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So let's figure out the bond order for our two molecules here that we figured out the electron configuration for.
让我们看看这里,两个分子的键序是多少。
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So the integers are totally ordered the reals are totally ordered lots of things are, the rationals are totally ordered.
对正数来说,这样的序列是排好序的,对实数也是如此,对很多事情都是如此。
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/2 So the bond order is going to be equal to 1/2, and then it will be 2 minus 2.
它的键序等于，然后乘以2减去2。
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So if you now reverse the story that we're telling here, what was the next line supposed to be after sorting left half?
回退到这个过程中来，在对左半部分排完序后，下一步该做什么呢？
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What bothers Socrates most about our democracy is a certain kind of instability, its tendency to be pulled between extremes of anarchy, between lawlessness and tyranny.
苏格拉底对我们民主最感困扰的是,不稳定性，倾向于失序两端,拉扯的特性，即介于目无法纪与专制之间。
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I haven't said how I'm going to get those sorted lists, but imagine I had two sorted lists like that.
我还没有说明我怎么才能得到已排序的列表,但是想像一下我现在已经有,两个已排序好序的列表了。
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And you can go ahead and tell me what you think the bond order is going to be for this molecule.
你们告诉我你觉得,这个分子的键序应该是怎样的。
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And if I'm walking down the list, this is probably order of the length of the list s because I'm looking at each element once.
这可能大概就是数组的长度，因为我会遍历数组中的每个元素一次,现在你可能会想，等等，数组已经排好序了。
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All right, so the bonding order, you're correct, should be 2, if we subtract the number of bonding minus anti-bonding electrons and take that in 1/2.
好，你们是对的，键序为，如果我们用成键数,减去反键数除以2。
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So what we see is a bond order of 0, and again, the bond is very, very weak.
我们看到键序是0，同样的，键非常非常的弱。
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So let me give you an example. Suppose I want to merge two lists, and they're sorted.
我想合并两个列表,并且这两个列表已经排好序了。
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- So again, if you're unwinding what's going on here, this-- we sorted the left half which meant sort the left half, then the right half then the merge.
同样，如果你展开正在进行的一切-,我们已对左半部分排好了序，接着右半部分，接着合并。
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We don't seem to be doing that just yet, certainly not as badly, alright, so at this point in the story I have a sorted list of size 4.
当然现在我们不需要那样做,此时此刻,我已对整个问题中大小为4的列表排好序了。
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PROFESSOR: And I didn't write up there but it is one, 1 and we can see that it's 1, 2 because it's 1/2 of 2, 4 minus 2, so 1/2 of 2, the bonding order is going to be equal to one.
教授：我没有写出来，但是是1，我们可以看到为什么是，因为它等于1/2的2，4减去，所以1/2的2，键序等于1。
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So we know that it's 1, because we have 1, 2, 3, 4 bonding, minus 2 anti-bonding, and 1/2 of that is a bond order of 1.
我们知道它是，因为我们有1，2，3，4个成键，减去2个反键，它的一半就是键序为1。
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Again. Basic premise of binary search, or at least we set it up was, imagine I have a sorted list of elements. We get, in a second, to how we're going to get them sorted, and I want to know, is a particular element in that list..
好，二分查找的基本前提，或者是我们建立二分查找的基础,我们已经有了一个排好序的元素列表，我们就需要知道如何来快速的排序,如何从列表中找到特定的元素。
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So let's just prove that to ourselves and figure out the bond order just using valence electrons.
让我们证明一下这一点,来看看只用价电子算出键序。
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Let's assume that I could somehow get to the stage where I've got two sorted lists.
让我们来假设目前的情况是：,我已经有两个已排好序的列表。
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I remind you, I know you're not really listening to me, but that's OK. I reminded you at the beginning of the lecture, I said, let's assume we have a sorted list, and then let's go search it.
没关系，我告诉过你在课程的开始,我们假设这是一个排好序的列表,然后才进行的搜索,那实际上有序列表从哪里来的呢？
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If I look for, say, minus 1, you might go, gee, wait a minute, if I was just doing linear search, I would've known right away that minus one wasn't in this list, because it's sorted and it's smaller than the first elements.
如果我要查找-1，你可能要怒了，呵呵，等一等，如果我用的是线性查找，我不会知道-1不在这个列表中,但是列表是排好序的,1又比第一个元素小。
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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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He also shows, however, that normal politics is, itself, dependent upon extraordinary politics, periods of crisis, anarchy, instability, revolution, where the normal rules of the game are suspended.
然而他也同时了解到常态政治，本身有赖于破例政治，诸如危机，失序，不稳定，革命等时期，当游戏的常态规范,遭搁置。
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And the basic idea was that we had some sort of a line and we knew the answer was somewhere between this point and this point.
去学习的二分搜索是有联系的,这种方法的基础思想,是我们有一个线性的序集,我们也明白答案在其中的某一段区间。
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All right? Because if I don't do any swaps on a pass through the algorithm, then it says everything's in the right order.
好么？因为如果我,这一遍没有做任何交换，那么就意味着已经排好序了。
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For hydrogen our bond order is going to equal 1/2, 2 minus 0.
对于氢原子键序等于1/2，2减去。
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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。
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/2 So this would mean the bond order is equal to 1/2, and in terms of valence electrons, how many bonding valence electrons do we have?
这意味着键序等于，对于价电子，有多少个成键价电子？
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So for the bond order we want to take 1/2 of the total number of bonding electrons, so that's going to be 4 minus anti-bonding is 4, so we end up getting a bond order that's equal to 0.
键序等于1/2乘以,总的成键电子数，也就是4，减去反键电子数，也就是4。,所以最后得到键序为0。
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What kind of a bond is a bond order of 1?
什么样的键键序是1？
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So now, I need to merge the sorted halves.
现在，我需要合并已排好序的部分。
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