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The experts at N.I.H.say you may have to repeat this process a few times.
VOA: special.2010.03.31
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Ah, n times, because for each value of i, I'm going to do that m thing, n*m so that is, close to what you said, right?
因此这就和你说的差不多了对不对？,这个问题的复杂度为，让我写下来，是-对不对，是？
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So I have n operations log n times, n log n there we go, n log n. Took us a long time to get there, but it's a nice algorithm to have.
所以我log，n遍的n次操作，就得到了,虽然花了不少时间得到了这个结论。
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But that merging process only takes N steps, N*log N so that's N times log of N. Now, it's a little tricky to reason through this perhaps the first time, let's just take a very simple example and see if we can do a little sanity check here.
但这个合并过程只需要N步，所以时间复杂度是,第一次对此进行推论可能会有点儿棘手，我们举一个简单的例子,看看我们能否做一些完整性的检查。
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So instead of being equal to negative z squared, now we're equal to negative z effective squared times r h all over n squared.
这里不再等于-z的平方，现在我们等于-有效的z的平方,乘以RH除以n的平方。
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Times the square of the elementary Z charge times n squared over Z.
乘以元电荷的平方,乘以n的平方除以。
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And in this case, we go from 8 to 4 to 2 to 1 three times and then on each iteration of this algorithm, each pass across the board I'm touching N numbers, so that means I'm doing N things, log N times.
在这个例子中，我们从8得到4，到2，再到1，是3次,在这个算法的每次迭代中，每一趟我都会操作N个数，也就是所我每次要做N步操作，一共要做，log，N，次。
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n So I've got to do n things n times.
所以就是做n次。
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The order complexity here, if I actually write it would be-- sorry, order n times m, and if m was equal to n, that would be order n squared, and this is quadratic.
如果m等于n的话,也就是n的平方，这是一个平方复杂度的问题，这是和前面不同类型的，好，我在做什么呢？
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So if I start off with a list of length n, how many times can I divide it by 2, until I get to something no more than two left?
我能够除以多少次2呢？,直到我得到的长度不超过2么？,对数次，对吧？就像刚才那位同学说的那样？
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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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k * n m plus k all times log n is in general going to be much better than k times n.
在普遍情况下要远远好于,实际情况要取决于n和k的取值。
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And that's going to be equal to negative z effective squared times r h over n squared.
有效的z的平方,乘以RH除以n的平方。
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So, we can get from these energy differences to frequency h by frequency is equal to r sub h over Planck's constant 1 times 1 over n final squared minus 1 over n initial squared.
所以我们通过不同能量,得到不同频率，频率等于R下标,除以普朗克常数乘以1除以n末的平方减去。
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In the linear case, meaning in the unsorted case what's the complexity of this? k times n, right? Order n to do the search, and I've got to do it k times, so this would be k times n.
复杂度是多少？k的n次方，对吧？,在序列n中做搜索，要做k次，所以是k的n次方次,如果先排序后搜索。
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So, if we just rearrange this equation, what we find is that z effective is equal to n squared times the ionization energy, IE all over the Rydberg constant and the square root of this.
我们可以发现有效的z等于n的平凡，乘以电离能除以里德堡常数,这些所有再开方，所以等于n乘以,除以RH整体的平方根。
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N 6 Sixteen, so that's 16 times log base 2 of 16 and though I'm writing small here, log base 2 of 16, 16 this gives me 4 'cause 2 to the 4 equals 16.
是多少呢？，Well，，N，is，what？,16，那就是16乘以以2为底16的对数6,在这儿我将2写小一些，以2为底16的对数是4，因为2^4等于。
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How many times you can divide N by 2 before you get down to 1?
在得到1之前需要将N除以2做几次呢？
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