If it's there, I'm done, if not, I keep walking down, and I only stop when I get to a place where the element I'm looking for is smaller than the value in the list., in which case I know the rest of this is too big and I can stop.

并且保持遍历,我只在当当前位置的数组元素,大于目标数时停止,这意味着剩下的元素都比目标元素大，但是其他的情况,我还是要遍历完整个数组。

麻省理工公开课 - 计算机科学及编程导论课程节选

And as long as that smaller computation reduces to another smaller computation, eventually I ought to get to the place where I'm down in that base case.

如果我不属于基础事件,那么我需要把它简化为更简单的计算，随着计算的不断简化,最终我能分解成基础事件。

麻省理工公开课 - 计算机科学及编程导论课程节选

I have sorted with the smaller problem 1 because that smaller problem right now is of size 1 and so it's sort of obviously the case that this cup is now sorted.

对这个较小的问题我已经排好序了,因为在这个小问题中只有1个元素1,那么很明显,这个杯子已经是有序的了。

哈佛公开课 - 计算机科学课程节选

So, for example, if we think of the fluorine minus case, would you expect fluorine minus to be larger or smaller than neutral fluorine? Okay. I heard mostly larger, but a little bit of a mix in there, and it turns out that larger is correct.

比如，如果我们来考虑一下负一价的氟离子的话，大家认为它大一些还是小一些？,对比于比中性氟原子，好的，我听到大部分人说大一些，但是也有一些不同意见，而正确答案应该是大一些。

麻省理工公开课 - 化学原理课程节选

But I want to stress again, as long as I do the base case right and my inductive or recursive step reduces it to a smaller version of the same problem, the code will in fact converge and give me out an answer.

就开心的去做吧,但是我想再次强调,只要基础事件处理正确而我的递归,或递推步骤能把它简化为更简单的同类问题,那么这段代码就可以收敛。

麻省理工公开课 - 计算机科学及编程导论课程节选