Short··0:21·Combinatorics
Ramsey Numbers r(s,t): OpenAI Claims the Sharp Log Exponent
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About this video
The Ramsey number r(s,t) is the fewest dots that force either s mutual friends or t mutual strangers, however the links are drawn. Since 1980 it has been known that r(s,t) grows at most like t^(s-1)/(log t)^(s-2), but the best lower bounds sat far below. OpenAI claims that for every fixed s ≥ 5, r(s,t) = t^(s-1)/(log t)^(s-2+o(1)), pinning down the logarithmic exponent and matching the classical upper bound at that scale; these are OpenAI's preprints, not yet peer reviewed.
Papers (OpenAI, catalogue result 170):
- The sharp logarithmic exponent of r(5,t): github.com/openai/math/blob/main/preprints/The-Sharp-Logarith…
- Sharp logarithmic exponents for fixed off-diagonal Ramsey numbers: github.com/openai/math/blob/main/preprints/Sharp-Logarithmic-…
The animation illustrates the result; it is not a proof.
Gradient Damage: mathematics you can see, equations you can hear.
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Every result explained: gradientdamage.com/p/170
About the problem
Combinatorics·OpenAI Math·Result 170
Sharp logarithmic exponents for off-diagonal Ramsey numbers
For every fixed integer s ≥ 5, proves r(s,t)=ts-1/(log t)s-2+o(1) as t → ∞, determining the logarithmic exponent and matching the classical upper bound at that scale. Here r(s,t) is the least number of vertices forcing an s-clique or a t-vertex independent set. (Lean)
Topics
Ramsey numbersoff-diagonal Ramsey numbersRamsey theorygraph theoryextremal combinatoricscliqueindependent setK5-free processrandom graphsAjtai Komlos SzemeredicombinatoricsOpenAIOpenAI mathAI mathematicsAI proofmathsmathematicsmath shorts3D animationRamsey numberparty problemfriends and strangersGradient Damage
