REVIEW 4 major objections 6 minor 1 cited by
Lower Bound for Online MMS Assignment of Indivisible Chores
T0 review · 4 major / 6 minor · reviewed 2026-08-06 · deepseek-v4-flash
Pith's one-line read This paper proves that no deterministic online algorithm for splitting chores can guarantee every agent less than n times her fair share, where n is the number of agents.
desk verdict Tight lower bound for deterministic online MMS chores; the base and two-agent cases are solid, but the general induction is sketched, not proved. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The carrying object is the recursive scenario sc(i, ε), an adaptive adversary that controls the costs of incoming chores. Its base case forces one agent to take n chores, and its inductive step is a geometric cost ladder for agent a1: chore costs increase exponentially, and any assignment to a1 costs at least 1/ε′ times the total cost, for a1, of the chores she did not take, while a stray run of chores to the other agents triggers the embedded sc(i) trap. The length parameter ℓ_i grows exponentially at each level—ℓ_1 = n and ℓ_2 ≈ n(1+1/ε′)^{ℓ_1}—so the full adversary is finite but tower-sized. The induction is what converts a trap for i agents into a trap for i+1 agents while keeping the MMS bound of the new backstop agent under control.
What would settle it
Run the explicit sc(2, ε) construction with small n (say n=2 or 3) and search deterministically for a strategy that keeps every agent's final cost below (n − nε) times her MMS; if such a strategy exists, the induction's embedding step is false, and if none exists, the base step can be inspected numerically to verify the claimed ratio.
Extended reading notes
Core claim
The paper's central claim is Theorem 3.1: for every ε > 0 there is an online instance with n agents on which any deterministic online algorithm leaves some agent with cost at least (n − nε) times her MMS, where MMS is the maximin share—the minimum, over all n-way partitions of the chores, of the maximum cost any bundle imposes on that agent. Since ε is arbitrary, no deterministic online algorithm can be (n − nε)-competitive; the best possible deterministic competitive ratio is exactly n. The proof constructs scenarios sc(i, ε) by induction. In sc(1, ε) a single trapped agent receives n chores whose costs dwarf her MMS, giving ratio about n. For sc(i+1, ε), the adversary generates chores whose costs for agent a1 grow geometrically, while an inner scenario sc(i, ε) is embedded on the remaining agents; if the algorithm gives too many consecutive chores to the inner agents the embedded trap fires, and if it gives one to a1, that single chore pays for all skipped chores by a factor of about 1/ε. At the end a1 has accumulated a stated fraction of the total generated cost while her MMS stays bounded, again forcing ratio n(1−ε).
Load-bearing premise
The whole proof rests on the unstated claim that the inner i-agent trap can be embedded inside the i+1-agent scenario and restarted whenever the fresh agent takes a chore, without changing that fresh agent's maximin-share bound; if that composition fails at some level, the theorem does not follow.
Editorial extensions
If this is right
- The deterministic competitive ratio for online MMS chore division is exactly n: the trivial algorithm that assigns every chore to one agent is optimal.
- The old lower bound of 2 is superseded; no constant-factor deterministic online algorithm exists for chores, in contrast to the offline setting where a 13/11 approximation is known.
- Any algorithm that wants a ratio below n must be randomized or must assume extra structure, such as a bound on the number of chores m or non-adversarial cost sequences.
- The adversary needs scenarios of length roughly exponential in n at the second level and tower-exponential deeper, so the result concerns worst-case instances with very long chore streams.
- If randomization or a known horizon m can break the bound, the paper's construction gives a concrete family of instances on which to test such algorithms.
Reading between the lines
- A natural next test is whether the bound survives with a bounded number of chores: since ℓ_2 is exponential in n, an algorithm that knows it will see at most poly(n) chores might beat n within that horizon, and this construction does not rule it out.
- The proof's reliance on adaptive cost revelation suggests that an oblivious adversary (all costs fixed in advance) may be strictly weaker; checking whether the same n lower bound holds for offline-known cost sequences would delimit the role of adaptivity.
- The same recursive ladder might transfer to other fairness benchmarks, such as proportional share or envy-free up to one chore, with the geometric ratio 1/ε playing the same role; testing that would show whether the n barrier is specific to MMS or generic to online chores.
- Because the construction forces a1 to accumulate cost while the other agents are trapped, it resembles a scheduler-exploiting adversary; a concrete follow-up could be to port the scenario to online load balancing and see whether the same ratio appears there.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper studies online assignment of indivisible chores to n agents under the maximin share (MMS) benchmark, where chores arrive one by one and must be irrevocably assigned. The main result (Theorem 3.1) claims that no deterministic online algorithm can be (n - nε)-competitive for any ε > 0, i.e., the competitive ratio is at least n, matching the trivial upper bound of assigning all chores to a single agent. The proof constructs a recursive adversarial scenario sc(i, ε) for i = 1, ..., n. The base scenario sc(1, ε) is written out: with prior cost x for agent a1, the adversary introduces n chores of cost w = x/ε, forcing cost nw against MMS at most w + x. The step from sc(1) to sc(2) is sketched with a geometric cost sequence and a bound on the total cost and MMS for a1. The transition from sc(i) to sc(i+1) is asserted to follow "the exact same blueprint" with exponentially growing ℓ_i, but is not formalized.
Significance. If the claimed lower bound is correct, it is a significant improvement over the previous best lower bound of 2 and closes the problem for deterministic online MMS chore assignment, since the trivial algorithm that assigns every chore to one agent is n-competitive. The base case argument is clean and the exponential-growth intuition for the recursive scenario is plausible. The paper is self-contained and does not rely on the result it proves. However, the proof as written contains a load-bearing unproven induction step; without a complete recursive construction, the theorem is not established for n > 2. The result is therefore potentially important, but the manuscript currently does not provide a complete proof.
major comments (4)
- [Section 3, paragraph beginning "The transition from each sc(i, eps) to sc(i+1, eps)"] The induction step from sc(i, ε) to sc(i+1, ε) is the core of the proof, but it is only asserted. The paper defines ℓ_1 = n and sketches ℓ_2, but does not define ℓ_i for general i, prove that the geometric sequences and restarting loops terminate, or show that the total number of chores generated remains finite. A universal lower bound for all n requires a fully specified recursion; as written, the proof only establishes the cases i = 1 and i = 2.
- [Section 3, sc(i+1) construction] The induction step requires composing an inner sc(i, ε) adversary on agents a2..a_{i+1} with an independent geometric cost sequence for a1, restarting the inner adversary after each abort. The proof never states the input-history hypothesis under which sc(i, ε) is invoked after an abort, nor proves that the inner scenario's guarantee (one of a2..a_{i+1} exceeds her MMS by the desired factor) is preserved when the inner scenario is run multiple times with interleaved chores assigned to a1. Without this composition lemma, the recursive construction is not well-defined.
- [Section 3, sc(2) MMS accounting] The bound "MMS1 ≤ nw/n + 3ε'w = (1+3ε')w" relies on an unstated scheduling-type inequality MMS ≤ T/n + (1 - 1/n)L, where T is total cost and L is the largest item cost. Even with that inequality, the displayed total cost bound from the previous line is T ≤ (n+2ε')w, not T = nw; the calculation as written drops the 2ε'w term. For the general induction, the paper gives no accounting of the accumulated total cost across all aborted restarts, which is needed to bound a1's final MMS. The sc(2) calculation therefore does not demonstrate the general invariant.
- [Section 3, sc(1) and sc(2) definitions of w] The definition w = x/ε (and similarly w = x/ε' in sc(2)) degenerates when the scenario is invoked with no prior chores: x = 0 gives w = 0 and all subsequent costs are zero, so the scenario never terminates and produces no positive cost. Since the online instance begins with an empty history, the construction as written does not apply to the initial invocation. The proof needs an explicit rule for the first chore (e.g., w = max(x, 1)/ε) or another way to seed a positive cost.
minor comments (6)
- [Section 1, MMS definition] In the introduction, MMS_i is defined as max over partitions of min over bundles, which is the goods form; Section 2 correctly defines it as min over partitions of max over bundles for chores. Please correct the introduction.
- [Section 3, sc(2) bound for ℓ_2] The inequality ℓ2 ≤ ℓ1 + n(1+1/ε')^{ℓ1} is stated without derivation; it would help to explain why each pattern can contribute at most nw to a1's cost and how the multiplicative factor n arises.
- [References] References [13] and [14] are duplicates of the same JACM paper; please remove the duplicate.
- [Section 3, notation] The symbol n appears both as the total number of agents in the theorem and as the index in sc(i); in the sc(2) calculation "nw/n" is confusing because the scenario has only a1 and a2 active. Please clarify that MMS is computed over n bundles, one per agent, even when only i agents ever receive chores.
- [Section 3, sc(2) proof] In the sc(2) proof, the phrase "if ℓ1 consecutive chores are given to agent a2 then the algorithm would not be (n - nε)-competitive by the guarantees of sc(1, ε)" should note that sc(1, ε) is invoked with the previous chores assigned to a2 as history; the costs of those chores for a2 need to be defined consistently.
- [Section 3, sc(1) proof] The proof of sc(1) states that costs for other agents "do not matter"; to make the online instance complete, specify that all other agents have sufficiently large costs for the constructed chores, or explicitly state that their values are irrelevant and can be set arbitrarily.
Circularity Check
No circularity: the recursive adversary construction is self-contained and does not assume the theorem.
full rationale
The derivation chain is a direct lower-bound adversary construction, not a cycle. The base case sc(1,eps) is specified from an arbitrary previous-history total x and produces chores with costs w = x/eps for agent a1; the bound MMS1 <= w(1+eps) follows from the same additive cost structure, not from the claimed competitive ratio. For sc(2,eps), the paper computes the ratio from the geometry of exponentially increasing costs and the accounting MMS1 <= (1+3eps')w, again using only the constructed costs. The passage from sc(i) to sc(i+1) is asserted as 'the exact same blueprint' rather than fully written out; this is a completeness or correctness gap in the proof, not circularity, because the induction hypothesis is a strictly weaker statement (failure on i agents) used to force a response on i+1 agents, and no step defines the target bound in terms of itself. Self-citations [9,10,17,18] appear only as literature context, not as evidence for Theorem 3.1. The prior lower bound [21] is external prior work used only as a baseline. No fitted parameter is relabeled as a prediction, no uniqueness theorem is imported from the authors, and the central claim is not equivalent by construction to its inputs.
Assumptions & free parameters
assumptions (3)
- domain assumption Additive cost valuations are the model for how agents evaluate bundles of chores.
- standard math For any list of item costs, the optimal n-bin makespan is at most total/n plus (1 - 1/n) times the largest item cost (Graham bound).
- ad hoc to paper The sc(i) adversary can be used as a black box inside sc(i+1), with a separate geometric cost sequence for the backstop agent, and aborted runs can be restarted without invalidating the backstop's MMS bound.
Cite this review
Pith. "Pith review of Lower Bound for Online MMS Assignment of Indivisible Chores." pith.science (2026). https://pith.science/paper/C5J3RKAH
@misc{pith2026250712984,
author = {Pith},
title = {Pith review of: Lower Bound for Online MMS Assignment of Indivisible Chores},
year = {2026},
howpublished = {\url{https://pith.science/paper/C5J3RKAH}},
note = {Machine review of arXiv:2507.12984}
}
abstract
We consider the problem of online assignment of indivisible chores under \MMS\ criteria. The previous work proves that any deterministic online algorithm for chore division has a competitive ratio of at least 2. In this work, we improve this bound by showing that no deterministic online algorithm can obtain a competitive ratio better than $n$ for $n$ agents.
Forward citations
Cited by 1 Pith paper
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Constant Weighted Maximin Share Approximations for Chores
Weighted maximin share for indivisible chores admits a 12-approximate allocation and cannot be approximated better than factor 2.
Reference graph
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Reviewed August 6, 2026 · model on record in the stance chip above.
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