Splitting a cake or a set of goods among three people sounds simple—yet the question of whether truly fair division is always possible stumped researchers for decades. It has now been answered, surprisingly, by a SAT solver.
The Problem: What Counts as "Fair"?
The core of fair division asks: given *n* agents and *m* indivisible goods, can we allocate them so that everyone is satisfied in the following sense—"no matter which single item you remove from my bundle, I do not prefer anyone else's bundle over mine"? This criterion is called EFX (envy-free up to any good).
Imagine you and two friends dividing three gifts. EFX requires that if any one item is taken away from your share, you no longer envy anyone else's bundle. It is a demanding standard—and exactly why theorists have been fascinated.
Research since the 1980s proved that EFX allocations always exist for two agents. But for three, four, or more agents, nobody knew. The question was widely regarded as *the* central open problem in discrete fair division.
The Method: Turning a SAT Solver into a Mathematician
The researchers encoded "does an EFX allocation exist?" as a SAT (Boolean satisfiability) instance:
- Unsatisfiable = no allocation satisfies EFX → EFX is always achievable
- Satisfiable = a counterexample has been found
- Title: A Counterexample to EFX n ≥ 3 Agents, m ≥ n+5 Items via SAT-Solving
- Authors: Hannaneh Akrami, Alexander Mayorov, Kurt Mehlhorn, Shreyas Srinivas, Christoph Weidenbach
- Preprint: arXiv:2604.18216 (cs.GT)
- Core contribution: First EFX counterexample discovered via SAT solving, formally verified in Lean
- Link: https://arxiv.org/abs/2604.18216
They set the instance to 3 agents and 8 items (≥ n+5). The SAT solver returned: satisfiable. This means there exists a valuation configuration for which *no* allocation meets the EFX criterion—the first-ever counterexample to EFX feasibility.
> The authors formally verified the correctness of the SAT encoding using the Lean proof assistant. The encoding has no bugs—the counterexample is a genuine counterexample. This means the chain of reasoning has been rigorously validated mathematically.
Results and Significance
Three key results:
1. An EFX counterexample exists. For n ≥ 3 and m ≥ n+5, an EFX allocation need not exist—refuting the conjecture that EFX always exists. 2. At the same time, 3 agents and 7 items is always feasible, sharply delineating the boundary. 3. Even where EFX fails, fallback guarantees exist—one of tEFX or EF1+EEFX always exists.
Using SAT solvers to settle open mathematical problems is not new, but applying a SAT solver to a decades-old conjecture *and* machine-verifying the encoding marks a promising new paradigm for theoretical computer science research.
> One caveat the author admits uncertainty about: is the +5 in m ≥ n+5 tight? Is m = 8 truly the smallest counterexample for n = 3, or could a counterexample only exist for larger m? Not 100% certain.
Paper Information
References
1. Akrami, H., et al. (2026). A Counterexample to EFX. arXiv:2604.18216. 2. Caragiannis, I., et al. (2019). Unifying Approximations of Envy-Freeness. 3. Plaut, B., & Roughgarden, T. (2020). Almost Envy-Freeness. *SODA 2020*.