(i 1)l, b+2il|1

i (b 1)/2} {1,2, . . . ,b + 1, b, b 1, . . . , 1, b l}, then L contains

2b b12

+ b = bb = r/2 positive integers and r/2 negative integers. Let 0

be the same set as in subcase 2.1 and j = {Cj + i, C j + i|0 i l 1} for

j = 1, 2, . . . , (b 1)/2. Thus, the collection

{0,1, . . . ,(b3)/2,(b1)/2}



Subcase 2.3. b 3 and a/a 3.

Since m > r and a 3a, we have 2eab > 2e1a+ 2 3 2e1a + 2, which

implies b > 1. We renew the antidirected path P0,0 by:

P0,0 : u0 u1 u1 u2 u2 ub+12 u b+1

2

ub32 u b+3

2 ub1 ub ub+1 ul.

Let P0,1 = P0,0 + l. Since l 2b + 3 implies that b + l > b + 1, the

vertices of P0,1 are distinct from the vertices of P0,0 except for u0, which is the

first vertex of P0,0 and the last vertex of P0,1. Similarly, the antidirected paths

P0,0, P0,1 = P0,0 + l, P0,2 = P0,0 + 2l, . . . , P0,2e1a1 = P0,0 + (2e1a 1)l

57

are vertex-disjoint except that the path P0,i ends at the first vertex of P0,i1

for 1 i 2e1a 1 and P0,2e1a1 ends at ul. Thus C0 = P0,0 P0,1

P0,2e1a1 is an antidirected m-cycle and the labels of the arcs of C0 are

1, 2, . . . , b, b + 2, b + 3, . . . , 2b 1, 2b + 1 and b + 1 + l. Let C 0 denote the

antidirected m-cycle obtained by reversing the orientation of all the arcs of C0.

If b > 3, then obtain the path Pi,0 by adding il to the subscripts of every

other vertex of P0,0 starting with u1, that is,

Pi,0 : u0 u1+il u1 u2+il u2

ub+12 u b+1

2+il ub3

2 u b+3

2+il

ub1+il ub ub+1+il ul,

for 1 i (b3)/2. Next, obtain the paths Pi,1, Pi,2, . . . , Pi,2e1a1 by letting

multiples of l act on Pi,0 in the same way they acted on P0,0. Thus, for each i

with 0 i (b 3)/2, we have that Ci = Pi,0 Pi,1 Pi,2 Pi,2e1a1 is

an antidirected m-cycle, where the labels of the arcs of Ci are 1+ il, 2+ il, 3+

il, . . . , b+il, b+2+il, b+3+il, . . . , 2b1+il, 2b+1+il and b+1+(i+1)l. The

largest label of arcs in Ci is b+1+( b1

2)l for 0 i (b3)/2. Since l 2b+3,

and since r < m, we have b = r2b

< m2b

= 2e1a, these inequalities imply that

b + 1 + ( b12)l < b + 1 + (2e2a 1

2)l = 2e2al l

2+ b + 1 < 2e2al = n2

2.

As before, let C i denote the antidirected m-cycle obtained by reversing the

orientation of all the arcs of Ci.

Lastly, define the path P b12

,0 as follows:

P b12

,0 : u0 u(2e2a1)l+1 u1 u(2e2a1)l+2 u2

u(2e2a1)l+ b1

2 ub+1

2 u2e2al+ b+1

2 ub1

2

u2e2al+ b+32 ub3

2 ub+1

u2e2al+b ul.

58

Then, obtain the paths P b12

,1, P b12

,2, . . . , P b12

,2e1a1 by letting multiples of l

act on P b12

,0 in the same way they acted on P0,0.

For convenient, let the (i+1)-th segment contains the vertices uil, uil+1, . . . ,

u(i+1)l1 for 0 i 2e1a 1. The path P b12

,0 use the second, third, . . .,

( b+12)-th vertices in the 2e2a-th segment, the ( b

+12

+ 1)-th, . . ., (b + 1)-th

vertices in the (2e2a+1)-th segment, and the last b1 vertices in the 2e1a-

th segment, and u0, ul. Since l 2b + 3, the vertices of the path P b12

,i are

distinct from the vertices of P b12

,j for i 6= j except the vertex uil which is the

first vertex of P b12

,i and the last vertex of P b12

,i1 for 1 i 2e1a 1 and

the last vertex of P b12

,2e1a1 is u0. Thus, C b12

= P b12

,0P b12

,1P b12

,2e1a1

is an antidirected m-cycle, where the labels of the arcs of C b12

are (2e2a

1)l+1, (2e2a 1)l+2, . . . , (2e2a 1)l+ b 1,2e2al+ b,2e2al+ b+

1, . . . ,2e2al + 2b 1 and (2e2a + 1)l + b. Since (2e2a + 1)l + b

n 2 (2e2a + 1)l + b = (2e2a 1)l + b > (2e2a 1)l + b 1, and

(2e2a 1)l+ b = n22 l+ b < n2

2, and 2e2al+ b = n2

2+ b > n2

2,

the labels of the arcs in C b12

are all distinct and with half positive and half

negative.

If b 2e1a 3, then the largest label of arcs in Ci for 0 i (b 3)/2

is b + 1 + ( b12)l b + 1 + (2e2a 2)l, which is less than (2e2a 1)l + 1,

the smallest label of arcs in C b12. Thus, the labels in C b1

2are distinct from

Ci, Ci for 0 i (b 3)/2. But if b = 2e1a 1, then b = r2b = 2

e1a 1

implies r = 2eab 2b, thus 2e1a + 2 = 2eab 2b. Divide 2e1a in both

side, we have

a

a= 2b 2b

2e1a 2

2e1a= b + (b b

+ 1

2e2a)

= b +2e2ab b 1

2e2a> b since e 3 and b > 1.

Hence we have l = 2b+a/a > 3b. The largest label of arcs in {Ci|0 i b32 }

is b+1+( b12)l = b+1+(2e2a1)l > b+(2e2a1)l (where b+(2e2a1)l

59

is the largest positive label of arcs in C b12

), and since l > 3b, we have

b+1+( b12)l = b+1+(2e2a1)l = 2e2al(lb1) < 2e2al2b+1(the

smallest negative label of arcs in C b12

in absolutely value). Now we check the

smallest positive label of arcs in C b12, which is (2e2a 1)l+1. Since l > 3b,

2b + 1 + (2e2a 2)l, the second largest label of arcs in {Ci|0 i b32 } is

less than (2e2a 1)l + 1. Thus the labels of arcs in C b12

are distinct from

{Ci, C i|0 i b32 } .

Let L = {1+ il, 2+ il, 3+ il, . . . , b+ il, b+2+ il, b+3+ il, . . . , 2b 1+

il, 2b+1+il, b+1+(i+1)l|0 i (b3)/2}{(2e2a1)l+1, (2e2a1)l+

2, . . . , (2e2a 1)l+ b,2e2al+ b,2e2al+ b+1, . . . ,2e2al+2b 1},

then L contains 2b b12

+ b = bb = r/2 positive integers and r/2 negative

integers. Let j = {Cj + i, C j + i|0 i l 1} for j = 0, 1, . . . , (b 3)/2;

and let (b1)/2 = {C(b1)/2 + i|0 i l 1}. Thus, the collection

{0,1, . . . ,(b3)/2,(b1)/2}



We are ready for the proof of Theorem 4.1.1.

Proof of Theorem 4.1.1.

Let Z = {1, 2, . . . , n42}. If r 0 (mod 4), by Lemma 4.2.3 and Lemma 4.2.4

together, we haveCm |

X (n 2;Z) K2; if r 2 (mod 4), by

Lemma 4.2.3 and Lemma 4.2.5 together, we haveCm |

X (n 2;Z) K2.

HenceCm |(Kn I). This completes the proof of Theorem 4.1.1.

60

Chapter 5

Caterpillar Factorization of

Crowns and Directed

Caterpillar Factorization of

Symmetric Crowns

5.1 Introduction and preliminaries

A caterpillar is a tree of order at least three which contains a path such that

each vertex not on the path is adjacent to a vertex on the path. For the

convenience of discussions, a caterpillar can also be defined alternatively as

follows. For a positive integer d, let Sd denote a star with d edges. For positive

integers d1, d2, , dv (v 1), S(d1, d2, , dv) denotes the graph obtained

from the stars Sd1 , Sd2 , , Sdv by identifying an endvertex of Sdi with the

center of Sdi+1 for i = 1, 2, , v1. As an illustration, S(5, 2, 1, 3) is exhibited

in Fig.1. It is easy to see that the graph S(d1, d2, , dv) is of order 1 + d1 +

d2+ +dv. A caterpillar can be also defined as a graph S(d1, d2, , dk) with

61

1 + d1 + d2 + + dk 3.

AAAA

@@

@@

HHHH@

@@

HHHH@

@@@

Fig.1. S(5, 2, 1, 3)

For positive integers k n, the crown Cn,k is a bipartite graph with bi-

partition (A,B) where A = {a0, a1, , an1} and B = {b0, b1, , bn1} and

edge set {aibj : 0 i n 1, j = i + 1, i + 2, , i + k (mod n)}. In the

following, the subscripts of ai and bj are always taken modulo n. Each edge

aibi+t (1 t k) in Cn,k has label t. Let Cn,k denote the digraph obtained

from Cn,k by replacing each edge by two arcs with opposite direction. We call

Cn,k a symmetric crown. Suppose G and H are graphs. An H-factor of G

is a spanning subgraph of G each component of which is isomorphic to H.

An H-factorization of G is a set of H-factors of G which partition the edges

of G. For digraphs G and H, H-factor of G and H-factorization of G are

similarly defined. In this chapter, we study caterpillar factorization of Cn,k

and directed caterpillar factorization of Cn,k. Here a directed caterpillar is a

digraph obtained from a caterpillar by orienting each edge arbitrarily.

Graph-factorization is a special type of edge decomposition of a graph.

Edge decompositions of crowns and multicrowns have been investigated for

isomorphic paths [71], isomorphic complete bipartite graphs [53], and isomor-

phic stars [51]. Ushio [84] proved that Kp,q has an Sl-factorization if and only

if p = q 0 (mod l(l + 1)). The Sl-factorization of Kp,q,r has been partially

solved by Ushio [85].

62

Lemma 5.1.1 Suppose G1 is a multigraph (directed or undirected) of order

n1, size e1 and G2 is a multigraph (directed or undirected) of order n2, size e2.

(1) If G1 has a G2-factor, then n1 0 (mod n2).

(2) If G1 has a G2-factorization, then e1n2 0 (mod n1e2).

Proof. Let t be the number of components in a G2-factor of G1.

(1)Since n1 = tn2, we have n1 0 (mod n2).

(2)Let r be the number of G2-factor in the factorization. Then e1 = rte2.

Thus e1 =rn1e2n2

, which implies r =e1n2n1e2

. Thus e1n2 0 (mod n1e2).

A connected bipartite graph with bipartition (C,D) is balanced if |C| = |D|.

It is easy to see that S(d1, d2, , dv) is balanced if and only if d1+d3+d5+ =

1 + d2 + d4 + .

Lemma 5.1.2 Suppose that H is a balanced bipartite graph of order 2n, and

G is a nonbalanced subgraph of H with order t. If H has a G-factor, then

n 0 ( mod t).

Proof. Suppose that (C,D) is a bipartition ofH with |C| = |D| = n. Suppose

(V1, V2) is a bipartition of G. Let t1 = |V1|, t2 = |V2|. Then t1 + t2 = t, t1 6= t2.

Suppose that in a G-factor of H, there are l copies of G with V1-part in C,

and m copies of G with V2-part in C. Then lt1 +mt2 = n and lt2 +mt1 = n.

Thus t2n t1n = m(t22 t21). Since t1 6= t2, we have n 0 (mod t1 + t2), i.e.,

n 0 (mod t).

Lemma 5.1.3 Suppose that n, p, q are positive integers such that pq n.

Let H be a graph. If Cn,p has an H-factorization, then Cn,pq has an H-

factorization.

Proof. It is easy to see that Cn,pq can be decomposed into q spanning subdi-

graphs of which each is isomorphic to Cn,p. Since, by assumption, Cn,p has an

H-factorization, so does Cn,pq.

63

5.2 Caterpillar factorization of crown

Lemma 5.2.1 Let G = S(d1, d2, , dv) be a caterpillar of order t. Let n

t 1. Then there exists a subgraph G(1) of the crown Cn,t1 such that

(i) G(1) = G,

(ii) the vertex set of G(1) is {ai : i = 0,1,2, ,d2 d4 d6 }

{bi : i = 1, 2, , d1 + d3 + d5 + }, and

(iii) the edges of G(1) have labels 1, 2, 3, , t 1.

Proof. Let G = S(d1, d2, , dv) be a caterpillar of order t. Then t =

1 + d1 + d2 + + dv. Let G(1) be the subgraph of Cn,t1 induced by the

following edge set

{a0bi : i = 1, 2, , d1}

{aibd1 : i = 1,2, ,d2}

{ad2bi : i = d1 + 1, d1 + 2, , d1 + d3}

{aibd1+d3 : i = d2 1,d2 2, ,d2 d4}

{ad2d4bi : i = d1 + d3 + 1, d1 + d3 + 2, , d1 + d3 + d5}

.

We see that

(i) G(1) = S(d1, d2, , dv) = G,

(ii) the vertex set of G(1) is {ai : i = 0,1,2, ,d2 d4 d6 }

{bi : i = 1, 2, , d1 + d3 + d5 + }, and

(iii) the edges of G(1) have labels 1, 2, , d1, d1 + 1, d1 + 2, ,

d1 + d2, d1 + d2 + 1, d1 + d2 + 2, , d1 + d2 + d3,

64

, d1 + d2 + d3 + + dv = t 1.

Lemma 5.2.2 Let G be a balanced-caterpillar of order t. Suppose that n 0

(mod t/2). Then Cn,t1 has a G-factorization.

Proof. Let G = S(d1, d2, , dv). Since G is balanced, d1 + d3 + d5 + =

1 + d2 + d4 + d6 + = t2 . Let n = n/ t

2. Let G(1) be the subgraph of Cn,t1

described in Lemma 5.2.1. Let F be a subgraph of Cn,t1 such that

F = G(1) (G(1) + t2) (G(1) +2 t

2) (G(1) +3 t

2) (G(1) +(n 1) t

2).

From Lemma 5.2.1(ii), we see that F is aG-factor of Cn,t1. From Lemma 5.2.1(iii),

we see that Cn,t1 can be decomposed into F, F + 1, F + 2, , F + ( t2 1).

Hence Cn,t1 has a G-factorization.


Theorem 5.2.3 Let G be a balanced caterpillar of order t. Then Cn,k has a

G-factorization if and only if n 0 (mod t2) and k 0 (mod t 1).

Proof. (Necessity) Note that Cn,k has 2n vertices, nk edges, and G has

t vertices, t 1 edges (where t is an even integer since G is balanced). By

Lemma 5.1.1, we have

2n 0 (mod t), t nk 0 (mod 2n(t 1)).

Thus n 0 (mod t2), and tk 0 (mod t 1). The latter implies k 0 (mod

(t 1)) since t and t 1 are relatively prime.

(Sufficiency) Since n 0 (mod t2), it follows from Lemma 5.2.2 that Cn,t1

has a G-factorization. Since k 0 (mod t 1), it follows from Lemma 5.1.3

that Cn,k has a G-factorization.

The nonbalanced-caterpillar factorization of crowns remains as an open

problem.

65

5.3 Directed caterpillar factorization of sym-

metric crowns

Recall that the symmetric crown Cn,k is the digraph obtained from Cn,k by

replacing each edge of Cn,k by two arcs with opposite directions. We assign the

arcaibj and the arc

bjai in C

n,k with the same label of aibj in Cn,k. Therefore

the arcaibi+t(1 t k) has label t, and the arc

bi+tai also has label t.

Suppose that H is a subdigraph of Cn,k and H has no isolated vertex.

The dual of H, denoted by Dual(H), is the subdigraph of Cn,k induced by

the arc set { biaj:aibj E(H)} {

ajbi :

bjai E(H)}. Note that

Dual(H) = H. Note also that the arc aibj in H, which is from A to B, and

the arcbiaj in D(H), which is from B to A, have the same label. Similarly,

the arcbjai in H, which is from B to A, and the arc

ajbi in Dual(H), which

is from A to B, have the same label.

Lemma 5.3.1 Let G be a caterpillar of order t. Suppose that G is a digraph

obtained from G by orienting each edge of G arbitrarily. Then we have the

following.

(1)If G is balanced, and n 0 (mod t2), n t, then Cn,t1 has a G-factorization.

(2)If G is nonbalanced, and n 0 (mod t), then Cn,t1 has a G-factorization.

Proof. Let G = S(d1, d2, , dv) be a caterpillar of order t. By Lemma 5.2.1,

there exists a subgraph G(1) of the crown Cn,t1 such that

(i) G(1) = G,

(ii) the vertex set of G(1) = {ai : i = 0,1,2, ,d2 d4 d6 }

{bi : i = 1, 2, , d1 + d3 + d5 + }, and

(iii) the edges of G(1) have labels 1, 2, 3, , t 1.

66

Since G(1) = G, we can orient the edges of G(1) so that the resulting digraph,

say G(2), is isomorphic to G. Hence G(2) can be viewed as a subdigraph of

Cn,t1 such that

(i) G(2) = G,

(ii) the vertex set of G(2) = {ai : i = 0,1,2, ,d2 d4 d6 }

{bi : i = 1, 2, , d1 + d3 + d5 + }, and

(iii) the arcs of G(2) have labels 1, 2, 3, , t 1.

Let G(3) = Dual(G(2)). Then

(i) G(3) = G,

(ii) the vertex set of G(3) is {ai : i = 1,2, ,d1 d3 d5 }

{bi : i = 0, 1, 2, , d2 + d4 + d6 + }, and


(1)Now G is balanced, and n 0 (mod t2), n t. We have d1+d3+d5+ =

1 + d2 + d4 + = t2 . Thus G(2) has t

2vertices in A and t

2vertices in B. Let

F1 = G(2) (G(2) + t

2) (G(2) +2 t

2) (G(2) +3 t

2) (G(2) +(2n

t 1) t

2),

and F2 = G(3) (G(3) + t

2) (G(3) + 2 t

2) (G(3) + 3 t

2) (G(3) + (2n

t

1) t2). Then F1 and F2 are G-factors of C

n,t1. We can see that C

n,t1 can be

decomposed into the following G-factors: F1, F1 + 1, F1 + 2, , F1 + ( t2 1)

and F2, F2 + 1, F2 + 2, , F2 + ( t2 1).

(2)Now G is nonbalanced, and n 0 (mod t). Let G(4) = G(3) + (1 + d1 +

d3 + d5 + ). Then

(i) G(4) = G,

(ii) the vertex set of G(4) is

{ai : i = 1, 2, ,1 + d1 + d3 + d5 + , d1 + d3 + d5 + }

67

{bi : i = 1 + d1 + d3 + d5 + , 2 + d1 + d3 + d5 + , ,

1 + d1 + d2 + d3 + d4 + }, and


Let G(5) = G(2) G(4).

Then the vertex set of G(5) =

{ai : d2 d4 d6 i d1 + d2 + d3 + }

{bi : 1 i 1 + d1 + d2 + d3 + d4 + }.

Let F = G(5) (G(5) + t) (G(5) + 2t) (G(5) + (nt 1) t). Then F

is a G-factor of Cn,t1, and Cn,t1 is decomposed into the following G-factors:

F, F + 1, F + 2, , F + (t 1).


Theorem 5.3.2 Let G be a caterpillar of order t, and let G be a digraph

obtained from G by orienting each edge of G arbitrarily.

(1) Suppose that G is balanced. Then Cn,k has a G factorization

if and only if n k, n 0 (mod t2), k 0 (mod t 1).

(2) Suppose that G is nonbalanced. Then Cn,k has a G factorization

if and only if n 0 (mod t), k 0 (mod t 1).

Proof. (1)(Necessity) Note that Cn,k has 2n vertices, 2nk arcs, and G has t

vertices, t1 arcs. No matter G is balanced or not, we have, by Lemma 5.1.1,

2n 0 (mod t), t 2nk 0 (mod 2n(t 1)). Hence n 0 (mod t2), tk

0 (mod t 1). The latter implies k 0 (mod t 1).

(Sufficiency) From the assumption, we have n 0 (mod t2), n t. By

Lemma 5.3.1(1), Cn,t1 has G-factorization. Since k 0 (mod t 1), Cn,k has

a G-factorization.

68

(2)(Necessity) As mentioned in the proof of the Necessity of (1), we have

k 0 (mod t 1). Since Cn,k is a balanced bipartite graph of order 2n, and

G is nonbalanced subgraph of Cn,k with order t, by Lemma 5.1.2 the existence

of a G-factor of Cn,k implies n 0 (mod t).

(Sufficiency) Since G is nonbalanced and n 0 (mod t), by Lemma 5.3.1(2),

Cn,t1 has a G-factorization. Since k 0 (mod t 1), by Lemma 5.1.1 Cn,khas a G-factorization.

The following result follows immediately from Theorem 5.3.2(2) by remov-

ing directions on the arcs.

Corollary 5.3.3 Let G be a nonbalanced caterpillar of order t. Suppose that

n 0 (mod t), k 0 (mod t 1). Then 2Cn,k has a G-factorization.

69

Documents

Path Cicle and Star Decomposition