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Negative 1 plus 0 should add up to negative 1, if in fact, we're correct for the c n anion.
负一加上零应该等于负一，如果是这样，我们对于氰离子的结果就是正确的。
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The other thing is that we can re-write our h c n in terms of bonds.
还有一件事是我们可以用键的形式来表示氰化氢。
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Let's have one mole, n equals one.
因为是一摩尔，所以n等于一。
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If I look at something that goes as n squared, if this is the edge of the nucleus here and if this is r1, 4 it says when n goes to two the radius goes to four.
如果在原子核外侧,我们发现某一半径和n的平方成正比，也就是说当n为2时半径等于。
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Log n Log n, because at each stage I'm cutting the problem in half. So I start off with n then it's n n/2 n/4 n/8 over two n over four n over eight.
因为总共有多少层？,因为在每一层，我都是把问题分解成两半,因此以n开始，然后是。
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So you have what are called escape characters like backslash N that is the shorthand notation of telling the computer put a new line character here.
幸亏我们有一些类似反斜杠n的转义字符,这只是一种简化方式,用来告知电脑要在这新添一行。
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You have n branches on the tree and you count the number of leaves and sum them up.
一共有n根树枝,你数了叶子的数量然后把他们加起来
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And,of course,it just comes from the Greek word canon, spelled with one "n," not two.
当然了，canon一词来源于希腊语，中间只有一个n，不是两个。
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So let's actually just simplify this to the other version of the Rydberg constant, since we can use that here.
除以n初始的平方，我们把它简化成,另一种形式的Rydberg常数。
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And the word degenerate simply means same energy, are of equal energy when they're degenerate.
简并“一词指的是，能量相同，你有n平方个等能轨道，是简并的。
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So just to show you how big a difference it makes, let's run a couple of numbers.
让我们试试一组数字吧,如果n是1000的话。
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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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So there are two electron configurations in the n equals one shell, if we follow according to the selection rules that we spelled out last day.
如果根据上次课,我们阐明的原子光谱选择定则，我们就会知道在n等于1的那一层,有两种电子图像构型。
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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 can tell you how to sort N elements by saying in kind of a snide way.
我可以告诉你一种方法,去对这N个元素进行排序。
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If you wrote only one policy, what's the probability distribution of x/n?
如果你只签了一份保单,那么x/n的概率分布是怎样的?
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Whatever the n number is, with the exception of helium, helium is the oddity because there's only two elements in n equals one shell.
无论n是多少，除了氦之外，氦是个特例,因为只有两个元素,在n为1的这一层。
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So if we go to the ground state, what you see is we're at that lowest energy level, and we only have one possibility for an orbital, because when n equals 1, that's all we can do.
如果我们在基态上，你可以看到，我们在能量最低的态上，只有一种，可能的轨道，因为n等于1，只有这种可能。
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Where n'th is somewhere between and 2 in this case.
而在这个例子中第n个元素,就是2和5之间的一个数。
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On the other hand, if I want to sort it first, OK, if I want to do sort and search, I want to sort it, it's going to take n log n time to sort it, and having done that, then I can search it in log n time.
我先排序，好的，如果我想排序再搜索，我要排序，这需要花n，log，n时间排序，然后做完了，我们能花log，n时间搜索,啊，哪一种更好呢？恩，呵呵。
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I was just finding very tunnel vision-like, the smallest elements at that moment in time which means I don't know anything about the other elements other than they are not the smallest and so no matter what with Selection Sort I had to repeat this again and again and again and if you do out the math it's roughly N squared steps in the worst case as well.
我只有一个狭窄的视野，只知道某时刻的最小元素,就意味着我并不知道其他元素的任何情况,只知道它们不是最小的,所以不管怎样，在选择排序中,我就得一遍一遍地重复选择过程,在最坏情况下,大概需要N的平方次比较。
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And then one of the things that I suggested was that if we could figure out some way to order it, and in particular, if we could order it in n log n time, and we still haven't done that, but if we could do that, then we said the complexity changed a little bit.
这就涉及到了排序,如果可以想出一种来将其进行排序,甚至可以在n，log，n的时间内完成,虽然目前我们没做这件事,但是一旦开始做这件事,那么复杂性就是发生一些变化。
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