New Math Finally Settles Who Takes Out the Trash
Could end arguments over chores, bills and bad shifts once and for all.
Researchers have resolved an open question about fairly dividing indivisible "bads" — allocations meant to guarantee every agent a bundle whose cost is no larger than a meaningful fairness benchmark. The canonical minimax share has been widely used, but it is not a simultaneous guarantee: one allocation that satisfies everyone. Hill (1987) initiated a complementary approach in which the share depends only on the number of agents and the largest possible single-item value. Li et al. (2024) gave the exact Hill formula and proved that its monotone closure is a simultaneous guarantee — but that closure treats the largest-item cost only as an upper bound, and can be strictly larger than the share conditioned on the actual largest item. They asked whether this smaller exact share is itself simultaneously guaranteed for three or more agents. This paper answers yes: for any number of agents and arbitrary heterogeneous largest-item costs, there is one allocation that satisfies every agent's exact Hill share. The authors also provide a polynomial-time algorithm to compute such an allocation, combining an ordered moving knife with a tail-domination invariant, plus a one-sided trimmed subset-sum routine for the two-agent endpoint without computing an exact minimax partition.
- "Bads" are things nobody wants — chores, fees, bad shifts — and splitting them fairly is mathematically trickier than splitting prizes.
- The team proved one single division can satisfy everyone's fair share at once, for any group size, ending a question open since 1987.
- They also gave a fast calculation method, so a computer could eventually turn this into real tools for roommates, managers or negotiators.
Why It Matters
Fairer, faster ways to divide unpleasant work and costs could reduce everyday arguments at home, at work and between countries.