And what he said was that when you get down to atomic length scales there are limits, there are constraints on our ability to observe.

讲的就是,当你去进入到原子长度的尺度中,我们的观察能力,有一定的限制条件。

麻省理工公开课 - 固态化学导论课程节选

And if I'm walking down the list, this is probably order of the length of the list s because I'm looking at each element once.

这可能大概就是数组的长度，因为我会遍历数组中的每个元素一次,现在你可能会想，等等，数组已经排好序了。

麻省理工公开课 - 计算机科学及编程导论课程节选

The result of this process if this replication went down the whole length of the DNA would be to form two identical, double stranded DNA molecules.

如果复制过程在整条链上持续进行的话,将生成两条完全一样的,双链DNA分子

耶鲁公开课 - 生物医学工程探索课程节选

A gig is 10 to the 9, so it does two operations in the length of time it takes light to get from one foot off the table down to the table. That's amazing.

因此光行进一英尺的,时间内电脑可以,进行两次基本运算,这太神奇了。

麻省理工公开课 - 计算机科学及编程导论课程节选

As I keep moving down, that part gets smaller, it's not always the initial length of the list, and you're right. But if you do the sums, or if you want to think of it this way, if you think about this more generally, it's always on average at least the length of the list.

等等，随着移动，剩下的部分越来越小，并不是初始那么长了，如果你算一算，或者你这么想，你考虑更一般的情况，平均下来至少是列表的长度。

麻省理工公开课 - 计算机科学及编程导论课程节选

I just doubled the indentation each time so you can see it. So each successive call, notice what's happening. The argument is getting reduced. And we're going another level in. When we get down to this point, we're calling it with just a string of length one.

因此每次成功调用注意它的过程,我们的命题不断简化，而且我们不断深入嵌套,当我们走到这一步的时候,我们就是在调用它处理，仅有一个元素的字符串了。

麻省理工公开课 - 计算机科学及编程导论课程节选

At each stage, no matter which branch, here or here, I take, I'm cutting down the length of the list that I'm searching in half. All right?

选的是这里还是那里,我总是把列表分成两半，对吧？,所以如果我处理一个长度为n的列表？

麻省理工公开课 - 计算机科学及编程导论课程节选

I've got to count my way down, which means that the access would be linear in the length of the list to find the i'th element of the list, and that's going to increase the complexity.

的位置并去访问，然后继续下去，也就意味着,找到数组中的第i个元素的方法,是关于数组的长度呈线性复杂度的，这回增加算法的复杂度。

麻省理工公开课 - 计算机科学及编程导论课程节选

It's going to let j run over the length of the list all right, so it's going to start at some point to move down, and then it's going to let i run over range, that's just one smaller, and what's it doing there?

就在这里，会做什么呢？,我们让j遍历这个列表,好的，它要从某一定开始移动,然后让i也跑遍整个范围，就是比之要小一，会发生什么？

麻省理工公开课 - 计算机科学及编程导论课程节选