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Given the code up there, if I want to move a tower of size n, what do I have to do?
鉴于这里的代码，如果我想移动N个圆盘，我该怎么做？
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And this is now consistent with my claim that I have sorted a list of size N equals 1.
这与我之前所说的是一致的,我已经将N为1的一个序列排好了序。
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Where you go from problem of size n to a problem of size n minus 1.
或者缩减了2，这都一样的,也就是把问题的规模从n变成了n-1。
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So if we're keeping n the same, we look and what we saw was that size actually decreases as we increase the value of l.
如果我们保持ｎ不变，我们看到随着ｌ值的,增大尺寸变小。
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So the running time of the problem where the input is T of size N as expressed here formulaically, T of N, the running time of an algorithm, given an input of size N. You know what?
因此一个输入为N的问题的运行时间,在这儿的公式表示为，如果输入为N，那么此算法的运行时间,是多少呢？
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N Well, here is a list of size N. How many times can you divide a list of size N by 2, right?
这是一个大小为N的列表，将一个大小为,的列表除以2需要几次呢？
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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。
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2 6 8 1 3 7 5 If I start off with fou, two, six, eight, one, three, seven, 8 five, so my list is of size N equals 8 at the moment.
顺序如下：,现在列表的大小N等于。
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If I want to move this stack here, I'm going to take a stack of size n minus 1 move it to the spare spot, now I can move the base disk over, and then I'm going to move that stack of size n minus 1 to there. That's literally what I did, OK?
如果我想移动这些圆盘，我先把从n-1个圆盘1,移动到多余的柱子上去，这样我就能把最底下的圆盘放到这儿了，然后再把从n-1个圆盘放到这儿来,这就是确切的我怎么做的对不对？
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And how about if the array is size N, and I say bracket N, where am I referring?
如果那个数组的大小为N，那会怎么样，我指明，涉及到了那个地方？
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And now I have a problem of size N minus 1.
现在是一个N-1大小的问题。
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And we said that was log rhythmic, took log n time where n is the size of the list.
当列表的长度为n的时候，整个算法耗时log，n的时间。
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So we say if n increases, the orbital size is also going to increase.
轨道的尺寸，也增大了。
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Down here, I've just got two things to merge, and then I've got things of size two to merge and then things of size four to merge. But notice a trade off. I have n operations if you like down there of size one.
但是n的大小是不同的，是吗？在这里我们只要合并两个元素,然后是合并长度为2的列表,接下来是合并长度为4的列表,但是观察一下之间的权衡关系。
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Up here I have n over two operations of size two. Up here I've got n over four operations of size four.
最下面有n个规模为1的操作,接着上面有n/2个规模为2的操作,再上面有n/4个规模为4的操作。
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If I'm running at nanosecond speed, 1000 n, the size of the problem, whatever it is, is 1000, and I've got a log algorithm, it takes 10 nanoseconds to complete.
如果这个问题的规模，也就是n，是，如果这个问题是对数级的，这将会占据10纳秒的时间，你一眨眼的时间。
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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。
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I now have a list of size 1 so N is, in fact, less than 2.
现在序列的大小是1，可见N小于。
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On the other hand, we've seen n that if the size of a is n, that's to say, we have n elements to choose from, then the number of possible subsets is 2 to the n.
另一方面，我们看到,如果a集合的大小是，也就是说我们有n个元素可供选择，而可能的子集的元素,个数就是2的n次方。
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