Xiao and Katona's clique-covering conjecture

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Let tk−1(n)t_{k-1}(n) denote the maximum number of edges in a Kk−1K_{k-1}-free graph on nn vertices, and let τk(G)\tau_k(G) be the minimum size of a vertex set meeting every copy of KkK_k in GG. For a balanced partition V1∪⋯∪Vk−1V_1\cup\cdots\cup V_{k-1} of [n][n], assume ∣V1∣≥⋯≥∣Vk−1∣|V_1|\geq\cdots\geq|V_{k-1}|. Xiao and Katona's conjecture. For fixed integers k≥4k\geq4 (and as stated in the source, s>t≥1s>t\geq1), every graph GG on nn vertices with tk−1(n)+1t_{k-1}(n)+1 edges and τk(G)≥2\tau_k(G)\geq2 contains at least

(∣V1∣+∣V2∣−2)∏i=3k−1∣Vi∣\left(|V_1|+|V_2|-2\right)\prod_{i=3}^{k-1}|V_i|

copies of KkK_k. The paper states that it gives a counterexample to one Xiao–Katona conjecture and proves a modified version, so this original conjecture is refuted.

References

Primary source

Xizhi Liu and Dhruv Mubayi, “On a generalized Erdős-Rademacher problem”, arXiv:2005.07224 (2020).

Progress summary

Refreshed
Claimed solved

A later paper claims to prove the conjecture’s predicted lower bound, but that claim has not been independently verified here.

Xiao and Katona conjectured that graphs just above the Turán threshold, with no single vertex meeting every KkK_k, must contain at least the number of KkK_k copies given by the balanced-partition construction.

Claimed verification in a later arXiv version (date not stated)

Theorem 2 of the later paper asserts that, for sufficiently large nn, every graph in the conjecture has at least (∣V1∣+∣V2∣−2)∏i=3k−1∣Vi∣(|V_1|+|V_2|-2)\prod_{i=3}^{k-1}|V_i| copies of KkK_k, identifying this as a verification of the Xiao–Katona conjecture. The same paper’s modified result concerns a different conjecture; the retrieved source reports no counterexample to the stated clique-covering conjecture. This theorem remains unverified in this scan.

Current status (as of September 2026): The conjecture is claimed proved in its intended sufficiently-large-nn form, while independent verification and any treatment of exceptional small nn are not recorded here.

Sources

Solutions 0

No solutions have been posted yet.