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arxiv: 1708.09777 · v1 · pith:RXCJ6ETLnew · submitted 2017-08-31 · 🧮 math.CO

Zero-sum K_m over mathbb{Z} and the story of K₄

classification 🧮 math.CO
keywords everyfunctionthereweightingzero-sumcopyequivfollowing
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We prove the following results solving a problem raised in [Y. Caro, R. Yuster, On zero-sum and almost zero-sum subgraphs over $\mathbb{Z}$, Graphs Combin. 32 (2016), 49--63]. For a positive integer $m\geq 2$, $m\neq 4$, there are infinitely many values of $n$ such that the following holds: There is a weighting function $f:E(K_n)\to \{-1,1\}$ (and hence a weighting function $f: E(K_n)\to \{-1,0,1\}$), such that $\sum_{e\in E(K_n)}f(e)=0$ but, for every copy $H$ of $K_m$ in $K_n$, $\sum_{e\in E(H)}f(e)\neq 0$. On the other hand, for every integer $n\geq 5$ and every weighting function $f:E(K_n)\to \{-1,1\}$ such that $|\sum_{e\in E(K_n)}f(e)|\leq \binom{n}{2}-h(n)$, where $h(n)=2(n+1)$ if $n \equiv 0$ (mod $4$) and $h(n)=2n$ if $n \not\equiv 0$ (mod $4$), there is always a copy $H$ of $K_4$ in $K_n$ for which $\sum_{e\in E(H)}f(e)=0$, and the value of $h(n)$ is sharp.

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