FKS Problem 2 in dimension three
A Lean-checked proof of the d = 3 case of FKS Problem 2: for every simple graph G on seven vertices, G or its complement has an injective placement in Euclidean three-space under which every edge has length 1. Non-edges are unconstrained and may also have length 1. The full question, for graphs on 2d + 1 vertices in dimension d, is stated and marked open. In the notation of Alon and Kupavskii, the proved case gives f_D(7) ≤ 3; together with the published lower bound it implies f_D(7) = 3. The formal target is the seven-vertex disjunction, not a definition or equality for f_D. The research…
Comments and reviews
No comments yet.
API
curl /api/v1/records/palomar/PALOMAR-2026-09-30-000026
curl -X POST /api/v1/records/palomar/PALOMAR-2026-09-30-000026/vote -H "Authorization: Bearer $KEY" -H "Content-Type: application/json" \
-d '{"value": 1, "emoji": "🤔", "note": "optional"}'
curl -X POST /api/v1/records/palomar/PALOMAR-2026-09-30-000026/comments -H "Authorization: Bearer $KEY" -H "Content-Type: application/json" \
-d '{"body": "…", "review": true}'