4

Okay, I've seen this question asked multiple ways, but I cannot get my head wrapped around it! With 9 people, I can have a schedule where each person participates in 4 meetings of 3 people per meeting, and every person will have been in a meeting with each other person in exactly one meeting. With 25 people, I can likewise have a schedule where each person participates in 6 meetings with 5 people per meeting, and every person will have been in a meeting with each other person exactly once.

Is there a schedule of 16 people where each person participates in 5 meetings of 4 people per meeting, and every person will meet each other exactly once?

Put another way: Is there a hypergraph with exactly $16$ vertices, such that each hyperedge has exactly $4$ vertices, each vertex is in exactly $5$ hyperedges, and for every pair of distinct vertices, there is exactly one hyperedge containing both vertices. Answer yes with an explicit construction, or no with a proof as to why there is no such hypergraph.

Mike
  • 23,484
Tom Rossi
  • 141
  • I added some additional tags. I don't know for sure myself, but I conjecture that it may be because $3$ and $5$ are primes whereas $4$ is not prime – Mike May 04 '22 at 23:17
  • @layabout for my question, the 16 people would each be in 5 different meetings with 4 people per meeting. – Tom Rossi May 04 '22 at 23:25
  • The subject you want to search for is "design theory". – Greg Martin May 04 '22 at 23:32
  • 1
    @Mike Look at the very bottom of saulspatz's answer. There are sixteen vertices, numbers 1 to 16. In each grid, the numbers are colored in four colors; each of those colors is a hyperedge. Since there are five grids, each vertex is contained in exactly five hyperedges. The hyperedges containing 0 (all white) are [0,1,2,3] in the first, [0,4,8,12] in the 2nd, [0,5,10,15] in the 3rd grid, [0,6,11,13] in the 4th, and [0,7,9,14] in the bottom. Note 0 is paired with everyone else exactly once (not at most once, as you claimed). – Mike Earnest May 05 '22 at 17:07
  • You are correct @MikeEarnest my apologies and thanks for pointing this out! – Mike May 05 '22 at 17:10

0 Answers0