{"id":"e3b14b92-b305-4b95-b352-7c0196a33572","arxiv_id":"2607.18139","paper_version":1,"verdict":"ACCEPT","confidence":"MODERATE","novelty_score":7.0,"correctness_risk":"low","formal_verification":"none","parameter_count":0,"one_line_summary":"Every four-agent instance with nonnegative additive valuations admits a complete 1-out-of-5 maximin-share allocation, and 5 is the smallest denominator for which a universal guarantee exists.","lead":"For four agents with additive valuations, a fair allocation meeting each agent's 1-out-of-5 maximin share always exists. This improves the known 1-out-of-6 guarantee and settles the four-agent question completely.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"No significant objection identified","rationale":"The reader correctly identifies Lemma 5's 2+2 case as the load-bearing intermediate, and I agree that this is where a hidden flaw would most likely live. However, after checking the inequalities and construction, the repair works. In the 2+2 case, the four cores have demands summing to 2, and the three unit cells U,V,W have total value 3. Splitting U and V via Lemma 4 produces shortfalls e and f, each at most m0/2, so e+f ≤ m0. Since m0 ≤ 1/2, the final repair inequality e+m0+f ≤ 1 holds, so W can be partitioned into Q and W\\Q that cover both shortfalls. The other subcases of Lemma 5 are simpler and correct: for k=1,2,3, the total-value accounting and trigger bounds suffice; for m0 ≤ 1/3, three minimal fills leave at least 2−3m0 ≥ 1 for the fourth anchor. The ordered-instance reduction and envy-free matching preserve the needed conditions. I found no circularity, missing case, or invalid numerical bound. The only weakness is the absence of the promised Lean sources in the submitted version, but the mathematical argument is self-contained and coherent. Therefore the reader's ACCEPT verdict need not change.","tokens_in":6637,"tokens_out":33599,"duration_ms":301182,"concrete_test":"Implement the Lemma 5 k=0, m0>1/3, pattern 2+2 construction in a small brute-force checker: enumerate small normalized additive valuations (e.g., 8 goods) that admit a unit witness partition with occupancy 2+2, and verify the described U,V,W split-and-repair procedure always yields four balanced bundles of value at least 1. If any valuation fails, the lemma is false; if all pass, this independently confirms the critical repair step.","verdict_should_be":"UNCHANGED","load_bearing_attack":"No significant objection identified. I stress-tested Lemma 5, especially the k=0, m0>1/3, pattern 2+2 case, which is the most intricate part of the proof. The accounting is consistent: Lemma 4 gives e,f ≤ m0/2, so e+f ≤ m0; equation (4) gives m0 ≤ 1/2; choosing Q ⊆ W with e ≤ v(Q) ≤ e+m0 leaves v(W\\Q) ≥ 1−e−m0 ≥ f, so both shortfalls are repairable. The other occupancy patterns and the k=1,2,3 cases rely on minimal-fill accounting that checks out, including the trigger-sum bound from Lemma 3. The ordered-instance reduction and envy-free matching are standard and preserve rejection correctly. The only caveat is that the claimed Lean formalization is not yet inspectable in the arXiv version; this is an external-verification concern, not a mathematical gap in the proof.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper proves Theorem 1: for every finite set of goods and every profile of four nonnegative additive valuations, there is a complete allocation in which each agent receives at least her 1-out-of-5 maximin share. The proof first normalizes each positive target to five unit witness cells, reduces the instance to a common ordered instance, and then establishes a Balanced Residual Lemma (Lemma 5) that allows rejected balanced bundles to be removed while preserving enough balanced unit-valued bundles for the remaining agents. An envy-free matching argument converts this single-agent lemma into the four-agent existence theorem. The paper also records Corollary 7, the full characterization for four agents: the universal guarantee holds exactly for d >= 5, using known exact-MMS counterexamples for d <= 4. The authors additionally claim a sorry-free Lean 4 formalization of the theorem.","tokens_in":6900,"tokens_out":26196,"duration_ms":198553,"significance":"If correct, the result closes the last open denominator for four-agent ordinal maximin share, improving the previous 1-out-of-6 guarantee to the optimal 1-out-of-5. The main technical contribution, Lemma 5, is a genuinely combinatorial balanced-residual partition lemma, and its intricate 2+2 occupancy case is handled by a neat nearest-crossing split argument. The proof is self-contained and uses only standard tools (Hall's theorem, envy-free matching). The claimed Lean 4 formalization is a potentially strong correctness certificate, but the artifacts are not yet available in the text; this is an external-verification concern rather than a mathematical gap. Overall, the paper is significant and, on reading, both the central theorem and its proof appear sound.","major_comments":[],"minor_comments":[{"comment":"The machine-verification claim is stated without the Lean source files being available in the arXiv text. Please include the Lean development, or a persistent link to it, in the final version so the sorry-free claim can be checked.","section":"Section 4"},{"comment":"The sentence 'would give v(R)<2' is correct but understated: the contradiction is with v(R)>3. Rewording to make this explicit would improve readability.","section":"Lemma 5, case k=2"},{"comment":"The sentence 'Use U for the two complementary demands x, 1-x' could be expanded to note that the two demands sum to exactly the value of U, which is why Lemma 4 is applicable to the unit cell U.","section":"Lemma 5, pattern 2+2"},{"comment":"The phrase 'All agents now agree that g1 is the highest rank' is potentially misleading because the agents still have different cardinal values. Consider clarifying: each agent's value is nonincreasing in the common rank order.","section":"Section 3.1"}],"recommendation":"accept","confidential_remarks":"The proof is sound and the result is a clean closing of the four-agent ordinal-MMS question. I recommend accepting. The only caveat is the unverifiable Lean claim; if the journal's policy allows, I would ask for the Lean sources to be deposited or linked in the final version, but this does not affect my recommendation."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Colleague,\n\nThis paper settles the last open ordinal-MMS question for four agents: 1-out-of-5 allocations always exist, and since exact MMS can fail, the guarantee is tight. It improves the previous 1-out-of-6 bound and completes the characterization. That's a real result, and the proof is worth your time.\n\nThe main technical work is the balanced-residual lemma. After removing rejected bundles that each contain one anchor (one of the four highest-ranked goods), the remaining goods can still be split into the required number of balanced unit-valued bundles. The hard case is the initial 2+2 occupancy pattern, where two witness cells each contain two anchors. The nearest-crossing split repair is a genuinely new idea: split two unit cells into pieces that almost meet the complementary demands, then use a third anchor-free cell to patch both shortfalls at once. I checked the accounting (e,f ≤ m0/2, e+f ≤ m0, m0 ≤ 1/2) and it holds. The k≥1 cases and the m0 ≤ 1/3 case are routine but correct. The envy-free matching step is standard and correctly preserves rejection.\n\nSoft spots, in proportion: the Lean 4 formalization is claimed but the sources are not yet available in the arXiv version, so the certificate is currently not inspectable. That's a verification gap, not a mathematical one; I'd want to see the code before fully relying on it. The 2+2 case is intricate, but I don't see a hole. The paper doesn't address computation, which is fine for an existence result. The result is specific to four agents; the discussion of what might generalize is honest rather than overstated.\n\nThe citation pattern looks right: the paper builds on Hosseini, Searns, and Segal-Halevi, and gives proper credit to the known counterexamples. The self-citation of prior work is appropriate.\n\nThis paper deserves a serious referee. I'd send it out, and I'd expect the main theorem to hold up. The Lean artifact should be made available during review so the formalization claim can be checked. If it verifies, this is a clean close of the four-agent case.\n\nFor your reading group: maybe—it's a focused proof paper, good for fair-division people but not a must-read beyond that.","headline":"A solid, self-contained proof of 1-out-of-5 MMS for four agents; the combinatorial lemma is genuinely new, and the only real caveat is that the Lean formalization isn't yet available to inspect.","tokens_in":7249,"tokens_out":6649,"would_cite":true,"duration_ms":62998,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["91B32"],"pacs":[],"model":"deepseek-v4-flash","headline":"For four agents, a 1-out-of-5 maximin-share allocation always exists.","keywords":["maximin share","ordinal MMS","fair division","four agents","additive valuations","balanced residual lemma","existence theorem","1-out-of-d"],"falsifier":"Search for a four-agent instance with no 1-out-of-5 maximin-share allocation. Since the theorem claims universal existence, a single counterexample would falsify it; an exhaustive check over small goods sets and discretized additive valuations is the natural test. More narrowly, one can test the 2+2 repair step: construct a normalized valuation with least anchor value above 1/2 and witness cells arranged as 2+2, and see whether the repair inequalities e+f≤m0 fail.","tokens_in":6602,"feed_emoji":"⚖️","tokens_out":4606,"duration_ms":47101,"temperature":0.7,"pith_summary":"The paper proves that for any four agents with nonnegative additive valuations over a finite set of goods, there is always an allocation in which each agent receives one of the five bundles she herself would have proposed; formally, a complete 1-out-of-5 maximin-share allocation exists. This improves the previous guaranteed denominator of 6 and, together with known counterexamples to exact maximin shares, shows that 5 is the smallest universal denominator for four agents. The proof works by reducing to a normalized ordered instance, showing that bundles rejected earlier can always be removed while leaving the remaining goods partitionable into equally valuable balanced bundles, and then applying an envy-free matching argument. If correct, the result fully settles the four-agent ordinal-MMS question.","feed_headline":"For 4 agents, a 1-out-of-5 maximin-share allocation always exists","feed_subtitle":"New proof shows denominator 5 is both always enough and the smallest possible for four-agent fair division.","key_machinery":"The balanced residual partition lemma (Lemma 5). Anchors are the four highest-ranked goods; a bundle is balanced if it contains exactly one anchor. The lemma guarantees that after removing k<4 rejected balanced bundles, the residual goods can be repartitioned into 4−k balanced bundles of value at least one. In its crucial 2+2 occupancy pattern, two witness cells each contain two anchors, and three anchor-free unit cells provide filler; Lemma 4 (nearest-crossing split) cuts a unit cell at a crossing point so the error is at most half the largest item value, and the third cell fixes both residual shortfalls because their combined size is bounded by the least anchor value.","core_discovery":"The central claim is Theorem 1: for every finite set of goods and every profile of nonnegative additive valuations over four agents, there exists a partition of all goods among the four agents such that each agent's bundle has value at least her 1-out-of-5 maximin share. The proof establishes a stronger single-agent structural lemma — the balanced residual lemma — which says that if up to three rejected balanced bundles (each containing exactly one of the four highest-ranked goods) have been removed, the remaining goods can always be repartitioned into the remaining number of balanced bundles with value at least one. The hardest case is the initial split, where two witness cells of an agent'","pith_inferences":["The 2+2 repair mechanism suggests a possible route to proving larger-denominator guarantees for more than four agents: find a multi-cell analogue of the nearest-crossing split that can handle many shortfalls with the same total budget.","Because the proof reduces to an ordered normalized instance, a practical algorithm might sort goods by each agent's value and then repeatedly apply the residual partition; the complexity of doing so remains open.","The four-agent characterization could serve as a test bed for conjectures about the smallest universal denominator in ordinal maximin-share guarantees for general n, with the 1/3 cutoff here hinting at a more general numerical obstruction."],"forward_implications":["For four agents, a complete 1-out-of-d maximin-share allocation is guaranteed exactly for d≥5; d=4 can fail, so the new result is the sharp threshold.","The proof is fully formalized and machine-checked in a proof assistant, giving the existence statement a high level of verification.","The balanced-residual argument is presented as the transferable core; extending 1-out-of-(n+1) existence to more agents would require a generalized analogue that can repair many simultaneous anchor collisions.","The construction is existential and does not yet address the computational complexity of finding such allocations."],"fun_headline_variants":["Four agents: 1/5 MMS always exists","Tight bound: 1/5 maximin share for 4 agents","Fair division for four: 1/5 share guaranteed","1-out-of-5 MMS proven for four agents","Machine-checked proof: 1/5 MMS for 4 agents"],"cache_read_input_tokens":2304,"weakest_assumption_plain":"The load-bearing premise is that in the 2+2 occupancy case, the three anchor-free unit cells can always be cut to repair both shortfalls, which requires the two shortfalls to sum to at most the least anchor value; if this fails for some valuation profile, the balanced residual lemma and hence the main theorem are not proven.","fun_headline_variants_meta":{"raw":{"variants":["Four agents: 1/5 MMS always exists","Tight bound: 1/5 maximin share for 4 agents","Fair division for four: 1/5 share guaranteed","1-out-of-5 MMS proven for four agents","Machine-checked proof: 1/5 MMS for 4 agents"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.00024,"raw_usage":{"total_tokens":1304,"prompt_tokens":640,"completion_tokens":664,"prompt_tokens_details":{"cached_tokens":256},"prompt_cache_hit_tokens":256,"prompt_cache_miss_tokens":384,"completion_tokens_details":{"reasoning_tokens":589}},"tokens_in":384,"tokens_out":664,"duration_ms":7155,"temperature":1.0,"reasoning_tokens":589,"cache_read_input_tokens":256,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-01T15:53:02.480150+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Search for a four-agent instance with no 1-out-of-5 maximin-share allocation. Since the theorem claims universal existence, a single counterexample would falsify it; an exhaustive check over small goods sets and discretized additive valuations is the natural test. More narrowly, one can test the 2+2 repair step: construct a normalized valuation with least anchor value above 1/2 and witness cells arranged as 2+2, and see whether the repair inequalities e+f≤m0 fail.","supporting_citations":[],"review_version":1}