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Seymour’s second-neighborhood conjecture
Proves Seymour's second-neighborhood conjecture: every nonempty finite oriented graph has a vertex with at least as many vertices at directed distance exactly two as at directed distance one. Oriented graphs may be arbitrary apart from the exclusion of loops and oppositely directed edge pairs.
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A proof of Seymour’s second-neighborhood conjecture
Sep 23, 2026·paper.pdfLean
We prove that every nonempty finite oriented graph has a vertex with at least as many vertices at directed distance two as at directed distance one. This resolves Seymour's second neighborhood conjecture positively.