A short proof of a conjecture on the Tr-choice number of even cycles Original Research Article

In this note we prove that the Tr-choice number of the cycle C2n is equal to the Tr-choice number of the path (tree) on 4n−1 vertices, i.e. Tr-ch(C2n)=⌊((4n−2)/(4n−1))(2r+2)⌋+1. This solves a recent conjecture of Alon and Zaks.