A note on the Alon-Saks-Seymour problem

Fuente: arXiv
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Main Author: Fox, Jacob
Format: Preprint
Published: 2026
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author Fox, Jacob
author_facet Fox, Jacob
contents Let $f(k)$ be the maximum possible chromatic number of a graph whose edge set can be partitioned into at most $k$ complete bipartite graphs. Alon, Saks, and Seymour conjectured that $f(k)=k+1$ for all $k$. While the conjecture was verified for $k \leq 9$ by Gao et al., it was disproved by Huang and Sudakov, and further Balodis et al. proved that $f(k) \geq 2^{\widetildeΩ((\log k)^2)}$. In this note, we give a simple proof of the recursive upper bound $f(k+1) \leq f(k)+f(\lfloor k/4 \rfloor)$. Consequently, $f(k) \leq 2^{(\log_2 (4k))^2/4}$ for $k \geq 1$. This improves the previous best known upper bound of Mubayi and Vishwanathan in the exponent by a factor which is asymptotically two. Note that these bounds are sharp up to a lower order factor in the exponent by the result of Balodis et al.
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institution arXiv
publishDate 2026
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spellingShingle A note on the Alon-Saks-Seymour problem
Fox, Jacob
Combinatorics
Discrete Mathematics
Let $f(k)$ be the maximum possible chromatic number of a graph whose edge set can be partitioned into at most $k$ complete bipartite graphs. Alon, Saks, and Seymour conjectured that $f(k)=k+1$ for all $k$. While the conjecture was verified for $k \leq 9$ by Gao et al., it was disproved by Huang and Sudakov, and further Balodis et al. proved that $f(k) \geq 2^{\widetildeΩ((\log k)^2)}$. In this note, we give a simple proof of the recursive upper bound $f(k+1) \leq f(k)+f(\lfloor k/4 \rfloor)$. Consequently, $f(k) \leq 2^{(\log_2 (4k))^2/4}$ for $k \geq 1$. This improves the previous best known upper bound of Mubayi and Vishwanathan in the exponent by a factor which is asymptotically two. Note that these bounds are sharp up to a lower order factor in the exponent by the result of Balodis et al.
title A note on the Alon-Saks-Seymour problem
topic Combinatorics
Discrete Mathematics
url https://arxiv.org/abs/2605.28915