Symmetric Hamilton cycle decompositions of complete multigraphs

Chitra, V.; Muthusamy, Appu

doi:10.7151/dmgt.1687

Cited by 8 publications

(7 citation statements)

References 7 publications

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“…Also, two times all cycles of

scriptH

together with all cycles of the orbit of R give the required decomposition of

3 K_{2 n} - I

. Case 2:

n \equiv 0

(mod 4), say

n = 4 m

. As observed in , the existence of a cyclic and symmetric HCS

(2 K_{2 n})

for any n is implicitly contained in Lemma 3.5 in . Thus, we have only to prove the existence of a cyclic and symmetric HCS

(3 K_{2 n} - I) .

Note that the sequence

α = (0, 2, - 2, 4, - 4, \dots, 2 m - 2, 2 - 2 m, 2 m, 2 m - 1, 1 - 2 m, \dots, 3, - 3, 1, - 1)

is n ‐transversal so that we can consider the cycle

A : = C_{italicrot} (α, n)

.…”

Section: Existence Of Cyclic Symmetric Hcss Of the Multigraphs λK2n Amentioning

confidence: 92%

“…So if

Γ = λ K_{v}

, then ψ fixes exactly one vertex or no vertex according to whether v is odd or even, respectively, while if

Γ = λ K_{v} - I

, then ψ is the permutation of

V (K_{v})

switching all pairs of endpoints of the edges of I . These symmetry requirements have been given by Akiyama, Kobayashi, and Nakamura for

Γ = K_{v}

, then Brualdi and Schroeder have considered the case

Γ = K_{v} - I

, and the conditions have furthermore been extended to multigraphs by Chitra and Muthusamy .…”

Section: Introductionmentioning

confidence: 99%

“…A symmetric HCS(K 2n − I ) exists if and only if n ≡ 1 or 2 (mod 4). Theorem 1.2 (Chitra and Muthusamy [10]). There exists a symmetric HCS(λK 2n − I ) for any odd λ > 1 and a symmetric HCS(λK 2n ) for any even λ.…”

Section: Introductionmentioning

confidence: 99%

“…An HCS( ) is cyclic if it is invariant under a cyclic permutation of all vertices of K v . Here are, for instance, the cycles of a cyclic HCS of K 12 − I with V (K 12 ) = Z 12 and I = {[0, 6], [1,7], [2,8], [3,9], [4,10], [5,11]}, invariant under the cyclic permutation x → x + 1 (mod 12):…”

Section: Introductionmentioning

confidence: 99%

“…(0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 1); (1,4,3,6,5,8,7,10,9,0,11,2); (0, 2, 7, 3, 5, 10, 6, 8, 1, 9, 11, 4); (1,3,8,4,6,11,7,9,2,10,0,5); (2,4,9,5,7,0,8,10,3,11,1,6).…”

Section: Introductionmentioning

confidence: 99%

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Hamiltonian Cycle Systems Which Are Both Cyclic and Symmetric

Buratti

Merola

2013

J of Combinatorial Designs

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The notion of a symmetric Hamiltonian cycle system (HCS) of a graph Γ has been introduced and studied by J. Akiyama, M. Kobayashi, and G. Nakamura [J Combin Des 12 (2004), 39–45] for Γ=Kv, by R. A. Brualdi and M. W. Schroeder [J Combin Des 19 (2011), 1–15] for Γ=Kv−I, and then naturally extended by V. Chitra and A. Muthusamy [Discussiones Mathematicae Graph Theory, to appear] to the multigraphs Γ=λKv and Γ=λKv−I. In each case, there must be an involutory permutation ψ of the vertices fixing all the cycles of the HCS and at most one vertex. Furthermore, for Γ=λKv−I, this ψ should be precisely the permutation switching all pairs of endpoints of the edges of I. An HCS is cyclic if it is invariant under some cyclic permutation of all the vertices. The existence question for a cyclic HCS of Kv−I has been completely solved by Jordon and Morris [Discrete Math (2008), 2440–2449]—and we note that their cyclic construction is also symmetric for v≡4 (mod 8). It is then natural to study the existence problem of an HCS of a graph or multigraph Γ as above which is both cyclic and symmetric. In this paper, we completely solve this problem: in the case of even order, the final answer is that cyclicity and symmetry can always cohabit when a cyclic solution exists. On the other hand, imposing that a cyclic HCS of odd order is also symmetric is very restrictive; we prove in fact that an HCS of λK2n+1 with both properties exists if and only if 2n+1 is a prime.

show abstract

“…Also, two times all cycles of

scriptH

together with all cycles of the orbit of R give the required decomposition of

3 K_{2 n} - I

. Case 2:

n \equiv 0

(mod 4), say

n = 4 m

. As observed in , the existence of a cyclic and symmetric HCS

(2 K_{2 n})

for any n is implicitly contained in Lemma 3.5 in . Thus, we have only to prove the existence of a cyclic and symmetric HCS

(3 K_{2 n} - I) .

Note that the sequence

α = (0, 2, - 2, 4, - 4, \dots, 2 m - 2, 2 - 2 m, 2 m, 2 m - 1, 1 - 2 m, \dots, 3, - 3, 1, - 1)

is n ‐transversal so that we can consider the cycle

A : = C_{italicrot} (α, n)

.…”

Section: Existence Of Cyclic Symmetric Hcss Of the Multigraphs λK2n Amentioning

confidence: 92%

“…So if

Γ = λ K_{v}

, then ψ fixes exactly one vertex or no vertex according to whether v is odd or even, respectively, while if

Γ = λ K_{v} - I

, then ψ is the permutation of

V (K_{v})

switching all pairs of endpoints of the edges of I . These symmetry requirements have been given by Akiyama, Kobayashi, and Nakamura for

Γ = K_{v}

, then Brualdi and Schroeder have considered the case

Γ = K_{v} - I

, and the conditions have furthermore been extended to multigraphs by Chitra and Muthusamy .…”

Section: Introductionmentioning

confidence: 99%

Section: Introductionmentioning

confidence: 99%

Section: Introductionmentioning

confidence: 99%

“…(0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 1); (1,4,3,6,5,8,7,10,9,0,11,2); (0, 2, 7, 3, 5, 10, 6, 8, 1, 9, 11, 4); (1,3,8,4,6,11,7,9,2,10,0,5); (2,4,9,5,7,0,8,10,3,11,1,6).…”

Section: Introductionmentioning

confidence: 99%

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