We revisit the problem of computing an optimal partial cover of points by intervals. We show that the greedy algorithm computes a permutation $\Pi = \pi_1, \pi_2,\ldots$ of the intervals that is $3/4$-competitive for any prefix of $k$ intervals. That is, for any $k$, the intervals $\pi_1 \cup \cdots \cup \pi_k$ covers at least $3/4$-fraction of the points covered by the optimal solution using $k$ intervals. We also provide an approximation algorithm that, in $O(n + m/\varepsilon)$ time, computes a cover by $(1+\varepsilon)k$ intervals that is as good as the optimal solution using $k$ intervals, where $n$ is the number of input points, and $m$ is the number of intervals (we assume here the input is presorted). Finally, we show a counter example illustrating that the optimal solutions for set cover do not have the diminishing return property -- that is, the marginal benefit from using more sets is not monotonically decreasing. Fortunately, the diminishing returns does hold for intervals.

翻译：我们重新审视了每隔一段时间计算最佳部分覆盖点的问题。 我们显示, 贪婪的算法计算了一个美元=\ pi=\ pi_ 1,\ pi_ 2,\ ldots 的间距为 3/4美元, 任何前缀为 美元, 任何前缀为 美元, 即, 对于任何一美元, 间隔 $\ pi_ 1\ cup\ cdots\ cdots 包括至少 3/4美元, 以 美元间隔计算最佳解决方案所涵盖的点。 我们还提供了一种近似算法, 以 $( n + m/\ varepsilon) 的时间计算, 以 $(1 ⁇ varepslon) 的间距为 3/4, 以 美元为, 以 美元 任何一美元 的间距为,, 美元为输入点数, 以 美元为 美元为 间隔数( 我们在此假设输入是预算 ) 。 最后, 我们展示了一个相反的例子, 说明设定的顶值解决方案的顶值解决方案没有 递减回产值, 的回产值不会减少 。