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

In the case of a tower of size 1, basically there are two things to do, right?

如果只有1个圆盘要移动，那么就只要做两件事就可以了对不对？

麻省理工公开课 - 计算机科学及编程导论课程节选

Should it be the date or the size of the document, pertinence to the case or another?

是文旦过的日期还是规格，与这个例子相关或其他的？

耶鲁公开课 - 心理学导论课程节选

Taking the problem, recognizing that you know what, 8 even though this is a pretty big problem size 8 in this case and last time it was size 8 or in the case of the papers in size of a thousand roughly with the phonebook, I assume these are in a perfectly straight line they won't quite fit.

以这个问题为例，你们要认识到，在这种情况下，这是个比较大的问题，其大小是,上次它的大小也是8，但在纸片那个问题中,电话簿的规模大概是上千的，现在假设这些,杯子完全在同一条直线上，虽然并不十分符合这个条件。

哈佛公开课 - 计算机科学课程节选

This case, I reduced the size of the problem in half.

这很好的表明了这是，对数级复杂度的问题，我马上就要解释。

麻省理工公开课 - 计算机科学及编程导论课程节选

the abuse is, you know, it's not quite right, it depends upon whether it's all ready, but you can see in either case, after 12 steps, 2 runs through this and down to a problem size b over 2.

在12步以后，两轮过后，这个问题的范围b被缩小了一倍，这为什么很不错呢？,这意味着再来12步。

麻省理工公开课 - 计算机科学及编程导论课程节选

So I propose this as a new algorithm for sorting N elements and being 8 in this case or really a thousand in the case of the phonebook, or anything of larger size.

所以我提出一种新的算法，来解决N个元素的排序问题,在这个问题中N是8，在电话簿的问题中N是一千,或者是大规模的任何问题。

哈佛公开课 - 计算机科学课程节选

I've got to test to see if I'm in the base case, and if I'm not, then I need to move a tower of size n minus 1, I need to move a tower of size 1, and I need to move a second-sorry about that a second tower of size n minus 1.

首先我看看我是不是在最基本的情况，如果不是的话，我得先做一个N-1个，圆盘的移动，移动一个圆盘,然后再做一次N-1个圆盘的移动。

麻省理工公开课 - 计算机科学及编程导论课程节选