Digital Object Identifier (DOI) 10.1007/s00373-006-0683-yGraphs and Combinatorics (2006) 22:515–525

Graphs andCombinatorics© Springer-Verlag 2006

⇀

C4-Decompositions of Dv\P and Dv ∪ P where P

is a 2-Regular Subgraph of Dv

Liqun Pu1, Hung-Lin Fu2 and Hao Shen1,�

1 Department of Applied Mathematics, Shanghai Jiao Tong University, Shanghai 200240,China. e-mail: liqunpu@yahoo.com.cn2 Department of Applied Mathematics, National Chiao Tung University, Hsin Chu 30050,Taiwan. e-mail: hlfu@math.nctu.edu.tw

Abstract. In this paper, we extend the study of C4-decompositions of the complete graphwith 2-regular leaves and paddings to directed versions. Mainly, we prove that if P is a ver-tex-disjoint union of directed cycles in a complete digraph Dv , then Dv \P and Dv ∪P can bedecomposed into directed 4-cycles, respectively, if and only if v(v − 1) − |E(P )| ≡ 0 (mod 4)and v(v − 1) + |E(P )| ≡ 0 (mod 4) where |E(P )| denotes the number of directed edges ofP , and v ≥ 8.

Key words. Directed 4-cycles, Complete digraph, Packing, Covering

1. Introduction

A packing of a graph G with 4-cycles is a set of edge-disjoint 4-cycles in G. Thegraph induced by the edges in G but not in any 4-cycle of the packing is called theremainder graph of the packing or the leave of the packing. If a packing has a leavewhich has the minimum number of edges, we call it a minimum leave. A maximumpacking of G (with 4-cycles) is a packing which has a minimum leave. Clearly, ifE(G) can be partitioned into sets which induce 4-cycles, then the leave is an emptygraph and we say that G has a 4-cycle decomposition.

A 4-cycle decomposition of a complete graph Kv is also known as a 4-cycle sys-tem of order v. It is folklore that a 4-cycle system of order v exists if and only if v ≡1 (mod 8) and the maximum packing of Kv [6] and Kv \P [2, 4] with 4-cycles whereP is a special subgraph of Kv are also known. When H is a 2-regular subgraph ofa complete graph K2m+1, H. L. Fu and C. A. Rodger give us the following result.

Theorem 1 [4]. Let H be a 2-regular subgraph of K2m+1 and |E(H)| be the numberof edges of H such that

(2m+12

) − |E(H)| is a multiple of 4. Then K2m+1 \ H has a4-cycle decomposition.

For convenience, we denote such a decomposition by C4|K2m+1 \ H .

� Present address: Insert the address here if needed.

516 L. Pu et al.

A covering of a graph G with 4-cycles is a collection of 4-cycles, P , such that eachedge of G occurs in at least one 4-cycle in P . So, if G(P) is the multigraph formedby joining each pair of vertices u and v with x edges if and only if P contains x

4-cycles that contain both u and v, then clearly C4| G(P). The multigraph G(P)\G

is called the excess graph of G; it is also known as the padding of the covering P ofG. A covering with smallest excess graph (in size) is called a minimum covering.

Similarly, packing and covering of complete digraph Dv with directed 4-cyclescan also be defined. In this paper, we extend the work of Theorem 1 and considerthe corresponding problem about packing and covering of a complete digraph withdirected 4-cycles.

2. Preliminaries

Let Dv be a complete digraph without loops of order v and for each vertex w in Dv,deg+(w) = deg−(w) = v − 1. Let Cli be a cycle of length li , P be a vertex-disjointunion of cycles in Dv and V (Cli ) be the set of vertices of Cli . Then Pn = ∪k

i=1Cli , if

n =k∑

i=1li , V (Cli ) ∩ V (Clj ) = ∅ (i �= j ).

Let A be an m-set, B be an n-set and A ∩ B = ∅. A complete bipartite directedgraph DA,B (Dm,n) contains 2mn directed edges. For example, D3,2 ={aibj , bj ai |1≤i ≤ 3, 1 ≤ j ≤ 2}, which has 12 edges. Note that aibj and bjai are two differentdirected edges.

Theorem 2 [5]. For an integer v, v ≥ 8, Dv has a⇀

C4-decomposition if and only ifv ≡ 0, 1 (mod 4).

We denote such a decomposition by⇀

C4 |Dv.The following lemma plays the most important role to prove our main theorem.

Since the proof is easy to see, we omit the proof.

Lemma 1. Let m, n be a positive integer such that min {m, n} ≥ 2 and mn is even.Then Dm,n can be decomposed into directed 4-cycles.

In fact, this result is a special case of the following theorem, the well-knownSotteau’s Theorem.

Theorem 3 [7]. Let m, n be two positive integers such that min {m, n} ≥ k and k|mn.Then Dm,n can be decomposed into directed 2k-cycles.

Lemma 2. If a digraph G can be decomposed into directed 4-cycles and X is a k-set,k ≡ 0 (mod 4) and k ≥ 8, such that V (G) ∩ X = ∅, then the graph G = G ∨ Dk

where V (G) = V (G) ∪ X and E(G) = E(G) ∪ E(X) ∪ E(DV (G),X) has a directed4-cycle decomposition.

⇀

C4-Decompositions of Dv \ P and Dv ∪ P where P is a 2-Regular Subgraph of Dv 517

Proof. It is a direct consequence of Theorem 2 that Dk can be decomposed intodirected 4-cycle and Lemma 1 that D|V (G)|,k can be decomposed into directed4-cycles. �

3. Packing Dt−P with Directed 4-Cycles

Before we give the proof of the first main theorem of this paper, we need a couple of

important facts. For convenience, we use k⇀

Ct to denote k vertex-disjoint directedt-cycles.

Lemma 3. D4, D4 \ ⇀

C4, D6 \ 2⇀

C3 and D7 \ 2⇀

C3 have no⇀

C4-decompositions.

Proof. The first two are easy to see and we prove the other two. Let D6 be definedon Z6 and (0, 1, 2) and (3, 4, 5) are the directed 3-cycles which are missing. LetA = {0, 1, 2} and B = {3, 4, 5}. Clearly, D6 can be written as D3 ∪ D3,3 ∪ D3where the first D3 is defined on A and the second one is defined on B. Now, we planto decompose (0, 2, 1)∪D3,3 ∪ (3, 5, 4) into directed 4-cycles. In order to use up thearcs in (0,2,1) and (3,5,4) respectively, each time we need two arcs from D3,3, onein (A, B) and the other one in (B, A). For convenience, we call a directed 4-cycleobtained this way a match-up. This implies that we shall have 3 match-ups in orderto use up all the arcs in (0,2,1)U(3,5,4). However, in total we have 9 arcs in (A, B)

and 9 arcs in (B, A) respectively. So we shall have an odd number of match-ups touse all the arcs in (0, 2, 1)∪(3, 5, 4), it is either 5 match-ups or 3 match-ups. By directconstructions, it is not possible for 5 match-ups in this case. We see that all of them

will leave at least two 2-cycles there. Therefore, a⇀

C4-decomposition of D6 \ 2⇀

C3

is not possible. Since D7 \ 2⇀

C3 = (D6 \ 2⇀

C3) ∪ D1,6, then D7 \ 2⇀

C3 has at

least two 2-cycles left by the result for D6 \ 2⇀

C3 and direct construction. �

Lemma 4. D4 \ 2⇀

C2, D5, D5 \ ⇀

C4 and D5 \ 2⇀

C2 have⇀

C4-decompositions.

Proof. Let D4 be defined on Z4 and (0,2), (1,3) are the directed 2-cycles which

are missing. Then D4 \ 2⇀

C2 = {(0, 1, 2, 3), (3, 2, 1, 0)}. Let D5 be defined on Z5

and (0,2), (1,3) be the directed 2-cycles which are missing. Then D5 \ 2⇀

C2 ={(4, 1, 0, 3), (4, 3, 2, 1), (4, 0, 1, 2), (4, 2, 3, 0)}. Let D5 be defined on Z5. Then D5 ={(i, 1 + i, 3 + i, 2 + i|i ∈ Z5)} and D5 \ ⇀

C4 can be easily obtained. �

Lemma 5. D6 \ ⇀

C2, D6 \ (⇀

C4 ∪ ⇀

C2), D6 \ 3⇀

C2 and D6 \ ⇀

C6 have⇀

C4-decomposi-tions.

Proof. Since D6 \3⇀

C2 = (D4 \2⇀

C2)∪D2,4, we can get the result by Lemma 1 andLemma 4. Let D6 be defined on Z6 where (5,4) and (3,2,1,0) are the missing cycles.

518 L. Pu et al.

Then D6 \ (⇀

C2 ∪ ⇀

C4) = {(4, 0, 1, 3), (5, 1, 2, 0), (5, 0, 4, 1), (4, 2, 3, 1), (5, 3, 0, 2),

(5, 2, 4, 3)}. If we add (3,2,1,0) to the 4-cycles set, we get D6\⇀

C2. If (0,1,2,3,4,5)

is the missing cycle, then D6\⇀

C6 = {(0, 3, 1, 4), (4, 1, 3, 0), (0, 2, 5, 1), (4, 3, 5, 2),

(0, 5, 3, 2), (2, 1, 5, 4)}. �

Lemma 6. D7\⇀

C2,D7 \3⇀

C2,D7\(⇀

C4 ∪ ⇀

C2),D7\⇀

C6,D8\ 2⇀

C2,D8\⇀

C4,D8\ 2⇀

C4,

D8 \ (⇀

C5 ∪ ⇀

C3), D8\⇀

C8, D8 \ 4⇀

C2, D8 \ (2⇀

C2 ∪ ⇀

C4) and D8 \ (⇀

C6 ∪ ⇀

C2) have⇀

C4-decompositions.

Proof. Since D7\⇀

C2 = D5 ∪ D2,5, D7 \3⇀

C2 = (D5 \2⇀

C2) ∪ D2,5, D7 \ (⇀

C4

∪ ⇀

C2) = (D5\⇀

C4) ∪ D2,5, D8 \ 2⇀

C2 = (D6\⇀

C2) ∪ D2,6, D8 \ 4⇀

C2 = (D4 \ 2⇀

C2) ∪ D4,4 ∪ (D4 \ 2⇀

C2), D8\⇀

C4 ∪ 2⇀

C2 = (D4 \ 2⇀

C2) ∪ D3,4 + (D5 \ C4),

D8\⇀

C6 ∪ ⇀

C2 = (D6\⇀

C6) ∪ D2,6, it is easy to decompose them into directed4-cycles by applying Lemma 4 and Lemma 5. By Theorem 2, D8 can be decomposed

into directed 4-cycles, so we delete a directed 4-cycle from it to get D8\⇀

C4 and delete

two vertex-disjoint directed 4-cycles from it to get D8 \ 2⇀

C4. Let D8 be defined on

Z4 ∪{a, b, c, d} where (0, 1, 2, 3, d) and (a, b, c) are the missing cycles. Then D8\(⇀

C5

∪ ⇀

C3) = {(a, 2, 0, 3), (1, a, 0, 2), (a, 3, 1, 0), (c, 2, b, 1), (d, 1, 3, 2), (d, 2, c, 1),

(d, b, 2, a), (d, a, 1, b), (3, 0, c, d), (d, c, b, 0), (3, c, 0, b), (3, b, a, c)}. Let D7 be de-

fined on Z6 ∪ {∞} and let (0, 1, 2, 3, 4, 5) be the missing cycle. Then D7\⇀

C6= {(∞, 3, 1, 4), (∞, 4, 0, 3), (∞, 1, 3, 0), (∞, 0, 4, 1), (0, 2, 5, 1), (4, 3, 5, 2), (∞, 5,

3, 2), (∞, 2, 0, 5), (2, 1, 5, 4)}. Let D8 be defined on Z8 and let (0, 1, 2, 3, 4, 5, 6, 7)

be the missing cycle. Then D8\⇀

C8 = {(7, 4, 3, 0), (3, 2, 1, 0), (2, 4, 6, 5), (2, 5, 1, 4),

(2, 7, 1, 6), (3, 1, 5, 7), (3, 7, 6, 1), (0, 2, 6, 4), (0, 4, 7, 2), (1, 7, 5, 4)} + D{0,3},{5,6}.�

Lemma 7. D8\(2⇀

C3 ∪ ⇀

C2), D9\(2⇀

C3 ∪ ⇀

C2), D9\2⇀

C2, D9\⇀

C4, D9\(2⇀

C2 ∪ ⇀

C4),

D9 \ (⇀

C6 ∪ ⇀

C2), D9 \ 2⇀

C4, D9 \ (⇀

C3 ∪ ⇀

C5) and D9\⇀

C8 have⇀

C4-decompositions.

Proof. Let D8 be defined on Z8 where (0, 1, 2), (3, 4, 5) and (6, 7) are the missing

cycles. Then D8 \ (2⇀

C3 ∪ ⇀

C2) = {(2, 1, 0, 3), (3, 0, 5, 4), (1, 3, 5, 6), (4, 0, 2, 6)∗,(7, 2, 3, 1)∗, (7, 5, 0, 4), (0, 7, 1, 6), (0, 6, 2, 7), (3, 7, 4, 6), (3, 6, 5, 7), (1, 4, 2, 5)∗,

(5, 2, 4, 1)∗}. Since D9 \ (⇀

C2 ∪2⇀

C3) = [D8 \ (2⇀

C3 ∪ ⇀

C2] ∪ D1,8, we can apply the

decomposition obtained above to find a⇀

C4-decomposition of D9 \ (⇀

C2 ∪ 2⇀

C3) by

matching the arcs in D1,8 with the directed 4-cycles in D8 \ (⇀

C2 ∪ 2⇀

C3) whichare marked with a “∗”. Here are the constructions: (5, 2, 4, 1) ∪ D{∞},{1,2} ={(∞, 2, 4, 1), (∞, 1, 5, 2)}. (1, 4, 2, 5) ∪ D{∞},{4,5} = {(∞, 4, 2, 5), (∞, 5, 1, 4)}.(4, 0, 2, 6) ∪ D{∞},{0,6} = {(∞, 0, 2, 6), (∞, 6, 4, 0)}. (7, 2, 3, 1) ∪ D{∞},{3,7} =

⇀

C4-Decompositions of Dv \ P and Dv ∪ P where P is a 2-Regular Subgraph of Dv 519

{(∞, 3, 1, 7), (∞, 7, 2, 3)}. This concludes the proof of this case. As for D9\⇀

C8, it isa special case of the proof of Theorem 4 Case (i). The remaining cases can be settledas those in Lemma 6. �

Lemma 8. D10\⇀

C2, D10\⇀

C6, D10 \ 3⇀

C2, D10 \ 2⇀

C3, D10 \ (⇀

C4 ∪ ⇀

C2), D10 \ (⇀

C4

∪ ⇀

C6), D10 \ (⇀

C2 ∪ ⇀

C8), D10 \ (⇀

C3 ∪ ⇀

C7), D10 \ 2⇀

C5, D10 \ (2⇀

C3 ∪ ⇀

C4) and

D10\⇀

C10 have⇀

C4-decompositions.

Proof. Let D10 be defined on Z7 ∪ {x, y, z} where (0, 1, 2, 3, 4, 5, 6) and (x, y, z)

are the missing cycles. Then D10 \ (⇀

C3 ∪ ⇀

C7) = {(0, 2, 5, 1), (4, 3, 6, 2), (0, 6, 3, 2),

(2, 1, 5, 4), (0, 3, 1, 4), (4, 1, 3, 0), (6, 1, z, 4), (6, 4, y, 1), (x, 0, y, 3), (5, 0, x, 3),

(5, 3, z, 0), (6, 5, z, y), (2, 6, y, x), (5, 2, x, z), (5, x, 1, y), (5, y, 4, x), (6, z, 1, x),

(6, x, 4, z), (2, y, 0, z), (2, z, 3, y)}. For D10\2⇀

C5, let D10 be defined on V = A∪B

where A = {1, 2, a, b, c, d} and B = {3, 4, 5, e}, ⇀

C5 = (1, 2, 5, 4, 3) and⇀

C5=(a, b, c, d, e). Let α = {(a, 3, 5, 2), (e, 3, 4, 5), (e, 5, 3, 2), (3, e, 4, d), (e, d, 1, 4)}and β = {(a, e, 1, 5), (d, 4, a, 5), (3, a, 4, 2), (3, d, 5, 1), (1, e, 2, 4), (b, e, c, 3),

(e, b, 3, c), (b, 4, c, 5), (4, b, 5, c)}. Then add α to use up all arcs in B and leave

a⇀

C6 = (a, b, c, d, 1, 2) in A. The rest of the directed edges of A can be partitionedinto 4-cycles by using Lemma 5. Now, it is left to complete the partition of DA,B intodirected 4-cycles β. Let D10 be defined on Z10 and let (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) be the

missing cycle. We can add two directed 4-cycles (0, 9, 8, 7) and (2, 5, 4, 3) to⇀

C10 and

divide⇀

C10 into seven cycles (0, 1, 2, 5, 6, 7), (3, 4) ∪ (8, 9), (4, 5), (0, 9), (7, 8) and

(2, 3). Then D10\⇀

C10 = {(0, 9, 8, 7), (2, 5, 4, 3), (0, 5, 1, 6), (6, 1, 5, 0), (4, 0, 2, 7),

(4, 7, 1, 0), (9, 5, 7, 2), (9, 2, 6, 5), (3, 0, 7, 5), (8, 5, 2, 0), (8, 0, 3, 5), (2, 1, 7, 6),

(9, 3, 8, 4), (4, 8, 3, 9)} ∪ D{3,9},{1,6,7} ∪ D{4,8},{1,2,6}. For D10\2⇀

C3 and D10\(2⇀

C3

∪ ⇀

C4), they are the special cases of Theorem 4. Case (iii). As for other cases, theycan be proved as those in Lemma 6. �

Lemma 9. For each integer t , t ≥ 3, D2t − ⇀

C2t has a⇀

C4-decomposition.

Proof. The proof is by induction. By Lemma 5, Lemma 6 and Lemma 8, it is truefor t = 3, 4, 5. Assume the assertion is true for all orders less than t , we shall provethat the assertion is true for t .

Let D2t be defined on V = A ∪ B ∪ M, M = Z2t−10, B = {bi |i ∈ Z4} and A ={ai |i ∈ Z6}. Let

⇀

C2t= (0, 1, , . . . , 2t −12, 2t −11, b0, b1, a0, a1, . . . , a5, b2, b3). Letα = {(a0, b1, b0, a5), (a5, b0, 2t−11, b2), (2t−11, 0, b3, b2)} and β={(b0, b2, b1, b3),

(b3, b1, b2, b0)}. Add α to⇀

C2t . The union of the above three cycles can be parti-

tioned into the following directed cycles⇀

C2t−10 = (0, 1, . . . , 2t − 12, 2t − 11),⇀

C6 = (a0, a1, . . . , a5),⇀

C2 ∪ ⇀

C2 = (b0, b1)∪ (b2, b3), (0, b3), (a0, b1), (2t − 11, b2),

520 L. Pu et al.

(2t − 11, b0), (a5, b2) and (a5, b0). Then

D2t\⇀

C2t = D2t \ (⇀

C2t ∪α) ∪ α

= [(DM\ ⇀

C2t−10) ∪ α] ∪ DM,A∪B ∪ [DA∪B \ (⇀

C6 ∪ ⇀

C2 ∪ ⇀

C2)] \ [(b0, a5)

∪(b0, 2t − 11) ∪ (b2, a5) ∪ (b2, 2t − 11) ∪ (0, b3) ∪ (b1, a0)]

= [(DM\ ⇀

C2t−10) ∪ α] ∪ DM,A∪B ∪ (DA\ ⇀

C6) ∪ DB \ (⇀

C2 ∪ ⇀

C2)

∪DA,B \ [(b0, a5) ∪ (b0, 2t − 11) ∪ (b2, a5) ∪ (b2, 2t − 11)

∪(0, b3) ∪ (b1, a0)]

= [(DM\ ⇀

C2t−10) ∪ α ∪ DB \ (⇀

C2 ∪ ⇀

C2)] ∪ DB,A∪M ∪ DA,M

∪(DA\ ⇀

C6) \ [(b0, a5) ∪ (b0, 2t − 11) ∪ (b2, a5) ∪ (b2, 2t − 11)

∪(0, b3) ∪ (b1, a0)]

= [(I) ∪ D{b0,b2},A∪M\{a5,2t−11}] ∪ D{b1,b3},A∪M

∪DA,M \ (0, b3) \ (b1, a0) ∪ (DA\ ⇀

C6)

= (I′) ∪ D{b1},A∪M\{a0} ∪ D{b3},A∪M\{0} ∪ DA,M ∪ (DA\ ⇀

C6)

= (I′) ∪ [(DA\ ⇀

C6) ∪ D{b3},A] ∪ D{b1},A∪M\{a0} ∪ D{b3},M\{0} ∪ DA,M

= (I′) ∪ (II) ∪ D{b1},(A\{a0})∪M ∪ D{b3},M\{0} ∪ DA,M

= (I′) ∪ (II) ∪ D{b1},A\{a0} ∪ D{b1},M ∪ D{b3},M\{0} ∪ DA\{a0},M ∪ D{a0},M= (I′) ∪ (II) ∪ D{b1},A\{a0} ∪ D{b1},M ∪ D{b3},M\{0} ∪ DA\{a0},M\{0}

∪DA\{a0},{0} ∪ D{a0},M= (I′) ∪ (II) ∪ [DM\{0},A∪{b3}\{a0} ∪ D{0,b1},A\{a0} ∪ DM,{a0,b1}]= (I′) ∪ (II) ∪ (III)

where (I)=(D2t−10\⇀

C2t−10)∪ α ∪(DB−⇀

C2 ∪⇀

C2), (I′) = (I)∪D{b0,b2},A∪M\{a5,2t−11},

(II) = Db3,A ∪ (DA\ ⇀

C6) = {(b3, a0, a3, a1), (b3, a4, a1, a3), (a0, a2, a5, a1),(b3, a1, a4, a0), (b3, a3, a0, a4), (a4, a3, a5, a2), (a0, a5, a3, a2), (b3, a2, a1, a5),(b3, a5, a4, a2)}. (III) = DM\{0},A∪{b3}\{a0} ∪ D{0,b1},A\{a0} ∪ DM,{a0,b1} �

With the above preparations, we are now in a position to prove the first main the-orem of this paper. For convenience, we shall denote the complete digraph definedon A by DA in what follows.

Theorem 4. Let v be an integer, v ≥ 8 and P be a vertex-disjoint union of directed

cycles in Dv. Then⇀

C4 |Dv \ P if and only if v(v − 1) − |E(P )| ≡ 0 (mod 4).

Proof. The necessity is obvious and we prove the sufficiency by induction on v. Notethat v(v−1)−(v−2)(v−3) is congruent to 2 modulo 4 while v(v−1)−(v−4)(v−5)

is congruent to 0 modulo 4. Therefore, the plan of our proof is reducing the order v

by 2 or 4. By Lemma 6 and Lemma 7, we conclude that⇀

C4 |D8 \ P and thus v = 8

⇀

C4-Decompositions of Dv \ P and Dv ∪ P where P is a 2-Regular Subgraph of Dv 521

is true. Assume the assertion is true for all orders smaller than v and we shall provethe assertion is also true for Dv \ P . Of course, if P contains no arcs, then Dv has a⇀

C4-decomposition by Theorem 2. Otherwise, P contains at least one directed cycle.

First, if P contains a⇀

C2, then Dv − P = [Dv−2 \ (P \ ⇀

C2)] ∪ D2,v−2. By the

induction process,⇀

C4 |Dv−2 \ (P \ ⇀

C2) while by Lemma 1,⇀

C4 |D2,v−2. Hence wefinish this case. So, in what follows, we consider the case where P contains directedcycles of length at least 3, that is,

(i) P contains directed cycles of length not less than 6;(ii) P contains a directed 4-cycle;

(iii) P contains two vertex-disjoint directed 3-cycles;(iv) P contains a directed 3-cycle and a directed 5-cycle;(v) P contains only directed 5-cycles;

Case (i). Let Dv be defined on Zv, A = {a0, a1, a3, a4}, B ′ = {at−1, a5, a2, x} and

B = Zv \ (A∪B ′). Suppose P contains a t-cycle⇀

Ct= {a0, a1, . . . , at−2, at−1} wherev > t ≥ 6. Add cycle set α = {(a0, a2, a5, a1), (a4, a3, at−1, a2), (a0, at−1, a3, a2),

(a2, a1, a5, a4), (x, a1, at−1, a4)∗, (x, a0, a5, a3)

∗} where the 4-cycles marked by “∗”

have double direction and x /∈ V (⇀

Ct). Then DZv \ P = DZv ∪ α∪ ⇀

Ct \(P \ ⇀

Ct

\α) = DZv \ [P \ ⇀

Ct ∪(a2, a5, . . . , at−1) ∪ (a1, a0) ∪ (a3, a4) ∪ DA,B ′ ] = (DZv\A \P ′) ∪ (DZv\A,A \ DA,B ′) ∪ [DA \ (a1, a0) \ (a3, a4)] = (DZv\A \ P ′) ∪ DZv\A∪B ′,A ∪[DA \ (a1, a0) \ (a3, a4)] = (DZv\A \ P ′) ∪ DA,B ∪ [DA \ (a1, a0) \ (a3, a4)] where

P ′ = P \ ⇀

Ct ∪(a2, a5, a6, . . . , at−2, at−1). By the induction process⇀

C4 |DZv\A \ P ′

and by Lemma 1 and Lemma 4, we have⇀

C4 |DA,B and⇀

C4 |DA \ (⇀

C2 ∪ ⇀

C2) respec-

tively. Therefore,⇀

C4 |Dv \ P . On the other hand, if v = t , then t must be even. ByLemma 9, we conclude the proof of this case.

Case (ii). Let P = P ′ ∪ (a0, a1, a2, a3) and x ∈ V (P ′). Then Dv\P = (Dv−4\P ′) ∪(D5 \ (a0, a1, a2, a3)) ∪ Dv−5,4 where D5 is defined on {a0, a1, a2, a3, x}. The prooffollows easily.

Case (iii). Let P be defined on Zv, (a, b, c) and (d, e, f ) be in P . Let x ∈ V \{a, b, c, d, e, f } and P ′ = P \ [(a, b, c)∪ (d, e, f )]. Let β = {(c, b, a, d), (d, a, f, e),

(d, b, x, c), (b, d, f, x), (e, a, c, x), (a, e, x, f )}. Add β to P to get P ′, D3,4, (b, c),(e, f ) and (a, d). Then DZv \ P = DZv \ [P ′ ∪ (a, b, c) ∪ (d, e, f ) ∪ β] ∪ β =DZv\{b,c,e,f } \ (P ′ ∪ (b, c) ∪ (e, f ) ∪ (a, d) ∪ D{a,d,x},{b,c,e,f }) ∪ β ∪DZv\{b,c,e,f },{b,c,e,f } ∪ D{b,c,e,f } = [DZv\{b,c,e,f } \ (P ′ ∪ (a, d))] ∪ β ∪ [D{b,c,e,f } \(b, c) \ (e, f )] ∪ (DZv\{b,c,e,f },{b,c,e,f } \ D{a,d,x},{b,c,e,f }) = [DZv\{b,c,e,f } \ (P ′ ∪(a, d))] ∪ β ∪ [D{b,c,e,f } \ (b, c) \ (e, f )] ∪ (DZv\{a,b,c,d,e,f,x},{b,c,e,f }). By induction,⇀

C4 |DZv\{b,c,e,f } \ (P ′ ∪ (a, d)) and by Lemma 4,⇀

C4 |D{b,c,e,f } \ (b, c) \ (e, f ), wehave the proof.

Case (iv). Let P ′ = P \ ⇀

C3 ∪ ⇀

C5. Then Dv \ P = (Dv−8 \ P′) ∪ D8,v ∪ (D8\

⇀

C3

∪ ⇀

C5). By Lemma 6 and the hypothesis, this case is proved.

522 L. Pu et al.

Case (v). Let Dv be defined on Zv and P ′ = P \ ⇀

C5 \ ⇀

C5. Then Dv \ P =(Dv−10 \P ′)∪Dv−10,10 ∪ (D10\

⇀

C5 \ ⇀

C5). By the assertion and Lemma 8, we finishthis case and the proof of this theorem. �

4. Packing Dt ∪ P with Directed 4-Cycles

In this section, we need the following Lemma.

Lemma 10. Let P be a vertex-disjoint union of directed cycles defined on V and allcycles in P have length not less than 3. Then for any a /∈ V , D{a},V ∪ 2P has a⇀

C4-decomposition.

The proof can be deduced from the following example immediately: D{a},{0,1,2} ∪(0, 1, 2) ∪ (0, 1, 2) = D{a},{0,1,2} ∪ (0, 1, 2) ∪ (0, 1, 2) = {(a, 0, 1, 2), (a, 1, 2, 0),

(a, 2, 0, 1)}.

Lemma 11. D4 ∪ ⇀

C4 , D6 ∪ ⇀

C2 and D7 ∪ ⇀

C2 have no⇀

C4-decompositions.

Proof. The first can be easily seen. For the second, we have D6 ∪ ⇀

C2 = D4 ∪D2,4 ∪ (D2∪

⇀

C2) where D2 ∪ ⇀

C2 = 2⇀

C2. By direct construction, we know there

exist at least two double edges left. D7 ∪ ⇀

C2 = D5 ∪ D2,5 ∪ (D2∪⇀

C2). By directconstruction, we know that there exist at least two 2-cycles left. �

Lemma 12. D4 ∪ 2⇀

C2, D5 ∪ 2⇀

C2, D5 ∪ ⇀

C4, D6 ∪ ⇀

C6 and D6 ∪ 3⇀

C2 have⇀

C4-decompositions.

Proof. Let D4 be defined on Z4 where 2⇀

C2= (0, 2) ∪ (1, 3). Then D4 ∪ 2⇀

C2={(0, 1, 3, 2)1, (1, 0, 2, 3)2, (0, 2, 1, 3), (2, 0, 3, 1)}. Let D5 be defined on Z5. Since

D5 ∪ 2⇀

C2 = (D4 ∪ 2⇀

C2) ∪ D1,4, choose the 4-cycles marked by “1” and “2”respectively from above and associate them with D{4},{0,3} and D{4},{1,2} respectively.Since D{4},{0,3} ∪ (0, 1, 3, 2) = (4, 0, 1, 3) ∪ (3, 2, 0, 4), D{4},{1,2} ∪ (1, 0, 2, 3) =(4, 1, 0, 2) ∪ (4, 2, 3, 1). We have the decomposition. Thus

⇀

C4 |D5 ∪ 2⇀

C2. Simi-

larly, D5 ∪ ⇀

C4 = {(i, 1 + i, 3 + i, 2 + i)|i ∈ Z5}∪⇀

C4. Let D6 be defined on Z6 and

C6 = (0, 1, 2, 3, 4, 5). Then D6 ∪ ⇀

C6 = (D6 \ (1, 4)) ∪ (0, 1, 4, 5) ∪ (1, 2, 3, 4).

Let D6 be defined on Z6 while 3⇀

C2 = (0, 1) ∪ (2, 3) ∪ (4, 5). D6 ∪ 3⇀

C2 = {(5, 1, 2, 3), (5, 3, 0, 1), (4, 0, 1, 3), (5, 0, 2, 4), (3, 2, 1, 0), (4, 2, 3, 1),(5, 4, 3, 2), (5, 2, 0, 4), (5, 4, 1, 0)}. �

Lemma 13. D8 ∪ ⇀

C3 ∪ ⇀

C5, D10 ∪ 2⇀

C5 and D10 ∪ ⇀

C3 ∪ ⇀

C7 have⇀

C4-decompositions.

⇀

C4-Decompositions of Dv \ P and Dv ∪ P where P is a 2-Regular Subgraph of Dv 523

Proof. Let D8 be defined on Z8, while (0, 1, 2) and (3, 4, 5, 6, 7) are the added cycles.

Then D8 ∪ ⇀

C3 ∪ ⇀

C5= {(0, 3, 4, 2), (2, 4, 7, 1), (3, 0, 1, 7), (4, 5, 6, 7), (5, 7, 6, 1),

(6, 5, 4, 1), (1, 2, 7, 5), (4, 6, 2, 1), (5, 2, 6, 4), (2, 5, 6, 7)}∪D{0,3},{1,2,4,5,6,7}. Let D10be defined on Z10, where (0, 1, 2, 3, 4) and (5, 6, 7, 8, 9) are the added cycles. Then

D10 ∪ 2⇀

C5= {(7, 0, 2, 6), (7, 6, 2, 0), (8, 2, 9, 0), (8, 0, 9, 2), (0, 3, 5, 4), (7, 3, 0, 4),

(7, 4, 5, 3), (3, 1, 6, 4), (4, 6, 1, 3), (8, 6, 3, 9), (5, 6, 8, 9), (5, 9, 3, 6), (7, 2, 1, 9),

(8, 1, 2, 7), (8, 7, 9, 1), (6, 7, 8, 9), (1, 2, 3, 4), (0, 1, 5, 6), (4, 0, 6, 9), (9, 5, 1, 4)} ∪D{1,5},{0,7} ∪D{3,4,5},{2,8}. Let D10 be defined on Z10 where (0,1,2) and (3,4,5,6,7,8,9)

are the added cycles. Then D10 ∪ ⇀

C3 ∪ ⇀

C7 = {(1, 2, 6, 4), (4, 6, 2, 1), (2, 7, 6, 9),

(9, 6, 7, 2), (5, 6, 7, 8), (4, 5, 7, 8), (4, 5, 8, 9), (3, 0, 1, 9), (3, 4, 2, 0), (2, 4, 9, 1),

(1, 6, 3, 7), (1, 7, 0, 6), (7, 4, 0, 9), (7, 9, 3, 4), (3, 5, 0, 8), (8, 0, 5, 3), (6, 0, 7, 3),

(9, 0, 4, 3), (4, 8, 7, 5)} + D{5,8,0,3},{1,2} + D{5,8},{6,9}. �

Lemma 14. D10 ∪ ⇀

C2, D11 ∪ ⇀

C2, D10 ∪ 3⇀

C2, D11 ∪ 3⇀

C2, D10 ∪ 5⇀

C2 and

D11 ∪ 5⇀

C2, have⇀

C4-decompositions.

Proof. Let D10 be defined on Z10 and⇀

C2 = (8, 9). Then D10 ∪ ⇀

C2 = {(8, 3, 1, 0)1,

(8, 1, 3, 2)2, (4, 5, 6, 7), (8, 7, 5, 4)3, (8, 5, 7, 6)4, (9, 1, 2, 3), (9, 3, 0, 1), (9, 8, 2, 0),

(9, 0, 3, 8), (9, 8, 0, 2), (9, 2, 1, 8), (9, 4, 6, 5), (9, 5, 8, 4), (9, 6, 4, 7), (9, 7, 8, 6)5} ∪D{0,1,2,3},{4,5,6,7}. Let D11 be defined on Z10 ∪{∞} and associate 4-cycles marked by“1” with D{∞},{3,0}, (8, 3, 1, 0)1 ∪D{∞},{3,0} = (∞, 3, 1, 0)∪ (∞, 0, 8, 3). As before,associate 4-cycles marked by “2, 3, 4” and “5” with D{∞},{1,2}, D{∞},{4,7}, D{∞},{5,6}and D{∞},{8,9}, respectively. Then,

⇀

C4 |D11 ∪ ⇀

C2. Since D10 ∪ 3⇀

C2 = D6 ∪ 3⇀

C2

∪D5,4 ∪ D5, D11 ∪ 3⇀

C2 = D6 ∪ 3⇀

C2 ∪D5,6 ∪ D5, D10 ∪ 5⇀

C2 = D6 ∪ 3⇀

C2

∪D4,6 ∪ (D4 ∪ 2⇀

C2), D11 ∪ 5⇀

C2 = D6 ∪ 3⇀

C2 ∪D5,6 ∪ (D5 ∪ 2⇀

C2), these casesare proved. �

Lemma 15. For integers t, l, t ≥ 3, l ≥ 3 and l ≡ 0 (mod 2), D2t ∪ ⇀

C2t and

D2l ∪ l⇀

Ci

2 where⇀

Ci

2 = (2i, 2i + 1) for 0 ≤ i ≤ l − 1, have⇀

C4-decompositions.

Proof. Let D2t be defined on Z2t and let⇀

Ci

2t = (0, 1, . . . , 2t − 2, 2t − 1). Then

D2t ∪ ⇀

C2t = {(i, i + 1, 2t − 2 − i, 2t − 1 − i)|0 ≤ i ≤ t − 2} ∪ [D2t \ ∑t−3i=0(1 +

i, 2t − 2 − i)]. Let D2l be defined on Z2l . A = {(j, 1 + j, 2l − 2 − j, 2l − 1 − j)|0 ≤j ≤ l − 1, j ≡ 0 (mod 2)} and B = {(j, 2l − 1 − j)|0 ≤ j ≤ l − 1}. Then

D2l ∪ l⇀

Ci

2 = A ∪ (D2l \ B). By Theorem 4, we have the proof. �

Lemma 16. For integers t1, t2, t1 ≥ 4, t2 ≥ 4, D2t1+2t2+2 ∪ ⇀

C2t1+1 ∪ ⇀

C2t2+1 has⇀

C4-decompositions.

524 L. Pu et al.

Proof. LetD2t1+2t2+2 be defined onV whereV = A∪B,V1 = V \{a0, a1, a2t1 , b0, b1,

b2t2}, A = {ai |i ∈ Z2t1+1} and B = {bi |i ∈ Z2t2+1}. Let⇀

C2t1+1 = (a0, a1, . . . , a2t1)

and⇀

C2t2+1 = (b0, b1, . . . , b2t2). We have⇀

C2t1+1 = {(a1+i , a2+i , a2t1−1−i , a2t1−i )|0 ≤i ≤ t1 − 2} \ {(a1+i , a2t1−i )|0 ≤ i ≤ t1 − 2} ∪ (a0, a1, a2t1) = A1 \ A2 \ (a1, a2t1) ∪(a0, a1, a2t1) where A1 = {(a1+i , a2+i , a2t1−1−i , a2t1−i )|0 ≤ i ≤ t1 − 2}, A2 ={(a1+i , a2t1−i )|1 ≤ i ≤ t1 − 2}. Let

⇀

C2t2+1 = {(b1+i , b2+i , b2t2−1−i , b2t2−i )|0 ≤i ≤ t2 − 2} − {(b1+i , b2t2−i )|0 ≤ i ≤ t2 − 2} + (b0, b1, b2t2) = B1 \ B2 \(b1, b2t1) ∪ (b0, b1, b2t2) where B1 = {(b1+i , b2+i , b2t2−1−i , b2t2−i )|0 ≤ i ≤ t2 − 2},B2 = {(b1+i , b2t2−i )|1 ≤ i ≤ t2 − 2}.

Then D2t1+2t2+2 ∪ ⇀

C2t1+1 ∪ ⇀

C2t2+1= D2t1+2t2−4 \ (A2 ∪ B2) ∪ (A1 ∪ B1) ∪[D6,2t1∪2t2−4 ∪D6 \ (a1, a2t1)\ (b1, b2t2)∪ (a0, a1, a2t1)∪ (b0, b1, b2t2)] = (I)∪ (II)∪(III).

(I) = D2t1+2t2−4 \ (A2 ∪ B2), (II) = A1 ∪ B1,(III) = D6,2t1∪2t2−4 ∪ D6 \ (a1, a2t1) \ (b1, b2t2) ∪ (a0, a1, a2t1) ∪ (b0, b1, b2t2) =

{(D{a2t1 ,b0,b1},V1\{e,f } ∪ D{a0,a1,b2t2 },V1) + D{a1,b0},{b1,b2t2 } ∪ {(e, a2t1 , f, b1),

(a2t1 , b0, e, b1), (a2t1 , b1, f, b0), (a0, b0, f, a2t1), (a0, a2t1 , e, b0), (a0, a1, b0, b1),

(a0, a1, a2t1 , b2t2), (a0, b2t2 , b0, a1), (b1, b2t2 , a2t1 , a0)} where e, f ∈ V1. By Theo-

rem 4,⇀

C4 | (I), then this case is proved. �

With the above preparations, we are now in a position to prove the second mainidea of this paper.

Theorem 5. Let v be an integer, v ≥ 8 and P be a vertex-disjoint union of directed

cycles in Dv. Then⇀

C4 |Dv ∪ P if and only if v(v − 1) + |E(P )| ≡ 0 (mod 4).

Proof. The necessity is obvious. We only need to prove the sufficiency.We divide the proof into three cases.

Case (i). P contains even number of 2-cycles. Let P2l = P2l1 ∪ P2l2 where 2l ≤ v,

l2 ≡ 0 (mod 2) , P2l2 = l2⇀

C2 and all the cycles in P2l1 have length longer than 3.Then Dv \P = Dv−2l2 ∪D2l2,v−2l2 ∪D2l2 ∪P2l1 ∪P2l2 = Dv−2l2−2l1 ∪D2l1,v−2l2−2l1 ∪(D2l1 \P2l1)∪(D1,2l1 ∪2P2l1)∪[D2l2,v−2l2−2l1 ∪D2l2−1,2l1 ∪(D2l2 ∪P2l2)]. By Lemma 1,Lemma 10, Lemma 15 and Theorem 4, we prove the cases.

Case (ii). P contains odd number of 2-cycles. Let P = P1∪⇀

C2 or P = P2 ∪ 3⇀

C2

or P = P3 ∪5⇀

C2. For the first cases, we have Dv ∪ P = (Dv−10 ∪ P1)∪D10,v−10 ∪(D10 ∪ ⇀

C2). By Lemma 1, Lemma 14 and Case (i) of this section, we finish theproof. For the other cases, we can proceed similarly.

Case (iii). All cycles in P have length longer than 2.If v = 2k + 1, P = P2l where l ≤ k, then we have D2k+1 ∪ P2l = D2k−2l ∪

(D2l+1 ∪ P2l ) ∪ D2k−2l,2l+1 = D2k−2l ∪ (D2l − P2l ) ∪ (D1,2l + 2P2l ) ∪ D2k−2l,2l+1.By Lemma 1, Lemma 10 and Theorem 4, we get the proof.

⇀

C4-Decompositions of Dv \ P and Dv ∪ P where P is a 2-Regular Subgraph of Dv 525

If v = 2k, P = P2l1∪⇀

C2l2 , then we have D2k ∪P = D2k−2l2 ∪D2k−2l2,2l2 ∪D2l2 ∪P2l1∪

⇀

C2l2= (D2k−2l2 ∪ P2l1) ∪ D2k−2l2,2l2 ∪ (D2l2∪⇀

C2l2) = [D2k−2l2−2l1 ∪ (D2l1 \P2l1) ∪ D2k−2l2−2l1,2l1 ∪ (2P2l1 ∪ D1,2l1)] ∪ (D2k−2l2−2l1,2l2 ∪ D2l1,2l2−1) ∪ (III). ByLemma 1, Lemma 10, Lemma 15 and Theorem 4, we get the proof.

If v = 2k, P = P2l0∪⇀

C2l1+1 ∪ ⇀

C2l2+1, then we have D2k ∪P = (D2k−2l1−2l2−2 ∪P2l0)∪D2k−2l1−2l2−2,2l1+2l2+2 ∪ (D2l1+2l2+2∪

⇀

C2l1+1 ∪ ⇀

C2l2+1)= [D2k−2l1−2l2−2−2l0∪(D2l0 \P2l0)∪D2k−2l1−2l2−2−2l0,2l0 ∪ (D1,2l0 ∪2P2l0)]∪ (D2k−2l1−2l2−2−2l0,2l1+2l2+2 ∪D2l0,2l1+2l2+1)∪(III). By Lemma 1, Lemma 10, Lemma 16 and Theorem 4, we getthe proof.

Thus we conclude the proof of Theorem 5. �

Acknowledgments. Thanks to the anonymous referees who carefully read the paper and givemany careful and valuable remarks which make the paper more readable.

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Received: June 4, 2005Final Version received: March 25, 2006