In the symmetric rendezvous problem two players follow the same (randomized) strategy to visit one of $n$ locations in each time step $t=0,1,2,\dots$. Their goal is to minimize the expected time until they visit the same location and thus meet. Anderson and Weber [J. Appl. Prob., 1990] proposed a strategy that operates in rounds of $n-1$ steps: a player either remains in one location for $n-1$ steps or visits the other $n-1$ locations in random order; the choice between these two options is made with a probability that depends only on $n$. The strategy is known to be optimal for $n=2$ and $n=3$, and there is convincing evidence that it is not optimal for $n=4$. We show that it is not optimal for any $n\geq 4$, by constructing a strategy with a smaller expected meeting time.

翻译：在对称汇合问题中，两个玩家遵循相同（随机化）策略，在每个时间步 $t=0,1,2,\dots$ 访问 $n$ 个位置中的一个。他们的目标是最小化访问同一位置从而相遇的期望时间。Anderson 和 Weber [J. Appl. Prob., 1990] 提出了一种策略，该策略以 $n-1$ 步为一轮进行操作：一个玩家要么在 $n-1$ 步内停留于一个位置，要么以随机顺序访问其他 $n-1$ 个位置；这两个选项之间的选择依据一个仅依赖于 $n$ 的概率进行。已知该策略对于 $n=2$ 和 $n=3$ 是最优的，且有令人信服的证据表明对于 $n=4$ 并非最优。我们通过构造一个具有更小期望相遇时间的策略，证明该策略对于任意 $n\geq 4$ 都不是最优的。