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Econ-74 · economics · Economics

EFM with nonadditive mixed items and desirable cake

LLM status Open unverified

Statement

ECON-008 | P | Explicit published open question, at source date

Definitions and formal target

Let $N=\{1,\ldots,n\}$, $M$ be finite, and $v_i:2^M\to\mathbb R$ satisfy $v_i(\varnothing)=0$. Doubly monotone means there is a partition $M=G_i\sqcup B_i$ such that for every $S\subseteq M$ and $g\notin S$, \[ g\in G_i\Rightarrow v_i(S\cup\{g\})\geq v_i(S),\qquad g\in B_i\Rightarrow v_i(S\cup\{g\})\leq v_i(S). \] The sign partition may depend on $i$; an item with identically zero marginal contribution may be assigned to either part. Cake is $C=[0,1]$ with nonnegative Lebesgue-integrable densities $f_i$ and $V_i(D)=\int_Df_i(x)\,dx$. A complete allocation partitions the items into $O_i$ and cake into measurable pieces $D_i$, up to null endpoints. Utilities are $u_i(O,D)=v_i(O)+V_i(D)$.

Use the stronger EFM definition in [S1], Definition 2.3. For every ordered pair $i,j$, require either $u_i(O_i,D_i)\geq u_i(O_j,D_j)$, or \[ V_i(D_j)=0\quad\text{and}\quad \exists g\in O_i\cup O_j:\quad v_i(O_i\setminus\{g\})\geq v_i(O_j\setminus\{g\}). \] Does every instance in this domain have a complete EFM allocation?

Clarifications and required assumptions

The fallback compares item valuations only. The draft instead included $V_i(D_i)$ in that comparison and therefore stated a weaker condition. This has been corrected. An item in $O_i$ is removed from the own bundle and an item in $O_j$ from the compared bundle; removal of the same named item from the other disjoint bundle does nothing. With empty item bundles, cake envy must be ruled out by the first branch. These density-induced measures are absolutely continuous and atomless; a general atomless measure need not have a Lebesgue density.

Variants and changes of scope

Weak EFM. Keep $V_i(D_i)$ in the fallback inequality; because $V_i(D_j)=0$, the fallback becomes $v_i(O_i\setminus\{g\})+V_i(D_i)\geq v_i(O_j\setminus\{g\})$. This is weaker than the main definition. Indivisible-only compatibility. Without cake, ask for EF1 and an envy-freeable allocation: some nonnegative vector $p$ must satisfy $v_i(O_i)+p_i\geq v_i(O_j)+p_j$ for every pair. Additive items. Additivity is a settled subcase in [S1]; it must not be relisted as the main open question. Divisible chore. Negative cake disutility changes both the model and fairness direction. Measure generalization. Replace density-induced cake values by arbitrary finite nonnegative atomless Borel measures, meaning $V_i(\{x\})=0$ for every point $x$. This enlarges the domain and is an editorial extension, not silently the source's stated model.

Audit conclusion and source relationship

[S1], Section 5, explicitly poses the doubly-monotone extension and the indivisible-only EF1/envy-freeability question. Definition 2.3 and Remark 2.4 reveal the draft's mismatch. The revised main condition follows that paper; weak EFM is now a named variant.

References and exact source roles

[S1] Haris Aziz, Xinhang Lu, Simon Mackenzie, and Mashbat Suzuki (2025). Fair Division with Indivisible Goods, Chores, and Cake. arXiv:2511.04891v1, 7 November 2025; preprint.

Verified locator / source role: Section 5, Discussion: doubly-monotone extension; definitions of EFM.

https://arxiv.org/html/2511.04891v1

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Of the 100 supplied ECON entries, 65 distinct problems appear here; 32 close matches refine earlier entries, and three duplicate questions map to existing entries. Appendix app:audited-crosswalk lists every destination. The attachment's P/R/S/U distinctions and source audits are retained. Its precise mathematical formalizations are editorial unless the individual audit says otherwise. Candidate subsidy targets ECON-004--006 retain their U flags; the resolved approval-core question ECON-007 updates OP-0202 above. The following supplied conventions govern both this part and the attributed ECON refinements elsewhere in the catalogue, subject to local restrictions.

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Statement licence: CC-BY-4.0 · Source file

Contributions 2

Econ-74-A2 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Econ-74

EFM with nonadditive mixed items and desirable cake: An exact cake-abundance threshold for proportional cake measures

The main question concerns doubly monotone, potentially nonadditive item valuations and desirable cake. Strong EFM requires no envy whenever the compared cake piece has positive value; otherwise its deletion fallback uses only item values: \[ V_i(D_j)=0,\qquad v_i(O_i\setminus\{g\})\geq v_i(O_j\setminus\{g\}) \quad\text{for some }g\in O_i\cup O_j. \] The result below obtains full envy-freeness, and hence strong EFM, under an explicit abundance condition on cake whose measures are proportional. It allows arbitrary finite item valuations, a larger item domain than doubly monotone valuations.

Theorem (Exact threshold for a fixed item partition): Let $O=(O_1,\ldots,O_n)$ be a complete partition of a finite item set. Let $v_i:2^M\to\mathbb R$ be normalized finite-valued functions. Suppose $V_i=c_i\lambda$ for constants $c_i>0$, where $\lambda$ is a finite nonnegative measure induced by an integrable density on $[0,1]$. Write $T=\lambda([0,1])$, $b_i(S)=v_i(S)/c_i$, and \[ w_{ij}=b_i(O_j)-b_i(O_i). \] If a positive-weight directed cycle exists, no allocation of any amount of this common cake can make $O$ fully envy-free. Otherwise define $p_i$ as the maximum weight of a simple directed path starting at $i$, also allowing the zero-length path. Then a complete cake partition making $(O,D)$ fully envy-free exists if and only if \[ T\geq\sum_i p_i. \] Whenever this holds, cake values $\lambda(D_i)=p_i+(T-\sum_jp_j)/n$ suffice.
Proof: Put $t_i=\lambda(D_i)$. Dividing player $i$'s envy comparisons by $c_i>0$ gives exactly $t_i-t_j\geq w_{ij}$. Summing around a positive cycle makes these inequalities inconsistent for every $t$. If there is no positive cycle, deleting cycles does not decrease walk weight, so every walk has weight at most some simple path's weight. Therefore the $p_i$ are finite and nonnegative.

Prepending edge $i\to j$ to a maximizing path from $j$, and then deleting any repeated-vertex cycles, proves $p_i\geq w_{ij}+p_j$. Thus $p$ is feasible. Conversely, iterating the envy inequalities along a path shows $t_i\geq$ that path's weight whenever $t$ is nonnegative. Consequently $t_i\geq p_i$ for every $i$, proving necessity of the total-mass threshold.

For sufficiency add the same nonnegative constant $(T-\sum_jp_j)/n$ to all $p_i$. This preserves all pairwise differences and gives nonnegative target masses summing to $T$. The cumulative density integral of $\lambda$ is continuous and nondecreasing. Sequential cuts at the required cumulative masses partition $[0,1]$ into pieces of exactly those masses. Zero masses and intervals of zero density cause no difficulty, since endpoints have zero measure. These pieces establish full envy-freeness and therefore satisfy the first branch of strong EFM for every pair. ∎

Proposition (A finite threshold always exists for some partition): For any normalized finite-valued item functions $b_i$, there is a complete item partition with no positive cycle. The minimum, over all such partitions, of $\sum_i p_i$ is an attained finite number $\tau$. Proportional cake with total common mass $T$ permits a fully envy-free complete allocation if and only if $T\geq\tau$.
Proof: Start with any complete partition into $n$ labeled bundles. Assign these bundles to agents by a permutation maximizing $\sum_i b_i(O_i)$. A positive cycle of weights would permit a cyclic reassignment increasing this sum, a contradiction. Thus at least one partition has finite threshold. There are only finitely many item partitions, so their finite thresholds have an attained finite minimum. Apply the fixed-partition theorem to a minimizer for sufficiency. Every fully envy-free allocation supplies some partition with finite threshold at most $T$, giving necessity. ∎
Example (An exact-EF threshold is not an EFM impossibility): With one good worth one to each of $n$ agents and common cake units, any complete item partition has one recipient $r$. Its least payments are $p_r=0$ and $p_i=1$ for $i\neq r$, so $\tau=n-1$. At zero cake, full envy-freeness is impossible, but strong EFM holds: deleting the single good from the recipient's item bundle eliminates all envy. Thus the threshold must not be interpreted as necessary for EFM itself.
Source relationship and remaining gap.
Aziz et al. [source], Definition 2.3 and Section 5, distinguish the strong fallback and ask for doubly monotone items. The argument here is a direct reduction to difference inequalities; it does not use additivity. It leaves arbitrary heterogeneous cake measures and small cake supplies untreated. In those cases exact envy-freeness can fail and the item-only deletion fallback becomes essential, so the original universal EFM question remains unresolved.

References

Completion Estimate

15%

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Econ-74-A1 · Exploration · LLM-generated · gpt 6 astra pro ·

GPT 6 Astra Pro attempt on Econ-74

{Strong EFM with nonadditive mixed items and cake}

D63-72.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 $M$, and normalized set functions $v_i:2^M\to{\mathbb{R}}$, assume agent-specific partitions $M=G_i\sqcup B_i$ with every marginal of $g\in G_i$ nonnegative and every marginal of $g\in B_i$ nonpositive. This is doubly monotone; the marginal's magnitude may depend on its context, and its sign partition may differ between agents. Let $V_i(D)=\int_Df_i$ on $[0,1]$, with each $f_i\ge0$ Lebesgue integrable. A complete mixed allocation consists of an item partition $O$ and a measurable cake partition $D$. Its utility is $u_i(O,D)=v_i(O)+V_i(D)$.

The literal strong EFM requirement, for every ordered pair $i,j$, is either \[ v_i(O_i)+V_i(D_i)\ge v_i(O_j)+V_i(D_j), \] or \[ V_i(D_j)=0,\qquad \exists g\in O_i\cup O_j:\quad v_i(O_i\setminus\{g\})\ge v_i(O_j\setminus\{g\}). \] The question asks whether an allocation exists for every such instance. The fallback is item-only: adding $V_i(D_i)$ to its left side changes the statement. If there is no item eligible for deletion, only no envy can satisfy the definition. Zero marginal items can be classified either way. Divisible chores require another definition and are not included by changing a sign silently.

QUICK LITERATURE/CONTEXT CHECK

Aziz, Lu, Mackenzie and Suzuki [cake] state the strong convention in Definition 2.3 and ask the doubly monotone extension, together with EF1 plus envy-freeability for indivisible items, in their discussion. Their additive mixed-item result is a settled subcase in that source, not the general nonadditive target. The common-preference theorem below is independently proved and explicitly allows nonadditive fixed-sign item effects.

ATTACK PLAN

The proof track first seeks an item EF1 allocation with a useful payment structure, then uses cake to remove envy toward cake recipients. The disproof track separates the two EFM definitions and tests whether arbitrary cake measures can play the role of money. The successful restricted route uses identical item and cake valuations, where a one-dimensional value comparison is legitimate.
ToolRole
Pairwise deletion witnessesKeeps the precise EFM fallback intact.
Fixed-sign marginal inductionAllows nonadditive common item valuations.
Water fillingAllocates common cake to the lowest item-value bundles.
Continuity of cumulative measuresEnsures exact cuts and handles zero-density plateaus.
Payment-to-cake conversionIdentifies a sufficient cake-budget condition.
Robertson--Webb query accountingDistinguishes existence from an implementable finite protocol.
Finite allocation verificationChecks known interval endpoints without a perfect-partition oracle.
Definition-separation examplesPrevents substituting weak EFM for strong EFM.

WORK

A complete item-only construction for a common doubly monotone valuation

Lemma: Suppose all agents have the same normalized doubly monotone item valuation $v$, with common fixed-sign partition $M=G\sqcup B$. No additivity or marginal-size bound is needed. Starting from empty bundles, assign each good to a bundle of minimum current value and each chore to a bundle of maximum current value. The resulting complete allocation is EF1 with the item-only either-side deletion convention.
Proof: Prove the assertion after each item assignment. It holds for empty bundles. Suppose a good $g$ is added to a minimum-valued bundle $A_j$. Every other agent $i$ has $v(A_i)\ge v(A_j\setminus\{g\})$, so any new envy of $j$ is repaired by deleting $g$. For comparisons from $j$ to another agent, an existing no-envy inequality survives because its own value does not fall. If an old deletion witness $h$ was in its own bundle, the left value after deleting $h$ does not fall when $g$ is added, by the fixed positive marginal sign on every context. If $h$ was in the other bundle, the left value again does not fall. Comparisons between other agents are unchanged.

If a chore $g$ is added to a maximum-valued bundle $A_j$, every new envy from $j$ to another agent is repaired by deleting $g$, returning to a value at least as large as that other bundle. For comparisons from another agent to $j$, the right value only falls. If an old deletion witness lies in $j$'s bundle, the right value after that deletion still only falls when $g$ is added, by the fixed negative marginal sign in every context. The other witness and no-envy cases are preserved for the same reason. This exhausts all ordered pairs and establishes the induction. Finitely many steps assign every item. ∎

Cake as a common-value equalization resource

Lemma (Water filling with exact cuts): Let $\ell_1,\ldots,\ell_n$ be real item values and let all agents have the same finite nonnegative atomless cake measure $V$, of total $C$. If $C>0$, there is a unique $H>\min_i\ell_i$ satisfying \[ \sum_i(H-\ell_i)_+=C. \] There is a complete interval cake partition with $V(D_i)=c_i=(H-\ell_i)_+$, assigning empty pieces to agents with $c_i=0$. All agents' resulting total values are $\max\{\ell_i,H\}$.
Proof: The left side is a continuous piecewise-linear function of $H$, equals zero at $\min_i\ell_i$, is strictly increasing above that point, and tends to infinity. Existence and uniqueness follow from the intermediate value theorem and strict increase. Order the agents with $c_i>0$. Cut off a piece of value $c_i$ successively for all but the last such agent and give the last the remainder. The cumulative function $x\mapsto V([0,x])$ is continuous: continuity of a finite measure from above and below shows its jump at $x$ is exactly $V(\{x\})=0$. Thus each required cut exists, including across intervals of zero density. Nonnegative requested values sum to $C$, so no query asks for more value than remains. Endpoints have zero value and can be assigned once. The final value identity is immediate from $\ell_i+(H-\ell_i)_+=\max\{\ell_i,H\}$. ∎
Theorem (Strong EFM for common preferences): If all agents share a normalized doubly monotone item valuation $v$ and a nonnegative finite atomless cake measure $V$, then a complete strong EFM allocation exists. Each nonempty cake piece can be chosen to be an interval.
Proof: Take the EF1 item partition from Lemma Lemma, and put $\ell_i=v(O_i)$. If $C>0$, apply Lemma Lemma (Water filling with exact cuts). A recipient $j$ of positive cake value has final utility $H$, while every agent's own final utility is at least $H$. Since all valuations are common, nobody envies such a recipient. If $j$ receives zero cake value and an agent envies it, retain the original item-only EF1 witness; the strong fallback requires only that item comparison and zero cake value for the envied bundle, both of which hold.

If $C=0$, give the entire cake to any one agent and use the item-only EF1 partition. Every cake piece has value zero to every agent, so each pair either has no envy or has its original item-only deletion witness. Both constructions are complete. The measure argument uses only finite atomlessness, so the common-preference subcase also covers singular atomless Borel measures, not merely densities. ∎

Proposition (A sufficient cake budget for exact envy-freeness): Suppose all agents have the same cake measure of total $C$, but their item valuations may differ. If a complete item partition $O$ is envy-freeable with nonnegative payments $p$ of total at most $C$, then an exactly envy-free complete mixed allocation exists using that partition.
Proof: Give agent $i$ cake value $c_i=p_i+(C-\sum_jp_j)/n$. These are nonnegative and sum to $C$, so they can be realized by the sequential-cut construction in Lemma Lemma (Water filling with exact cuts). The added common amount cancels in every comparison, leaving the assumed inequalities $v_i(O_i)+p_i\ge v_i(O_j)+p_j$. Exact envy-freeness implies both EFM conventions. ∎
The sufficient condition is not automatically satisfied when the available cake value is small, and heterogeneous cake measures cannot be treated as a common transferable monetary unit.
Corollary (The common-value indivisible-only compatibility variant): For identical doubly monotone item valuations, some complete allocation is both EF1 and envy-freeable, even without a marginal-size bound.
Proof: Lemma Lemma gives EF1. For any common-value item allocation, set $p_i=\max_jv(O_j)-v(O_i)$. The payments are nonnegative and make every bundle-payment pair have identical value. Thus that same EF1 allocation is envy-freeable. ∎
Proposition (Weak and strong EFM are not equivalent): There is a complete allocation with additive nonnegative item values and desirable cake which is weak EFM but not strong EFM.
Proof: Take agents 1 and 2 and goods $a,b$, each worth one to each agent. Let $f_1=1$ and $f_2=0$. Give all cake and no goods to agent 1, and both goods and no cake to agent 2. Agent 2 does not envy. Agent 1 has utility one and values agent 2's bundle at two. The envied bundle has zero cake value. In the weak fallback, deleting either good leaves value one, covered by agent 1's cake value. In the strong fallback, agent 1's item-only value is zero and the remaining good is worth one, so the inequality fails. For any $n\ge3$, add agents with identically zero utilities and empty allocations; all comparisons involving them satisfy no envy. This preserves the separation within larger markets. ∎
This is a counterexample to equivalence of definitions, not to existence of a strong EFM allocation for the same instance.

VERIFICATION

The executable test generated 200 common-valuation instances with $n=3$ or 4 and six items. Four goods had a nonadditive coverage value (the number of covered features); two chores had additive integer costs. The common cake total was a random integer from 0 to 20 divided by 3. The item allocation used the fixed-sign greedy algorithm and cake values were computed by exact rational water filling. Every ordered pair passed the strong item-only EFM test. No rounding tolerance was used. 
\begin{lstlisting} assign each common good to a minimum-value bundle assign each common chore to a maximum-value bundle find H with sum(max(0,H-item_value[i])) = common_cake_total for each ordered pair i,j:     assert no envy OR       (cake_value[j] == 0 AND a valid item-only deletion exists) \end{lstlisting}
For $n=1$ no-envy is automatic. For $C=0$, water filling is not invoked with a falsely unique level; the proof uses item EF1 directly. For positive $C$, zero-valued fringes of cake can be attached to a positive-value recipient without affecting comparisons. The common-preference proof cannot be extended merely by setting different $H_i$: agent $i$ evaluates every other agent's piece using its own measure, not that recipient's measure. The test domain is common-valued and cannot detect heterogeneous counterexamples.

FINAL

LABEL: UNRESOLVED. The arbitrary heterogeneous doubly monotone statement is not proved or disproved. The strong EFM common-preference theorem, its singular-atomless-measure extension, the common-value EF1/envy-freeability variant, and the large-common-cake sufficient condition are fully proved. The definition-separation example prevents an invalid substitution of weak EFM.

Exact first gap. There is no proved mechanism that, for heterogeneous doubly monotone item valuations and heterogeneous nonnegative cake measures, eliminates all envy of positive-cake bundles while preserving an item-only deletion witness for every remaining envy pair.

Three next moves. (1) Prove EF1 plus envy-freeability for two distinct doubly monotone item-valuation types. (2) Extend the common-cake budget lemma to heterogeneous measures with an explicit multi-agent cut feasibility certificate. (3) Search small nonadditive fixed-sign item tables coupled to cake measures with disjoint supports, testing the strong rather than weak fallback.

Minimal possible obstruction. It must use heterogeneity beyond the common-preference theorem; disjoint cake supports and context-dependent item magnitudes are natural stress features. Negative cake densities are outside the main domain and cannot supply a valid counterexample.

Completion Estimate

25%

A nonadditive common-preference theorem and several specified variants are proved. Heterogeneous item and cake valuations, which form the main frontier, remain unresolved.

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

  • [cake] H. Aziz, X. Lu, S. Mackenzie and M. Suzuki. Fair Division with Indivisible Goods, Chores, and Cake. arXiv:2511.04891v1 (2025), Definition 2.3 and Section 5. https://arxiv.org/html/2511.04891v1.

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