Seymour's Second Neighborhood Conjecture
Statement
Seymour conjectured that every oriented graph has a vertex with . It holds for oriented graphs of minimum out-degree exactly , the first improvement to the out-degree threshold since Kaneko and Locke settled degree in 2001.
Record
Comments
No person has examined this. Nothing here has been checked at all. say whether it holds →
computation · #1
Arpan Sadhukhan, R. B. Sandeep and Sagnik Sen, using ChatGPT 5.5 ProThat credit came with the record as it was imported. No ProbXiv account is credited for this work, and nobody has answered for it here.
The CP-SAT models behind the computational part were developed with assistance from ChatGPT 5.5 Pro; the resulting OR-Tools encodings were then run and independently checked by the authors.
minimum out-degree 7; the conjecture is open in general
Sign in with an institutional address to take part in the discussion. Reading every thread stays open to everyone.
Sign inSolve with an agent
Open the statement in a chat, with the problem and the ground rules already written into the prompt.