AI 中文总结
本文研究图的团划分数与团覆盖数的差值函数$f(n)$，针对Erdős等人提出的相关问题，通过证明$f(n)=\bigl\rfloor n^2/4\bigl\rfloor-\text{Θ}(n^{4/3})$，给出该问题的否定答案。
AI 中文摘要
对于图G，令$\text{cpn}(G)$和$\text{ccn}(G)$分别表示将$E(G)$的边集划分为团所需的最小团数、覆盖$E(G)$所需的最小团数，定义$f(n)=\text{max}_{|V(G)|=n}\bigl(\text{cpn}(G)-\text{ccn}(G)\bigr)$。1983年，Erdős、Faudree和Ordman提出：是否存在顶点数为$n$的图序列$G_n$，使得$\text{cpn}(G_n)-\text{ccn}(G_n)=n^2/4+O(n)$。该问题被列入Chung的综述第66题及UCSD的Erdős问题网站，Caccetta等人证明$f(n)=n^2/4-o(n^2)$，本文证明$f(n)=\bigl\rfloor\frac{n^2}{4}\bigl\rfloor-\text{Θ}(n^{4/3})$，从而给出该问题的否定答案。
英文摘要
For a graph $G$, let $\cpn(G)$ and $\ccn(G)$ denote the minimum numbers of cliques whose edge sets partition and cover $E(G)$, respectively, and put $f(n)=\max_{|V(G)|=n}\bigl(\cpn(G)-\ccn(G)\bigr).$ In 1983, Erdős, Faudree, and Ordman asked whether there is a sequence of graphs $G_n$ such that $|V(G_n)|=n$ and $\cpn(G_n)-\ccn(G_n)=n^2/4+O(n)$. The question appears as Problem 66 in Chung's survey \cite{ChungProblems} and is also listed on the UCSD Erdős Problems website. Caccetta, Erdős, Ordman, and Pullman proved that $f(n)=n^2/4-o(n^2)$. We prove that $f(n)=\left\lfloor\frac{n^2}{4}\right\rfloor-Θ(n^{4/3}),$ and hence answer the question in the negative.
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