The Honeymoon Oberwolfach Problem: small cases

The Honeymoon Oberwolfach Problem HOP (2m₁, 2m₂, , 2mₜ) asks the following question. Given n=m₁+m₂+ +mₜ newlywed couples at a conference and t round tables of sizes 2m₁, 2m₂, , 2mₜ, is it possible to arrange the 2n participants at these tables for 2n-2 meals so that each participant sits next to their spouse at every meal, and sits next to every other participant exactly once? A solution to HOP (2m₁, 2m₂, , 2mₜ) is a decomposition of K₂₍+ (2n-3) I, the complete graph K₂₍ with 2n-3 additional copies of a fixed 1-factor I, into 2-factors, each consisting of disjoint I-alternating cycles of lengths 2m₁, 2m₂, , 2mₜ. The Honeymoon Oberwolfach Problem was introduced in a 2019 paper by Lepine and Sajna. The authors conjectured that HOP (2m₁, 2m₂, , 2mₜ) has a solution whenever the obvious necessary conditions are satisfied, and proved the conjecture for several large cases, including the uniform cycle length case m₁==mₜ, and the small cases with n 9. In the present paper, we extend the latter result to all cases with n 20 using a computer search.