In communication complexity the input of a function $f:X\times Y\rightarrow Z$ is distributed between two players Alice and Bob. If Alice knows only $x\in X$ and Bob only $y\in Y$, how much information must Alice and Bob share to be able to elicit the value of $f(x,y)$? Do we need $\ell$ more resources to solve $\ell$ instances of a problem? This question is the direct sum question and has been studied in many computational models. In this paper we focus on the case of 2-party deterministic communication complexity and give a counterexample to the direct sum conjecture in its strongest form. To do so we exhibit a family of functions for which the complexity of solving $\ell$ instances is less than $(1 -\epsilon )\ell$ times the complexity of solving one instance for some small enough $\epsilon>0$. We use a customised method in the analysis of our family of total functions, showing that one can force the alternation of rounds between players. This idea allows us to exploit the integrality of the complexity measure to create an increasing gap between the complexity of solving the instances independently with that of solving them together.

翻译：在通信复杂度理论中，函数$f:X\times Y\rightarrow Z$的输入被分配给两名参与者Alice和Bob。若Alice仅知晓$x\in X$而Bob仅知晓$y\in Y$，Alice和Bob需要共享多少信息才能确定$f(x,y)$的值？我们是否需要$\ell$倍的资源来解决$\ell$个问题实例？此即直接和问题，已在多种计算模型中得到研究。本文聚焦于两方确定性通信复杂度情形，针对最强形式的直接和猜想给出了反例。为此，我们构造了一个函数族，对于该族函数，求解$\ell$个实例的复杂度小于求解单个实例复杂度的$(1 -\epsilon )\ell$倍（其中$\epsilon>0$足够小）。我们在分析该全函数族时采用了定制化方法，证明了可以强制参与者之间进行轮次交替。这一思路使我们能够利用复杂度度量的整数特性，在独立求解实例与联合求解实例的复杂度之间构造出递增的差距。