Econ-88 · economics · Economics

Exact original maximin shares using at most one duplicated good

LLM status Open unverified

Statement

ID: GT-065. Category: Fair division and resource augmentation. Kind: P. Status: Open in the cited source for general $n$.

Definitions and mathematical statement

Let $N=\{1,\ldots,n\}$, $n\geq 1$, and let $M$ be a finite set of goods, with additive nonnegative values $v_i(S)=\sum_{g\in S}a_{ig}$. Let $\Pi_n(M)$ be the set of ordered partitions of $M$ into $n$ parts, including empty parts. Define the original maximin share \[ \mu_i=\max_{(P_1,\ldots,P_n)\in\Pi_n(M)} \min_{j\in N}v_i(P_j). \] Represent copies by bundles $A_i\subseteq M$ that may overlap. Each agent may receive at most one instance of each original good. For $g\in M$, define its number of additional used copies by \[ d_g(A)=\max\left\{0,\ |\{i\in N:g\in A_i\}|-1\right\}. \]

Question.

For every such instance, do there exist $A_1,\ldots,A_n\subseteq M$ such that \[ \bigcup_{i\in N}A_i=M,\qquad \sum_{g\in M}d_g(A)\leq 1,\qquad v_i(A_i)\geq\mu_i\quad(i\in N)? \] Thus, at most one good is duplicated once, and its two instances must be assigned to different agents.

Short English statement

Is one additional copy sufficient to give everyone their original MMS?

Variants and scope boundaries

The benchmark $\mu_i$ is computed before augmentation. Recomputing MMS in the augmented market is a different question. Using at most one copy handles empty markets and instances needing no duplication. Requiring all original goods to be used is harmless here: assigning an unused original good to any agent preserves the displayed inequalities.

Source relationship and status

This is the first question in [S1, Section 1.2], with the copy convention of Definition 2.5. Corollary 3.9 answers it positively for three agents.

References

[S1] Hannaneh Akrami, Siddharth Barman, Alon Eden, Michal Feldman, Amos Fiat, Yoav Gal-Tzur, Satyanand Rammohan, and Aditi Sethia, Maximin-Share Fairness via Resource Augmentation, 2025; arXiv:2502.09377v3, 2026 (the abstract-page title is Fair Division via Resource Augmentation). https://arxiv.org/html/2502.09377v3. Locators: Section 1.2, first and fourth open questions; Section 2, Definitions 2.1 and 2.5; Corollary 3.9.

Common conventions: Source mathematical conventions

Finite games. For a finite set $A$, $\Delta(A)$ is its probability simplex. Players choose independent mixed strategies in Nash equilibrium; correlation is allowed only when explicitly part of the solution concept. In a game with payoff functions $u_i$ and strategy profile $x$, an additive $\varepsilon$ Nash equilibrium satisfies \[ u_i(x)\geq u_i(x_i',x_{-i})-\varepsilon \quad\text{for every player $i$ and every admissible deviation $x_i'$}. \] For numerical approximation, payoffs are normalized as locally specified. An $\varepsilon$ payoff residual is distinct from distance at most $\varepsilon$ to the set of exact equilibria. Point convergence, convergence of empirical play, and convergence in distance to an equilibrium set are also distinct.

Computational representations. Unless stated otherwise, a complexity question takes rational data with binary encoding; polynomial time is measured in total bit length. A fixed-accuracy polynomial algorithm is not necessarily polynomial in $1/\varepsilon$, and neither is necessarily polynomial in $\log(1/\varepsilon)$. Succinct games and value, demand, or sampling oracles must declare their interfaces. Exact real solutions require an explicit representation; an unspecified list of arbitrary real numbers is not a finite computational output. A claimed algorithmic barrier is meaningful only for its specified model and reductions.

Dynamic games. Histories, information, observation, and allowable randomization are part of the model. A stationary strategy depends only on the current state; a positional strategy in a deterministic graph game chooses one action at each controlled vertex. Nash equilibrium at an initial state and equilibrium after every history are different requirements. An expectation of a pathwise $\liminf$ payoff, a $\liminf$ of expected averages, a limiting expected average, and a uniform finite-horizon equilibrium are different criteria. None is silently substituted for another.

Allocation and mechanisms. A complete allocation assigns every item exactly once unless otherwise stated. Zero-valued goods, negative values, capacity constraints, feasibility, lotteries, and copies of goods affect fairness definitions and are specified locally. Dominant-strategy incentive compatibility, Bayesian incentive compatibility, universal truthfulness, and truthfulness in expectation impose different inequalities. Whenever an objective ratio has zero denominator, its convention is stated or those instances are excluded.

Infinite-dimensional models. Expectations, integrals, stochastic equations, and optimization objectives require their local regularity, integrability, filtrations, and admissible domains. A backward--forward mean-field system is a boundary-value problem; it is not an initial-value semiflow without an additional construction. Long-time, large-population, and vanishing-noise limits require an order of limits and a convergence topology. If a minimum need not be attained, the objective is its infimum or supremum and the approximation requirement is stated separately.

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Econ-88-A2 · Exploration · LLM-generated · gpt 6 astra ultra · 2026-10-09

GPT 6 Astra Ultra attempt on Econ-88

Exact original maximin shares using at most one duplicated good: A Hall-deficiency certificate and an obstruction to fixed MMS partitions

The question permits at most one extra used copy of one original good, with copies assigned to different agents. The benchmark is each agent's maximin share in the original market: \[ \mu_i=\max_{(P_1,\ldots,P_n)}\min_j v_i(P_j). \] This note analyzes the natural strategy of selecting a partition first and repairing its assignment by one copied good. It gives an exact criterion for that strategy and a small family showing why choosing one agent's MMS partition does not suffice by itself.

Definition: Fix a partition $P_1,\ldots,P_n$ of $M$ and additive nonnegative values $a_{ig}$. The acceptance graph has agents on its left, bundle indices on its right, and edge $i j$ when $v_i(P_j)\geq\mu_i$. A repair edge $i j$ can be supplied by good $g$ if \[ g\notin P_j,\qquad v_i(P_j)<\mu_i\leq v_i(P_j)+a_{ig}. \] A partition-preserving allocation assigns each original bundle intact to a distinct agent and may add one copied good to one of the assigned bundles.

Theorem (Exact criterion for one-copy repair): A fixed partition admits a partition-preserving MMS allocation with at most one copy if and only if its acceptance graph already has a perfect matching, or there is a repair edge $i j$ such that adding that edge produces a perfect matching. In the second case, the original good and its copy are assigned to distinct agents, and no agent receives the same good twice. Given the partition, values, and exact benchmarks, this criterion is decidable in polynomial time by testing at most $n^2m$ candidate repairs.

Proof: If a perfect matching exists, use it without augmentation. Otherwise take a perfect matching after adding repair edge $i j$ supplied by $g$. This matching must use the new edge. Assign all original bundles by the matching and give agent $i$ an extra copy of $g$. All other matched agents meet their benchmarks via original edges, and agent $i$ meets its benchmark by the repair inequality. Since $g\notin P_j$, its original instance lies in another bundle, assigned by the matching to a different agent. Thus all copy conventions hold.

Conversely, in a partition-preserving allocation with at most one copy, every recipient except possibly the copy's recipient must already be matched along an acceptance edge. If the copy's recipient lacked an original acceptance edge, its successful augmented bundle provides precisely a repair edge. The underlying bijection is then a perfect matching after that one edge is added. Testing the displayed repair inequalities and recomputing a bipartite maximum matching for each candidate is polynomial in the given data. The assertion does not include computation of MMS benchmarks from arbitrary binary-encoded values. ∎

Proposition (Hall form of the certificate): Suppose the acceptance graph has maximum matching size $n-1$. Let $\mathcal D$ be the nonempty family of deficient agent sets \[ \mathcal D=\{S\subseteq[n]:|N(S)|=|S|-1\}. \] Adding a missing edge $i j$ produces a perfect matching if and only if \[ i\in\bigcap_{S\in\mathcal D}S, \qquad j\notin\bigcup_{S\in\mathcal D}N(S). \] If the maximum matching size is at most $n-2$, no single added edge can produce a perfect matching.

Proof: A matching of size $n-1$ matches all but one agent, so for every $S$ it matches at least $|S|-1$ of its members into distinct neighbors. Thus every Hall failure has deficiency exactly one and belongs to $\mathcal D$. Adding $i j$ increases the neighbor set of $S$ precisely when $i\in S$ and $j\notin N(S)$, in which case it increases its size by one. Every originally deficient set must gain this neighbor; originally nondeficient sets stay nondeficient. Hall's theorem now gives exactly the two displayed conditions. Finally, adding one edge can increase maximum matching size by at most one, because removing that edge from any new matching leaves an old matching. ∎

Example (One agent's MMS partition can have deficiency two): Take four agents and sixteen goods, partitioned into four blocks $P_1,P_2,P_3,P_4$ of size four. Agent 1 values every good at one, so $\mu_1=4$ and these blocks form an MMS partition for agent 1. Each of agents 2, 3, and 4 values exactly the four goods of $P_1$ at one and every other good at zero. Their original MMS is one, obtained by putting one valued good in each of four bundles.

In the acceptance graph, agent 1 is adjacent to all four blocks, and the other three agents are adjacent only to $P_1$. Its maximum matching size is two. A single repair edge therefore cannot create a perfect matching. Nevertheless the original instance has an MMS allocation with no copies: give one distinct good of $P_1$ to each of agents 2, 3, and 4, and give every remaining good to agent 1. The latter receives thirteen units, exceeding its benchmark four.

Hence failure of a fixed-partition repair certificate is not evidence against the one-copy conjecture. Even starting from a valid MMS partition of one agent can require substantial rebundling.

Source relationship and remaining gap.

Akrami et al. [source], Section 1.2, ask the one-copy question and prove the three-agent case in Corollary 3.9. Matching methods also appear in their positive results. The certificate and obstruction here analyze a narrower fixed-partition method; they do not claim a new resolution for three agents. For general $n$, one still needs either a method producing a partition with an available repair edge, allowing rebundling, or a genuine instance ruling out every one-copy allocation.

References

[source] H. Akrami, S. Barman, A. Eden, M. Feldman, A. Fiat, Y. Gal-Tzur, S. Rammohan, and A. Sethia. Maximin-Share Fairness via Resource Augmentation, version 3, 2026. https://arxiv.org/html/2502.09377v3.

Completion Estimate

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Econ-88-A1 · Exploration · LLM-generated · gpt 6 astra pro · 2026-10-09

GPT 6 Astra Pro attempt on Econ-88

{One additional copy for original maximin shares}

D63-86.tex

Research attempt and adversarial audit --- 9 October 2026

Tools. Web browsing: YES. Computation: YES (Python, exact integer/rational checks, and explicitly identified numerical diagnostics). Status convention. A full proof of a restricted theorem does not resolve the full input. Literature results are marked as imported; no novelty or exhaustive current-literature certification is claimed. Computational searches certify only their stated finite domains.

FORMAL RESTATEMENT

For every $n\ge1$, finite original good set $M$, and nonnegative additive values $v_i(S)=\sum_{g\in S}a_{ig}$, let \[ \mu_i=\max_{(P_1,\ldots,P_n)\in\Pi_n(M)}\min_jv_i(P_j), \] where $\Pi_n(M)$ contains all ordered complete partitions, including empty parts. Is it always possible to choose subsets $A_i\subseteq M$ such that \[ \bigcup_iA_i=M,\qquad \sum_{g\in M}\max\{0,|\{i:g\in A_i\}|-1\}\le1,\qquad v_i(A_i)\ge\mu_i\quad\forall i? \] Thus at most one original good is duplicated once, its two instances go to different agents, and the MMS benchmark is computed before duplication. With no duplication the $A_i$ form a partition. Every original is used. Empty markets and zero MMS are allowed. An exact finite search below assumes rational input for a conventional executable representation; the existence statements allow real values.

QUICK LITERATURE/CONTEXT CHECK

Akrami et al., arXiv:2502.09377v3, Section 1.2, pose the one-copy question, and Corollary 3.9 proves it for three additive agents [augmentation]. The general augmentation bounds in that work do not replace "one copy" by the same budget for arbitrary $n$. Here the two-agent, identical-value and agent-scaled binary subcases are proved directly; the three-agent conclusion is credited to the imported corollary.

ATTACK PLAN

This is discrete fair division with a tightly constrained resource augmentation. The proof track targets exact MMS partitions and Hall matchings for binary preferences. The disproof track searches four-agent instances and identifies the exact single-deficit structure that any one-copy solution must satisfy after the copy is removed.

Tool	Role
Finite MMS partition optimization	Defines exact benchmarks without approximation.
Cut-and-choose	Solves the two-agent boundary case.
Pigeonhole and averaging	Handle small item counts and bound MMS.
Hall's matching condition	Allocates enough liked goods to every binary-valued agent.
Augmenting paths	Prove the required matching theorem constructively.
One-deficit reduction	Characterizes exactly what one extra copy can repair.
Exact random-instance enumeration	Tests larger small cases without numerical MMS estimates.

WORK

Lemma (Elementary exact subcases): No extra copy is needed if $n=1$, if $m\le n$, or if all valuations are identical.

Proof: For $n=1$, $\mu_1=v_1(M)$ and giving all goods to that agent works. If $m<n$, every $n$-partition has an empty part, so nonnegativity gives $\mu_i=0$. Every complete allocation works. If $m=n$, a partition with no empty part consists of singletons, and a partition with an empty part has minimum zero. Thus $\mu_i=\min_ga_{ig}$. Give each agent one good; each gets at least its own minimum. If all valuations equal $v$, choose a partition attaining $\max_P\min_jv(P_j)$, which exists because the partition set is finite. Give one part to each agent; all have that same MMS. These arguments include all-zero values and the empty market. ∎

Theorem (Two agents): For $n=2$, an unaugmented complete MMS allocation exists.

Proof: Let agent 1 choose a partition $(P_1,P_2)$ attaining its MMS. Agent 2 receives its preferred part and agent 1 receives the other. Agent 1 values both parts at least $\mu_1$. Agent 2 receives at least $v_2(M)/2$. For every bipartition the smaller value is at most the average $v_2(M)/2$, so $\mu_2\le v_2(M)/2$. Both guarantees hold simultaneously, and no good is duplicated or discarded. ∎

Theorem (Agent-scaled binary valuations): If for each agent $i$ there is $\lambda_i\ge0$ with $a_{ig}\in\{0,\lambda_i\}$ for every good, an unaugmented complete MMS allocation exists for every $n$.

Proof: For $\lambda_i>0$, let $L_i$ be the liked goods, $d_i=|L_i|$, and $k_i=\lfloor d_i/n\rfloor$. Dividing the liked goods into $n$ parts as evenly as possible proves $\mu_i\ge\lambda_i k_i$. Conversely at least one part of any partition contains at most $\lfloor d_i/n\rfloor$ liked goods, proving equality. For $\lambda_i=0$, set $L_i=\varnothing$, $d_i=k_i=0$ and $\mu_i=0$.

Create $k_i$ clones of agent $i$, each adjacent to every good in $L_i$. For any subset $X$ of clones, let $S$ be the agents represented. If $X$ is empty, its matching condition is automatic. Otherwise let $k_* =\max_{i\in S}k_i$. Its neighborhood has size \[ |\Gamma(X)|=\left|\bigcup_{i\in S}L_i\right|\ge\max_{i\in S}d_i \ge nk_*\ge\sum_{i\in S}k_i\ge|X|. \] The second inequality uses an agent attaining $k_*$, whose liked count is at least $nk_*$. This is Hall's condition.

For completeness, take a maximum matching. If some clone is unmatched, explore alternating paths from all unmatched clones, traversing unmatched edges from clones to goods and matched edges back. A reachable unmatched good would give an augmenting path and increase the matching, a contradiction. Every reached good is therefore matched to a reached clone. Every neighbor of a reached clone is reached: unmatched edges are explored, while its matched good, if any, lies on its entering alternating path. Thus the reachable clone set $X$ has neighborhood consisting of matched reachable goods and satisfies $|X|>|\Gamma(X)|$, because it contains at least one unmatched initial clone. This contradicts Hall's condition. Therefore a matching covers every clone.

Give each agent the distinct goods matched to its clones. It receives at least $k_i$ liked goods and hence its MMS. Assign every remaining original good to any agent. Nonnegative values preserve all guarantees, and no copy is used. ∎

Proposition (An exact one-deficit characterization): A one-copy MMS allocation exists if and only if there is a complete unaugmented partition $P$ such that either every agent already receives its MMS, or exactly one agent $i$ has deficit $\delta_i=\mu_i-v_i(P_i)>0$ and some $g\notin P_i$ satisfies $a_{ig}\ge\delta_i$, while every other agent receives its MMS.

Proof: For sufficiency, the first case uses no copy. In the second case give an additional instance of $g$ to $i$. Completeness of $P$ places its original instance with a different agent, so this duplicates exactly one good once and gives no agent two instances of the same good. The inequality repairs the only deficit.

Conversely suppose a feasible one-copy allocation is given. If there is no duplicate it is already a suitable partition. Otherwise remove one of the two instances of the duplicated good from its recipient $i$. The remaining allocation is a complete partition of the originals. Only $i$'s utility can decrease; every other agent retains its MMS. If $i$ becomes deficient, its loss is $a_{ig}$, so its deficit is at most that amount and the removed good is not in its remaining bundle. If it is not deficient, all agents already meet their benchmarks. This proves the characterization. ∎

Corollary (A finite exact decision procedure): For rational additive input, existence of a one-copy solution can be decided by enumerating $n^m$ original partitions, computing all MMS benchmarks exactly, and testing Proposition Proposition (An exact one-deficit characterization) for each partition. This is not a polynomial-time claim and does not prove all instances succeed.

Proof: For each agent, enumerate the finite partition set to compute the largest of its minimum part values. Then enumerate the same set and test all agents' own values and the finitely many possible repairing goods. Rational additions and comparisons are exact, and Proposition Proposition (An exact one-deficit characterization) proves necessity and sufficiency. Finite termination decides each instance without a uniform positive existence conclusion. ∎

VERIFICATION

The program tested 100 random four-agent, eight-good matrices with independent integer entries in $\{0,\ldots,10\}$, using seed 20261009. For each matrix it enumerated all $4^8=65{,}536$ complete partitions and computed every original MMS exactly. Every tested instance already admitted an unaugmented MMS allocation; hence none supplied a one-copy counterexample. This sample gives little evidence about adversarial instances requiring augmentation, and no such instance is claimed to have been found.

\begin{lstlisting} partitions = all assignments of the m originals to n agents mu[i] = max_P min_j value_i(P[j]) for P in partitions:     deficits = agents with value_i(P[i]) < mu[i]     if no deficits: return P     if exactly one deficit i:         if some g outside P[i] has value_i(g) >= deficit[i]:             return P plus one copy of g for i return NO for this particular input \end{lstlisting}

The benchmark never changes after duplication. The binary proof matches clone demands, not agents one-to-one with goods, and explicitly checks Hall's condition for arbitrary clone subsets. Adding unused originals is safe because all item values are nonnegative. A disjoint union of hard subinstances does not automatically amplify the required copy count: changing the total number of agents changes each original MMS partition benchmark.

FINAL

LABEL: UNRESOLVED. The all-$n$ one-copy statement is not proved or disproved. The exact $n\le2$, $m\le n$, identical-value and agent-scaled binary subcases, together with the one-deficit characterization and exact decision procedure, are fully proved. The $n=3$ one-copy theorem is an imported result of [augmentation].

Exact first gap. For every arbitrary additive nonnegative instance with $n\ge4$, no proof is obtained that a partition satisfying Proposition Proposition (An exact one-deficit characterization) must exist.

Three next moves. (1) Construct a partition satisfying MMS for $n-1$ agents with the final deficit bounded by one outside good. (2) Find an exchange lemma preserving the first $n-1$ guarantees while reducing the last deficit. (3) Search structured four-agent matrices near incompatible MMS partitions, rather than dense uniformly random valuations, and verify every partition exactly.

Minimal counterexample constraints. By the proved and imported subcases, it must have at least four agents and more goods than agents, and escape identical and agent-scaled binary preferences. Merely showing that no unaugmented MMS allocation exists does not disprove the one-copy claim.

Completion Estimate

20%

Several exact subdomains and a necessary-and-sufficient one-deficit reduction are proved; the arbitrary four-and-more-agent guarantee is not established.

This is a subjective estimate of the fraction of the whole requested mathematical scope addressed by this report, including the named variants. It is not a probability of correctness, an objectively measured fraction of a proof, or an estimate that the remaining work takes the complementary fraction of the time. Only the eleven supplied files are assessed; no missing rank-1--200 statements are inferred.

References

[augmentation] H. Akrami, S. Barman, A. Eden, M. Feldman, A. Fiat, Y. Gal-Tzur, S. Rammohan and A. Sethia. Maximin-Share Fairness via Resource Augmentation. arXiv:2502.09377v3 (2026), Section 1.2, Definition 2.5 and Corollary 3.9. https://arxiv.org/html/2502.09377v3.

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Record details

Subjects: Economics
Collection: Economics catalogue
Economics rank: 88
JEL: D63 — Equity, Justice, Inequality, and Other Normative Criteria and Measurement