Turán’s tetrahedron problem

From four points A, B, C and D, there are four triples: ABC, ABD, ACD and BCD. Choosing any three of them avoids choosing the complete collection of four. On a much larger set, impose that same rule on every group of four points.

How large a fraction of all possible triples can be chosen? Turán's conjecture predicts that the best achievable fraction approaches $5/9$ as the number of points grows. The three-out-of-four example explains the forbidden pattern; its fraction is not the predicted large-size answer.[2] [3]

How many triples can be packed into a finite set while every four-point subset misses at least one of its four possible triples? The open target is the largest limiting fraction achievable without any restriction on how the selected triples are distributed. Such a collection is a three-uniform hypergraph: each selected triple is an edge. Write $K_{4}^{(3)}$ for the forbidden four-triple configuration.

If $M(n)$ is the largest permitted number of triples on $n$ vertices, the ordinary density is $$\pi(K_{4}^{(3)})=\lim_{n\to\infty}\frac{M(n)}{\binom{n}{3}}.$$ The denominator counts all choices of three vertices. The limit exists; the conjectured value is $5/9$.

The proposed value comes from a concrete construction. Split the vertices into three nearly equal groups, cyclically ordered A, B, C. Choose every triple taking one vertex from each group. Also choose triples with two vertices in A and one in B, two in B and one in C, or two in C and one in A. This avoids the forbidden four-point pattern.

With exactly $m$ vertices in each group, the chosen count is $$m^{3}+3m\binom{m}{2}.$$ Dividing by $\binom{3m}{3}$ tends to $5/9$ as $m$ grows. This establishes an achievable limiting density; proving no other arrangement exceeds it is the missing direction.[2] [3]

The conjectured limiting fraction is $5/9$; the exact answer remains open.[2] [3]

The September 23, 2026 preprint reports an ordinary-density upper bound below 0.557808, still above $5/9$. Its exact certificate and Lean formalization concern that bound, not equality with $5/9$.

The September uniform-density paper concerns a different parameter: it requires a lower density throughout every sufficiently large vertex subset. Its reported value $1/2$ does not apply to unrestricted ordinary density. Results imposing additional forbidden configurations likewise address narrower targets.[1] [2] [3]

This is a question about when abundance forces a pattern. For ordinary links joining pairs of points, many such thresholds are understood. Triples can overlap in more ways, so even forbidding the complete pattern on four points exposes structure that pair-based graph theory does not determine.[2]

An exact answer would identify the point where a local four-vertex obstruction becomes unavoidable in every sufficiently large collection, regardless of its arrangement. The contrast with the uniform variant shows why the total triple count and the way triples are distributed carry different information.

The problem also connects combinatorial structure with exact computation. Small-subgraph inequalities can give universal density bounds, and numerical searches for those inequalities can be converted into rational certificates. The cited preprint describes that conversion and formal verification. Better numerical evidence alone does not identify the optimal structure or prove the proposed equality.[1] [3]