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

Step 5 of 8: Method M2: absorbing the missing piece with leftover colours
In plain words

After Step 4, most of the tree sits rainbow inside K2n+1K_{2n+1}, but exactly n−e(T^′)n - e(\hat{T}') colours and a matching number of vertices remain unused, and the small removed part of TT (its trimmed leaves in Case A, or the far ends of its bare paths in Case B) still needs to be attached using precisely those leftover colours -- no more, no fewer. This is a delicate matching problem: absorption designs a flexible structure among the leftover random vertices that can complete the embedding using exactly the unused colours, whatever they happen to be.

n−e(T^′)n - e(\hat{T}')
Detailed analysis

Sections 4 and 5 of Montgomery, Pokrovskiy and Sudakov (2021) prove the "finishing lemmas" for Case A and Case B respectively (Method M2). The idea, adapted from the general absorption method of Rodl, Rucinski and Szemeredi, is to design an absorption structure among the leftover random vertex set VV that can be completed using exactly the unused colour set CC, regardless of which specific colours happen to be in CC. Case A (many separated leaves) and Case B (many long bare paths) each need their own tailored absorption gadget, because the missing pieces have different shapes (single pendant edges versus long paths), but both rely on the same underlying randomness guarantees from Theorem 2.2 (Step 4).

Terms in this step
Absorption method
A technique, introduced by Rodl, Rucinski and Szemeredi, that first builds a small flexible "absorbing" structure capable of incorporating almost any leftover piece, then embeds the bulk of the target structure separately, and finally uses the absorber to mop up whatever remains.
Knowledge used in this step