Abstract
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.
Original language | English |
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Pages (from-to) | 243-246 |
Journal | Discrete Applied Mathematics |
Volume | 92 |
Issue number | 2-3 |
DOIs | |
Publication status | Published - 1999 |