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

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

Now if you lived in a culture where the only meat that got served was the size of a deck of cards, and the only pasta that got served was the size of a tennis ball, there wouldn't be such problems, but that's not the case.

如果我们生活在一个,肉类每日摄入量是一副牌大小,或者意大利面只有网球大小的环境中,问题就不会存在,可惜这不是现实

耶鲁公开课 - 关于食物的心理学、生物学和政治学课程节选

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

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