In standard number-in-hand multi-party communication complexity, performance is measured as the total number of bits transmitted globally in the network. In this paper, we study a variation called local communication complexity in which performance instead measures the maximum number of bits sent or received at any one player. We focus on a simple model where $n$ players, each with one input bit, execute a protocol by exchanging messages to compute a function on the $n$ input bits. We ask what can and cannot be solved with a small local communication complexity in this setting. We begin by establishing a non-trivial lower bound on the local complexity for a specific function by proving that counting the number of $1$'s among the first $17$ input bits distributed among the participants requires a local complexity strictly greater than $1$. We further investigate whether harder counting problems of this type can yield stronger lower bounds, providing a largely negative answer by showing that constant local complexity is sufficient to count the number $1$ bits over the entire input, and therefore compute any symmetric function. In addition to counting, we show that both sorting and searching can be computed in constant local complexity. We then use the counting solution as a subroutine to demonstrate that constant local complexity is also sufficient to compute many standard modular arithmetic operations on two operands, including: comparisons, addition, subtraction, multiplication, division, and exponentiation. Finally we establish that function $GCD(x,y)$ where $x$ and $y$ are in the range $[1,n]$ has local complexity of $O(1)$. Our work highlights both new techniques for proving lower bounds on this metric and the power of even a small amount of local communication.
翻译:在标准数字- 手边多党通信复杂度中, 业绩是用网络中全球传输的比特总数来测量的。 在本文中, 我们研究一个叫做本地通信复杂性的变异, 即业绩衡量每个玩家发送或收到的比特的最大数目。 我们关注一个简单的模型, 即美元玩家, 每个输入一个位数, 执行协议, 交换信息以计算输入点的函数 $n 。 我们问在这种环境下, 可以用一个小的本地通信复杂度来解答什么可以和无法解答什么( 本地通信复杂度较小 ) 。 我们首先对本地的复杂度设定一个非3级的比值, 具体功能是本地的复杂度。 我们进一步调查更难计算这种类型的问题是否能产生更强的下限, 通过显示本地的复杂度足以计算整个输入点的1美元, 并且因此计算任何对本地的复杂度功能。 除了计算外, 我们还显示, 本地的排序和搜索的1美元是本地的复杂度, 也能够算出本地的复杂度。 我们的复杂度, 最终计算出本地的复杂度, 。