1
$\begingroup$

I'm looking to schedule 16 players to play a round robin tournament with each other such that there are 4 players at each table. I'd like for each player to play with each other player exactly once over the course of 5 rounds.

I just took a look at this post: Rearrangement of groups such that no two members meet again , and I was wondering if anyone knows if the case n=m=4 has been solved, and if so what that solution is.

$\endgroup$
  • $\begingroup$ looks like a variant of the social golfer problem. mathworld.wolfram.com/SocialGolferProblem.html I do think that this only works with 20 players, 5 foursomes in 5 rounds. $\endgroup$ – Doug M Jun 15 '16 at 20:08
  • $\begingroup$ @DougM, note, the social golfer problem says "no more than once," which in this case turns out to be the same as "exactly once." $\endgroup$ – Barry Cipra Jun 16 '16 at 20:01
1
$\begingroup$

This should work:

Round robin

Please verify. (Each row is a table and the 4 tables for each round is enclosed by a square).

$\endgroup$
0
$\begingroup$

I believe this can be done with $16$ players over $4$ rounds. Start by labeling the players according to the first table at which they play, so at table $0$, the first round will see players $A_0,B_0,C_0,D_0$, and the other tables $1,2,$ and $3$ have similar players $A,B,C,D$.

For the next three rounds, keep the $A_j$ players at table $j$. At the tables of even index, bring in a $B$, a $C$, and a $D$ player whose indices are all distinct and start with the next available number counting up, with the restriction that you start over at $0$ after reaching $3$, and skip the label of the table (the players with the same index have already seen each other). At the tables of odd index, start with the next available number counting down.

For example, at table $2$, the second round will see $A_2,B_3,C_0$, and $D_1$. The next round will see $A_2,B_0,C_1$, and $D_3$.

In other words, at table even $j$, the indices of the $B$, $C$, and $D$ players will cyclically progress through rounds $2$ through $4$ respectively (where everything is reduced modulo $4$) as

$$(j+1,j+2,j+3)\\(j+2,j+3,j+1)\\(j+3,j+1,j+2).$$

At table odd $k$, the indices of the $B$, $C$, and $D$ players will cyclically progress through rounds $2$ through $4$ respectively (where everything is reduced modulo $4$) as

$$(k-1,k-2,k-3)\\(k-2,k-3,k-1)\\(k-3,k-1,k-2).$$

For the final round, put all the $A$ players at a table, $B$ players at another, $C$ at another, and $D$ at the final table.

I imagine there's some nice way to extend this to other specific cases of the general problem, but it sounds like you are trying to organize an actual tournament of 16 players rather than solve a maths problem in the abstract.

$\endgroup$
  • $\begingroup$ Unless I'm doing this wrong, this doesn't quite work. B0 will play with C3 twice, at table 1 (labeling tables 0,1,2,3) in round 2 (0+1=1, 3+2 = 1) and in round 3 at table 2 (0+2=2, 3+3=2) (All the B/C pairs will repeat). Alternatively if B moves 2 up after the second round, and C moves 3 from where they were, and D moves 1 from where they were, then D plays with the same A in rounds 1 and 3. Am I missing something here? $\endgroup$ – Ian Jun 22 '16 at 18:51
  • $\begingroup$ Fixed, modulo typos committed in my haste to cover my shame. $\endgroup$ – grapher Jun 23 '16 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.