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 » Rotation Chart for 42 people, 7 tables, 6 people to table
Topic review (newest first)
 bobbym
 20121123 02:30:32
Hi;
Would you consider allowing me to see your solution?
 1975rachel
 20121123 01:26:46
Hi Bob,
You are right and at least it means I wasn't being that stupid!
Thank you ever so much for your time and help, it is very kind of you.
I have devised a scheme which sort of works, there is some cross over, but it will do.
Thank you again
 bob bundy
 20121123 00:08:59
hi 1975rachel
Welcome to the forum.
I agree with bobbym. I'll try to show you why this is impossible.
The first time it doesn't matter who sits with whom, so let's get them seated and then give everyone a number as follows.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 18 20 21 22 23 24 25 26 27 28 29 30
Now who will person 1 sit with, the next time around?
No one from 26, so I'll choose someone from the next table, let's say person 7. As those six folk can shuffle their numbers around 712, I'm not losing any generality by saying 7.
Now who will 1 and 7 sit with? It'll have to be someone who wasn't on either of the first tables so let's choose person 13. Aagin I'm not losing generality by picking that number (it could equally be 14 or 15 ...)
So who will 1, 7 and 13 sit with?
I'll have to choose from a table that hasn't been picked yet so I'll have person 18, ... and to speed things up I'll pick 25 next.
So I've got 1, 7, 13, 18, 25.
Now you might think I've fiddled it by choosing these numbers. But my point is, to avoid putting people with folk they've already met, you have to choose one person from each table.
So, for example, 1, 8, 15, 22, 30 is another way of picking five to go on the first table in round two.
But I have only picked five people and the table must seat six. So who else can I choose?
As I have already picked one person from each of the tables, there's no one left who I can pick without putting someone with someone they've already met.
Conclusion: It is impossible.
Bob
 bobbym
 20121123 00:06:25
Hi;
Sorry that I can not help but some of them I just do not know the answer to or even if there is an answer.
 1975rachel
 20121123 00:02:43
Hi bobbym
Many thanks for coming back to me.
Unfortunately we are stuck with this seating plan!
Nevermind, thanks for your help.
 bobbym
 20121122 22:29:18
Hi 1975rachel;
I am sorry I do not know of any structure of 5 tables of 6 people for any number of courses. Not all structures are possible. The best I can do is 6 tables of 5 people each for 3 courses.
 1975rachel
 20121122 21:59:56
Hello, I really hope you can help me.
I have the same problem but it is for 5 tables of 6 people and I need just to rotate it for 3 courses. The idea being that no one sits with the same person twice.
Please please could you help me as I am losing the will to live.
Thanking you in advance.
 bobbym
 20121109 03:34:34
Hi adlady;
Welcome to the forum and save me some chicken!
 adlady
 20121109 02:55:49
Thank you very much. I don't know how you did it. It is appreciated.
 bobbym
 20121108 17:56:37
 adlady
 20121108 17:32:49
I can't figure out how to do this. We don't want to sit at a table with someone we have already visited with. We know we won't get to see all the people. Just need 34 rounds. Can someone help?
