On the rod placement theorem of Rybko and Shlosman

AbstractGiven n−1 points $$x_1 \le x_2 \le \cdots \le x_{n-1}$$ on the real line and a set of n rods of strictly positive lengths $$\lambda_1 \le \lambda_2 \le \cdots \le \lambda_n$$, we get to choose an n-th point xn anywhere on the real line and to assign the rods to the points according to an arbitrary permutation π. The rod $$\lambda_{\pi(k)}$$ is thought of as the workload brought in by a customer arriving at time xk into a first in -first out queue which starts empty at − ∞. If any xi equals xj for i < j, service is provided to the rod assigned to xi before the rod assigned to xj.Let $$Y_{\pi} (x_n)$$ denote the set of departure times of the customers (rods). Let $$N_{\pi}(x_1, \ldots, x_{n-1}; \lambda_1, \ldots, \lambda_n)$$ denote the number of choices for the location of xn for which $$0 \in Y_{\pi}(x_n)$$. Rybko and Shlosman proved that $$ \sum_\pi N_\pi (x_1, \ldots, x_{n-1}; \lambda_1, \ldots, \lambda_n) = n ! $$ for Lebesgue almost all $$(x_1, \ldots, x_{n-1}; \lambda_1, \ldots, \lambda_n)$$.Let $$y_{\pi,k}(x_n)$$ denote the departure point of the rod λk. Let Nπ, k(y) denote the number of choices for the location of xn for which $$y_{\pi,k}(x_n) = y$$ and let $$N_k(y) = \sum_\pi N_{\pi, k}(y)$$. In this paper we prove that for every $$(x_1, \ldots, x_{n-1}; \lambda_1, \ldots, \lambda_n)$$ and every k we have $$N_k (y) = (n-1) !$$ for all but finitely many y. This implies (and strengthens) the rod placement theorem of Rybko and Shlosman.