Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
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Statement
Definition 8.5 (p. 40). An -edge-coloring of is a balanced -coloring when, for each color , every set of vertices contains a monochromatic in color . Example 8.6 (p. 41) shows a balanced -coloring of : every vertices induce an edge of both colors.
Lemma 8.7 (p. 41), quoted: "If a finite projective plane of order exists, then has a balanced -coloring. In other words, there is an -edge-coloring of so that for any any vertices induce an edge in color ."
The two sentences agree because . The remark after the lemma (p. 41) notes that a projective plane of order exists for every prime power , and that existence for other orders is open.
Proof pointer
No proof is given in the thesis. The lemma is credited in its heading to Erdős and Gyárfás, Theorem 5 of P. Erdős and A. Gyárfás, Split and balanced colorings of complete graphs, Discrete Math. 200 (1999), 79--86, whose corpus home is the Theorem 5 page of the Erdős--Gyárfás card. That theorem is printed as under the same projective-plane hypothesis, being the least order of a complete graph with a balanced -coloring; the thesis restates it as the existence of a balanced -coloring of and is a secondary statement of it.
Read depth
Claims checked: Definition 8.5, Example 8.6, Lemma 8.7 and the remark after it were read clause by clause on the printed pages. The lemma is not proved in the thesis and no proof was checked here.
Dependencies
None in the thesis.
Source. N. Almási, The Ramsey Turnaround Numbers, master's thesis, Karlsruhe Institute of Technology, 2023; the edition read is named on the source card.
Bears on
- Problem 617: the problem asks whether, for , every -coloring of the edges of has vertices whose induced misses a color, that is, whether has no balanced -coloring, since . The lemma concerns the larger order , where the tested sets have vertices: for every with a projective plane of order , it gives a balanced -coloring there. It says nothing about itself, and restricting its coloring to vertices does not give sets of vertices that see every color.