We revisit the direct sum theorems in communication complexity which askes whether the resource to solve $n$ communication problems together is (approximately) the sum of resources to solve these problems separately. Our work starts with the observation that Meir and Dinur's fortification lemma for protocol size over rectangles can be generalized to a general fortification lemma for a sub-additive measure over set. By applying this lemma to the case of cover number, we obtain a dual form of cover number, called ``$\delta$-fooling set'' which is a generalized fooling set. Given a communication problem $S\subseteq (X\times Y) \times Z$, let $\Lambda \subseteq X\times Y$ be a $\delta$-fooling set of $S$, then given any subset $\tilde{\Lambda} \subseteq \Lambda$ such that $|\tilde{\Lambda}|/{|\Lambda|} > \delta$, there is no monochromatic rectangle that covers the subset $\tilde{\Lambda}$. Particularly, there is a $\frac{16\log|X| |Y|}{\mathsf{ Cov}(S)}$-fooling set of communication problem $S$. With this fact, we are able to reprove the classic direct sum theorem of cover number with a simple double counting argument. And we prove a new direct sum theorem about protocol size which imply a better direct sum theorem for two functions in terms of protocol size. Formally, let $\mathsf{L}$ denote the protocol szie, given a communication problem $F:A \times B \rightarrow \{0,1\}$, $ \log\mathsf{L}\left(F\times F\right)\geq \log \mathsf{L}\left(F\right) +\Omega\left(\sqrt{\log\mathsf{L}\left(F\right)}\right)-\log\log|A||B| -4$.We also prove a tight cover number lower bound for the agree problem introduced by Amos Beimel et al.
翻译:我们重新审视通信复杂性中的直接和理学 {Fright {Fright} complication $(约) 用于单独解决这些问题的资源总和。我们的工作始于观察Meir 和 Dinur 在矩形上的协议大小的堡垒 emma, 在设定的子添加度度测量中, 可以普遍化为普通的堡垒 lemma 。 通过对覆盖数的案例中应用这个 lemmma, 我们获得了双倍的封面号, 叫做 $\del$- folton 集合。 鉴于通信问题 $Sdelem(Xtime Y), 让 $\lambda\ a committed $( delta$- fol) 。 然后给任何子集 $(lax labda) 问题\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\