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 |
Number of pages | 4 |
Journal | Discrete Applied Mathematics |
Volume | 92 |
Issue number | 2-3 |
DOIs | |
Publication status | Published - Jun 1999 |