A Recolouring Version of a Conjecture of Reed
Lucas De Meyer, Clément Legrand-Duchesne, Jared León, Tim Planken, Youri Tamitegama
TL;DR
The paper investigates a recolouring analogue of Reed's conjecture using Kempe changes, defining k-recolourability and frozen colourings as the main obstructions. It establishes a general upper bound ε ≤ 1/3, proves its optimality up to colour-permutation uniqueness, and conjectures ε = 1/3 for all graphs. It then sharpens the picture for special graph classes: for odd-hole-free graphs, ε = 1/2 (tight up to one colour); for triangle-free graphs, ε = 4/9 (tight); and it provides precise thresholds η⋆ that separate frozen non-uniqueness from uniqueness in these classes. The results rely on inductive recolouring schemes with faithful extensions of colourings and careful analysis of Kempe chains, yielding tight obstructions and guiding future attempts to resolve the recolouring Reed-type conjectures. The work clarifies the landscape of recolouring obstructions and highlights the distinct behaviours across general, triangle-free, and odd-hole-free graphs, with potential implications for constructive approaches to Reed-type colouring questions.
Abstract
Reed conjectured that the chromatic number of any graph is closer to its clique number than to its maximum degree plus one. We consider a recolouring version of this conjecture, with respect to Kempe changes. Namely, we investigate the largest $\varepsilon$ such that all graphs $G$ are $k$-recolourable for all $k \ge \lceil \varepsilon ω(G) + (1 -\varepsilon)(Δ(G)+1) \rceil$. For general graphs, an existing construction of a frozen colouring shows that $\varepsilon \le 1/3$. We show that this construction is optimal in the sense that there are no frozen colourings below that threshold. For this reason, we conjecture that $\varepsilon = 1/3$. For triangle-free graphs, we give a construction of frozen colourings that shows that $\varepsilon \le 4/9$, and prove that it is also optimal. In the special case of odd-hole-free graphs, we show that $\varepsilon = 1/2$, and that this is tight up to one colour.
