When is round-robin scheduling possible and with in minimal time? Suppose that you have six teams $x_0, x_1, x_2, x_3, x_4, x_5$. Can you schedule round-robin games between them so that if one game is played each day, the series of games can be completed in five weeks?
A naive attempt fails:
$$(x_0,x_1), (x_2, x_3), (x_4, x_5)$$
$$(x_0,x_2), (x_1, x_3) .. oops$$ 
Some care is needed to get this to work. A solution:
$$(x_0,x_3), (x_1, x_5), (x_2, x_4)$$
$$(x_1,x_4), (x_2, x_0), (x_3, x_5)$$
$$(x_2,x_5), (x_3, x_1), (x_4, x_6)$$
$$(x_0,x_1), (x_2, x_3), (x_4, x_5)$$
$$(x_0,x_5), (x_2, x_1), (x_4, x_3)$$
This solution certainly is not easy to generalize. Questions:


*

*Is this in fact always possible for an even number of teams?

*Is there a convenient algorithm for producing the schedule?

*Can this problem be easily expressed in terms of graph theory, combinatorics, or abstract algebra?


Since you only ask for even $n$, you could use the fact that the complete graph $K_n$ is Hamilton decomposable for odd $n$ and decomposable into a perfect matching plus Hamiltonian cycles for even $n$.
There is one single permutation that generates the Hamiltonian cycles.
For even $n$ each Hamiltonian cycle gives you 2 rounds: one where player $2i$ plays player $2i+1$ and one where he plays player $2i-1$ (all $\pmod{2n}$).
The perfect matching gives you the last round.
The fact that this is a decomposition guarantees that every 'edge' is played.
Details are here.
