On the honeymoon Oberwolfach problem

The honeymoon Oberwolfach problem HOP ( 2 m 1 , 2 m 2 , … , 2 m t ) asks the following question. Given n = m 1 + m 2 + ⋯ +m t newlywed couples at a conference and t round tables of sizes 2 m 1 , 2 m 2 , … , 2 m t , is it possible to arrange the 2 n participants at these tables for 2 n − 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 ( 2 m 1 , 2 m 2 , … , 2 m t ) is a decomposition of K 2 n + ( 2 n − 3 ) I , the complete graph K 2 nwith 2 n − 3 additional copies of a fixed 1‐factor I , into 2‐factors, each consisting of disjoint I ‐alternating cycles of lengths 2 m 1 , 2 m 2 , … , 2 m t . It is also equivalent to a semi‐uniform 1‐factorization ofK 2 nof type( 2 m 1 , 2 m 2 , … , 2 m t ) ; that is, a 1‐factorization { F 1 , F 2 , … , F 2 n − 1} such that for all i ≠ 1 , the 2‐factor F 1 ∪ F i consists of disjoint cycles of lengths 2 m 1 , 2 m 2 , … , 2 m t . In this paper, we first introduce the honeymoon Oberwolfach problem and then present several results. Most notably, we completely solve the case with uniform cycle lengths, that is, HOP ( 2 m , 2 m , … , 2 m ) . In addition, we show that HOP ( 2 m 1 , 2 m 2 , … , 2 m t ) has a solution in each of the following cases: n ≤ 9 ; n is odd and t = 2 ; as well as m i ≡ 0( mod 4 ) for all i . We also show that HOP ( 2 m 1 , 2 m 2 , … , 2 m t ) has a solution whenever n is odd and the Oberwolfach problem with tables of sizes m 1 , m 2 , … , m t has a solution.