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For a graph let denote the minimal number of vertices that include at least one from each maximal clique of on at least two vertices (sometimes called the clique transversal number).
Let be maximal such that every triangle-free graph on vertices contains an independent set on vertices.
If is a graph on vertices then is
Source: erdosproblems.com/151
No claim settles this problem.
Open. No proof, disproof or proof claim for the inequality was found in the search whose scope the Current assessment records. The best explicit upper bound valid for every is Theorem 3 of Erdős, Gallai and Tuza (Discrete Math. 108 (1992), refereed): for every graph on vertices, from a linear-time algorithm; their Theorem 1 gives by averaging two lemmas. The best asymptotic bound is for some and all large , Corollary 2 of Joret, Micek, Reed and Smid (2021) with the one-line transfer recorded on Problem 610. The conjectured value is with of order : the 1992 paper records from Ajtai, Komlós and Szemerédi (1980, refereed; their Theorem 3, , the bound on being its elementary rewriting) and Erdős (1961), and Kim's Theorem 1.1 (1995) gives for large . The 1992 authors write that "so far we could not construct examples worse than triangle-free ones" ([EGT92], p. 280), and Erdős that the conjecture "is perhaps completely wrongheaded". The asymptotic bound matches the order of but not its constant, so it leaves this problem open. This is a bounded negative finding, not a certificate of openness. The inequality holds for every chordal graph, by Tuza's bound for chordal graphs (Discrete Math. 86 (1990), Theorem 2(a), refereed), an accepted partial claim recorded on its claim page (Tuza, 1990). A partial proof claim of 2026-09-28, the inequality for every graph on at most 39 vertices, is on the site's proof-claim tab; it is recorded as claimed on its claim page (veljjanoski, 2026) and derives nothing for the standing, which stays open.