Plan 4 nights of dinners for 12 families. Each fam hosts 3 others 1 night. Can each fam eat with every other?

Question

I would like to make sure that every family gets to dine with every other family during a family vacation week. In reality on one night there will only be two families cooking for five others. Grandma and Grandpa don't have to cook anymore. This simplifies the problem some, but I am still not sure if each family can dine with everyother family at least once that week. Any suggestions?

I would like to make sure that every family gets to dine with every other family during a family vacation week. Each family cooks for themselves and three other families one night, and then is the guest of another family the other three nights. In reality on one night there will only be two families cooking for five others. Grandma and Grandpa don't have to cook anymore. This simplifies the problem some, but I am still not sure if each family can dine with every other family at least once that week. Any suggestions?

Answer 1

Well, each family can't host 3 others in only 4 nights, if they have one dinner per night and there are 12 families. (you need at least 12 nights to host one a piece). But that isn't what is important.

The families can't eat together in this situation (no matter who does the hosting).

This is why? (it will most likely be very hard to understand since it is a strange situation).

Call the families a, b, c, d, e, f, g, h, i , j, k, and l

Each family eats 4 dinners with three others (so a total of 12 others). Therefore each family eats with one family twice and the rest once.

Let's say that a eats with b twice (thus b eats with a twice). And shuffle the days around so that this happens on the first and second nights.

Now let's just say that a eats with b, c, and d the first night (doesn't matter who serves); e,f,g,h eat together; i,j,k,l eat together.

The second night a eats with b and two others. There are two cases for who the other two will be: either they have eaten together on the first night, or haven't eaten together.

Case 1) they have eaten together. Since they are still general we can just assume that they are e and f. So a,b,e,f are together. That means that c,d,g,h,i,j,k,l must be split into two groups. Since i,j,k,l have already eaten together, they must be split into groups of two. Let's say i,j and k,l. Now we have another case (either c,d and g,h eat with each other again, or they are split).

Case 1.1)c,d eat together again and g,h do too. Thus we can assume that the groups are split up like a,b,e,f; c,d,i,j; and g,h,k,l for the second night. Therefore every family has eaten with one family twice, and can't repeat a dinner again.
On the third night a can not eat with b,c,d,e, of f. Therefore a must eat with three out of g,h,i,j,k,l. But there is no way to choose three out of this group so that they don't repeat. Thus case 1.1 doesn't work.

Case 1.2)c and d don't eat with each other. Therefore we can split up the second night: a,b,e,f; c,g,i,j; and d,h,k,l. In this case a,b,e,f,i,j,k, and l have all eaten with a family more than once.
Night 3 a must eat with three from g,h,i,j,k,l. Since g,h are the only ones that can still eat together from this group, a must eat with g,h and one from i,j,k,l. But i,j have already eaten with g, and k,l have already eaten with h, so this is not possible. Therefore Case 1 is not possible.

Case 2) the two families that join a and b on night 2 have not eaten together. We can say that they are e and i. Therefore a,b,e,i eat together and c,d,f,g,h,j,k,l must be split up. f,g,h and j,k,l must be split up so that only 2 from each group eat together. we can say that f,g eat together and j,k eat together, and then just split the rest up generally. Thus for the second night, we have a,b,e,i; c,f,g,l; and d,h,j,k.

For the third night a can not eat with b,c,d,e, or i, and thus must eat with three from f,g,h,j,k,l. We can assume that f is one of them. f can not eat with g,h. f cannot eat with g,h or l, but j can't eat with k, so f,j,k can't eat together. Thus a can't eat with f the second night, but since the fourth night will be even more controlled, a can't eat with f then, and thus can't eat with f at all. Therefore Case 2 doesn't work either.

Therefore it can't happen.


But if one night there are only 2 groups and not 3, that could change things. Let me think about that one for a bit.