How to spread adversarial nodes?: rotate!

In this paper we study the problem of how to keep a dynamic system of nodes well-mixed even under adversarial behavior. This problem is very important in the context of distributed systems.More specifically, we consider the following game: There are n white pebbles and ε n black pebbles for some fixed constant ε < 1. Initially, all of the white pebbles are laid down in a ring, and the adversary has all of the black pebbles in its bag. In each round, the adversary can look at the entire ring and can select to add a black pebble to the ring (if its bag is not empty) or to take any black pebble from the ring and put it back into its bag (i.e. we consider adaptive adversaries). However, the adversary cannot place a black pebble into any position it likes. This is handled by a join strategy to be specified by the system. The goal is to find an oblivious join strategy, i.e. a strategy that cannot distinguish between the white and black pebbles in the ring, that integrates the black pebbles into this ring and may do some further rearrangements so that for a polynomial number of rounds the adversary will not manage to include its black pebbles into the ring so that there is a sequence of s=Θ(log n) consecutive pebbles in which at least half of the pebbles are black. If this is achieved by the join strategy, it wins. Otherwise, the adversary wins.Of course, the brute-force strategy of rearranging all of the pebbles in the ring at random after each insertion of a black pebble will achieve the stated goal, with high probability, but this would be a very expensive strategy. The challenge is to find a join strategy that needs as little randomness and as few rearrangements as possible in order to win with high probability. In this paper, we present and analyze a very simple strategy called k-rotation that chooses k-1 existing positions uniformly at random in the ring, creates a new position uniformly at random in the ring, and then rotates the new pebble and the k-1 old pebbles along these positions. Interestingly, even if the adversary has just $s$ pebbles, it can still win for k=2. But the k-rotation rule wins with high probability for k=3 as long as ε<2/3, demonstrating that there is a sharp threshold for keeping pebbles in a sufficiently perturbed state.