延迟的力量：在线维持恒定竞争的Steiner树
The Power of Deferral: Maintaining a Constant-Competitive Steiner Tree Online
摘要：
在在线Steiner树问题中，一个点序列被一个接一个地显示出来：当一个点到达时，我们只有时间添加一条连接这个点和前面的点的边，并且我们希望最小化添加的边的总长度。二十年来，我们知道贪心算法维护的树的成本是O（log n）乘以Steiner树的成本，这是最好的可能。但是假设，除了我们添加的新边之外，我们还可以从之前的一组边中改变一条边：我们能做得更好吗？我们能维持一棵持续竞争的树吗？我们对这个问题的回答是肯定的。我们给出了一个原始-对偶算法，和一个新的基于对偶的分析，每一步只进行一次交换（除了添加连接新点到先前点的边），并且使得树的成本仅为最优成本的常数倍。对于这个问题，以前的结果给出了一个执行摊销常数交换次数的算法：对于每个n，前n步的交换次数为O（n）。我们也给出了一个更简单的严密的分析，这种摊销的情况。
英文摘要：
In the online Steiner tree problem, a sequence of points is revealed one-by-one: when a point arrives, we only have time to add a single edge connecting this point to the previous ones, and we want to minimize the total length of edges added. For two decades, we know that the greedy algorithm maintains a tree whose cost is O(log n) times the Steiner tree cost, and this is best possible. But suppose, in addition to the new edge we add, we can change a single edge from the previous set of edges: can we do much better? Can we maintain a tree that is constant-competitive? We answer this question in the affirmative. We give a primal-dual algorithm, and a novel dual-based analysis, that makes only a single swap per step (in addition to adding the edge connecting the new point to the previous ones), and such that the tree's cost is only a constant times the optimal cost. Previous results for this problem gave an algorithm that performed an amortized constant number of swaps: for each n, the number of swaps in the first n steps was O(n). We also give a simpler tight analysis for this amortized case.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
