Worked solution: A rainbow-tree proof of Ringel's conjecture (2020)
Rather than trying to place the whole tree at once, the authors first place a slightly smaller version (missing only the special leaves or bare-path ends singled out by the case division) as a rainbow copy chosen at random among many valid placements. Randomness is the key trick: instead of tracking exactly which vertices and colours end up unused, the authors only need to know that a large, well-behaved random leftover set of vertices and colours remains, ready to absorb the missing piece.
Montgomery, Pokrovskiy and Sudakov (2021, Section 2.1, building on their earlier paper [Montgomery-Pokrovskiy-Sudakov, prior work]) prove Theorem 2.2: for a -factorised ND-colouring of and a forest on vertices, there is a randomised rainbow copy of , together with random subsets , such that is rainbow with high probability, and both and are provably "nicely random" (each element appears independently with a fixed probability). This lets the argument avoid tracking exact leftover vertices and colours, working instead with their statistical distribution.
- 2-factorised colouring
- An edge-colouring of in which every vertex is adjacent to exactly edges of each colour; the ND-colouring has this property.
- q-random set
- A random subset of a ground set in which each element appears independently with the same fixed probability , making its statistical behaviour easy to control.