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Online MMS Allocation for Chores

Jiaxin Song, Biaoshuai Tao, Wenqian Wang, Yuhao Zhang

TL;DR

The paper resolves a central question in online fair division of chores by showing a tight $n-\varepsilon$-MMS impossibility for any fixed $n$ and $\varepsilon>0$, while simultaneously presenting a practical online algorithm that achieves $\min\{n, O(k), O(\log D)\}$-MMS, where $k$ is the maximum number of disutility types per agent and $D$ the max per-agent ratio of disutilities. The proposed method builds on a generalized round-robin via per-type pressure and a greedy, type-aware allocation rule, augmented by a value-rounding step and a novel Stacking Game discrepancy framework that reduces the online-MMS analysis to a structured adversarial game. The paper also proves a tight bound in the personalized bi-valued case, attaining $(2+\sqrt{3})$-MMS, and discusses how the results extend to broader settings via the discrepancy machinery. These contributions clarify the limits of online MMS allocations and offer practical guarantees under realistic conditions (small $k$ or small $D$), with a rigorous connection between online allocation and discrepancy minimization through the Stacking Game.

Abstract

We study the problem of fair division of indivisible chores among $n$ agents in an online setting, where items arrive sequentially and must be allocated irrevocably upon arrival. The goal is to produce an $α$-MMS allocation at the end. Several recent works have investigated this model, but have only succeeded in obtaining non-trivial algorithms under restrictive assumptions, such as the two-agent bi-valued special case (Wang and Wei, 2025), or by assuming knowledge of the total disutility of each agent (Zhou, Bai, and Wu, 2023). For the general case, the trivial $n$-MMS guarantee remains the best known, while the strongest lower bound is still only $2$. We close this gap on the negative side by proving that for any fixed $n$ and $\varepsilon$, no algorithm can guarantee an $(n - \varepsilon)$-MMS allocation. Notably, this lower bound holds precisely for every $n$, without hiding constants in big-$O$ notation, thereby exactly matching the trivial upper bound. Despite this strong impossibility result, we also present positive results. We provide an online algorithm that applies in the general case, guaranteeing a $\min\{n, O(k), O(\log D)\}$-MMS allocation, where $k$ is the maximum number of distinct disutilities across all agents and $D$ is the maximum ratio between the largest and smallest disutilities for any agent. This bound is reasonable across a broad range of scenarios and, for example, implies that we can achieve an $O(1)$-MMS allocation whenever $k$ is constant. Moreover, to optimize the constant in the important personalized bi-valued case, we show that if each agent has at most two distinct disutilities, our algorithm guarantees a $(2 + \sqrt{3}) \approx 3.7$-MMS allocation.

Online MMS Allocation for Chores

TL;DR

The paper resolves a central question in online fair division of chores by showing a tight -MMS impossibility for any fixed and , while simultaneously presenting a practical online algorithm that achieves -MMS, where is the maximum number of disutility types per agent and the max per-agent ratio of disutilities. The proposed method builds on a generalized round-robin via per-type pressure and a greedy, type-aware allocation rule, augmented by a value-rounding step and a novel Stacking Game discrepancy framework that reduces the online-MMS analysis to a structured adversarial game. The paper also proves a tight bound in the personalized bi-valued case, attaining -MMS, and discusses how the results extend to broader settings via the discrepancy machinery. These contributions clarify the limits of online MMS allocations and offer practical guarantees under realistic conditions (small or small ), with a rigorous connection between online allocation and discrepancy minimization through the Stacking Game.

Abstract

We study the problem of fair division of indivisible chores among agents in an online setting, where items arrive sequentially and must be allocated irrevocably upon arrival. The goal is to produce an -MMS allocation at the end. Several recent works have investigated this model, but have only succeeded in obtaining non-trivial algorithms under restrictive assumptions, such as the two-agent bi-valued special case (Wang and Wei, 2025), or by assuming knowledge of the total disutility of each agent (Zhou, Bai, and Wu, 2023). For the general case, the trivial -MMS guarantee remains the best known, while the strongest lower bound is still only . We close this gap on the negative side by proving that for any fixed and , no algorithm can guarantee an -MMS allocation. Notably, this lower bound holds precisely for every , without hiding constants in big- notation, thereby exactly matching the trivial upper bound. Despite this strong impossibility result, we also present positive results. We provide an online algorithm that applies in the general case, guaranteeing a -MMS allocation, where is the maximum number of distinct disutilities across all agents and is the maximum ratio between the largest and smallest disutilities for any agent. This bound is reasonable across a broad range of scenarios and, for example, implies that we can achieve an -MMS allocation whenever is constant. Moreover, to optimize the constant in the important personalized bi-valued case, we show that if each agent has at most two distinct disutilities, our algorithm guarantees a -MMS allocation.

Paper Structure

This paper contains 22 sections, 14 theorems, 54 equations, 5 figures, 1 algorithm.

Key Result

Lemma 1

For any agent $i\in [n]$, it holds that $\sum_{\theta=1}^k ({\rm MMS}\xspace_i^\theta - V_i^\theta) \le {\rm MMS}\xspace_i \le \sum_{\theta=1}^k {\rm MMS}\xspace_i^\theta$, where $k$ is the number of distinct disutility of agent $i$.

Figures (5)

  • Figure 1: An example operation of the stacking game with $k=1$.
  • Figure 2: An example of mapping from pressures to subintervals when $n=3$ and $k=2$.
  • Figure 3: An example showing $\tilde{F}^{t+1}(y) \ge F^{t+1}(y)$ with contiguous interval $\tilde{A}\cup\tilde{B}$.
  • Figure 4: An example of $(a, b, A, B)$ and $f^t$, where $x$ is to the left of $B$.
  • Figure 5: An example for the upgrading interval and the downgrading interval.

Theorems & Definitions (31)

  • Definition 1: Minmax share (MMS)
  • Definition 2: MMS for type-$\u$ items
  • Lemma 1
  • proof
  • Theorem 1
  • Definition 3: Stacking game
  • Example 2: Running example for Stacking game
  • Lemma 2
  • proof
  • Proposition 1
  • ...and 21 more