Cycles of weight divisible by $k$

TL;DR

This paper investigates the maximum size of weighted complete directed graphs avoiding cycles with weights divisible by k, providing new bounds and extending results to undirected graphs with group weights.

Contribution

It establishes new upper bounds for the size of such graphs and proves a minimum degree condition for undirected graphs ensuring cycles with weights divisible by k.

Findings

Proves f(k) < k + 2Ω(k), where Ω(k) is the number of prime factors of k.

Shows that undirected graphs with minimum degree 2k-1 contain cycles with weights divisible by k.

Extends the cycle divisibility result to graphs with weights from a finite abelian group.

Abstract

A weighted (directed) graph is a (directed) graph with integer weights assigned to its vertices and edges. The weight of a subgraph is the sum of weights of vertices and edges in the subgraph. The problem of determining the largest order $f(k)$ of a weighted complete directed graph that does not contain a directed cycle of weight divisible by $k$, for an integer $k \ge 2$, was raised by Alon and Krivelevich [J. Graph Theory 98 (2021) 623-629]. They showed that $f(k)$ is $O(k\log k)$ and $f(k) \le 2k-2$ if $k$ is prime. The best bounds known to us are $f(k) \le 2k-2$ for all $k$ and $f(k) < (3k-1)/2$ for prime $k$. It is also known that $f(k) \ge k$ and this is believed to be the correct value. We prove that $f(k) < k+2\Omega(k)$, where $\Omega(k)$ is the number of prime factors, not necessarily distinct, in the prime factorization of $k$. We also show that any weighted undirected graph of minimum degree $2k-1$ contains a cycle of weight divisible by $k$. This result is proved in the more general setting in which the weights are from a finite abelian group of order $k$, and the cycle has weight equal to the group identity. We conjecture that this holds for undirected graphs with minimum degree $k+1$.