16
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I'm throwing an event where every individual is suppose to meet every other individual so I'm trying figure out how to rotate them. My friends say its easy but they have yet to come up with an answer and our event looms closer and closer.

We are shooting for n = 20 but my gut says n has to be a power of 2 for this to work. The first half is easy, you just have the odds stay in their seats and rotate evens. Then take the 10 odds, renumber them, repeat. Then split to 5... oops. you've got an odd number. but thats ok, you've got 4 groups with 5 each so create 2 pairs and you've got 5 sets of 2 pairs.

At this point, my head hurts and it's taking more time to tell my guests who to meet than they spend meeting them.

Is there a simpler answer for n=20?

(edit: lots of questions about the table setup and who they are suppose to meet. Assume whatever table arrangement works, we have a variety. Regardless, i think the long narrow solution below works.)

  • 0
    "Keep one person stationary and rotate the rest" (Jesse Phillips's answer) is the right solution. See also Wikipedia's article on [Round-robin tournament](http://en.wikipedia.org/w/index.php?title=Round-robin_tournament&oldid=432980993#Scheduling_algorithm) for an illustration. This is *exactly* what you want.2011-06-29

3 Answers 3

0

For sixteen people, sort them into four groups of four: A,B,C,D. Then seat them at two tables of eight:
ABAB CDCD
ABAB CDCD
then
ACAC BDBD
ACAC BDBD
then
ADAD BCBC
ADAD BCBC

Rotate the people within A, and those within B, etc, so they meet everyone within their own group.
Then, counting neighbours as next-to, opposite, and opposite's next-to, some people can meet all 15 others, but everyone else meets only 11 others.

--

For twelve people, number the chairs 1 to 6 on one side of the table, and 7 to 12 on the other side, so 1 faces 7, 2 faces 8 and so on. Everyone moves from seat $n$ to seat $3n (\mod 13)$ after each course.
ABCDEF
GHIJKL
then
IEAJFB
KGCLHD
then
CFILBE
HKADGJ
In this arrangement, everyone meets everyone else (next-to, opposite or opposite's next-to) except for BK, EH and FG.

11

I think the answer is:
Make one person stationary, rotate all the folks. This actually works.

Example with N=6. 1 is stationary. (Each person in the top row "meets" the corresponding person just below in the bottom row.)

1) (#1:4) (#2:5) (#3:6) (#4:1) (#5:2) (#6:3)

1 2 3 4 5 6 

2) (#1:4,5) (#2:5,3) (#3:6,2) (#4:1,6) (#5:2,1) (#6:3,4)

1 4 2 5 6 3 

3) (#1:4,5,6) (#2:5,3,4) (#3:6,2,5) (#4:1,6,2) (#5:2,1,3) (#6:3,4,1)

1 5 4 6 3 2 

4) (#1:4,5,6,3) (#2:5,3,4,6) (#3:6,2,5,1) (4:1,6,2,5) (5:2,1,3,4) (6:3,4,1,2)

1 6 5 3 2 4 

5) (#1: 4,5,6,3,2) (#2:5,3,4,6,1) (#3:6,2,5,1,4) (4:1,6,2,5,3) (5:2,1,3,4,6) (6:3,4,1,2,5)

1 3 6 2 4 5 
  • 0
    I changed the formatting in the answer from "vertical" (which wasn't working) to "horizontal", but feel free to revert. BTW, this solution which you independently found is also present on [Wikipedia](http://en.wikipedia.org/w/index.php?title=Round-robin_tournament&oldid=432980993#Scheduling_algorithm).2011-06-29
1

Seat your guests at a long banquet table, but don't seat anyone at the heads of the table. So like this:

A B C D E F L K J I H G 

then have them rotate:

L A B C D E  K J I H G F   K L A B C D  J I H G F E 

And so on.

Done!

(Only you will have 10 on each side.)

  • 0
    With this solution, every person will only meet half the other people (like speed-dating). By keeping one person fixed (see below), everyone will meet everyone else (like speed-networking).2015-03-18