Short··0:40·Combinatorics
Barnette's Conjecture: Does This Loop Always Exist?
Watch on YouTubeAbout the problem →
About this video
Can you walk along the edges of a shape, visit every corner exactly once, and end where you started?
In 1969 David Barnette conjectured that you always can when the shape has 3 edges at every corner and every face has an even number of sides (a cubic, bipartite, 3-connected planar graph). OpenAI says its AI has now proved it. The proof is OpenAI's claim and has not yet been peer reviewed.
Paper: github.com/openai/math/blob/main/preprints/Paired-states-and-…
The animation illustrates the statement; it is not the proof. Every loop shown is a real Hamiltonian cycle, found by computer search and checked: cube (8 corners), hexagonal prism (12), truncated octahedron (24), truncated cuboctahedron (48), truncated icosidodecahedron (120). Pink and cyan mark the two colour classes, so the loop always alternates.
Visuals and sound: original, made with three.js and synthesized audio driven by the same events. Voice (B): ElevenLabs.
Narration is an AI voice (ElevenLabs).
About the problem
Combinatorics·OpenAI Math·Result 180
Barnette’s Hamiltonian-cycle conjecture
Proves that every finite simple cubic bipartite planar 3-vertex-connected graph has a Hamiltonian cycle, resolving Barnette's conjecture. Equivalently, every three-edge path in a finite simple cubic 3-vertex-connected bipartite Pfaffian graph lies in a Hamiltonian cycle. (Lean)
Topics
Barnette's conjectureHamiltonian cycleHamiltonian pathgraph theorycubic graphbipartite graphplanar graphpolyhedrontruncated icosidodecahedronDavid BarnetteTutteOpenAI mathAI proofmaths shorts
