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OPEN This is open, and cannot be resolved with a finite computation. - $500
What is $\mathrm{ex}_3(n,K_4^3)$? That is, the largest number of $3$-edges which can placed on $n$ vertices so that there exists no $K_4^3$, a set of 4 vertices which is covered by all 4 possible $3$-edges.
Disclaimer: The open status of this problem reflects the current belief of the owner of this website. There may be literature on this problem that I am unaware of, which may partially or completely solve the stated problem. Please do your own literature search before expending significant effort on solving this problem. If you find any relevant literature not mentioned here, please add this in a comment.
A problem of Turán. Turán observed that dividing the vertices into three equal parts $X_1,X_2,X_3$, and taking the edges to be those triples that either have exactly one vertex in each part or two vertices in $X_i$ and one vertex in $X_{i+1}$ (where $X_4=X_1$) shows that\[\mathrm{ex}_3(n,K_4^3)\geq\left(\frac{5}{9}+o(1)\right)\binom{n}{3}.\]This is probably the truth. The current best upper bound is\[\mathrm{ex}_3(n,K_4^3)\leq 0.5611666\binom{n}{3},\]due to Razborov [Ra10].

See also [712] for the general case.

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This page was last edited 05 October 2025.

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Related OEIS sequences: A140462
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When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:

T. F. Bloom, Erdős Problem #500, https://www.erdosproblems.com/500, accessed 2026-01-16