{"id":574,"date":"2026-05-04T19:20:51","date_gmt":"2026-05-04T19:20:51","guid":{"rendered":"https:\/\/sites.cs.queensu.ca\/ocw2026\/?post_type=tribe_events&#038;p=574"},"modified":"2026-05-04T19:21:06","modified_gmt":"2026-05-04T19:21:06","slug":"the-honeymoon-oberwolfach-problem-a-recursive-approach","status":"publish","type":"tribe_events","link":"https:\/\/sites.cs.queensu.ca\/ocw2026\/event\/the-honeymoon-oberwolfach-problem-a-recursive-approach\/","title":{"rendered":"The Honeymoon Oberwolfach Problem: A Recursive Approach"},"content":{"rendered":"<p><strong>Mary Rose Jerade, University of Ottawa<\/strong><\/p>\n<p>The Honeymoon Oberwolfach Problem (HOP) is a variant of the Oberwolfach Problem, which was first introduced by Mateja \u0160ajna. The problem asks whether it is possible to seat n couples attending a conference for 2(n \u2212 1) consecutive nights at l round tables of sizes 2m1, 2m2, . . . , 2ml such that mi \u2265 2 for all i = 1, . . . , l, \u03a3 i=1 mi = 2n, and each participant must be seated next to their spouse every night, but next to each other person exactly once. This problem is denoted by HOP(2m1, 2m2, . . . , 2ml).<\/p>\n<p>This seating arrangement problem smoothly translates into a graph theory problem. It is equivalent to finding a cycle decomposition of the graph K2n + (2n \u2212 3)I (that is, the complete graph K2n with 2n \u2212 3 additional copies of a fixed 1-factor I) into 2-factors such that each 2-factor consists of disjoint I-alternating cycles of lengths 2m1, . . . , 2ml. So far, solutions to HOP have been established for various cases, among which: for all n \u2264 20; when all tables are of the same size; when the table sizes are divisible by 8; and when n is odd and the Oberwolfach Problem OP(m1, . . . , ml) has a solution.<\/p>\n<p>In this talk, we present a recursive approach to the problem. In particular, we show that if m1, . . . , ml are positive even integers, t is an integer such that &amp;Sigma i=1 mi &lt; t, and HOP(2m1, . . . , 2ml) has a solution, then HOP(2m1, . . . , 2ml, 2t) has a solution as well. Moreover, we will briefly touch on the case when m1, . . . , ml are not all even.<\/p>\n<p>This is joint work with Mateja \u0160ajna.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Mary Rose Jerade, University of Ottawa The Honeymoon Oberwolfach Problem (HOP) is a variant of the Oberwolfach Problem, which was first introduced by Mateja \u0160ajna. The problem asks whether it 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