Given a set of $n$ labeled points in general position in the plane, we remove all of its points one by one. At each step, one point from the convex hull of the remaining set is erased. In how many ways can the process be carried out? The answer obviously depends on the point set. If the points are in convex position, there are exactly $n!$ ways, which is the maximum number of ways for $n$ points. But what is the minimum number? It is shown that this number is (roughly) at least $3^n$ and at most $12.29^n$.

翻译：给定平面上处于一般位置的 $n$ 个标记点，我们逐一移除所有点。每一步中，从剩余点集的凸包中删除一个点。整个过程有多少种执行方式？答案显然取决于点集。若点处于凸位置，则恰好有 $n!$ 种方式，这是 $n$ 个点的最大方式数。但最小方式数是多少？研究表明，该数（大致）至少为 $3^n$，至多为 $12.29^n$。