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 $(1 -ε)\ell$ times the complexity of solving one instance for some small enough $ε>0$. 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：在通信复杂度问题中，函数$f:X\times Y\rightarrow Z$的输入分布在两位玩家爱丽丝和鲍勃之间。若爱丽丝仅知$x\in X$，鲍勃仅知$y\in Y$，则他们需共享多少信息才能计算出$f(x,y)$的值？解决$\ell$个问题实例是否需额外消耗$\ell$倍的资源？这一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，并以最强形式给出了直接和猜想的反例。为此，我们构造了一族函数，使得对于足够小的$ε>0$，解决$\ell$个实例的复杂度低于$(1 -ε)\ell$倍单实例复杂度。在分析全函数族时，我们采用定制化方法，证明玩家间可以迫使轮流交替轮次。此思路使我们能利用复杂度度量的整数特性，在独立求解与联合求解实例的复杂度之间制造递增差距。