TITLE: Least number of rounds to find all hidden pairs
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: This post is on the necessity (not sufficiency) side, and so I decided to start a new answer.  Hope this is acceptable etiquette.
Claim: The game cannot be won in $\lfloor (n+1)/2 \rfloor$ rounds.  In fact, stronger claim: For the first $\lfloor (n+1)/2 \rfloor$ rounds, no matter what pairings the players tried, the producer can say every such tried pair is a non-match (and still have a valid answer).
Obviously, this bound is extremely loose.  It is offered here only as a sharing of ideas.  In particular I hope the stronger claim above may be the basis for a much better lower bound.

The proof is an application of Hall's marriage theorem: https://en.wikipedia.org/wiki/Hall%27s_marriage_theorem
The $n$ women and $n+1$ men form a bipartite graph, with an edge between every woman and every man.  Let $E = $ the set of edges.  If a pair $(w,m)$ is tried and revealed to be a non-match, then the edge $(w,m)$ is removed from $E$.
In $k$ rounds, each woman could have tried at most $k$ pairings, and have at most $k$ edges removed (out of the $n+1$ edges she started with).  Similarly, a man could have tried at most $k$ pairings, and have at most $k$ edges removed (out of the $n$ edges he started with).
Now consider a subset $W$ of women, and consider the subset of men $M(W)$ that at least some woman $w \in W$ is still connected to, i.e. $M(W) = \{m: \exists w \in W, (w,m) \in E\}$.  Hall's marriage theorem states that a matching exists (using the edges still in $E$) iff $\forall W, |W| \le |M(W)|$.
To exclude a man $m$ from $M(W)$, i.e. $m \notin M(W)$, the $|W|$ edges from $m$ to  all $w \in W$ must be removed.  This has 2 consequences:

In $k$ rounds, the women in $W$ have removed at most $k|W|$ edges.  It takes $|W|$ removed edges to exclude a man, so the number of excluded men $\le k|W| / |W| = k$, or equivalently, $|M(W)| \ge n+1 - k$.
Meanwhile, it takes at least $|W|$ rounds to remove $|W|$ edges.  So: $|W| > k \Rightarrow$ no man can be excluded $\Rightarrow |M(W)| = n+1 \ge |W|$.

So only the "small" subsets where $|W| \le k$ remains to be checked.  For such subsets:
$$k \le (n+1)/2 \Rightarrow n+1 - k \ge k \Rightarrow |W| \le k \le n+1-k \le |M(W)|$$
I.e., when $k \le (n+1)/2$, Hall's theorem applies and a matching $F$ can be found in the remaining $E$ (even if every tried pair is a non-match and the corresponding edge has been removed).  In this matching $F$, one of the men is not involved.  However, he (like every man) has $n$ edges to start with, and at most $k < n$ have been removed, so he still has an edge $e$ left, and $F \cup \{e\}$ represents a valid answer to the original problem setting.

Further thoughts: the $(n+1)/2$ bound is tight for the stronger claim.  In $k$ rounds, a subset of $k$ women could have exhaustively tried all pairings with $k$ men (e.g. the concentric circles idea).  If all $k^2$ pairs are non-matches, these $k$ women must match with the remaining $n+1-k$ men, which is impossible if $k > n+1 - k,$ i.e. $k > (n+1)/2$.  In other words, if the players try the maximum number of pairings each round, the producer must reveal some match(es) within the first $\lfloor (n+1)/2 \rfloor +1$ rounds.