I.e.: could the gentlemen circle a rectangular table in a clockwise fashion, then rearrange themselves and continue in another fashion such that given any number of men, every man would be paired with every other man in the smallest number of iterations and without pairing two men together twice.

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 mathematically...@Qiaochu 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 $3+3=6$ 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 $2n$ vertices - see, e.g., en.wikipedia.org/wiki/Graph_factorization The odd case is the same as math.stackexchange.com/questions/54846/…

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.

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Please check out our SCHEDULE of upcoming events to register for our next Gay speed dating party.If there are no upcoming Gay speed dating events, EMAIL US to request one."Hi Pre Dating: Myself and Theresa (pictured) were the first couple to chat at a Honolulu Pre Dating event (venue All-Star Hawaii) in March of last year.After a successful "match" and quick exchange of e-mails, I asked Theresa to a local concert.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).With straight speed dating, (assuming the number of men and women are equal) the number of iterations that need to be made before every man has chatted with every woman is N/2, where N is the total number of people. Counting combinations in gay speed dating manually, I see the following numbers: These numbers suggest that gay speed dating can be done with N or N-1 iterations (albeit in a much more chaotic pattern). Also, if it is N iterations, would there be a pattern that could be followed?