MathLabs

Worked solution: A rainbow-tree proof of Ringel's conjecture (2020)

Step 4 of 8: Method M1: randomly embedding almost all of the tree
In plain words

Rather than trying to place the whole tree TT at once, the authors first place a slightly smaller version T′T' (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.

(1−ϵ)n(1-\epsilon)n
Detailed analysis

Montgomery, Pokrovskiy and Sudakov (2021, Section 2.1, building on their earlier paper [Montgomery-Pokrovskiy-Sudakov, prior work]) prove Theorem 2.2: for a 22-factorised ND-colouring of K2n+1K_{2n+1} and a forest T′T' on (1−ϵ)n(1-\epsilon)n vertices, there is a randomised rainbow copy T^′\hat{T}' of T′T', together with random subsets V⊆V(K2n+1)∖V(T^′), C⊆C(K2n+1)∖C(T^′)V \subseteq V(K_{2n+1})\setminus V(\hat{T}'), \ C \subseteq C(K_{2n+1})\setminus C(\hat{T}'), such that T^′\hat{T}' is rainbow with high probability, and both VV and CC 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.

Terms in this step
2-factorised colouring
An edge-colouring of K2n+1K_{2n+1} in which every vertex is adjacent to exactly 22 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 qq, making its statistical behaviour easy to control.
Knowledge used in this step