Turán’s tetrahedron problem

For the tetrahedron $K_4^{(3)}$ , determine whether its Turán density satisfies $\pi(K_4^{(3)})=\frac{5}{9}$ . Equivalently, if $\operatorname{ex}(n,K_4^{(3)})$ denotes the maximum number of edges in a $K_4^{(3)}$ -free $3$ -uniform hypergraph on $n$ vertices, determine whether $\lim_{n\to\infty}\frac{\operatorname{ex}(n,K_4^{(3)})}{\binom{n}{3}}=\frac{5}{9}$ , where $\pi(K_4^{(3)})$ is this limiting density.

References

Primary source

arXiv

Additional references

A New Upper Bound for the Turán Density of the Tetrahedron — arXiv — Gyeongwon Jeong, Seonghun Park, Seonghyuk Im, Joonkyung Lee, Hongseok Yang

Posted 1 week ago · edited 1 week ago

Progress summary

Refreshed 1 week ago

Claimed progress

A new machine-checked result lowers the best known upper bound, but the conjectured exact answer remains unproved.

Turán’s tetrahedron problem asks whether the ordinary density of tetrahedron-free $3$ -uniform hypergraphs equals the conjectured value $\pi(K_4^3)=5/9$ . The classical problem remains open.

Known results

Baber’s earlier upper bound was approximately $0.5615$ .

September 23, 2026 upper-bound improvement

Gyeongwon Jeong, Seonghun Park, Seonghyuk Im, Joonkyung Lee, and Hongseok Yang report an improved upper bound, $\frac{312372062889819}{560000000000000}$ , with the certificate formalized in Lean $4$ . This is claimed progress toward $5/9$ , not a proof of the conjectured value.

Current status (as of September 2026): The reported upper bound improves Baber’s bound and has a Lean $4$ certificate, but the conjecture $\pi(K_4^3)=5/9$ remains open.

Sources

No solutions have been posted yet.