Create unique table rotation for an event I am planning to host an event with 14 people participating in it. There will be five tables in total. (4 tables with 3 people each and 1 with 2).
I want to rotate the tables 5-6 times and to ensure that everyone gets to meet as many people as possible and minimize (if not completely avoid) that some people see each other twice.
I figured out how to do tables of 2, but I struggle with tables of 3.
I would very much appreciate your help!
 A: Here's a 7-round schedule with every pair covered exactly once, obtained via integer linear programming:
1  {6,8,11}  {10,12,14} {4,5,7}   {3,9,13}   {1,2}
2  {3,7,11}  {1,4,10}   {5,8,13}  {2,6,14}   {9,12}
3  {2,7,8}   {5,9,10}   {1,11,12} {3,4,14}   {6,13}
4  {1,3,8}   {4,6,9}    {2,5,12}  {10,11,13} {7,14}
5  {2,9,11}  {6,7,10}   {4,8,12}  {1,13,14}  {3,5}
6  {2,3,10}  {8,9,14}   {1,5,6}   {7,12,13}  {4,11}
7  {5,11,14} {3,6,12}   {2,4,13}  {1,7,9}    {8,10}

A: The best that I could come up with was this seven round schedule. Out of the $14\times 13/2=91$ pairs of people, this schedule causes all but six pairs of people to meet each other, with only six pairs of people meeting twice.





Table 1
Table 2
Table 3
Table 4
Table 5




Round 1
$4, 6, 7$
$1, 2, 5$
$9, 12, 14$
$3, 10, 13$
$8, 11$


Round 2
$1, 6, 11$
$8, 9, 10$
$7, 12, 13$
$2, 3, 14$
$4, 5$


Round 3
$3, 4, 8$
$5, 10, 12$
$1, 7, 14$
$2, 11, 13$
$6, 9$


Round 4
$4, 13, 14$
$5, 9, 11$
$3, 6, 12$
$2, 7, 10$
$1, 8$


Round 5
$5, 6, 13$
$8, 12, 14$
$2, 4, 9$
$3, 7, 11$
$1, 10$


Round 6
$6, 10, 14$
$3, 9, 13$
$5, 7, 8$
$1, 4, 12$
$2, 11$


Round 7
$3, 5, 14$
$4, 10, 11$
$1, 7, 9$
$2, 6, 12$
$8, 13$



