Least number of rounds to find all hidden pairs In my country there's a TV reality show in which $n+1$ men and $n$ women live in a house and, over the course of the show, they have to, as a group, find out who their 'match' is. $($In the actual TV show, $n=10)$.
Each man is assigned to a single woman such that every woman save for one has a single man assigned to her; this exceptional woman therefore has two men assigned to her.
At the start of the game, no assignment is known to any player.
Each round lasts a week, and on each round the players assemble into pairs (hence, one man is left out).
Then, each pair that is a match is revealed to be a correct pair.
The players win when $n$ correct pairs are assembled in a round.
What is the least number $r(n)$ of rounds that guarantees a win?

More generally, what kind of theory/bibliography is available to deal with questions like this?
Information theory? Combinatorial group theory? I confess I'm at a loss here, and devising a winning strategy here by 'trial and error' seems like an approach whose feasibility decreases very fast as $n$ increases.

A very slight improvement on the circle idea of @saulspatz makes $n$ rounds sufficient for a win, as follows.
Terminology: There is one woman matched with 2 men.  We'll call her the "special" woman, and her 2 men, the "special" men.
Algorithm: randomly set one man aside.  This leaves $n$ men and $n$ women.  Sit them in 2 concentric circles.  In each round pair off each man and woman facing each other.  Between each round the women move clockwise 1 space.  (This happens even if a match is found.)  After $n-1$ rounds, each man (except the one set aside) has met $n-1$ different women, and therefore he knows which woman he is matched with, because he has either met her, or, she is the only one he hasn't met.


*

*If the set-aside man is one of the "special" men, then the $n$ men in the circle match with different women.  Just pair them up.

*If both "special" men are among the $n$ (i.e. not set aside), then they both know who the "special" woman is.  Pair her up with one of them (doesn't matter which).  For the other $n-2$ men in the circle, pair each up with his match.  This leaves one woman whom none of the $n$ men are matched with.  Pair her up with the set-aside man.
I'm not sure this is the best solution though...
