Directed 3-cycle decompositions of complete directed graphs with quadratic leaves

(1)

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Discrete Mathematics

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Directed 3-cycle decompositions of complete directed graphs with

quadratic leaves

Liqun Pu

a,∗

, Hung-Lin Fu

b

, Hao Shen

c

a_{Department of Mathematics, Zhengzhou University, Zhengzhou 450052, China}

b_{Department of Applied Mathematics, National Chiao Tung University, Hsin Chu, 30050, Taiwan} c_{Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China}

a r t i c l e i n f o

Article history:

Received 3 January 2007 Accepted 28 May 2008 Available online 2 July 2008

Keywords:

Mendelsohn triples Complete directed graph Packing

a b s t r a c t

Let G be a vertex-disjoint union of directed cycles in the complete directed graph Dt, let

|E(G)|be the number of directed edges of G and suppose G6= *C2∪

* C3or * C5if t=5, and G6= *C3∪ *

C3if t=6. It is proved in this paper that for each positive integer t, there exist

*

C3-decompositions for Dt−G if and only if t(t−1) − |E(G)| ≡0(mod 3).

1. Introduction

A Steiner triple system of order t, denoted STS

(

t

)

, is a pair

(

V

,

B

)

where V is a t-set andBis a collection of 3-element subsets (called triples) of V such that each pair of elements occurs in a unique triple. It is well-known that an STS

(

t

)

exists if and only if t

≡

1 or 3 (mod 6). In terms of graph decompositions, an STS

(

t

)

can also be viewed as a partition of the edges of Kt, each element of which induces a triangle C3; we denote such a decomposition by C3

|

Kt.

A packing of a graph T with triangles is a partition of the edge set of a subgraph G of T , each element of which induces a triangle; the remainder graph of this packing, also known as the leave, is the subgraph T

−

G formed from T by removing

the edges in G. If the leave is minimum in size (that is, has the least number of edges among all possible leaves of T ), then the packing is called a maximum packing. The following result is well-known.

Theorem 1.1 ([5]). The leaves G for any maximum packing of Ktwith triangles are as follows:

t

(

mod 6

)

0 1 2 3 4 5

G F

∅

F

∅

F1 C4

F is a 1-factor, F1is an odd spanning forest with₂t

+

1 edges (tripole), and C4is a cycle of length 4.

It is natural to ask for which subgraphs G of Kt, C3

|

(

Kt

−

G

)

. When t is odd and G is a 2-regular graph, the following result

has been obtained by Colbourn and Rosa [2].

Theorem 1.2 ([2]). Let t be an odd positive integer. Let G be a 2-regular subgraph of Kt. If t

=

9, then suppose that G

6=

C4

∪

C5.

Then C3

|

(

Kt

−

G

)

if and only if the number of edges in Kt

−

G is a multiple of 3.

∗_{Corresponding author.}

(2)

In this paper, we consider the corresponding problem about packings of directed graphs.

A Mendelsohn triple system of order t, denoted MTS

(

t

)

, is a pair

(

V

,

B

)

where V is a t-set,Bis a collection of cyclically ordered 3-subsets of V (called Mendelsohn triples) such that each ordered pair of V appears in exactly one Mendelsohn triple ofB. In terms of graph decomposition, the existence of an MTS

(

t

)

is equivalent to partitioning the directed edges (or edges in short) of Dtinto a collection of directed 3-cycles. We denote such a decomposition by

*

C3

|

Dt. It is well-known that an

MTS

(

t

)

exists if and only if t

≡

0

,

1 (mod 3), t

6=

6 [1].

In this paper, we shall extend the work ofTheorem 1.2to directed graphs. We consider packings of Dtwith Mendelsohn

triples and prove the following result.

Theorem 1.3. Let t be a positive integer. Let G be a vertex-disjoint union of directed cycles in the complete directed graph Dt, and

suppose G

6=

*C2

∪

* C3or * C5if t

=

5, and G

6=

* C3

∪

* C3if t

=

6. Then *

C3

|

(

Dt

−

G

)

if and only if t

(

t

−

1

)−|

E

(

G

)| ≡

0

(

mod 3

)

.

We can get 2Ktfrom Dtwhere 2Ktis the multigraph in which each pair of vertices is joined by exactly two edges. Thus

we haveTheorem 1.4.

Theorem 1.4. Let t be a positive integer. Let G be a vertex-disjoint union of cycles in 2Kt, and suppose G

6=

C2

∪

C3if t

=

5, and

G

6=

C3

∪

C3if t

=

6. Then C3

|

(

2Kt

−

G

)

if and only if t

(

t

−

1

) − |

E

(

G

)| ≡

0

(

mod 3

)

.

Note here that the exceptional case 2K5

−

C5can be obtained by direct construction and the other two cases remain

impossible.

2. Preliminaries

In this section, we will give some notation, symbols and lemmas which are useful in the proof of our main theorem. Let Dtbe a complete directed graph of order t containing no loops so that for each vertex

v

in Dt

,

deg+

(v) =

deg−

(v) =

t

−

1. Let DV1 denote the directed complete subgraphs of Dt induced by V1 where V1

⊆

V

(

Dt

)

. Let V1 be an m-set,

V2 be an n-set and V1

∩

V2

= ∅

. A complete bipartite directed graph DV1,V2

(

Dm,n

)

contains 2mn directed edges; e.g.

D3,2

= {

aibj

,

bjai

|

1

≤

i

≤

3

,

1

≤

j

≤

2

}

which has 12 edges. (For convenience, we call them edges instead of directed

edges.)

The join of two directed graphs G1and G2is denoted by G1

∨

G2where G1and G2are vertex-disjoint. Then E

(

G1

∨

G2

) =

E

(

G1

) ∪

E

(

G2

) ∪

E

(

D|V(G1)|,|V(G2)|). For convenience, we use G1

+

G2to denote E

(

G1

) ∪

E

(

G2

)

. Obviously, Dm+n

=

Dm

∨

Dn

=

Dm

+

Dn

+

Dm,n. In the following, we denote a directed cycle of length l by *

Cl.

Definition 2.1. Let t be a positive integer. If H is a spanning subgraph of G, then H is a t-factor of G if deg

(v) =

t for each

vertex

v

in H.

Definition 2.2. Let H be a 2-factor of G. If G is oriented such that for each vertex

v

in H

,

deg+

(v) =

deg−

(v) =

1, then H is a directed 2-factor of G.

Lemma 2.1. Let C be a directed 2-factor. Then*C3

|

(

C

∨

K1

)

where V

(

C

) ∩

V

(

K1

) = ∅

.

The lemma can be deduced from the following example immediately.

Example 1. Let V1

=

Z5, V2

= {∞}

and

*

C5

=

(

0

,

1

,

2

,

3

,

4

)

. The graph T contains a single point

∞

. Then T

∨

*

C5

=

DV1,V2

+

*

C5

= {

(

0

,

1

, ∞), (

1

,

2

, ∞), (

2

,

3

, ∞), (

3

,

4

, ∞), (

4

,

0

, ∞)}

.

Theorem 2.1 ([6]). Let G be a t-regular graph with t an even number. Then G can be decomposed into₂t 2-factors.

An analog of Petersen’s 2-factor theorem (Theorem 2.1), in the case of directed graphs, is also needed in our proof. For clarity, we present a proof here.

Theorem 2.2. Let G be a t-regular digraph, i.e., deg+

(v) =

deg−

(v) =

t for any

v ∈

V

(

G

)

. Then G can be decomposed into t directed 2-factors (or t 1-regular directed spanning subgraphs).

Proof. Construct a bipartite graph H

=

(

V1

,

V2

)

from G by letting V1

=

V2

=

V

(

G

)

and a vertex a of V1is adjacent to a

vertex b of V2if and only if

(

a

,

b

)

is an arc in G. By assumption, we conclude that H is t-regular and thus by Hall’s condition

H has a perfect matching M

= {

a1b1

,

a2b2

, . . . ,

anbn

}

where n

= |

V

(

G

)|

.

From M, we obtain a 1-regular directed spanning subgraph

(

a1

,

b1

,

ai1

,

bi1

,

ai2

,

bi2

, . . .)

of G which is the desired

subgraph. Then, the proof follows by considering the perfect matchings one at a time (t of them).

Since we shall use induction on the order t to prove our main results, the following lemmas which show the direct construction for small orders are essential. We start at t

=

5 since the cases when t

≤

4 are easy to be seen.

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Lemma 2.2. If G

=

*C2

∪

* C3or * C5, then D5

−

G has no * C3-decomposition.

Proof. The first case is easy to see. Now, if G

=

*C5, then, without loss of generality, we may let

*

C5

=

(

0

,

1

,

2

,

3

,

4

)

where

V

(

D5

) =

Z5. Therefore, there are 15 edges in D5

−

*

C5which includes all

(

a

,

b

)

with a

,

b

∈

Z5and a

−

b

≡

2 or 3 (mod

5). Since these 10 edges themselves cannot form a directed 3-cycle, we shall need one edge of difference 4 to combine with two of the 10 edges with difference 2 or 3. But, this is impossible by direct checking.

Lemma 2.3. If G

=

*_C₃

∪

*_C₃, then D6

−

G has no

*

C3-decomposition.

Proof. Let the two directed 3-cycles be defined on A

= {

a

,

b

,

c

}

and B

= {

d

,

e

,

f

}

respectively. Then there are 18 edges in

DA,Bwhich cannot form a directed 3-cycle themselves. Since there are only 6 edges left in A and B, no *

C3-decompositions

can be obtained for D6

−

G.

Lemma 2.4. Suppose G

6=

*C3

∪

*

C3if t

=

6. For 6

≤

t

≤

12, if G contains a

*

C2(or D2), then Dt

−

G has a *

C3-decomposition.

Proof. The proof follows by adding directed 3-cycles to G

−

*C2defined on V

(

Dt

)\

V

(

*

C2

)

to obtain a

(

t

−

5

)

-regular subgraph

H of Dt−2(Dt−2

−

H is 2-regular) and then applyTheorem 2.2.

Note that the number of directed 3-cycles (denoted by N) we add can be seen in the following table. Since they are easy to calculate, we omit the details.

t 6 7 8 9 10 11 12

N 0 2 4 7 10 15 20

Lemma 2.5. For 7

≤

t

≤

12, if G contains a directed 3-cycle, then Dt

−

G has a *

C3-decomposition.

Proof. The idea is similar to the proof of Lemma 2.4for 8

≤

t

≤

12. Instead of making the subgraph H

(

t

−

5

)

-regular, we need a

(

t

−

7

)

-regular graph H. Therefore, for 8

≤

t

≤

12, we have a similar proof. For completeness, we include a*C3-decomposition of D7

−

(

*

C3

∪

*

C3

)

in what follows. Let V

(

D7

) = {

a

,

b

,

c

,

d

,

e

,

f

,

g

}

and let the two

directed 3-cycles be defined on

{

a

,

b

,

c

}

and

{

d

,

e

,

f

}

respectively. Then D7

−

({(

a

,

b

,

c

)} ∪ {(

d

,

e

,

f

)}) = {(

a

,

e

,

b

),

(

b

,

f

,

c

), (

c

,

d

,

a

), (

a

,

f

,

e

), (

b

,

d

,

f

), (

c

,

e

,

d

), (

g

,

a

,

d

), (

g

,

f

,

a

), (

g

,

c

,

f

), (

g

,

e

,

c

), (

b

,

e

,

g

), (

d

,

b

,

g

)}

.

Lemma 2.6. There exist*C3-decompositions for D3

−

* C3, D6

−

* C6, D7

−

* C6, D8

−

* C8, D9

−

* C9, D11

−

* C11, D12

−

(

* C6

∪

* C6

)

, and D12

−

* C12.

Proof. We give the proof by direct construction which can be found in theAppendix.

Lemma 2.7. Let G contain two directed cycles with one of them*C4or

*

C5. Then, for t

=

8

,

9

,

10

,

11

,

12, the graph Dt

−

G has

a*C3-decomposition if and only if t

(

t

−

1

) − |

E

(

G

)| ≡

0

(

mod 3

)

.

Proof. The proof can also be found in theAppendix.

Note that if we need the case t

=

13 in our main proof, we can decompose D13

−

G into

*

C3-decomposition by finding

four vertex-disjoint 3-cycles in B0which is the set of directed 3-cycles of D13

−

G.

For example, let G

=

*C5

∪

*

C7. Let the set of directed 3-cycles be B00

= {

(

e

,

i

,

l

), (

h

,

c

,

f

), (

j

,

d

,

a

), (

b

,

k

,

g

)} ⊆

B0. Let a new

vertex be m, then D13

−

G

=

B0

−

B00

+{

(

m

,

e

,

i

), (

m

,

i

,

l

), (

m

,

l

,

e

), (

m

,

h

,

c

), (

m

,

c

,

f

), (

m

,

f

,

h

), (

m

,

j

,

d

), (

m

,

d

,

a

), (

m

,

a

,

j

),

(

m

,

b

,

k

), (

m

,

k

,

g

), (

m

,

g

,

b

)}

.

On the other hand, if G contains a D2, then the proof follows by a similar idea as inLemma 2.4.

With the above preparations, we are now in a position to prove our main results. For readers’ convenience, we start with a special case.

Lemma 2.8. Let t be a positive integer such that t

≡

4

(

mod 6

)

. Let G be a 1-regular directed subgraph of Dt such that

t

(

t

−

1

) − |

E

(

G

)|

is a multiple of 3 and*C2is not a component of G. Then Dt

−

G can be decomposed into directed 3-cycles. Proof. For t

=

4 and t

=

10, the proof follows by direct constructions. Assume that t

≥

16. By counting,

|

E

(

G

)| ≡

0

(

mod 3

)

. For convenience, let t

=

6k

+

4, V

(

Dt

) = {v

i

|

i

∈

Z6k+4

}

and V

(

G

) = {v

i

|

3l

+

1

≤

i

≤

6k

+

3

}

. Now, let

e

G

=

G

+

B₀where

B0

= {

(v

1

, v

2

, v

3

), (v

4

, v

5

, v

6

), . . . , (v

3l−2

, v

3l−1

, v

3l

)}

. Let

e

G0be the graph obtained by reversing the direction of the arcs on

e

G. Then, the following results are easy to see:

(i) ByTheorem 1.2, Dt−1

−

(

e

G

+

e

G0

)

has a *

C3-decomposition where Dt−1is defined on V

(

Dt

) \ {v

0

}

;

(ii) byLemma 2.1,

e

G0

∨ {

v

₀

}

has a *

(4)

Let the collection of directed 3-cycles obtained from (i) and (ii) be denoted by B1and B2respectively. Then Dt

−

G can be

decomposed into the directed 3-cycle collection B0

∪

B1

∪

B2.

To simplify arguments, we also use the following equalities to see the process of obtaining the desired decomposition.

Dt

−

G

=

Dt

−

(

G

+

B0

) +

B0

=

Dt

−

e

G

+

B0

=

Dt−1

+

DV(Dt)\{v0},{v0}

−

e

G

+

B0

=

Dt−1

−

e

G

−

e

G 0

₊

₍

DV(Dt)\{v0},{v0}

+

e

G 0

_{) +}

B0

=

B1

∪

B2

∪

B0 where B1

=

Dt−1

−

e

G

−

e

G 0 and B2

=

DV(Dt)\{v0},{v0}

+

e

G 0

_.

In order to prove our main theorem, we first consider the case when G is a directed cycle of size very close to t, i.e., either

t or t

−

1.

Theorem 2.3. For each positive integer t,*C3

|

(

Dt

−

*

Ct

)

when t

≡

0

,

2

(

mod 3

)

and *

C3

|

(

Dt

−

*

Ct−1

)

when t

≡

1

(

mod 3

)

.

Proof. We give the proof by induction. For small cases, the proof can be found inLemma 2.6. Assuming inductively that the assertion is true for values less than t, we shall prove the assertion is true for t. Note here that when t

≡

4 (mod 6), we can obtain the result byLemma 2.8independently. The proof can be divided into two cases.

Case 1. t

≡

0

,

2

(

mod 3

)

.

Case (1.1). t

≡

3

(

mod 6

)

.

For t

≡

3

(

mod 6

)

, let t

=

6k

+

3 and let V

(

Dt

) =

A

∪

B where A

= {∞

i

|

i

∈

Z3k

}

and B

= {

i

|

i

∈

Z3k+3

}

.

Furthermore, let*Ct

=

(∞

0

, ∞

1

, . . . , ∞

3k−1

,

0

,

1

, . . . ,

3k

+

2

)

. Now, by adding

α = {(

k

+

2

, ∞

3k−1

, ∞

0

), (∞

0

,

3k

+

2

,

k

+

2

), (∞

3k−1

,

k

+

2

,

0

), (

0

,

k

+

2

,

3k

+

2

)}

to

*

Ctwhere

α

is a set of four Mendelsohn triples, we obtain * Ct

+

α =

* C (1)

+

*_C (2)

+

β + {(

k

+

2

,

3k

+

2

), (

0

,

k

+

2

)}

where*C (1)

=

(∞

₀

, ∞

₁

, . . . , ∞

_3k−1

)

, * C (2)

=

(

0

,

1

, . . . ,

3k

+

2

)

and

β = {(∞

0

,

3k

+

2

), (∞

0

,

k

+

2

), (∞

3k−1

,

k

+

2

), (∞

3k−1

,

0

)}

. First, considering Dt

=

DA

+

DB

+

DA,B, we can easily get

DA

−

*

C

(1)

which has a*C3-decomposition by induction. Second, since DA,B

=

DA\{∞0,∞3k−1},B

+

D{∞0},B

+

D{∞3k−1},B, we

associate it with

β + {(

k

+

2

,

3k

+

2

), (

0

,

k

+

2

)}

to obtain DA\{∞0,∞3k−1},B

+ {

D{∞0},B\{k+2,3k+2}

+ [

f1

−

(

k

+

2

,

3k

+

2

)]}

I

₊

{

D{∞3k−1},B\{k+2,0}

+ [

f2

−

(

k

+

2

,

0

)]}

II

₋

_f

1

−

f2where f1and f2are two different, edge-disjoint directed 2-factors of DB

−

*

C

(2)

such that f1contains a 2-cycle

(

k

+

2

,

3k

+

2

)

and f2contains a 2-cycle

(

k

+

2

,

0

)

. In addition, f1and f2do not contain any edges

of difference k

+

1. ByLemma 2.1, there exist*C3-decompositions for (I) and (II). Finally,

(

DB

−

*

C

(2)

−

f1

−

f2

)

III

+

DA\{∞0,∞3k−1},B

is left. Since the edges of difference k

+

1 is not used, let F1

= {

(

i

,

i

+

k

+

1

,

i

+

2k

+

2

) |

0

≤

i

≤

k

}

. Then

(

DB

−

*

C

(2)

−

f1

−

f2

)−

F1

is a

(

3k

−

2

)

-regular directed graph and can be decomposed into a set of

(

3k

−

2

)

directed 2-factors, say H1

, . . . ,

H3k−2by

applyingTheorem 2.2. For each i

=

1

,

2

, . . . ,

3k

−

2, Hi

+

D∞i,Bcan be decomposed into a set of directed 3-cycles (denoted

by T₂i) byLemma 2.1. Thus,

(

DB

−

* C (2)

−

f1

−

f2

) +

DA\{∞0,∞3k−1},B

=

F1

+

S

3k−2 i=1 T i

2. We give an example inFig. 1.

For clarity, the*C3-decomposition of Dt

−

*

Ctcan be obtained by the following steps.

Dt

−

* Ct

=

Dt

−

(

* Ct

+

α) + α

=

Dt

−

(

* C (1)

+

*C (2)

+

β + {(

k

+

2

,

3k

+

2

), (

0

,

k

+

2

)}) + α

=

DA

+

DB

+

DA\{∞0,∞3k−1},B

+

D{∞0},B\{k+2,3k+2}

+

D{∞3k−1},B\{k+2,0}

+ {

(∞

0

,

3k

+

2

), (∞

0

,

k

+

2

)}

+ {

(∞

_3k−1

,

k

+

2

), (∞

3k−1

,

0

)} − (

* C (1)

+

*C (2)

+

β + {(

k

+

2

,

3k

+

2

), (

0

,

k

+

2

)}) + α

=

(

DA

−

* C (1)

) + [(

DB

−

* C (2)

−

f1

−

f2

) +

DA\{∞0,∞3k−1},B

]

+ {

D{∞0},B\{k+2,3k+2}

+ [

f1

−

(

k

+

2

,

3k

+

2

)]} + {

D{∞3k−1},B\{k+2,0}

+ [

f2

−

(

k

+

2

,

0

)]} + α.

Case (1.2). t

≡

0

(

mod 6

)

.

For t

≡

0

(

mod 6

)

, let t

=

6k and V

(

Dt

) =

A

∪

B where A

= {∞

i

|

i

∈

Z3k−1

}

and B

= {

i

|

i

∈

Z3k+1

}

. Furthermore,

let*Ct

=

(∞

0

, ∞

1

, . . . , ∞

3k−2

,

0

,

1

, . . . ,

3k

)

. Now, by adding

α = {(

k

+

2

, ∞

3k−2

, ∞

0

)(∞

0

,

3k

,

k

+

2

), (∞

3k−2

,

k

+

2

,

0

), (

0

,

k

+

2

,

3k

)}

to*Ct where

α

+

α =

* C (1)

+

*C (2)

+

β +

{

(

k

+

2

,

3k

), (

0

,

k

+

2

)}

where*C (1)

=

(∞

₀

, ∞

₁

, . . . , ∞

_3k−2

)

, * C (2)

=

(

0

,

1

, . . . ,

3k

)

and

β = {(∞

0

,

3k

), (∞

0

,

k

+

2

),

(∞

3k−2

,

k

+

2

), (∞

3k−2

,

0

)}

. Moreover, let f1and f2be two different, edge-disjoint directed 2-factors of DB

−

*

C

(2)

(5)

Fig. 1. t=21 for k=3.

f1contains a 2-cycle

(

k

+

2

,

3k

)

(

k

+

2

,

0

)

. Then, the main steps of

* C3-decomposition of Dt

−

* Ct are as follows. Dt

−

* Ct

=

α + (

DA

−

* C (1)

) + [(

DB

−

* C (2)

−

f1

−

f2

) +

DA\{∞0,∞3k−2},B

]

I

+ {

D{∞0},B\{k+2,3k}

+ [

f1

−

(

k

+

2

,

3k

)]}

II

_{+ {}

_D {∞3k−2},B\{k+2,0}

+ [

f2

−

(

k

+

2

,

0

)]}

III

_.

Now, by induction, DA

−

* C (1)

has a*C3-decomposition and the last two parts (II and III) also have

*

C3-decompositions (by Lemma 2.1), so it is left to show that (I) has a*C3-decomposition.

(

DB

−

*

C

(2)

−

f1

−

f2

)

is a

(

3k

−

3

)

-regular directed graph

and by applyingTheorem 2.2andLemma 2.1,*C3

|

(

DB

−

*

C

(2)

−

f1

−

f2

+

DA\{∞0,∞3k−2},B

)

.

Case (1.3). t

≡

2

(

mod 6

)

.

For t

≡

2 (mod 6), let t

=

6k

+

2 and V

(

Dt

) =

A

∪

B where A

= {∞

i

|

i

∈

Z3k

}

and B

= {

i

|

i

∈

Z3k+2

}

. Furthermore,

let*Ct

=

(∞

0

, ∞

1

, . . . , ∞

3k−1

,

0

,

1

, . . . ,

3k

+

1

)

. Now, by adding

α = {(

k

+

2

, ∞

3k−1

, ∞

0

), (∞

0

,

3k

+

1

,

k

+

2

),

(∞

3k−1

,

k

+

2

,

0

), (

0

,

k

+

2

,

3k

+

1

)}

to

*

Ct where

α

+

α =

* C (1)

+

* C (2)

+

β + {(

k

+

2

,

3k

+

1

), (

0

,

k

+

2

)}

where *C (1)

=

(∞

₀

, ∞

₁

, . . . , ∞

_3k−1

)

, * C (2)

=

(

0

,

1

, . . . ,

3k

+

1

)

and

β =

{

(∞

₀

,

3k

+

1

), (∞

0

,

k

+

2

), (∞

3k−1

,

k

+

2

), (∞

3k−1

,

0

)}

. Moreover, let f1and f2be two different, edge-disjoint directed

2-factors of DB

−

*

C

(2)

where f1contains a 2-cycle

(

k

+

2

,

3k

+

1

)

(

k

+

2

,

0

)

. Then, the

*

C3-decomposition

of Dt

−

*

Dt

−

* Ct

=

α + (

DA

−

* C (1)

) + [(

DB

−

* C (2)

−

f1

−

f2

) +

DA\{∞0,∞3k−1},B

]

I

+ {

D{∞0},B\{k+2,3k+1}

+ [

f1

−

(

k

+

2

,

3k

+

1

)]}

II

_{+ {}

_D {∞3k−1},B\{k+2,0}

+ [

f2

−

(

k

+

2

,

0

)]}

III

_.

−

* C (1)

has a*C3-decomposition and the last two parts: (II) and (III) also have

*

C3-decompositions

(byLemma 2.1), it is left to show that the part (I) has a*C3-decomposition.

(

DB

−

*

C

(2)

−

f1

−

f2

)

is a

(

3k

−

2

)

-regular directed

graph and by applyingTheorem 2.2andLemma 2.1,*C3

|

(

DB

−

*

C

(2)

−

f1

−

f2

+

DA\{∞0,∞3k−1},B

)

.

Case (1.4). t

≡

5

(

mod 6

)

.

For t

≡

5 (mod 6), let t

=

6k

+

5 and let V

(

Dt

) =

A

∪

B where A

= {∞

i

|

i

∈

Z3k

}

and B

= {

i

|

i

∈

Z3k+5

}

. Furthermore,

let*Ct

=

(∞

0

, ∞

1

, . . . , ∞

3k−1

,

0

,

1

, . . . ,

3k

+

4

)

. Now, by adding

α = {(

k

+

2

, ∞

3k−1

, ∞

0

), (∞

0

,

3k

+

4

,

k

+

2

),

(∞

3k−1

,

k

+

2

,

0

), (

0

,

k

+

2

,

3k

+

4

)}

to * Ct, we obtain * Ct

+

α =

* C (1)

+

*C (2)

+

β + {(

k

+

2

,

3k

+

4

), (

0

,

k

+

2

)}

where * C (1)

=

(∞

₀

, ∞

₁

, . . . , ∞

_3k−1

)

, * C (2)

=

(

0

,

1

, . . . ,

3k

+

4

)

and

β = {(∞

0

,

3k

+

4

), (∞

0

,

k

+

2

), (∞

3k−1

,

k

+

2

), (∞

3k−1

,

0

)}

.

Moreover, let f1and f2be two different, edge-disjoint directed 2-factors of DB

−

*

C

(2)

(

k

+

2

,

3k

+

4

)

(

k

+

2

,

0

)

. In addition, f1and f2do not contain any edges of difference 1

,

2

,

3k

+

2. Then, the

*

(6)

-decomposition of Dt

−

*

Dt

−

* Ct

=

α + (

DA

−

* C (1)

) + [(

DB

−

* C (2)

−

f1

−

f2

) +

DA\{∞0,∞3k−1},B

]

I

+ {

D{∞0},B\{k+2,3k+4}

+ [

f1

−

(

k

+

2

,

3k

+

4

)]}

II

_{+ {}

_D {∞3k−1},B\{k+2,0}

+ [

f2

−

(

k

+

2

,

0

)]}

III

_.

−

* C (1)

has a*C3-decomposition and the two parts (II) and (III) also have

*

C3-decompositions by Lemma 2.1. It is left to show that the part (I) has a*C3-decomposition. Since the edges of differences 1

,

2

,

3k

+

2 have not

been used, let F1

= {

(

i

,

i

+

1

,

i

+

3

) |

i

∈

Z3k+5

}

. Then

(

DB

−

*

C

(2)

−

f1

−

f2

)−

F1is a

(

3k

−

2

)

-regular directed graph and when

associated with DA\{∞0,∞3k−1},Bcan be decomposed into a set of directed 3-cycles (denoted by F2) by applyingTheorem 2.2

andLemma 2.1. Thus,

(

DB

−

*

C

(2)

−

f1

−

f2

) +

DA\{∞0,∞3k−1},B

=

F1

+

F2.

Case 2. t

≡

1

(

mod 3

)

.

Since the case t

≡

4

(

mod 6

)

has been settled by Lemma 2.8, it suffices to consider the case t

≡

1

(

mod 6

)

. Let t

=

6k

+

1 and let V

(

Dt

) =

A

∪

B where A

= {∞

i

|

i

∈

Z3k−2

}

and B

= {

i

|

i

∈

Z3k+3

}

. Furthermore, let

* Ct−1

=

(∞

0

, ∞

1

, . . . , ∞

3k−4

,

0

,

1

, . . . ,

3k

+

2

)

. Now, by adding

α = {(

k

+

2

, ∞

3k−4

, ∞

0

), (∞

0

,

3k

+

2

,

k

+

2

), (∞

3k−4

,

k

+

2

,

0

), (

0

,

k

+

2

,

3k

+

2

)}

to*Ct−1, we obtain * Ct−1

+

α =

* C (1)

+

*C (2)

+

β + {(

k

+

2

,

3k

+

2

), (

0

,

k

+

2

)}

where * C (1)

=

(∞

₀

, ∞

₁

, . . . , ∞

_3k−4

)

, * C (2)

=

(

0

,

1

, . . . ,

3k

+

2

)

and

β = {(∞

0

,

3k

+

2

), (∞

0

,

k

+

2

), (∞

3k−4

,

k

+

2

), (∞

3k−4

,

0

)}

.

Moreover, let f1and f2be two different, edge-disjoint directed 2-factors of DB

−

*

C

(2)

(

k

+

2

,

3k

+

2

)

(

k

+

2

,

0

)

. In addition, f1and f2do not contain any edges of difference 1

,

2

,

3k.

Similar to Case (1.1), we can get DA

−

*

C

(1)

. When k is odd, by induction, DA

−

*

C

(1)

has a*C3-decomposition. When k is

even, byLemma 2.8, DA

−

*

C

(1)

has a*C3-decomposition. We also associate DA,Bwith

β + {(

k

+

2

,

3k

+

2

), (

0

,

k

+

2

)}

and get

two parts, one is:

{

D{∞0},B\{k+2,3k+2}

+ [

f1

−

(

k

+

2

,

3k

+

2

)]}

I

_{+ {}

_D

{∞3k−4},B\{k+2,0}

+ [

f2

−

(

k

+

2

,

0

)]}

II

₋

_f

1

−

f2. ByLemma 2.2,

there exist*C3-decompositions for (I) and (II) respectively. Finally, we have

(

DB

−

*

C

(2)

−

f1

−

f2

) +

DA\{∞0,∞3k−4},Bleft. Since

the edges of differences 1

,

2 and 3k have not been used, let F1

= {

(

i

,

i

+

1

,

i

+

3

) |

i

∈

Z3k+3

}

. Then

[

(

DB

−

*

C

(2)

−

f1

−

f2

)−

F1

]

III

is a

(

3k

−

4

)

-regular directed graph. ByLemma 2.1andTheorem 2.2, we can get a set of directed 3-cycles, denoted by F2by

associating it with DA\{∞0,∞3k−4},B. Thus

(

DB

−

*

C

(2)

−

f1

−

f2

) +

DA\{∞0,∞3k−4},B

=

F1

+

F2.

Then, the*C3-decomposition of Dt

−

*

Dt

−

* Ct

=

(

DA

−

* C (1)

) + [(

DB

−

* C (2)

−

f1

−

f2

) +

DA\{∞0,∞3k−4},B

]

+ {

D{∞0},B\{k+2,3k+2}

+ [

f1

−

(

k

+

2

,

3k

+

2

)]} + {

D{∞3k−4},B\{k+2,0}

+ [

f2

−

(

k

+

2

,

0

)]} + α.

This concludes the proof ofTheorem 2.3.

3. Decomposing Dt

−

G

Finally, we will proveTheorem 3.1by induction.

Theorem 3.1. Let t be a positive integer and let G be a vertex-disjoint union of directed cycles in Dt while G

6=

*

Ct when

t

≡

0

,

2

,

3

,

5

(

mod 6

)

and G

6=

*Ct−1when t

≡

1

(

mod 3

)

. Suppose G

6=

* C2

∪

* C3or * C5if t

=

5, and G

6=

* C3

∪

* C3if

t

=

6. Then*C3

|

(

Dt

−

G

)

if and only if t

(

t

−

1

) − |

E

(

G

)| ≡

0

(

mod 3

)

.

Proof. The necessity is obvious. We prove the sufficiency by induction on t. Assuming inductively that the assertion is true

for values less than t, we shall prove that it is true for t. First, we take the following partition of t.

(1) t

≡

0

(

mod 6

),

t

=

6k

=

(

3k

+

1

) + (

3k

−

1

)

; (2) t

≡

3

(

mod 6

),

t

=

6k

+

3

=

(

3k

+

3

) +

3k; (3) t

≡

2

(

mod 6

),

t

=

6k

+

2

=

(

3k

+

3

) + (

3k

−

1

)

; (4) t

≡

4

(

mod 6

),

t

=

6k

+

4

=

(

3k

+

4

) +

3k; (5) t

≡

1

(

mod 6

),

t

=

6k

+

1

=

(

3k

+

3

) + (

3k

−

2

)

; (6) t

≡

5

(

mod 6

),

t

=

6k

+

5

=

(

3k

+

5

) +

3k.

(7)

In the following, we will discuss these cases one by one.

Case (1): t

≡

0 (mod 6).

Let t

=

6k. By counting,

|

E

(

G

)| ≡

0 (mod 3). Let

|

E

(

G

)| <

6k or

|

E

(

G

)| =

6k. For the former, we can add a directed 3-cycle

T to G, where V

(

G

) ∩

V

(

T

) = ∅

. This process can be repeated until

|

E

(

G

)| =

6k. Therefore, it is enough to consider the case

|

E

(

G

)| =

6k.

First, if G has a component G1such that

|

E

(

G1

)| =

3k

−

1, then let G2

=

G

−

G1and

|

E

(

G2

)| =

3k

+

1. We denote V

(

G1

)

by A and V

(

G2

)

by B. Then we have

Dt

−

G

=

DA

+

DB

−

(

G1

+

G2

) +

DA,B

=

(

DA

−

G1

) + (

DB

−

G2

) +

DA,B

.

By induction, DA

−

G1has a

*

C3-decomposition. Then DB

−

G2is a

(

3k

−

1

)

-regular directed graph and byTheorem 2.2

andLemma 2.1,*C3

|

(

DB

−

G2

+

DB,A

)

.

Second, if we cannot get G1and G2satisfying the condition of the first case, we can rearrange our leave by adding three

directed 3-cycles to get G∗₁and G∗₂satisfying

|

G∗₁

| =

3k

−

1 and

|

G∗₂

| =

3k

+

1. Even if all the components of G have cardinality 0 (mod 3), we can also rearrange our leave by adding three directed 3-cycles to get G∗

1and G ∗ 2. Suppose min

{

3k

−

1

− |

E

(S

* C∈G * C

)|} =

l

>

0, then let G1

=

S

* C∈G *

C where 3k

−

1

− |

E

(

G1

)| =

l. Furthermore, let

G2

=

G

−

G1. In the following, we will choose a directed cycle

*

C from G2and divide

*

C into two parts: G∗₁and G∗

2such that

|

G∗₁

| =

3k

−

1 and

|

G∗₂

| =

3k

+

1. The details are as follows. For any*C

∈

G2,

|

E

(

*

C

)| >

l holds. Otherwise, if there exists one cycle denoted by*C

0

∈

G2and

|

E

(

* C 0

)| ≤

l, we can get 3k

−

1

− |

E

(

G1

S

* C 0

)| <

l, a contradiction to the construction of G1. Choose

*

C

∈

G2where

*

C

=

(

x1

,

x2

, . . . ,

xj

)

and

j

≥

l

+

1. If l

=

1, we can choose j

≥

4. Let x0

∈

V

(

G2

−

* C

)

and

α = {(

x0

,

xl+1

,

xl

), (

x1

,

x0

,

xl

), (

xj

,

x0

,

x1

)}

. Then we have * C

+

α = β +

*Cl

+

* Cj−l+1 where

β = {(

x1

,

xj

), (

x1

,

x0

), (

xl

,

xl+1

), (

xl

,

x0

)},

* Cl

=

(

x1

,

x2

, . . . ,

xl

)

and * Cj−l+1

=

(

x0

,

xl+1

, . . . ,

xj

).

* Cj−l+1

=

(

x0

,

xl+1

, . . . ,

xj

) = (

xl+1

,

xl+2

, . . . ,

xj

) −

xjxl+1

+

x0xl+1

+

xjx0

=

*Cj−l

−

xjxl+1

+

x0xl+1

+

xjx0 where * Cj−l

=

(

xl+1

,

xl+2

, . . . ,

xj

).

Let G∗ 1

=

G1

+

* Cland G∗2

=

G2

−

* C

+

*Cj−l. Obviously,

|

G∗1

| =

3k

−

1 and

|

G ∗ 2

| =

3k

+

1. We denote V

(

G ∗ 1

)

by A and V

(

G ∗ 2

)

by B. Then we have Dt

−

G

=

DB

+

DA

+

DA,B

+

α − [

G1

+

(

G2

−

* C

) + (

*C

+

α)]

=

DB

+

DA

+

DA,B

− [

G1

+

(

G2

−

* C

) + β +

*Cl

+

* Cj−l+1

] +

α

= {

DB

− [

(

G2

−

* C

+

*Cj−l

) −

xjxl+1

+

x0xl+1

+

xjx0

]} + [

DA

−

(

G1

+

* Cl

)] + (

DA,B

−

β) + α

= {

DB

−

G∗2

+

xjxl+1

−

x0xl+1

−

xjx0

− [

f1

−

(

xj

,

x0

)] − [

f2

−

(

xl+1

,

x0

)]} + (

DA

−

G∗1

)

+ {

D{x1},B\{xj,x0}

+ [

f1

−

(

xj

,

x0

)]} + {

D{xl},B\{xl+1,x0}

+ [

f2

−

(

xl+1

,

x0

)]} +

DA\{x1,xl},B

+

α

= {[

DB

−

G∗2

−

f1

−

f2

+

(

x0

,

xj

,

xl+1

)] +

DA\{x1,xl},B

} +

(

DA

−

G ∗ 1

)

+ {

D{x1},B\{xj,x0}

+ [

f1

−

(

xj

,

x0

)]} + {

D{xl},B\{xl+1,x0}

+ [

f2

−

(

xl+1

,

x0

)]} + α.

Note that f1and f2are two different, edge-disjoint directed 2-factors where f1contains

(

xj

,

x0

)

and f2contains

(

xl+1

,

x0

)

.

Moreover, f1and f2are defined on B.

−

G∗1has a

*

C3-decomposition. ByLemma 2.1,

{

D{x1},B\{xj,x0}

+ [

f1

−

(

xj

,

x0

)]}

and

{

D{xl},B\{xl+1,x0}

+

[

f2

−

(

xl+1

,

x0

)]}

have

*

C3-decompositions. It is left to show that

(

DB

−

G∗2

−

f1

−

f2

) +

DA\{x1,xl},Bhas a

*

C3-decomposition.

DB

−

G∗2

−

f1

−

f2is a directed

(

3k

−

3

)

-regular graph and byTheorem 2.2andLemma 2.1,

*

C3

|

(

DB

−

G∗2

−

f1

−

f2

+

DA\{x1,xl},B

)

.

We give an example inFig. 2as follows. The upper ovals represent cycles in G2, the lower ovals represent the cycles in

G1andFig. 2shows how to get G∗1and G

∗

2.

Case (2). t

≡

3 (mod 6).

Let t

=

6k

+

3, by counting,

|

E

(

G

)| ≡

0 (mod 3). Similar to Case 1, we only consider the case

|

E

(

G

)| =

6k

+

3.

|

E

(

G1

)| =

3k, G2

=

G

−

G1and

|

E

(

G2

)| =

3k

+

3. We denote V

(

G1

)

by A and

V

(

G2

)

by B. Then we have

(8)

Fig. 2. t=6k.

By induction, DA

−

G1has a

*

C3- decomposition. The difference triple

(

k

+

1

,

k

+

1

,

k

+

1

)

(short orbit) whose corresponding

set of directed 3-cycles is F1

= {

(

i

,

i

+

k

+

1

,

i

+

2k

+

2

) |

0

≤

i

≤

k

}

from DB

−

G2. Then DB

−

G2

−

F1is a 3k-regular

directed graph and byTheorem 2.2andLemma 2.1,*C3

|

(

DB

−

G2

−

F1

+

DB,A

)

.

Second, if we cannot get G1and G2satisfying the condition of the first case, we can rearrange our leave by adding three

directed 3-cycles, as follows, to get G∗

1and G ∗ 2satisfying

|

G ∗ 1

| =

3k and

|

G ∗ 2

| =

3k

+

3. Suppose min

{

3k

− |

E

(S

* C∈G * C

)|} =

l

>

0, then let G1

=

S

* C∈G *

C where 3k

− |

E

(

G1

)| =

l. Further, let G2

=

G

−

G1. In

the following, we will choose a directed cycle*C from G2and divide

*

C into two parts so that we can get G∗₁and G∗₂,

|

G∗₁

| =

3k and

|

G∗₂

| =

3k

+

3. The detail can be found as follows.

For any*C

∈

G2,

|

E

(

*

C

)| >

l holds. Otherwise, if there exists one cycle denoted by*C

0

∈

G2and

|

E

(

* C 0

)| ≤

l, we can get 3k

− |

E

(

G1

S

* C 0

)| <

l, a contradiction to the construction of G1. Choose

* C

∈

G2where C

=

(

x1

,

x2

, . . . ,

xj

)

and j

≥

l

+

1. Let x0

∈

V

(

G2

−

* C

)

and

α = {(

x0

,

xl+1

,

xl

), (

x1

,

x0

,

xl

), (

xj

,

x0

,

x1

)}

. Then we have * C

+

α = β +

*Cl

+

* Cj−l+1 where

β = {(

x1

,

xj

), (

x1

,

x0

), (

xl

,

xl+1

), (

xl

,

x0

)},

* Cl

=

(

x1

,

x2

, . . . ,

xl

)

and * Cj−l+1

=

(

x0

,

xl+1

, . . . ,

xj

).

* Cj−l+1

=

(

x0

,

xl+1

, . . . ,

xj

) = (

xl+1

,

xl+2

, . . . ,

xj

) −

xjxl+1

+

x0xl+1

+

xjx0

=

*Cj−l

−

xjxl+1

+

x0xl+1

+

xjx0 where * Cj−l

=

(

xl+1

,

xl+2

, . . . ,

xj

).

Let G∗ 1

=

G1

+

* Cland G∗2

=

G2

−

* C

+

*Cj−l. Obviously,

|

V

(

G∗1

)| =

3k and

|

V

(

G ∗ 2

)| =

3k

+

3. We denote V

(

G ∗ 1

)

by A and V

(

G∗₂

)

by B.

Similar to Case 1, we have

Dt

−

G

= {[

DB

−

G∗2

−

f1

−

f2

+

(

x0

,

xj

,

xl+1

)] +

DA\{x1,xl},B

} +

(

DA

−

G

∗

1

)

+ {

D{x1},B\{xj,x0}

+ [

f1

−

(

xj

,

x0

)]} + {

D{xl},B\{xl+1,x0}

+ [

f2

−

(

xl+1

,

x0

)]} + α.

Note f1and f2 are two different, edge-disjoint directed 2-factors where f1 contains

(

xj

,

x0

)

and f2 contains

(

xl+1

,

x0

)

.

Moreover, f1and f2are defined on B.

−

G∗1has a

*

C3-decomposition. ByLemma 2.1,

{

D{x1},B\{xj,x0}

+ [

f1

−

(

xj

,

x0

)]}

and

{

D{xl},B\{xl+1,x0}

+

[

f2

−

(

xl+1

,

x0

)]}

have a

*

C3-decomposition.

It is left to show

(

DB

−

G∗2

−

f1

−

f2

) +

DA\{x1,xl},Bhas a

*

C3-decomposition. The difference triple

(

k

+

1

,

k

+

1

,

k

+

1

)

whose

corresponding set of directed 3-cycles is F1

= {

(

i

,

i

+

k

+

1

,

i

+

2k

+

2

) |

0

≤

i

≤

k

}

can be obtained from DB

−

G∗2

−

f1

−

f2.

Then DB

−

G∗2

−

f1

−

f2

−

F1is a

(

3k

−

2

)

-regular directed graph. We associate it with DA\{x1,xl},Bto get a set of directed

3-cycles (denoted by F2) by applyingTheorem 2.2andLemma 2.1. Thus, DB

−

G∗2

−

f1

−

f2

+

DA\{x1,xl},B

=

F1

+

F2.

Case (3). t

≡

2 (mod 6).

By counting,

|

E

(

G

)| ≡

2 (mod 3), let t

=

6k

+

2. Similar to Case 1, we only consider the case

|

E

(

G

)| =

6k

+

2. First, if G has a component G1such that

|

E

(

G1

)| =

3k

−

1, G2

=

G

−

G1and

|

E

(

G2

)| =

3k

+

3.

(9)

Then we have

Dt

−

G

=

(

DA

−

G1

) + (

DB

−

G2

) +

DB,A

.

By induction, DA

−

G1has a

*

C3- decomposition. The difference triples

(

k

+

1

,

k

+

1

,

k

+

1

)

and

(

2k

+

2

,

2k

+

2

,

2k

+

2

)

from DB

−

G2can form a set of directed 3-cycles F1

= {

(

i

,

i

+

k

+

1

,

i

+

2k

+

2

), (

i

,

i

+

2k

+

2

,

i

+

4k

+

4

) |

0

≤

i

≤

k

}

. Then

DB

−

G2

−

F1is a

(

3k

−

1

)

-regular directed graph and when associated with DB,Acan be decomposed into a set of directed

3-cycles (denoted by F2) by applyingTheorem 2.2andLemma 2.1. Thus, DB

−

G2

+

DB,A

=

F1

+

F2.

Second, suppose min

{

3k

−

1

− |

E

(S

*

C∈G *

C

)|} =

l

>

0, then let G1

=

S

*

C∈G *

C where 3k

−

1

− |

E

(

G1

)| =

l. The remainder

of the proof of this case is similar Case 2.

Case (4). t

≡

4

(

mod 6

)

.

By counting

|

E

(

G

)| ≡

0

(

mod 3

)

, let t

=

6k

+

4. Similarly, as that in Case 1, we only consider the case

|

E

(

G

)| =

6k

+

3. First, if G has a component G1such that

|

E

(

G1

)| =

3k, G2

=

G

−

G1,

|

E

(

G2

)| =

3k

+

3. We denote V

(

G2

)

by B and

V

(

D6k+4

) \

V

(

G2

)

by A. Note

|

V

(

DA

)| =

3k

+

1.

Then we have

Dt

−

G

=

(

DA

−

G1

) + (

DB

−

G2

) +

DB,A

.

By induction, DA

−

G1has a

*

C3- decomposition. The difference triples

(

k

+

1

,

k

+

1

,

k

+

1

)

and

(

2k

+

2

,

2k

+

2

,

2k

+

2

)

(short orbit) from DB

−

G2can form a collection of directed 3-cycles F1

= {

(

i

,

i

+

k

+

1

,

i

+

2k

+

2

), (

i

,

i

+

2k

+

2

,

i

+

4k

+

4

) |

i

=

0

,

1

, . . . ,

k

}

. Then DB

−

G2

−

F1is a 3k-regular directed graph and when associated with DB,Acan be decomposed into a

set of directed 3-cycles (denoted by F2) by applyingTheorem 2.2andLemma 2.1. Thus, DB

−

G2

+

DB,A

=

F1

+

F2.

Second, suppose min

{

3k

− |

E

(S

*

C∈G * C

)|} =

l

>

0, then let G1

=

S

* C∈G *

C where 3k

− |

E

(

G1

)| =

l. The remainder of the

proof of this case is similar to Case 2.

Case (5). t

≡

1 (mod 6).

By counting,

|

E

(

G

)| ≡

0 (mod 3), let t

=

6k

+

1. Similar to Case 1, we only consider the case

|

E

(

G

)| =

6k.

|

E

(

G1

)| =

3k

−

3, G2

=

G

−

G1,

|

E

(

G2

)| =

3k

+

3. We denote V

(

G2

)

by B and

V

(

D6k+1

) \

V

(

G2

)

by A. Note

|

V

(

DA

)| =

3k

−

2.

Then we have

Dt

−

G

=

(

DA

−

G1

) + (

DB

−

G2

) +

DB,A

.

By induction, DA

−

G1has a

*

C3-decomposition. The difference triple

(

1

,

2

,

3k

)

from DB

−

G2can form a collection of

directed 3-cycles F1

= {

(

i

,

i

+

1

,

i

+

3

) |

i

∈

Z3k+3

}

. Then DB

−

G2

−

F1is a

(

3k

−

2

)

-regular directed graph. We associate with

DB,Ato get a set of directed 3-cycles (denoted by F2) by applyingTheorem 2.2andLemma 2.1. Thus, DB

−

G2

+

DB,A

=

F1

+

F2.

Second, suppose min

{

3k

−

3

− |

E

(S

*

C∈G * C

)|} =

l

>

0, then let G1

=

S

* C∈G *

C where 3k

−

2

− |

E

(

G1

)| =

l. The remainder

of the proof of this case is similar to Case 2.

Case (6). t

≡

5 (mod 6).

Similar to Case 1, we only consider t

=

6k

+

5 and

|

E

(

G

)| =

6k

+

5. Then, the proof follows by a similar argument, we omit the details.

4. Conclusion

Now, by combiningLemma 2.8,Theorems 2.3and 3.1, we have proved our main result:Theorem 1.3. We also get

Theorem 1.4as corollary ofTheorem 1.3.

Theorem 1.3. Let t be a positive integer. Let G be a vertex-disjoint union of directed cycles in Dtand suppose G

6=

* C2

∪

* C3or * C5if t

=

5 and G

6=

* C3

∪

* C3if t

=

6. Then *

C3

|

(

Dt

−

G

)

if and only if t

(

t

−

1

) − |

E

(

G

)| ≡

0

(

mod 3

)

.

Theorem 1.4. Let t be a positive integer. Let G be a vertex-disjoint union of cycles in 2Ktand suppose G

6=

C2

∪

C3or G

6=

C5if

t

=

5 and G

6=

C3

∪

C3if t

=

6. Then C3

|

(

2Kt

−

G

)

if and only if t

(

t

−

1

) − |

E

(

G

)| ≡

0

(

mod 3

)

.

The covering of Ktwith triangles was first considered by Colbourn and Rosa [3] and then by Fu, Fu and Rodger [4]. Mainly,

they prove the following.

Theorem 4.1. Let G be a 2-regular (not necessarily spanning) subgraph of Ktwhere t is odd. Then C3

|

(

Kt

∪

G

)

if and only if the

number of edges in Kt

∪

G is a multiple of 3.

Now, by using the results we obtain in this paper, we are able to prove a digraph version. The details are omitted here.

Theorem 4.2. Let t be a positive integer. Let G be a vertex-disjoint union of directed cycles in Dtand

|

E

(

G

)|

be the number of