Yet another proof of Brooks' theorem
Combinatorics
2014-09-25 v1
Abstract
Arguably the simplest variation of this style of proof as we avoid reducing to the cubic case entirely.
Cite
@article{arxiv.1409.6812,
title = {Yet another proof of Brooks' theorem},
author = {Landon Rabern},
journal= {arXiv preprint arXiv:1409.6812},
year = {2014}
}
Comments
A referee for "Brooks' Theorem and Beyond" asked to make this proof available on the arXiv