(1 + Ω(1))-approximation to MAX-CUT requires linear space

We consider the problem of estimating the value of MAX-CUT in a graph in the streaming model of computation. We show that there exists a constant ϵ* > 0 such that any randomized streaming algorithm that computes a (1 + ϵ*)-approximation to MAX-CUT requires Ω(n) space on an n vertex graph. By contrast, there are algorithms that produce a (1 + ϵ)-approximation in space O(n/ϵ2) for every ϵ > 0. Our result is the first linear space lower bound for the task of approximating the max cut value and partially answers an open question from the literature [2]. The prior state of the art ruled out (2 − ϵ)-approximation in [EQUATION] space or (1 + ϵ)-approximation in n1−O(ϵ) space, for any ϵ > 0. Previous lower bounds for the MAX-CUT problem relied, in essence, on a lower bound on the communication complexity of the following task: Several players are each given some edges of a graph and they wish to determine if the union of these edges is ϵ-close to forming a bipartite graph, using one-way communication. The previous works proved a lower bound of [EQUATION] for this task when ϵ = 1/2, and n1−O(ϵ) for every ϵ > 0, even when one of the players is given a candidate bipartition of the graph and the graph is promised to be bipartite with respect to this partition or ϵ-far from bipartite. This added information was essential in enabling the previous analyses but also yields a weak bound since, with this extra information, there is an n1−O(ϵ) communication protocol for this problem. In this work, we give an Ω(n) lower bound on the communication complexity of the original problem (without the extra information) for ϵ = Ω(1) in the three-player setting. Obtaining this Ω(n) lower bound on the communication complexity is the main technical result in this paper. We achieve it by a delicate choice of distributions on instances as well as a novel use of the convolution theorem from Fourier analysis combined with graph-theoretic considerations to analyze the communication complexity.