Then again, I figure if everyone in the dating schematic is gay, the two-gender case reduces to a maximum of the separate one-gender cases, so it's not really a problem [email protected] Yuan: I think he means counting the number of maximal sets of disjoint edges in a complete graph on $n$ vertices. With 3 people, you need 3 rounds (because everyone has to sit out a round), not 2.I think I encountered something like this problem (including the proper terminology) in a graph theory textbook once, but I don't have easy access to it at the moment. With 6 people with your method, you seem to require +3=6$ rounds, but it is possible with 5 rounds, as the OP says. The technical term for what Ross has done (in the even case) is finding a 1-factorization of the complete graph on n$ vertices - see, e.g., en.wikipedia.org/wiki/Graph_factorization The odd case is the same as math.stackexchange.com/questions/54846/…Here's an interesting problem that I came up with the other night. Assuming in straight speed dating, the men stay at their tables, the "sitting" men in gay speed dating won't meet one another (nor will the "standing" men).

I think the question can be rephrased more formally as: Given a set $S$ of $n$ elements, what is the shortest sequence $C_i$ of sets of unordered pairs in $S$ such that each unordered pair occurs in exactly one $C_i$ and no pairs in a given $C_i$ "overlap"? Imagine a long table with a seat at one end and $\frac{N-1}{2}$ seats along each long side. After each round, each person moves one seat clockwise. This gets us N-1 rounds in the even case, which is optimal. I run gay speed dating events and have the seating charts for 12 participants up to 22.

You should be able to convince yourself that each person meets each other after N rounds, but you can't do better as each person needs to meet N-1 others and has to sit out once. It's a little complex, but essentially you split the room into 2 parts, and then have 1/2 the room meet the other 1/2.

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