Hi, I need some help with this mathmatical problem I have. I would also need it to be explained or perhaps programmed for future use. Ok this is basically what I need: - I have an event that 72 people will be attending - In this event, there will be 9 different tables - Each table will have an assigned captain who will always stay on the same table - Each person except the table captain will need to switch tables twice and have completely different people on the other tables. - So each person will sit on 3 different tables and have every single person be different, and can't sit on the same table. Table captains can't sit with anyone twice also. - So there will be 8 people on each of the 9 tables. The number of people, number of tables and number of people per table could change per event. I am not sure if this is mathmatically possible with any number. I need to know how this can be possible and perhaps programmed to be a table plan. Let me know what kind of work is involved to make this possible, or is it something quite simple? I need some kind of formula to do this. I have attached a spreadsheet that I have put together which almost solves it, but I had 3-5 people who were sitting on same table or same people more than once. I put this together by guessing so just take it as an example. Please let me know if you need more info.
Nope not yet. I've spent hours trying to find a away of doing it but I've given up now. It looks impossible.
Hi ahmasa, I was just shortly looking at it, and I believe it's not possible with conditions you have. Small proof using your schema from xls file : person | tables 10 1,2,3 19 1,4,5 28 1,6,7 37 1,8,9 => you can never find table for person 46, 55, 64 other than the previous ones. So they can never sit with completely different people. To make it possible you would need at least the number of tables as : (number of moving persons(per table) * number of switches) - (number of moving persons(per table) + 1) Which would be 15 => doesn't work with placing all 72 people by same groups. So you would probably need 12 tables (groups by 6). Then it would work, I guess. Hope I'm not totally missing something..... Of course I am missing something. Sorry for confusion.
Thanks for your help guys. darkmessiah, the spreadsheet looks correct! That is for 9 tables x 8 people per table; will that pattern work for 7x7 or 8x7 for example? or will that need to change for different numbers? Thanks
Honestly, it took longer to edit the SS than it did to come up with the solution LOL My solution was to insert a row at the top below the captains.. take the last row and put it in the inserted row, but shifted (right) over by 1, move the overflow into the starting position. continue to insert a new row at the top and increasing the shift by 1, so the second insert will start in the 3rd positions. etc.. Hope that makes sense.
Hmm... this looks fun. Helping each other with homework. Just cross fingers your teacher isn't logged in, too. LOL.
No need to login , if your teacher searches in google about the homework he had given this post might appear in first page.
The solution is all over the internet and was solved over a hundred years ago. Just last week I was perusing the net and found it, but I can't remember the google search criteria to find it again.