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Erdős #901 · combinatorics / hypergraphs

Let m(n)m(n) be minimal such that there is an nn-uniform hypergraph with m(n)m(n) edges which is 33-chromatic. Estimate m(n)m(n).

MSC 05 · Combinatorics

Inferred from the community field; finer classification has not been reviewed.

Source reports OPENAs of 2026-09-08Source & evidence ↓
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Problem statement

Let m(n)m(n) be minimal such that there is an nn-uniform hypergraph with m(n)m(n) edges which is 33-chromatic. Estimate m(n)m(n).

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Background

In other words, the hypergraph does not have Property B. Property B means that there is a set SS which intersects all edges and yet does not contain any edge.

It is known that m(2)=3m(2)=3, m(3)=7m(3)=7, and m(4)=23m(4)=23. Erdős proved

2n≪m(n)≪n22n2^n \ll m(n) \ll n^2 2^n

(the lower bound in [Er63b] and the upper bound in [Er64e]). Erdős conjectured that m(n)/2n→∞m(n)/2^n\to \infty, which was proved by Beck [Be77], who proved m(n)≫(log⁡n)2nm(n)\gg (\log n)2^n, and later [Be78] improved this to

n1/3−o(1)2n≪m(n).n^{1/3-o(1)}2^n \ll m(n).

Radhakrishnan and Srinivasan [RaSr00] improved this to

nlog⁡n2n≪m(n).\sqrt{\frac{n}{\log n}}2^n \ll m(n).

Pluhar [Pl09] gave a very short proof that m(n)≫n1/42nm(n) \gg n^{1/4}2^n.

In [ErLo75] Erdős and Lovász speculate that n2nn2^n is the correct order of magnitude for m(n)m(n).

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The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #901 link

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