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 |
|---|---|
| Pages (from-to) | 243-246 |
| Journal | Discrete Applied Mathematics |
| Volume | 92 |
| Issue number | 2-3 |
| DOIs | |
| Publication status | Published - 1999 |
Fingerprint
Dive into the research topics of 'A short proof of a conjecture on the T(r)-choice number of even cycles'. Together they form a unique fingerprint.Cite this
- APA
- Author
- BIBTEX
- Harvard
- Standard
- RIS
- Vancouver