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Let's just first fill in this for the b 2 case.
让我们先填完这个B2的情况。
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So first, I choose a volatility randomly, from some distribution of possible volatilities 2 from to, in this case, 0.2.
来决定的一个值，所以首先我先随机选择一个浮动值，从可能的浮动值中的分布进行选择,在这个例子中就是0。
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If we have a higher z effective, it's pulled in tighter, we have to put in more energy in order to eject an electron, so it turns out that that's why case 2 is actually the lowest energy that we need to put in.
而如果有效核电量更高，原子核的束缚也就更紧，我们不得不输入更多的能量来打出一个电子，这就是第二种情况,所需要输入的,能量更少的原因。
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It's going to be /2 in the case where Coke and Pepsi cost exactly the same.
当可口可乐和百事可乐的价格一样时,销量将是/2
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And so, in this case, l could equal zero, 2 l could equal one or l could equal two.
所以这样的话，l可以等于，也可以等于1或。
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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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I have the case where p = .2, so the probability of an accident is 20%.
假设这里的p等于0.2,也就是一次事故发生的概率为20%
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In that case, y will be -gt^/2+c+bt.
在这个情况下y=-gt^/2+c+bt
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So in this case choosing 2 is better than choosing 1.
在这种情况下选择立场2比1好
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And if we do that calculation, what we find out is that the binding energy, in this case where we have no shielding, 72× is negative 8 . 7 2 times 10 to So, let's compare what we've just seen as our two extremes.
我们会发现结合，能在这个情况中，没有屏蔽，等于-8。,所以我们来对比一下,我们在两个极端的案例中看到了什么。
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So what Christine is arguing is, even though it's the case that 2 is not a dominated strategy, if we do the process of iterative deletion of dominated strategies and we delete the dominated strategies, then maybe we should look again and see if it's dominated now.
克里斯汀说的是,即使选择立场2不是劣势策略,如果我们迭代剔除劣势策略,然后我们剔除掉了劣势策略,然后再来回头看看还有没有劣势策略了
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the abuse is, you know, it's not quite right, it depends upon whether it's all ready, but you can see in either case, after 12 steps, 2 runs through this and down to a problem size b over 2.
在12步以后，两轮过后，这个问题的范围b被缩小了一倍，这为什么很不错呢？,这意味着再来12步。
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In an extreme case b, we had a z effective of 2, so essentially what we had was no shielding at all.
我们有效的z是，所以本质上我们完全没有屏蔽。
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So if N is not less than 2, that is I have two elements or more in which case there's definitely some sorting to be done.
因此如果N不小于2，也就是有2个或更多的元素,那当然是需要做排序的。
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And lower case q sub 2 is the charge on the electron.
小写q加个下标2是指电子所带电量。
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So, in fact, it's not the case that 2 is dominated by 3.
因此说立场2劣于立场3
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So 50% in this case, and once again, 2 did better than 1.
综上所述,选择离场2确实比1好
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Good, so even if everyone in the number-- everyone in the room didn't choose randomly but they all chose a 100, a very unlikely circumstance, but even if everyone had chosen 100, the highest, the average, sorry, the highest two-thirds of the average could possibly be is 66 2/3, hence 67 would be a pretty good choice in that case.
好的,如果每个人,教室里的每个人不是随机选择数字,而是全选了100,这好像不太可能,但如果每个人真的都选了100,就是最大的书,那平均数,抱歉,平均数的2/3会是66又2/3,此情况下67应该是个不错的选项
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So if that's the case doing a quick little calculation, what would the ionization energy be for a 2 p electron in neon?
么请稍微计算一下，氖原子的，2，p，电子的,电离能是多大？
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So let's look at the first case 2px where we have either the 2 p x or 2 p y type of orbitals that we're combining.
让我们看看第一个情形,我们要组合,或者2py轨道。
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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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So this code is identical functionally nonswitch c to the last implementation we saw, nonswitch.c, but I'm just ever-more emphatically saying, "In case 1," that is when n equals 1 -- or when case 2 applies -- when n equals 2 or when n equals 3 do what?
所以这些代码到最后的实现上，功能是,完全相同的，但是我想要再次强调一下，“在case，1“，那是当n等于1时1,或者当case，2适用-,当n等于2或者当n等于3，它将做什么？
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Yeah. You're jumping slightly ahead of me, but basically, I'm done when this is equal to 1 right? Because I get down to the base case, so I'm done when b u is over 2 to the k is equal to 1, and you're absolutely right, that's when k is log base 2 of b.
因为这就是最基本的情况了，因此当b/2的k次方等于1的时候就停止了1,你说的太对了，就是k等于b的以2为底的对数的时候,你们都坐的挺靠后的啊，我不知道是不是我讲的不太明白？
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So it'll be 1 plus 1/4 times 4, 1/4 times 4 is 1, so 1 plus 1 is 2, so Player II's best response in that case will be 2.
这回就是1+4/4,即1+1=2,所以此时参与人II的最佳对策是2
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But there's something you'll note here also when I point out the case of the 2 s versus the 2 p, which is what I mentioned that I would be saying again and again, which is when we look at the hydrogen atom, the energy of all of the n equals 2 orbitals are exactly the same.
但是这里有一些事情你们也会注意到,当我指出2s和2p的情况，我之前提过,我会一次又一次的说，我们在观察氢原子的时候，2层轨道的所有n的能量,是完全相同的。
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oh, it's kind of hard to compare case 2 and 3 when we can't see it anymore.
噢，我们不太好比较第二和第三选项3,这里看不清了。
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Let's compare case 2 and 3, since this where some people seem to have gotten confused.
让我们来比较一下第二和第三种情况，因为不少人好像在这里有些困惑。
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In case 2, we're taking the 3 p out of the neutral atom, whereas in case 3, we're taking it out of the ion.
在第二种情况中，我们要从中性原子中拿走，3，p，电子，而在第三种情况中，我们要从这个离子中拿走它。
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All right. So let's look now at the case where we do have 2 p z orbitals that we're talking about.
好的，让我们现在看一个,需要2pz轨道的例子。
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If we look at our last structure here where we have nitrogen the middle, we can also figure out all those formal charges, and in this case we have plus 1 on the nitrogen, we have minus 2 on the carbon, and then we end up with a 0 on the sulfur there.
如果我们来看看最后一个结构,在中间的原子是氮，我们同样可以计算出所有的形式电荷，而在这种情况下，氮为正一，碳为负二，而最后硫为零。
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