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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 $\frac$ 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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Считаю, что свою роль сыграла цель моего пребывания там, то есть я не шла искать себе мужа!

Поэтому в тот вечер была естественной, общительной и очень веселой! Оказалось — реальность, Лариса и Евгений прямое этому подтверждение.

Обоих на вечер знакомств в Москве от Fast Life 15 августа пригласили друзья. Осенью прошлого года я посетила один из вечеров свиданий Fast Life и это вылилось в серьёзные отношения!

И пусть, я буквально настояла на своём участии в быстрых знакомствах :-) именно 22 сентября, огромное вам спасибо!

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).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?С 7 по 9 ноября 2014 года Fast Life провел самое массовое романтическое свидание на высоте 337м.В честь дня Рождения Останкинской телебашни более сотни человек встретились на самой высокой смотровой площадке Европы.Мы познакомились здесь, в Fast Life, на мгновенных свиданиях. Я тоже так думала, но сейчас могу смело заявить: Еще как бывает!

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