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Elections in which every Condorcet winning set has four candidates
Construct an election in which every Condorcet winning set has at least four candidates, refuting the conjecture that three always suffice.
| 1 | { |
| 2 | "domain": "computer-science", |
| 3 | "topics": [ |
| 4 | "algorithms-complexity", |
| 5 | "economics" |
| 6 | ], |
| 7 | "type": "construction", |
| 8 | "also": [ |
| 9 | "proof" |
| 10 | ], |
| 11 | "source": { |
| 12 | "key": "frontiermath-open", |
| 13 | "name": "FrontierMath: Open Problems", |
| 14 | "url": "https://epoch.ai/frontiermath/open-problems/condorcet-winning-sets", |
| 15 | "id": "condorcet-winning-sets", |
| 16 | "curator": "Epoch AI", |
| 17 | "poser": "Moses Charikar, Alexandra Lassota, Prasanna Ramakrishnan, Adrian Vetta and Kangning Wang", |
| 18 | "status": "unsolved", |
| 19 | "as_of": "2026-10-20", |
| 20 | "catalog": "0521c19bbd38" |
| 21 | }, |
| 22 | "review": true, |
| 23 | "approval": "first-time" |
| 24 | } |