Erdős Problem #1091 — Let GG be a K4K_4-free graph with chromatic number 44. Must GG contain an odd cycle with at least two diagonals?

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Let GG be a K4K_4-free graph with chromatic number 44. Must GG contain an odd cycle with at least two diagonals? More generally, is there some f(r)→∞f(r)\to \infty such that every graph with chromatic number 44, in which every subgraph on ≤r\leq r vertices has chromatic number ≤3\leq 3, contains an odd cycle with at least f(r)f(r) diagonals?

References

Progress summary

Refreshed
Claimed solved

The original question has a positive answer, while a 2026 preprint says its stronger version is false.

Erdős posed the question in 1976: every K4K_4-free graph with chromatic number 44 should contain an odd cycle with at least two diagonals. The stronger local-colourability formulation asks whether the number of diagonals must become arbitrarily large.

Known results

  • Larson, 1979: a K4K_4-free graph with no odd cycle having a diagonal is bipartite, has a cut vertex, or has a vertex of degree at most 22.
  • Voss, 1982: every K4K_4-free 44-chromatic graph contains an odd cycle with at least two diagonals.
  • The pentagonal wheel shows that three diagonals are not always guaranteed.

April 9, 2026 counterexample

Alexeev, Putterman, Sawhney, Sellke, and Valiant give explicit K4K_4-free graphs GmG_m with 20m+3120m+31 vertices, chromatic number 44, and every proper subgraph 33-colourable, indeed 22-degenerate. Every cycle has at most 1010 chords, so the preprint claims to disprove the proposed function f(r)→∞f(r)\to\infty; this claim is unverified.

Current status (as of September 2026): Voss’s two-diagonal theorem settles the original question, while the stronger unbounded-diagonal assertion is refuted by an explicit but not independently verified 2026 preprint.

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