The Hamilton–Waterloo problem for cycle sizes 3 and 4

Abstract

The Hamilton-Waterloo problem seeks a resolvable decomposition of the complete graph K n , or the complete graph minus a 1-factor as appropriate, into cycles such that each resolution class contains only cycles of specified sizes. We completely solve the case in which the resolution classes are either all 3-cycles or 4-cycles, with a few possible exceptions when n = 24 and 48. q

The Hamilton-Waterloo Problem for C3-Factors and Cn-Factors

“…Proof: Let the vertex set be Γ = Z 10 × Z 3 and the 1-factor be Cay(Γ, {5} × {0}). For (α, β) = (10, 4), we get the conclusion by using Construction 2.6 with an HW(10; 3, 5, 0, 4) and an HW(K 3 [10]; 3, 5, 10, 0) from Theorem 1.1. For all the other cases, the methods of generating the required α C 3 -factors and β C 5 -factors are listed in Table 1.…”

The Hamilton—Waterloo problem: The case of triangle‐factors and one Hamilton cycle

“…The first paper on this topic, [1], settled the problem for all odd n ≤ 17 and in addition proved that the necessary conditions for the existence of an HW(r, s; m, k) are sufficient when (m, k) ∈ { (3,5), (3,15), (5, 15)} except that an HW(6, 1; 3, 5) does not exist and the case HW( v−3 2 , 1; 3, 5) is unresolved for n ≡ 0 (mod 15) with n > 15 . In a recent paper [6] it is shown that the necessary conditions for the existence of an HW(r, s; 3, 4) are sufficient with 7 possible exceptions. It seems somewhat fitting that the seating arrangement in Hamilton should be one big cycle (a Hamilton cycle).…”

“…In this paper we will require the existence of 3-frames of the following types: 2 4 , 6 1 12 4 , 6 1 12 5 , 6 4 12 2 , 6 5 12 1 , 6 6 , 12 4 , 12 5 , and 12 6 . Existence of all of these frames (except for the one of type 6 4 12 2 ) is given in Section IV.5.2 of [5].…”

“…Additionally add three infinite points {x, y, z}. Now, in each inflated block of D (except b) put the blocks of a 3-frame of type either 12 4 , 12 5 or 12 6 (respecting the groups). On the first and second groups put the blocks of an HW(12k 1 + 3 : 12k 1 , 3) and a HW(12k 2 + 3 : 12k 2 , 3), respectively where the 3-cycle in the non-triangle 2-factor is (x, y, z).…”

A note on the Hamilton–Waterloo problem with C8-factors and CmWang1, Cao2 2018Discrete Mathematics90100View full textAdd to dashboardCite

“…Many authors have considered the Hamilton-Waterloo problem for small values of m and n. A complete solution for the existence of an HW(v; 3, n; α, β) in the cases n ∈ {4, 5, 7} is given in [1,7,14,21,24]. For the case (m, n) ∈ {(3, 15), (5,15), (4,6), (4,8), (4,16), (8,16)}, see [1].…”