Shaking hands in the minimal number of turns. There are $2n$ persons shaking hands in turns. In every turn every person shakes hands with exactly one other person. After the turn all of them may choose another partner.
I'm searching for:


*

*The minimal number of turns needed so that every person has shaken hands with all of the $2n-1$ other persons.

*An algorithm that calculates a minimal sequence of turns.

 A: would take $2n-1$ turns, starting in alphabetical order of names, the person would shake hands with the first person in alphabetical order that they have not shaken hands with yet and is not selected to shake hands with anyone in this turn - then the next unselected person in alphabetical order would have the same applied to them - this process would result in n handshakes per turn and would go through n-1 turns
for n = 2 : 4 people
if you had A,B,C,D
round 1
A-B
C-D
round 2
A-C  (has shaken B's hand)
B-D   (C is shaking hands)
round 3
A-D (only person left)
B-C
done!
A: If there are $2n$ people then there must be $1+ \ldots + (2n-1) = n(2n-1)$ handshakes. Each turn we can do $n$ handshakes so we want to solve for $x$ in the following equation: $$ n(2n-1) - xn  = 0. $$ If we do this we get $$ x = 2n-1 .$$
A: From wikipedia: https://en.wikipedia.org/wiki/Round-robin_tournament#Scheduling_algorithm

If $n$ is the number of competitors, a pure round robin tournament
  requires $\begin{matrix} \frac{n}{2} \end{matrix}(n - 1)$ games.  If
  $n$ is even, then in each of $(n - 1)$ rounds, $\begin{matrix} \frac{n}{2} \end{matrix}$ games can be run concurrently, provided
  there exist sufficient resources (e.g. courts for a  tennis 
  tournament).  If $n$ is odd, there will be $ n$ rounds, each with
  $\begin{matrix} \frac{n - 1}{2} \end{matrix}$ games, and one
  competitor having no game in that round.

The wikipedia entry goes on to provide a scheduling algorithm.
