I want to set up a rotation of sixteen couples with four couples per house so that all couples eventually have dinner together, no repetition. Each couple is to host one dinner. Meetings are monthly over four or five months. I have done it twice now with great difficulty. There must be an easy way to figure this out. The group may expand to twenty couples this year. Can someone please help me with this. Thanks so much.

  • $\begingroup$ Observation: Each couple can meet at most three new couples per dinner, so you'll certainly need at least five dinners... $\endgroup$ – Ben Millwood Oct 4 '13 at 12:06
  • $\begingroup$ "There must be an easy way to figure this out." Typically false for this kind of problem. I think what you have is a variant on the "social golfer problem"; do a search for that, and let us know whether it helps. $\endgroup$ – Gerry Myerson Oct 4 '13 at 12:48

First number is the host couple;

spring 1,14,15,16; 2,9,11,12; 3,6,7,10; 4,5,8,13; summer 5,1,10,12; 6,11,13,14; 7,2,4,15; 8,3,9,16; fall 9,1,4,6; 10,2,8,14; 11,3,5,15; 12,7,13,16; winter 13,1,2,3; 14,5,7,9; 15,6,8,12; 16,4,10,11;


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