I first tested it on the square root of 4 and in one iteration it found 2.

我首先测了下求4的平方根4,它只迭代了一次并返回了2。

麻省理工公开课 - 计算机科学及编程导论课程节选

So what Christine is arguing is, even though it's the case that 2 is not a dominated strategy, if we do the process of iterative deletion of dominated strategies and we delete the dominated strategies, then maybe we should look again and see if it's dominated now.

克里斯汀说的是,即使选择立场2不是劣势策略,如果我们迭代剔除劣势策略,然后我们剔除掉了劣势策略,然后再来回头看看还有没有劣势策略了

耶鲁公开课 - 博弈论课程节选

And in this case, we go from 8 to 4 to 2 to 1 three times and then on each iteration of this algorithm, each pass across the board I'm touching N numbers, so that means I'm doing N things, log N times.

在这个例子中，我们从8得到4，到2，再到1，是3次,在这个算法的每次迭代中，每一趟我都会操作N个数，也就是所我每次要做N步操作，一共要做，log，N，次。

哈佛公开课 - 计算机科学课程节选