Ringel's Conjecture
The complete graph on 2n+1 vertices can be cut into 2n+1 identical copies of any given tree with n edges.
Resolved
Proved for all sufficiently large n by Richard Montgomery, Alexey Pokrovskiy and Benny Sudakov (2020), using an absorption method combined with a randomised near-perfect embedding of the tree.
Formal statement
Obstruction
Why the direct approaches fail
Resolved for large n, and retained here as the memorisation control. Its proof is recent and well documented, so a model can “solve” it by recall — which makes it the one place we can distinguish reconstruction from recitation. A run scores as reconstruction only if it rebuilds the absorption-plus-random-embedding strategy and explains why a purely greedy embedding fails, rather than quoting the headline.
Attack surface
Registered entries
A run commits to exactly one of these and states why it chose it.
- 01Reconstruct the absorption argument and explain what the absorber structure must guarantee.
- 02Explain why a naive greedy edge-disjoint embedding stalls, and where randomness rescues it.
- 03Identify the role of the tree’s bipartition and why 2n+1 (not 2n) copies is the right count.
Ringel has its own lane in the solver, running continuously alongside every other problem.