Q:

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:

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.

