I've been trying to think through an extension with n roommates and n+1 potential rooms, with the last necessarily being $0. Restricting us to the outer faces of the n+1 simplex. There will be n envy free solutions - one for each of the rooms being assigned empty - but it's unclear to me which they should go with
The new house has more rooms than roommates, which i'm not sure how to handle and neither is spliddit. The current plan is to use one as an office. we COULD just assign smallest room to be office and run the algo with equal numbers... but we also signed the lease and are moving in before we found an occupant for the fourth and final room. So, a heuristic for now and then run it when we find the last person?
I've been trying to think through an extension with n roommates and n+1 potential rooms, with the last necessarily being $0. Restricting us to the outer faces of the n+1 simplex. There will be n envy free solutions - one for each of the rooms being assigned empty - but it's unclear to me which they should go with
Did you use this to choose rooms in your new house?
The new house has more rooms than roommates, which i'm not sure how to handle and neither is spliddit. The current plan is to use one as an office. we COULD just assign smallest room to be office and run the algo with equal numbers... but we also signed the lease and are moving in before we found an occupant for the fourth and final room. So, a heuristic for now and then run it when we find the last person?