Cyclically decomposing the complete graph into cycles

Note

Cyclically decomposing the complete graph into cycles

Hung-Lin Fu

_{, Shung-Liang Wu}

a_{Department of Applied Mathematics, National Chaio Tung University, Hsin Chu, Taiwan, ROC} b_{National United University, Miaoli, Taiwan, ROC}

Received 9 July 2001; received in revised form 8 July 2003; accepted 28 December 2003

Abstract

Let m1; m2; : : : ; mk be positive integers not less than 3 and let n =
ki=1mi. Then, it is proved that the complete graph

of order 2n + 1 can be cyclically decomposed into k(2n + 1) cycles such that, for each i = 1; 2; : : : ; k, the cycle of length mi occurs exactly 2n + 1 times.

c

2003 Elsevier B.V. All rights reserved.

Keywords: Complete graph; Cycle system; Skolem sequence; Hooked Skolem sequence; Near Skolem sequence

1. Introduction

A Steiner triple system (STS) is an ordered pair (V; B), where V is a <nite nonempty set of elements, and B is a collection of 3-element subsets of V called triples, such that each pair of distinct elements of V occurs together in exactly one triple of B. The order of a Steiner triple system (V; B) is the size of V , denoted by |V |.

From “graph decomposition” point of view, the existence of a Steiner triple system of order v (STS(v)) is equivalent to the existence of a decomposition of the complete graph Kvof order v into edge-disjoint triangles, denoted by C3. It is not

diBcult to see the necessary condition for such a decomposition to exist is that v ≡ 1 or 3 (mod 6). In fact, this condition was proved to be suBcient around 150 years ago by Kirkman [4]. An automorphism of a STS (V; B) is a bijection  : V → V such that {x; y; z} ∈ B if and only if {(x); (y); (z)} ∈ B. A STS(v) is cyclic if it has an automorphism that is a permutation consisting of a single cycle of length v, for example (1; 2; 3; : : : v).

Cyclic Steiner triple systems do exist. In 1939, Peltesohn used the so-called diIerence method to settle the existence problem.

Theorem 1.1 (Peltesohn [7]). For all v ≡ 1 or 3 (mod 6) except v = 9, there exists a cyclic STS(v).

We move on to consider an analog of Steiner triple systems. An m-cycle system of order v is a pair (V; C), where V = V (Kv) and C is a collection of edge-disjoint m-cycles which partition the edge set of Kv. Let J be an automorphism

groupof the m-cycle system (V; C) (i.e., a groupof permutations on v vertices leaving the collection C of cycles invariant). If there is an automorphism  ∈ J of order v, then the m-cycle system (V; C) is said to be cyclic. For an m-cycle system of Kv, the vertex set V can be identi<ed with Zv. It is easy to see the necessary conditions for such a decomposition are

(i) v is odd and (ii) m|v 2

 .

The study of “existence problem” of m-cycle systems started around 40 years ago. Recently, Alspach and Gavlas [2] and LSajna [10] proved that an m-cycle system exists as long as the above conditions are met. Thus, we have all the m-cycle systems for each m ¿ 3.

1_{Research Supported by NSC 89-2115-M-009-041.}

E-mail address:hlfu@math.nctu.edu.tw(H.-L. Fu).

0012-365X/$ - see front matter c
2003 Elsevier B.V. All rights reserved.

Conjecture. Let m1; m2; : : : ; mh be positive integers not less than 3 such that _i=1mi = ₂ for odd n (respectively,

_h

i=1 mi=n₂− n=2 for even n). Then Kn (respectively, Kn− F) can be decomposed into cycles C1; C2; : : : ; Ch such that

the length of Ci is mi for i = 1; 2; : : : ; h.

In this paper, we prove a special case of the conjecture, namely, we prove that if m1; m2; : : : ; mk are positive integers

all at least 3, then the complete graph K2n+1, where n =
ki=1mi, has a cyclic decomposition into k(2n + 1) cycles such

that for each i = 1; 2; : : : ; k, there are exactly 2n + 1 cycles of length mi.

2. The main results

Throughout this paper, we shall use diIerence methods. The di5erence between two vertices x and y in the complete graph Kn with V (Kn) = Zn is |x − y| or n − |x − y|, whichever is smaller. We will say that the edge xy has diIerence

min{|x − y|; n − |x − y|}. Thus, the set of diIerences possible in Kn is {1; 2; : : : ; n=2} and each diIerence induces a

2-factor except the diIerence n=2 induces a 1-factor whenever n is even. For convenience, we shall use G[D] to denote the subgraph of G induced by the set of diIerences D ⊆ {1; 2; : : : ; ‘}. It is easy to check that K2‘+1[i] is a disjoint union

of cycles of length (2‘ + 1)=(2‘ + 1; i), where (2‘ + 1; i) denotes the greatest common divisor of 2‘ + 1 and i. Clearly, if (2‘ + 1; i) = 1, then K2‘+1[i] is a Hamiltonian cycle in K2‘+1. It should be mentioned that if cycles Ci (1 6 i 6 k) have

diIerence sets Ai which partition {1; 2; : : : ; ‘}, then there exists a cyclic decomposition of K2‘+1 into cycles Ci.

Notice that if H is a subgraph of K2‘+1such that each edge of H has a distinct diIerence, then the graph H +i obtained

from H by adding i (mod 2‘ + 1) to each vertex of H is an isomorphic copy of H. The following results are given in [13] and will be used in the proof of Theorem 2.5.

Lemma 2.1 (Wu [13]). For positive integers b and s, there exists a cycle C of length 4s with di5erence set {b; b + 1; : : : ; b + 4s − 1}

in Kn where n is odd with n ¿ 2(b + 4s − 1) + 1.

Lemma 2.2 (Wu [13]). Let b and s be positive integers. (1) There exists a cycle C of length 4s + 2 with di5erence set

{b; b + 1; : : : ; b + 4s; b + 4s + 2}

in Kn where n is odd with n ¿ 2(b + 4s + 2) + 1.

(2) There exists a cycle C of length 4s + 2 with di5erence set {b; b + 2; b + 3; : : : ; b + 4s + 2}

in Kn where n is odd with n ¿ 2(b + 4s + 2) + 1.

Note that one may use a consecutive block of integers to construct cycles of length congruent to 0 modulo 4 and/or an even number of cycles of length congruent to 2 modulo 4. For example, if m1= 4s + 2 and m2= 4t + 2, then applying (1)

and (2) of Lemma2.2give cycles C1 and C2 of lengths m1 and m2 with diIerence sets {b; b + 1; : : : ; b + 4s; b + 4s + 2}

and {b + 4s + 1; b + 4s + 3; : : : ; b + 4s + 4t + 3}, respectively, for any positive integer b. For convenience, in the following lemmas, we use a typical odd cycle as in Fig. 1.

Lemma 2.3. For positive integers a; b; c, and r, with c = a + b and r ¿ c, and a nonnegative integer s, there exists a cycle C of length 4s + 3 with di5erence set {a; b; c; r; r + 1; : : : ; r + 4s − 1} in Kn where n is odd and n ¿ 2(r + 4s − 1) + 1.

Fig. 1.

Proof. The proof is divided into two cases. Case 1: Either a or b is odd, say b.

The cycle C of length 4s + 3 is de<ned as the following:

An easy veri<cation shows that the vertices of the cycle C are: for i = 0; 1; : : : ; s, v2i+1= a + 2i, v2i+1
= c + 2i,

v2i= a − r − 2(i − 1), v
2i = c + r + 4s − 2i + 1, where all indices are taken modulo n, and the diIerence set is

{a; b; c; r; r + 1; : : : ; r + 4s − 1}. Observe that since c = a + b and b is odd, it follows that a and c have opposite parity. Thus a; a + 2; : : : ; a + 2s and c; c + 2; : : : ; c + 2s have opposite parity and hence are distinct. Also, c + r + 4s − 1, c + r + 4s − 2; : : : ; c + r + 2s + 1 and a − r ; a − r − 2; : : : ; a − r − 2s − 2 have opposite parity when considered modulo n and thus are distinct. Therefore, the vertices of C are distinct.

Case 2: Both a and b are even.

(i) r is even: Let e1= a, e
1= r + 4s − 2, e2= r + 4s − 1, e
2= c, e2s+2= b, and for i = 3; 4; : : : ; 2s + 1, let ei= r +

4s − 2i + 3, e

i = r + 2i − 6. Now, we de<ne the vertices accordingly. Let v0= 0, v1= a, v
1= r + 4s − 2 and for

i = 1; 2; : : : ; s; v2i= a + r + 4s − 2i + 1, v
2i= c + r + 4s + 2i − 4, v2i+1= a + 2i, and v
2i+1= c + 4s − 2i.

(ii) r is odd: In this subcase, we let e1= a, e
1= r + 4s − 1, e2= r + 4s − 2, e2
= c; e2s+2= b, and for i = 3; 4; : : : ; 2s +

1; ei= r + 4s − 2i + 2, e
i = r + 2i − 5. Then according to the diIerences, we de<ne the vertices for the cycle. Let

v0= 0, v1= a, v1
= r + 4s − 1, and for i = 1; 2; : : : ; s, v2i= a + r + 4s − 2i, v2i+1= a + 2i, v
2i= c + r + 4s + 2i − 3,

and v

2i+1= c + 4s − 2i.

Remark. In Lemma 2.3, if c = a + b ± 2, then we can use a similar method mentioned above to construct a cycle of length 4s + 1 with diIerence set {a; b; c; r; r + 1; : : : ; r + 4s − 4; r + 4s − 2}. Notice that this construction will be used in Theorem2.5. See, for example, Fig.2, where a = 2; b = 3; c = 7; s = 3, and r = 9.

Notice that the value of a + b is even (odd) and c is odd (even). By routine computation, it follows that the diIerence set is {a; b; c; r; r + 1; : : : ; r + 4s − 3}, and the distinct vertices of C are: v1= a, v
1= c, v2s= c + 2s − 3, v
2s= c + r + 2s − 2,

and for i = 1; 2; : : : ; s − 1, v2i= c + 2i − 3, v
2i= c + r + 4s − 1 − 2i, v2i+1= c − r − 2i − 1, and v2i+1
= c + 2i.

In order to prove the main theorem, we also need to use Skolem sequences, hooked Skolem sequences, and near Skolem sequences.

A Skolem sequence of order n is a sequence (s1; s2; : : : ; s2n) such that for each j ∈ {1; 2; : : : ; n}, there exists a unique

i ∈ {1; 2; : : : ; 2n} such that si= si+j= j. It is proved by Skolem [12] that such a sequence exists if and only if n ≡ 0 or 1 (mod 4).

A hooked Skolem sequence of order n is a sequence (s1; s2; : : : ; s2n+1) such that s2n= 0 and for each j ∈ {1; 2; : : : ; n},

there exists a unique i ∈ {1; 2; : : : ; 2n − 1; 2n + 1} such that si= si+j= j. These sequences are known to exist if and only

if n ≡ 2, 3 (mod 4) (see [6]).

An m-near Skolem sequence of order n (m 6 n) is a sequence (s1; s2; : : : ; s2n−2) of 2n − 2 integers which satis<es for

every j ∈ {1; 2; : : : ; n} \ {m}, there exists a unique i ∈ {1; 2; : : : ; 2n − 2} such that si= si+j= j. It is proved by Shalaby

[11] that an m-near Skolem sequence of order n exists if and only if either (1) m is odd and n ≡ 0 or 1 (mod 4) or (2) m is even and n ≡ 2 or 3 (mod 4).

Remark that a Skolem sequence (s1; s2; : : : ; s2n) of order n gives a partition of {1; 2; : : : ; 3n} into triples {j; n + i; n + i +

j|si= si+j= j} for j = 1; 2; : : : ; n. Similarly, a hooked Skolem sequence gives a partition of {1; 2; : : : ; 3n − 1; 3n + 1} into

triples {j; sj; tj} satisfying j+sj=tj (1 6 j 6 n) and an m-near Skolem sequence gives a partition of {1; 2; : : : ; 3n−2}\{m}

into triples {j; sj; tj} satisfying j + sj= tj for each j ∈ {1; 2; : : : ; n} \ {m}.

Now, we are ready for the proof of our main result.

Theorem 2.5. Let m1; m2; : : : ; mk be positive integers not less than 3 such that n =
ki=1mi. Then there exists a cyclic

(m1; m2; : : : ; mk)-cycle system of order 2n + 1.

Proof. For convenience, let m1; m2; : : : ; mi1 denote the integers which are congruent to 3 modulo 4, mi1+1; mi1+2; : : : ; mi2

denote the integers which are congruent to 1 modulo 4, mi2+1; mi2+2; : : : ; mi3 denote the integers which are congruent to

0 modulo 4, and thus mi3+1; mi3+2; : : : ; mk will be the integers which are congruent to 2 modulo 4. It suBces to partition

the set {1; 2; : : : ; n} into sets A1; A2; : : : ; Ak such that:

• |Ai| = mi for each i with 1 6 i 6 k;

• each of the sets A1; A2; : : : ; Ai1 satis<es the conditions for the diIerence set given in Lemma2.3;

• each of the sets Ai1+1; Ai1+2; : : : ; Ai2 satis<es the conditions for the diIerence set given in Lemma2.4or the remark after Lemma 2.3;

• each of the sets Ai2+1; Ai2+2; : : : ; Ai3 satis<es the conditions for the diIerence set given in Lemma2.1; and

• each of the sets Ai3+1; Ai3+2; : : : ; Ak satis<es the conditions for the diIerence set given in either (1) or (2) in

Case 1: Suppose that i2≡ 0, 1 (mod 4). Clearly, if i2= 0, then it is easy to de<ne the sets A1; A2; : : : ; Ak by choosing

the diIerences for the cycles of length congruent to 0 modulo 4 <rst followed by choosing the diIerences for those cycles of length congruent to 2 modulo 4 last, using n + 1 for n as necessary. In fact, after de<ning the sets A1; A2; : : : ; Ai2,

if we left with a set {b; b + 1; : : : ; n} for some positive integer b, then we can easily choose the diIerences for sets Ai2+1; Ai2+2; : : : ; Ak.

Since i2 ≡ 0, 1 (mod 4), there exists a Skolem sequence of order i2 such that the set {1; 2; : : : ; 3i2} can be partitioned

into triples {i; si; ti} with i + si= ti for i = 1; 2; : : : ; i2. Suppose <rst that i2= i1 so that there are no cycles of length

congruent to 1 modulo 4. Then, the sets A1; A2; : : : ; Ai1 are de<ned as follows:

• 1, s1; t1∈ A1,

• 2, s2; t2∈ A2,

.. .

• i1, si1, ti1∈ Ai1, and

• starting with 3i1+ 1, assign the next m1− 3 consecutive integers to A1, the next m2− 3 consecutive integers to A2 and

so on until assigning mi1− 3 consecutive integers to Ai1.

Observe that the diIerences left are
i1 i=1 mi+1,

_i1

i=1 mi+2; : : : ; n, and as remarked earlier, the sets Ai2+1; Ai2+2; : : : ; Ak

are easily found.

Now suppose that i2¿ i1. Suppose <rst that i2− i1 is even. De<ne the sets A1; A2; : : : ; Ai1 as follows:

• 1, s1, t1∈ A1, • 2, s2, t2∈ A2, .. . • i1, si1, ti1∈ Ai1, • i1+ 2, si1+1, ti1+1∈ Ai1+1, • i1+ 1, si1+2, ti1+2∈ Ai1+2, .. . • i2, si2−1, ti2−1∈ Ai2−1, • i2− 1, si2, ti2∈ Ai2, and

• starting with 3i2+ 1, assign the next m1− 3 consecutive integers to A1, the next m2− 3 consecutive integers to A2 and

so on until assigning mi2− 3 consecutive integers to Ai2.

The diIerences remaining are
i2

i=1 mi+ 1,

_i2

i=1 mi+ 2; : : : ; n and the sets Ai2+1; Ai2+2; : : : ; Ak are easily found.

Now assume that i2 − i1 is odd. Then, by [11], there exists a 1-near Skolem sequence of order i2 so that the set

{2; 3; : : : ; 3i2− 2} can be partitioned into triples {i; si; ti} with i + si= ti for i = 2; : : : ; i2. If there are no cycles of length

congruent to 2 modulo 4, then we de<ne A1; A2; : : : ; Ai2 as follows:

• 2, s2, t2∈ A1, • 3, s3, t3∈ A2, .. . • i1+ 1, si1+1, ti1+1∈ Ai1, • i1+ 3, si1+2, ti1+2∈ Ai1+1, • i1+ 2, si1+3, ti1+3∈ Ai1+2, .. . • i2− 1, si2, ti2∈ Ai2−1, • 1, n − 1, n + 1 ∈ Ai2, and

• starting with 3i2− 1, assign the next m1− 3 consecutive integers to A1, the next m2− 3 consecutive integers to A2 and

• i1+ 2, si1+3, ti1+3∈ Ai1+2, .. . • i2− 1, si2, ti2∈ Ai2−1, • 1, 3i2− 1, 3i2+ 1 ∈ Ai2, • Ai3+1= {3i2; 3i2+ 2; 3i2+ 3; : : : ; 3i2+ mi3+ 1}, and

• starting with 3i2+ mi3+1+ 1, assign the next m1− 3 consecutive integers to A1, the next m2− 3 consecutive integers

to A2 and so on until assigning mi2− 3 consecutive integers to Ai2.

In either case, the diIerences remaining form a set of consecutive integers, and the sets Ai2+1; Ai2+2; : : : ; Ai3−1; Ai3+1; : : : ; Ak

are easily found.

Now, we have completed the partition of {1; 2; : : : ; n} into the sets A1; A2; : : : ; Ak. By Lemmas 2.1–2.4, we are able to obtain a cyclic cycle decomposition for K2n+1. This concludes the proof of the case when i2≡ 0 or 1 (mod 4).

Case 2: Suppose that i2≡ 2, 3 (mod 4). The proof can be obtained by using a procedure similar to those in Case 1,

and so we omit the details.

For clearness, we present two examples to show the idea of partition.

Example 1. n = 51. (m1; m2; : : : ; m10) = (3; 3; 7; 5; 9; 4; 4; 4; 6; 6), i2= 5. A1= {1; 6; 7}, A2= {2; 12; 14}, A3= {3; 8; 11; 16; 17, 18; 19}, A4 = {5; 9; 13; 20; 21}, A5 = {4; 10; 15; 22; 23; 24; 25; 26; 27}, A6 = {28; 29; 30; 31}, A7 = {32; 33; 34; 35}, A8 = {36; 37; 38; 39}, A9= {40; 41; 42; 43; 44; 46}, and A10= {45; 47; 48; 49; 50; 51}. Example 2. n = 51. (m1; m2; : : : ; m10) = (3; 3; 3; 7; 5; 5; 9; 4; 6; 6), i2= 7. A1= {1; 8; 9}, A2= {3; 14; 17}, A3= {4; 11; 15}, A4= {5; 13; 18; 22; 23; 24; 25}, A5= {7; 10; 16; 26; 27}, A6= {6; 12; 19; 28; 29}, A7= {2; 20; 21; 30; 31; 32; 33; 34; 35}, A8= {36; 37; 38; 39}, A9= {40; 41; 42; 43; 44; 46}, and A10= {45; 47; 48; 49; 50; 51}.

With the theorem we proved, the following known result can be obtained easily.

Corollary 2.6. For each m ¿ 3 there exists a cyclic m-cycle system of order v ≡ 1 (mod 2m).

Proof. Since v = 2km + 1 for some integer k ¿ 1, by Theorem2.5where we choose n = mk, we conclude the proof. Remark. By an independent eIort, we can also obtain the consequence, that is, for each odd prime p, there exists a cyclic p-cycle system.

Acknowledgements

The authors wish to thank the referees for their valuable comments and suggestions, and in particular, one of the referees for his/her considerable eIort to rewrite this paper into a more readable form.

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