Hamilton Cycle Decomposition of6-Regular Circulants of Odd Order

Matthew DeanCentre for Discrete Mathematics and Computing, Department of Mathematics,The University of Queensland, Queensland 4072, Australia,E-mail: mdean@maths.uq.edu.au

Received June 23, 2005; revised March 16, 2006

Published online 23 October 2006 in Wiley InterScience (www.interscience.wiley.com).DOI 10.1002/jcd.20118

Abstract: The circulant G = C(n, S), where S ⊆ Zn \ {0}, is the graph with vertex set Zn andedge set E(G) = {{x, x + s}|x ∈ Zn, s ∈ S}. It is shown that for n odd, every 6-regular connectedcirculant C(n, S) is decomposable into Hamilton cycles. © 2006 Wiley Periodicals, Inc. J CombinDesigns 15: 91–97, 2007

Keywords: Hamilton cycle decomposition; circulant; graph decomposition; graph factorization

1. INTRODUCTION

Given an Abelian group � and a subset S ⊆ � \ {0}, the Cayley graph G = Cay(�, S) isdefined to have vertex set V (G) = � and edge set E(G) = {{x, x + s}|x ∈ �, s ∈ S}. Theset S is called the connection set. Note that the graph Cay(�, S) is connected if and onlyif S generates the group �. In this paper all graphs are simple, without loops. If s ∈ S hasorder two in �, then Cay(�, {s}) is a 1-factor in G. If s ∈ S has order greater than two, thenCay(�, {s}) = Cay(�, {−s}) and is a 2-factor in G. It is well known [4] that every connectedCayley graph on a finite Abelian group contains a Hamilton cycle. Alspach [1] conjecturedthat every connected 2k-regular Cayley graph on a finite Abelian group has a Hamilton cycledecomposition. That is, a partition of the edge set into k edge-disjoint Hamilton cycles.

Liu [8] has shown this conjecture is true when S is a strongly minimal generating set of �

(S is a strongly minimal generating set if for each si ∈ S, 2si does not belong to the subgroupgenerated by S \ {si}). This implies that the conjecture is true for all Cayley graphs of odd

Contract grant sponsor: Australian Postgraduate Award.

© 2006 Wiley Periodicals, Inc.

91

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order with S a minimal generating set (S is a minimal generating set if each element si ∈ S

does not belong to the subgroup generated by S \ {si}).The conjecture is trivial for the case k = 1. Bermond et al. [3] proved that the conjecture

is true for k = 2. Liu and Fan et al. have also found partial results for the case k = 3. Theseinclude Theorems 2.7 and 2.8 in [7], Theorems 3.15 and 3.17 in [6], and Theorem 32 in[4]. Another partial result for the case k = 3 was proven in [5]. This is Theorem 2.5 below.There are no known exceptions to Alspach’s conjecture, but the general problem remainslargely unsolved for k ≥ 3.

A circulant is a Cayley graph on a cyclic group. In this paper we denote a circulantCay(Zn, S) by C(n, S). For 2k-regular circulants C(n, S), we assume without loss ofgenerality, that si < n

2 for each si ∈ S. In this article we settle the case k = 3 for circulantsof odd order.

Main Theorem. Every 6-regular connected circulant of odd order has a Hamilton cycledecomposition.

2. PRELIMINARY RESULTS

Fan et al. [6] found Hamilton cycle decompositions of connected 4-regular Cayley graphsby representing them as pseudo-cartesian products of cycles and then applying switchingconfigurations. We review these concepts.

A. Pseudo-Cartesian Products

Given integers a ≥ 1, b ≥ 1, and r ∈ Zb, the r-pseudo-cartesian product Ca ×r Cb, is thesimple graph with vertex set Za × Zb and edge set

{(i, j), (i, j + 1)} for i ∈ Za, j ∈ Zb

{(i, j), (i + 1, j)} for i ∈ Za\{a − 1}, j ∈ Zb and{(a − 1, j), (0, j + r)} for j ∈ Zb

Following the convention in Reference [6], we draw a pseudo-cartesian product sothat vertex (i, j) is in the ith column and the jth row (not the ith row and jth col-umn). The variable r is referred to as the jump number since the edges from verticesin column a − 1 jump down r rows (modulo b) to connect to vertices in column 0.Note that an ordinary cartesian product of cycles is an r-pseudo-cartesian product withr = 0. Figure 1 shows C5 ×3 C6. For clarity of the figure, some of the edges are not fullydrawn.

The labeling of the vertices in Figure 1 illustrates that C(30, {5, 9}) is isomorphic toC5 ×3 C6. This is an example of the following well-known result, see [3,6].

Theorem 2.1. Any 4-regular connected circulant C(n, {s, t}) is isomorphic to a pseudo-cartesian product Ca ×r Cb where a = gcd(t, n), b = n/gcd(t, n), and 0 ≤ r < b. The iso-morphism maps vertex is +jt (modulo n) to vertex (i, j) for 0 ≤ i < a, and 0 ≤ j < b.

B. Switching configurations

Suppose that F1 and F2 are two 2-factors in a 2-factorizationF of a graph G and (α, β, γ, δ)is a 4-cycle in G such that {α, β}, {γ, δ} ∈ E(F1) and {β, γ}, {α, δ} ∈ E(F2). Let F ′

1 be the

HAMILTON CYCLE DECOMPOSITION 93

FIGURE 1. C (30, {5, 9}) ∼= C5 X3 C6

2-factor obtained from F1 by replacing {α, β} and {γ, δ} with {β, γ} and {α, δ}, and let F ′2

be the 2-factor obtained from F2 by replacing {β, γ} and {α, δ} with {α, β} and {γ, δ}. It isclear that F ′ = (F \ {F1, F2}) ∪ {F ′

1, F′2} is a new 2-factorization of G. We shall refer to

this process as a switch on the 4-cycle (α, β, γ, δ). For a given 2-factorizationF, a switchingconfiguration S is a set of switches involving edge disjoint 4-cycles. If we apply each ofthe switches in a switching configuration S to F, then we get a new 2-factorization whichwe denote by S(F). We also define S(Fi) for i = 1, 2, . . . , k in the obvious way so thatS(F) = {S(F1),S(F2), . . . ,S(Fk)}. Note that the order in which the switches are appliedmakes no difference to the resulting 2-factorization.

Figure 2 illustrates the use of a switching configuration S1 to find a decompositionof C(n, {s, t}) ∼= Cτ ×r Cm into Hamilton cycles. The initital 2-factorization {F1, F2} is{C(n, {s}), C(n, {t})}. The Hamilton cycle decomposition is {S1(F1),S1(F2)}. In the figure,each switch of S1 is designated by the symbol �, S1(F1) is shown by dashed lines andS1(F2) is shown by solid lines. This example is taken from the proof of Theorem 2.2 ([6]).It is for the cases where τ and m are both odd and greater than 1.

Theorem 2.2 (Fan et al [6]). Any pseudo-cartesian product can be decomposed into twoHamilton cycles.

The following results are also used in our proof.

Theorem 2.3 (Fan et al [6]). Suppose that � is an Abelian group of odd order and S ={s1, s2, s3} is a generating set of � such that ord(s1) ≥ ord(s2) > ord(s3), then Cay(�, S)can be decomposed into Hamilton cycles.

Theorem 2.4 (Aubert and Schneider [2]). If G can be decomposed into two Hamiltoncycles and C is a cycle, then G × C can be decomposed into three Hamilton cycles.

Theorem 2.5 ([5]). Every 6-regular circulant C(n, S) where ord(si) = n for some si ∈ S,has a Hamilton cycle decomposition.

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FIGURE 2. C(n, {s, t}) represented as Cτ ×r Cm.

3. NEW RESULTS

As we shall see later, we can now completely settle the Hamilton cycle decompositionproblem for 6-regular circulants of odd order if we settle the case where S has exactly twoelements of the same order. First we need the following technical result.

Lemma 3.1. Given odd integers σ ≥ 3, z ≥ 3 and m satisfying m = zσ, and a pos-itive integer q < m/2 coprime to m, then there is a triple {e − q(modulo m), e, e +q(modulo m)} ⊂ Zm \ {0, 1, 2, . . . , σ − 1}.Proof. If q < σ then σ + 2q < 3σ ≤ m so we may choose e = σ + q. If q > σ thenσ − q < 0 so we may choose e = σ. Note that q = σ since σ|m but gcd(q, m) = 1 andσ > 1. �Lemma 3.2. If G is a 6-regular connected circulant C(n, S) where n is odd and S hasexactly two elements of the same order, then G has a Hamilton cycle decomposition.

Proof. Since a 6-regular circulant has 3 generators, let our 6-regular circulant beC(n, {s, t, u}). We may assume that 0 < s, t, u < n/2. Let t and u be the two elementsof equal order. Thus gcd(t, n) = gcd(u, n) = τ for some integer τ. Also let σ = gcd(s, n).

HAMILTON CYCLE DECOMPOSITION 95

FIGURE 3. C(n, {s, u}) represented as Cτ ×r′ Cm.

Since C(n, {s, t, u}) is connected, this implies gcd(n, s, t, u) = 1 and gcd(σ, τ) = 1. Thus,C(n, {s, t}) and C(n, {s, u}) are both connected and 4-regular, and may be represented aspseudo-cartesian products Cτ ×r Cm (Fig. 2) and Cτ ×r′ Cm (Fig. 3) respectively. Alson = zστ for some positive integer z. The 2-factor F1 = C(n, {s}) consists of σ disjointcycles, each of length n/σ = zτ. In both Figures 2 and 3 each of these cycles con-sists of z horizontal rows and z diagonal edges. The 2-factor F2 = C(n, {t}) consists ofτ disjoint cycles, each of length m = n/τ = zσ. There is one cycle on each coset of〈t〉, the subgroup generated by t, in Zn. These are the vertical cycles in Figure 2. Simi-larly the 2-factor F3 = C(n, {u}) consists of τ disjoint m-cycles, one cycle on each cosetof 〈u〉 = 〈t〉 in Zn. These are the vertical cycles in Figure 3. The fact that 〈u〉 = 〈t〉also implies that each row in Cτ ×r′ Cm (Fig. 3) is also a row in Cτ ×r Cm (Fig. 2).In other words, the sequence of rows in Cτ ×r′ Cm (Fig. 3) is a permutation of thesequence of rows in Cτ ×r Cm (Figure 2). Note that {F1, F2, F3} is a 2-factorizationof C(n, {s, t, u}).

If either σ or τ equals 1, then at least one of s, t, or u has order n and Theorem 2.5shows that C(n, {s, t, u}) has a Hamilton cycle decomposition. If z = 1, then n = στ andC(n, {s, t, u}) is isomorphic to the ordinary cartesian product Cτ × C(m, {t/τ, u/τ}). Thegraph C(m, {t/τ, u/τ}) is 4-regular and connected, so by Theorems 2.1 and 2.2 it can bedecomposed into two Hamilton cycles. By Theorem 2.4 it follows that C(n, {s, t, u}) has a

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FIGURE 4. Switching configurations L (above) andR (below).

Hamilton cycle decomposition. Therefore, since n is odd, for the following we may assumeσ, τ, and z are all odd and greater than 1.

To construct a Hamilton cycle decomposition of C(n, {s, t, u}) we shall use two switchingconfigurations. The first is S1 shown in Figure 2. This acts on F1 and F2. As noted in the dis-cussion preceding Theorem 2.2, F1

′ = S1(F1) and F2′ = S1(F2) are both Hamilton cycles.

Note F3′ = S1(F3) = F3 and F ′ = {F1

′, F2′, F3

′} is a 2-factorization of C(n, {s, t, u}).A second switching configuration S2 applied to F1

′ and F3′ will give the sought decom-

position. For S2, we intend to use either L or R as shown in Figure 4 on three rows d, e,and f which are consecutive in Cτ ×r′ Cm (Fig. 3) but are not any of the rows involved inswitches in S1.

Three rows are consecutive in Cτ ×r′ Cm (Fig. 3), if and only if they are spaced q rowsapart in Cτ ×r Cm (Fig. 2), where q = min{q′, m − q′} and q′ = u/t (mod m). Note thatq is coprime to m. By Lemma 3.1, we can find the required three rows d = e − q, e, andf = e + q. Thus we may applyS2 = L orR as a switching configuration exchanging edgesbetween F1

′ (horizontal edges) and F3′ (vertical edges).

Regardless whether S2 is L or R, the resulting 2-factor S2(F3′) is a Hamilton cycle,

since the τ vertical cycles of F3′ are connected by the τ − 1 switches. However for S2(F1

′)to be a Hamilton cycle, sometimes we need S2 = L and sometimes S2 = R. Notice thatF1

′ can contain the three rows d, e, and f as subpaths in at most eight possible ways, listedas cases 1–8 below.

1. (d0T , . . . , e0T , . . . , f0T , . . .) 5. (d0T , . . . , f0T , . . . , e0T , . . .)2. (d0T , . . . , e0T , . . . , fT0, . . .) 6. (d0T , . . . , f0T , . . . , eT0, . . .)3. (d0T , . . . , eT0, . . . , f0T , . . .) 7. (d0T , . . . , fT0, . . . , e0T , . . .)4. (d0T , . . . , eT0, . . . , fT0, . . .) 8. (d0T , . . . , fT0, . . . , eT0, . . .)

where d0T = dt, dt + s, . . . , dt + (τ − 1)s, dT0 = dt + (τ − 1)s, . . . , dt + s, dt ande0T , eT0, f0T , and fT0 are defined similarly.

Figure 5 illustrates Case 3. The dashed line in the top figure shows how rows d, e, and f aresubpaths of F1

′. The bottom figure shows that if S2 = L then S2(F1′), indicated by a dashed

line, is also a Hamilton cycle. For cases 1,3,4, and 7, we letS2 = L and for cases 2,5,6, and 8letS2 = R. The other seven cases may be verified by examining similar diagrams, which weomit. Thus {S2(F1

′), F2′,S2(F3

′)} is a Hamilton cycle decomposition of C(n, {s, t, u}). �

HAMILTON CYCLE DECOMPOSITION 97

FIGURE 5. F1′ (above) and F1

′′ (below), for case 3.

Proof of Main Theorem. The generating set S for a 6-regular circulant C(n, S) has threeelements. Call them s1, s2, and s3. If one, say s3 has smallest order, Theorem 2.3 guaranteesthat the circulant is Hamilton cycle decomposable. If exactly two, say s2 and s3 have equalsmallest order then Lemma 3.2 applies. Finally, suppose all three of s1, s2, and s3 have thesame order. Since the graph is connected, gcd(s1, s2, s3, n) = 1 and so gcd(si, n) = 1 foreach i. Thus {C(n, {s1}), C(n, {s2}), C(n, {s3})} is a Hamilton cycle decomposition. �

ACKNOWLEDGEMENT

The author thanks Darryn Bryant for careful supervision. This research was funded by anAustralian Postgraduate Award.

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