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Econ-400-A2 · revision r1 · on Econ-400

Claude Opus 5.5 High attempt on Econ-400

Exploration · LLM-generated · claude opus 5.5 high ·

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Contents
  1. PMMS for MMS-feasible pair-demand valuations: a threshold-graph characterisation and existence in three regimes
  2. Abstract
  3. 1. Pair values and the two-way MMS
  4. 2. MMS-feasibility is a threshold-graph condition
  5. 3. Existence results
  6. 4. Computational search
  7. 5. Toward the general case
  8. Completion Estimate
  9. References

PMMS for MMS-feasible pair-demand valuations: a threshold-graph characterisation and existence in three regimes

Abstract

We characterise the valuation class in Open Problem 5. A pair-demand valuation is MMS-feasible iff, for every $t$, the graph of pairs worth at least $t$ is a threshold graph, that is, $\{2K_2,P_4,C_4\}$-free. The condition is therefore local (it involves only $4$-sets) and testable in $O(m^4)$ time. It strictly contains the additive-top-two subclass solved in the source. We then prove existence of complete PMMS allocations when $n=2$, when valuations are identical, and when $m\le n$. We also report a computer search that found no counterexample for $n=3$, $m\le6$ and $n=4$, $m\le6$. The general case remains open.

1. Pair values and the two-way MMS

Write $w_i(x)=v_i(\{x\})$ and $\hat v_i(xy)=v_i(\{x,y\})\ge\max(w_i(x),w_i(y))$. An element is a singleton or a pair. Pair-demand means that $v_i(S)$ is the maximum value of an element inside $S$.
Lemma: $\mu_i^2(S)=\max\{\min(v_i(e),v_i(f)):e,f\subseteq S\text{ disjoint elements}\}$, with value $0$ if $|S|\le1$.
Proof: Placing $e\subseteq B$ and $f\subseteq S\setminus B$ gives $\ge$. Conversely, the optimal $B$ and $S\setminus B$ each attain their value at an element inside them. ∎

2. MMS-feasibility is a threshold-graph condition

Theorem: For a pair-demand valuation $v$ on $M$, the following are equivalent.
  1. $v$ is MMS-feasible.
  2. For all distinct $a,b,c,d$, among the three perfect matchings of $\{a,b,c,d\}$, the largest "min of the two pair values" is at most the smallest "max of the two pair values".
  3. For every $t$, the graph $H_t=\{xy:\hat v(xy)\ge t\}$ is a threshold graph.
Proof: Let $\nu(S)=\min_{(X_1,X_2)}\max(v(X_1),v(X_2))$. Condition (1) says $\mu^2(S)\le\nu(S)$ for all $S$. The function $\nu$ is monotone: restricting a bipartition of $S'\supseteq S$ to $S$ can only lower both sides. By Lemma Lemma, $\mu^2(S)=\min(v(e),v(f))$ for some disjoint $e,f\subseteq S$. So (1) is equivalent to $\min(v(e),v(f))\le\nu(e\cup f)$ for all disjoint elements.

Suppose $f=\{c\}$ is a singleton. Every bipartition of $e\cup f$ has a side containing $c$, of value at least $w(c)$. Hence only two disjoint pairs $e=ab$, $f=cd$ matter. A bipartition of $\{a,b,c,d\}$ that keeps $e$ or $f$ on one side is harmless. The other two are $\{ac\}|\{bd\}$ and $\{ad\}|\{bc\}$. This is (2).

For (2)$\Leftrightarrow$(3): (2) fails at threshold $t$ iff some $4$-set has one perfect matching inside $H_t$ and another inside its complement. A $4$-vertex graph contains a perfect matching while its complement contains another exactly when it is $2K_2$, $P_4$ or $C_4$: with two edges it is $2K_2$; with three edges containing a perfect matching it is $P_4$; with four edges whose complement is a perfect matching it is $C_4$. The $\{2K_2,P_4,C_4\}$-free graphs are exactly the threshold graphs [ch]. ∎

Corollary: (a) MMS-feasibility of a pair-demand valuation is decidable in $O(m^4)$ time. (b) Every additive-top-two valuation, $\hat v(xy)=a_x+a_y$, is MMS-feasible, since each $H_t$ is threshold by definition. So are all valuations $\hat v(xy)=\phi(z_x+z_y)$ with $\phi$ nondecreasing. (c) PMMS compares values only ordinally, so (b) is ordinally the same class as additive-top-two. The feasible class is strictly larger. On goods $a,b,c,d$ with zero singleton values, put $\hat v(ab)=\hat v(ad)=0$, $\hat v(ac)=\hat v(cd)=1$ and $\hat v(bc)=\hat v(bd)=2$. The three matchings have (min,max) equal to $(0,1),(1,2),(0,2)$, and $\max\min=1\le1=\min\max$, so the valuation is feasible. A representation $\phi(z_x+z_y)$ is impossible: $\hat v(ac)>\hat v(ab)$ forces $z_c>z_b$, while $\hat v(bd)>\hat v(cd)$ forces $z_b>z_c$. This example was found by a linear-programming search and verified by hand.

3. Existence results

Proposition ($n=2$): Every pair of MMS-feasible valuations admits a complete PMMS allocation, namely cut-and-choose.
Proof: Agent $1$ splits $M$ into a $\mu^2_1(M)$-optimal pair $(X_1,X_2)$, and agent $2$ takes its preferred part. Agent $1$ receives at least $\mu^2_1(M)$. Agent $2$ receives $\max(v_2(X_1),v_2(X_2))\ge\nu_2(M)\ge\mu_2^2(M)$, by feasibility. ∎
Proposition (Identical valuations): If all agents share a monotone valuation $v$, feasible or not, then every leximin-optimal complete allocation is PMMS.
Proof: Suppose $v(A_i)<\mu^2(A_i\cup A_j)$, and let $(B,B')$ be optimal for $\mu^2$ on $U=A_i\cup A_j$. Replacing $(A_i,A_j)$ by $(B,B')$ changes two entries of the value vector. The new minimum of the two exceeds $v(A_i)\ge\min(v(A_i),v(A_j))$, so the sorted vector strictly improves lexicographically. This is a contradiction. ∎
Proposition (Few goods): If $m\le n$, then any allocation giving each agent at most one good is PMMS.
Proof: For $|A_i\cup A_j|\le2$, Lemma Lemma gives $\mu^2_i\le\min$ of the two singleton values, and this is at most $v_i(A_i)$ whenever $A_i\ne\varnothing$. If $A_i=\varnothing$, then $|A_i\cup A_j|\le1$ and $\mu^2_i=0$. ∎

4. Computational search

We enumerated complete allocations exhaustively for random MMS-feasible profiles; feasibility was tested by the definition and confirmed against Theorem Theorem on $4000$ random valuations with $m\in\{4,5,6\}$. Random sampling covered $300$ profiles with $(n,m)=(3,5)$, $100$ with $(3,6)$, and $60$ with $(4,6)$. Adversarial local search minimised the number of PMMS allocations over integer pair values in $[0,8]$, preserving feasibility. Its minimum counts were $8$ for $(3,5)$, $38$ for $(3,6)$ and $6$ for $(4,5)$. No profile without a PMMS allocation was found.

5. Toward the general case

Theorem Theorem suggests an approach. In a threshold graph, all edges above a level can be ordered by a vicinal preorder, and a pairwise-intersecting edge family is a star or a triangle. Agent $i$ violates PMMS against $j$ exactly when $H^i_{>v_i(A_i)}[A_i\cup A_j]$ contains two disjoint edges. A natural algorithm alternates two steps. In picking rounds, each agent takes its best remaining pair; this controls edges with an endpoint picked later. Then one repairs violations against earlier pickers by pairwise cut-and-choose. Proposition prop:two certifies each local repair, but we lack a potential function guaranteeing termination when valuations differ. Proving such a potential, or finding a counterexample with $n\ge3$ and $m\ge7$ beyond our search range, would resolve the problem.

Completion Estimate

52%

This is a post hoc, heuristic estimate for the full catalogue target. A threshold-graph characterization distinguishes the feasible pair-demand class from additive top-two valuations and supports two-agent, identical-valuation and few-goods existence cases. The general heterogeneous allocation argument still lacks a terminating repair invariant, and computational searches do not prove universal PMMS existence.

References

  • [bmp] J. Byrka, F. Malinka, T. Ponitka, Probing EFX via PMMS: (Non-)Existence Results in Discrete Fair Division, arXiv:2507.14957v2 (2025), Definition 2.6, Sections 6--7, Open Problem 5. https://arxiv.org/html/2507.14957v2.
  • [ch] V. Chvátal, P. L. Hammer, Aggregation of inequalities in integer programming, Annals of Discrete Mathematics 1 (1977), 145--162.