The geometric transportation problem takes as input a set of points $P$ in $d$-dimensional Euclidean space and a supply function $\mu : P \to \mathbb{R}$. The goal is to find a transportation map, a non-negative assignment $\tau : P \times P \to \mathbb{R}_{\geq 0}$ to pairs of points, so the total assignment leaving each point is equal to its supply, i.e., $\sum_{r \in P} \tau(q, r) - \sum_{p \in P} \tau(p, q) = \mu(q)$ for all points $q \in P$. The goal is to minimize the weighted sum of Euclidean distances for the pairs, $\sum_{(p, q) \in P \times P} \tau(p, q) \cdot ||q - p||_2$. We describe the first algorithm for this problem that returns, with high probability, a $(1 + \varepsilon)$-approximation to the optimal transportation map in $n\varepsilon^{-O(d)}\log^{O(d)}{n}$ time. In contrast to the previous best algorithms for this problem, our near-linear running time bound is independent of the spread of $P$ and the magnitude of its real-valued supplies.
翻译:几何运输问题需要输入一组美元P$, 以美元为单位的 Euclidean 空间和供应功能$\mu: P\ to\\\ mathb{R}$。 目标是为所有点找到一个运输地图, 一个非负派任务$\ tau: P\ times P\to \ mathb{R ⁇ geq 0} 到对点, 所以每个点的总任务离开等于其供应量, 即 $\ sum ⁇ r\ in P}\ tau( q, r) - sum_p\ p} p}\ tau( p)\ tau( p)\ sum\ p\ p}\ tau( p, q) q) =\ mu(q) p$( q) = p$( q) =( qu) 美元。 目标是最大限度地减少双对点的 Euclidean距离, $( p=) p times =n___ levelyal_ pral___ pral__ pal_ pal_ pal____ pal_ pal_____ pal_ pral__ pr____________________ pr___________________________________________________________