Math Is Fun Forum

Doug Allender

Member

Registered: 2011-01-05

Posts: 1

Lunch rotation for 18 people in a group, 6 groups of 3 each month

I am in a group of 18 members who would like to start meeting once a month for lunch.

Each month, we want 3 people to meet for lunch, so there would be 6 groups of 3 people meeting each month. 

The goal is to have each member eat with two people they have not met for lunch in previous months until they have eaten lunch with every member of the group at least once before repeating the cycle.  This should take about 9 months to complete a cycle.

If there is a formula to figure this out, that would be great.  If you have a solution, or know of other ways to solve this lunch schedule, please share with me.

Your suggestions are welcome.

Regards,

Doug Allender

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Lunch rotation for 18 people in a group, 6 groups of 3 each month

Hi Doug Allender;

Your problem is called a progressive dinner problem. I worked on two of them here already. I am working on yours. They are part of the study of incomplete block diagrams ( designs?) and latin squares. The problem begins with Reverend Kirkman, J.J.Sylvester and Arthur Cayley. It is generally a very difficult problem and some arrangements are not possible.

This is the official answer, call your people A,B,C,D,E,F,G,H,I,a,b,c,d,e,f,g,h,i and arrange them like this. Each row is one month.

The first problem is obvious. Each person needs to eat with 17 others. He eats with 2 at a time. 17 is not divisible by 2! The above answer should be missing someone for each person. Your problem does not have a better solution.

My suggestion is to add more people or take a few away. Or go with the above table which arranges it that you eat with everyone else but your spouse.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.