Daykin–Häggkvist partial Latin square completion conjecture

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A partial Latin square of order nn is an n×nn\times n array whose entries are symbols from [n][n], with no symbol repeated in any row or column, while some cells may be empty. A completion is a Latin square obtained by filling all empty cells while preserving the existing entries.

Daykin–Häggkvist conjecture. Given a partial Latin square LL of order nn in which each row, column, and symbol is used at most n/4n/4 times, it is possible to complete LL into a Latin square of order nn.

The conjecture is a completion form of triangle-decomposition theory for complete tripartite graphs. The source discusses later results giving a weaker asymptotic bound of roughly n/25n/25, so the stated n/4n/4 threshold remains open.

References

Primary source

Stefan Glock, Daniela Kühn and Deryk Osthus, “Extremal aspects of graph and hypergraph decomposition problems”, arXiv:2008.00926 (2021).

Progress summary

Refreshed
Claimed progress

A September 2026 manuscript improves the best known completion guarantee but does not reach the conjectured one-quarter threshold, so the problem remains open.

Daykin and Häggkvist conjectured in 1983 that every partial Latin square whose rows, columns, and symbols each occur at most n/4n/4 times can be completed. The conjectured threshold is sharp: larger densities can fail.

Known results

  • Bartlett proved completability at density 9.8×10−59.8\times 10^{-5}.
  • Bowditch and Dukes, and independently Barber, Kühn, Lo, Osthus, and Taylor, raised the guarantee to roughly 0.04n0.04n.
  • A 2016 result established an asymptotic guarantee of (1/25−ε)n(1/25-\varepsilon)n.
  • Yu and Feng later raised the stated bound to 2n/252n/25.

September 2026 improvement

Allsop, Bowtell, Lesgourgues, and Petrova report a sufficient threshold of 0.231n0.231n, substantially closer to n/4n/4. Their manuscript is unrefereed and does not prove the conjecture.

Current status (as of September 2026): The n/4n/4 conjecture remains open; a new unrefereed manuscript claims completability up to 0.231n0.231n.

Sources

Solutions 0

No solutions have been posted yet.