For a meeting, how do I divide 30 people into 4 groups, x 4 sessions, with maximum diversity each session?

Question

To clarify: we divide into four small groups four times. Each time, the groups will be divided differently and I want to ensure the maximum diversity/least overlap possible.

Answer 1

How about this:

First round:
1) ABCDEFGH
2) IJKLMNOP
3) QRSTUVW
4) XYZ1234
The naming of people is completely arbitrary, so I just used alphabetical order (and the first four digits when I ran out of letters).

Second round:
1) AB IJ QR XY
2) CD KL ST Z1
3) EF MN UV 2
4) GH OP W 34
(You can't split up all of ABCDEFGH, since there are eight members and only four groups to distribute them to. So they have to be distributed in pairs.)

Then, for the third round, and the fourth round as well, you'd want to ensure that the pairs you kept together between sessions 1-2 were separated between sessions 3-4. You don't want anyone to see any other person more than twice (which is unavoidable as explained above).

Third round:
1) AE KO S Z 3
2) BF LP TW 14
3) CG IM QU X
4) DH JN RV Y2

For example: "KL" were together in both the first two sessions, so they now need to be split up. Since they were paired with CD in the second session, and CD were also split up this time, I distributed KL to the two sessions that CD didn't get into. Similarly, AB are split up, and then also separated from all non ABCDEFGH people that they were with in the second round.

For the fourth round, you want to be sure that people are not with their first retained pairing (e.g. A-B separated) and also not with the pairing they got in the third around (e.g., A-E separated).

Fourth round:
1) AH LM Q 23
2) BG KN RW Z
3) CF JO SV X1
4) DE IP TU Y4

I distributed the second group (IJKLMNOP) so that they were with members of the first group (ABCDEFGH) that they hadn't previously seen. For example, LM were the only two from the second group that neither A nor H had been with before.

This left us with two people (from the first two groups) that every third-group member had seen, and two that they hadn't, in each of the four teams, so the distribution of the third group was pretty arbitrary. I just used the same pattern.

Then for each member of the fourth (original) group, I counted how many of the the members of groups 1-3 they'd seen previously in each of the four teams, and distributed them to the one that contained the fewest people they'd already seen.

I think this does as good a job as possible of mixing people up. It's not possible for every person to see every other person, and there must be some number of repeats. But it is spread out fairly well:

A, for example, sees:
B(x2), C, D, E(x2), F, G, H(x2), I, J, K, L, M, O, Q(x2), R, S, X, Y, Z, 2, 3 (x2).
He sees five different people twice, and misses only N, P, T, U, V, W, 1, and 4.

X, for example, sees:
A, B, C(x2), F, G, I, J(x2), M, O, Q(x2), R, S, U, V, Y(x2), Z, 1(x2), 2, 3, 4
He also sees five different people twice, and misses only D, E, H, K, L, N, P, T, W, and X.

Answer 2

You have to start with assigning people to one of the 4 teams. The teams will either have 7 or 8 people...two teams of 7 and two teams of 8. (7 + 7 + 8 +8 = 30.)

You will have overlap, as each team member will likely be on a new team with one of their old members. In the second round, assign 1/4 of each team to a new team. (2 people go to Team 1, 2 go to Team 2, 2 go to Team 3 and 1/2 goes to Team 4.)

Repeat this process for the next two rounds, but ensure that the person who is "traveling" with someone has not "traveled" with that person before.

You can do all of this by building a "tree" type structure. An Excel spreadsheet is the best way to set up the teams, but be aware that there will be overlap. The idea is to minimize the number of times that two people share one team.

Answer 3

Since your group together is not a multiple of four, it'll be a bit difficult. The only thing I can come up in mind without making the process excruciatingly difficult is to give everyone a number 1-4, and then in those groups, give each group two cards of each number and tell them to pick a number, and rearrange into those groups. You're always going to have two groups of seven and two groups of eight, however.

Answer 4

1st season divide 4 group with 2 people in each group
2nd season divide 4 group with 1 people in each hgroup
3rd season divide 4 group with 3 people in each group
4th season divide 3 group with 2 people in each group