问题
A coworker came to me with an interesting problem, a practical one having to do with a "new people in town" group she's a part of.
18 friends want to have dinner in groups for each of the next 4 days. The rules are as follows:
- Each day the group will split into 4 groups of 4, and a group of 2.
- Any given pair of people will only see each other at most once over the course of the 4 days.
- Any given person will only be part of the size 2 group at most once.
A brute force recursive search for a valid set of group assignment is obviously impractical. I've thrown in some simple logic for pruning parts of the tree as soon as possible, but not enough to make it practical.
Actually, I'm starting to suspect that it might be impossible to follow all the rules, but I can't come up with a combinatorial argument for why that would be.
Any thoughts?
回答1:
16 friends can be scheduled 4x4 for 4 nights using two mutually orthogonal latin squares of order 4. Assign each friend to a distinct position in the 4x4 grid. On the first night, group by row. On the second, group by column. On the third, group by similar entry in latin square #1 (card rank in the 4x4 example). On the fourth, group by similar entry in latin square #2 (card suit in the 4x4 example). Actually, the affine plane construction gives rise to three mutually orthogonal latin squares, so a fifth night could be scheduled, ensuring that each pair of friends meets exactly once.
Perhaps the schedule for 16 could be extended, using the freedom of the unused fifth night.
EDIT: here's the schedule for 16 people over 5 nights. Each row is a night. Each column is a person. The entry is the group to which they're assigned.
[0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3]
[0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3]
[0, 1, 2, 3, 1, 0, 3, 2, 2, 3, 0, 1, 3, 2, 1, 0]
[0, 2, 3, 1, 1, 3, 2, 0, 2, 0, 1, 3, 3, 1, 0, 2]
[0, 3, 1, 2, 1, 2, 0, 3, 2, 1, 3, 0, 3, 0, 2, 1]
来源：https://stackoverflow.com/questions/12586587/combinatorial-algorithm-for-assigning-people-to-groups