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We have 24 people, and 4 tables. Over 4 sessions, each person will visit each table once. What system can I use to make sure each person gets to each table once, while making sure they meet as many other people as possible (so dont go from table to table with the same people).

Is there something that can help with this?

Apologises if this doesnt belong in Maths, not sure what kind of problem it is?

Caleb Stanford
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Mike Langlois
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  • have all 24 people sit at the same table? Are you assuming that each table can only have 6 people? Or are you assuming no two people ever sit at the same table twice and they all sit four times? What are your restrictions? It's not possible that no two people never sit at the same table twice. – fleablood Sep 08 '16 at 22:21
  • Sorry I should have specificed. Each table can only sit 6 at a time. Each person will visit each table once and only once. People can sit at a table with people they have sat with before, however would need to try and minimise this as much as possible, if that makes sense? – Mike Langlois Sep 08 '16 at 22:23
  • Hmm, well the must "average" would be: After the first seating each of the 3 pairs at each table to a different table. Then each of those pairs to a different third table. and then finally to the last table. So at table 1 person 1a meets 1b-1f. That's five people. Then at table 2 person 1a again is sitting with 1b but meet 3c and 3d and 4e and 4f. That's nine people. Then at table 3, he meets 2b and 2f, is seated with 1c and 4e again but meets 4d. So that's 12 people At table four he meets 2 more people for 14 people. Can we beat that? I don't know. – fleablood Sep 08 '16 at 22:52

1 Answers1

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Most average:

Turn 1:

Table x(1-4) seats people (xa-xf). Everybody meets 5 people.

Example: Table 1 seats 1a,1b,1c,1d,1e,1f

Turn 2:

xa and xb go to table x+1; xc and xd go to table x+2, xe and xf go to table x-1

Example: Table 2 seats 1a,1b,4c,4d,3e,3f. Everybody meets 4 new people.

Turn 3: xa,xc,xe got to the next table they haven't been to and xb,xd,xf go to the previous table they havent been to.

So Table 3 seats 1a,4b,4c,2d,1e,2e. Everybody meets 3 new people.

Turn 4: everybody goes to the table they haven't visited.

So Table 4 seats 1a,2b,3c,1d,2e,3f. Everybody meets 3 new people. Everybody met 15 people.

I don't know if that can be beat.

fleablood
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