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Statement

Printed p. 280 (PDF p. 2 of the publisher's scan, whose text layer drops exponents; read on the page image). The problem is motivated by the heuristic that when every clique is large, fewer vertices should suffice to meet them all, and is posed in these words:

Problem 2. "Suppose that each clique of GG has at least k=k(n)k=k(n) vertices. Which value of k(n)k(n) insures that τC(G)\tau_C(G) is less than n−cnn-cn (for some absolute constant cc), or is o(n)o(n) or O(nα)O(n^\alpha) for a given α\alpha, 0<α<10<\alpha<1?"

The paragraph that follows makes these points. Theorem 5 shows that requiring every clique to have at least nc′/log⁡log⁡nn^{c'/\log\log n} vertices, for a suitable constant c′c', does not force τC(G)≤n−cn\tau_C(G)\le n-cn, so k(n)k(n) must be at least of that order. In the other direction no threshold k(n)k(n) was known to force a small clique-transversal number, and the authors suggest studying k(n)=nαk(n)=n^\alpha, 0<α<10<\alpha<1. For constant k(n)=kk(n)=k, "the constructions of Section 3 [sic]" give graphs with τC(G)≥n−O(nslog⁡tn)\tau_C(G)\ge n-O(n^s\log^tn), where s=s(k)=1−1/ks=s(k)=1-1/k (for kk a power of 2) and t=t(k)=k/2t=t(k)=k/2. Whether these bounds are sharp is left open, in particular whether the exponent s(k)s(k) is optimal, as it is for k=2k=2.

Here a clique is an inclusion-maximal complete subgraph with at least two vertices and τC(G)\tau_C(G) the least size of a set meeting every clique (p. 279); n=∣V(G)∣n=|V(G)|. The paper's constructions are in its Section 5 (the substitution G⟨F⟩G\langle F\rangle and Theorem 5, pp. 287--288); "Section 3" is the printed text (Section 3 is the linear-time algorithm), recorded as printed. The site's Problem 611 asks two specializations: whether cliques of at least cncn vertices force τC(G)=oc(n)\tau_C(G)=o_c(n) (the o(n)o(n) clause at k(n)=cnk(n)=cn), and the least kc(n)k_c(n) forcing τC(G)<(1−c)n\tau_C(G)<(1-c)n (the n−cnn-cn clause). The paper's Note added in proof (p. 288) reports the threshold for τC(G)=1\tau_C(G)=1, "Motivated by Problem 2" (note_added_in_proof).

Source. P. Erdős, T. Gallai and Zs. Tuza, Covering the cliques of a graph with vertices, Discrete Math. 108 (1992), 279--289, doi:10.1016/0012-365X(92)90681-5; printed p. 280 = PDF p. 2, read on the page image. The edition is identified in the source digest.

Read depth. Claims checked: the passage was read clause by clause on the page image on 2026-09-19. It poses a question and summarizes Theorem 5 and the constant-kk constructions; the latter's bound n−O(n1−1/klog⁡k/2n)n-O(n^{1-1/k}\log^{k/2}n) was not checked against the proof of Theorem 5, which the paper states only in the form n−o(n)n-o(n).

Proof pointer

None; a question. Its known side is Theorem 5 (the necessary condition) and Theorem 2 (the bound n−knn-\sqrt{kn} for cliques of more than kk vertices).

Dependencies

None.

Bears on

  • Problem 611: the primary formulation of both of the problem's questions, with the authors' statement that no upper bound on k(n)k(n) was known to insure a small clique-transversal number.