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So this first little piece of code right here says, ok you give me 2 points, I'll create another 1 of these lists and I'll simply take the x, sorry I shouldn't say x, I'm going to assume it's the x, the x-values are the two points, add them together, just right there, the y-values, add them together and return that list.
好，为了来认识到这一点,让我们来看一个简单的小例子，在你们的课堂手册上，你可以看到我写了一个小程序,它假设我得到了，这些点中的一些，我想对它们做一些操作，例如我想把它们加到一起，那么这里的第一小片，代码的意思是，好给我两个点，我会再创建一个数组。
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And so everything's kind of new for me, again, coming back to London after 2 years.
所以对于两年后回到伦敦的我来说，再次地，所有的事物对我来是都是新鲜的。
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And frankly it would be incredibly time-consuming and tedious for me, to count this room full of people old school style-- 1, 2, 3 and so forth.
坦白说，按学校的老办法一个人一个人的数,1个，2个，3个……,对我来说极其费时费力。
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I go up to the green line and go across and this tells me the best response for Firm 2.
根据这条绿线的对应法则,就可以对应出公司2的最佳对策
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*10^-18 1.6 times 10 to the minus 18, excuse me, 6*10^-19 we have 1.6 times 10 to the minus 19, 2.18 times 10 to the minus 18, etc.
。,1。,2。18*10^-18，等等。
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This was a huge number for me up until 2 weeks ago Besides the 100 billion neurons in the brain . there are 12-15 thousand cochlear hair cells.
直到两周前，这对我来说还是个庞大的数字,除了这一兆亿个神经元外,还有12，000到15，000个毛细胞。
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And I can obviously do the other direction which is I can say skip to index 2 and all the remaining pieces. This lets me slice out, if you like, the front part or back part or a middle part of the tuple as I go along.
很明显我可以从索引2开始切片,所有剩下的元素，这可以让我2,如果你喜欢我这么说的话,取得元素的前面的或者后面的,或者中间的部分。
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So, why don't you take a look at this and tell me which are possible for a 2 s electron in a lithium atom where z 3 is going to be equal to three?
你们为什么不看一下这个然后告诉我对,于一个锂原子中的2s电子哪些是可能,的？它的有效电荷量,可能等于？
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As I say, the syllabuses should be accessible on the Classsv2 server; however, I've had problems with that in the past and you should please let me know if it's not.
你们可以在2号课程服务器上下载大纲;,不过我曾经下载不了,如果现在还不能下的话，一定要告诉我。
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The sum of path number 2 and path number 3 get me to the same place, so the energy change by going through this time path, this intermediate point here back all the way to final state should be the same the red path.
而经过路径2和3可以3,到达同样的末态，因此经过路径,2和3带来的能量的变化，与路径1带来的,能量变化相同。
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So let me go back now to this expression, x=at^/2+c+bt.
让我们再回到这个式子,x=at^/2+c+bt
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Good, so what I have here is an equation that tells me Player 1's best response for each possible choice of Player 2.
这个方程表示,参与人2不同策略下参与人1的最佳对策
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Let me divide this into 2 lists of size 1 and now done, right?
那么我将其划分为只有1个元素的2个列表？
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Two, so now let me drop down 2.
，把2写下来。
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Let's just... We'll get there, just to remind you, the way we read this is you give me a quantity of Firm 2, I find Firm 1's best response by going across to the pink line and dropping down.
我们当然也能算出来,提醒一下各位,这个图像的个意思是任意给出公司2产量,然后通过这条粉色的线,就可找出与之对应的公司1的最佳对策
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4 So 2 times 24 plus 16 gives me 64.
*24再加16等于。
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That's 1800 times faster. But the real one that blows me away is, it has 2 gig of memory inside of it. That's 12 thousand times more memory. Oh, and by the way?
比以前快了1800倍,但是真让我感到吃惊的是,它有2G的内存，这是以前的12000倍,顺便说一句我的？
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So 2 against 1 gives me 90% of the votes, 3 against 1 gives me what?
我选立场2他选1,我得90%选票,要是我选立场3对手选1呢
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6 Well again, we can go to our ones column, twos, four, eight, 16, 2 damn, 32. So we can go to 32 128 and then we can go to 64, and then 128, 1 2 3 4 5 6 7 8 and this gives me 1, 2, 3, 4, 5, 6, 7, 8.
好吧，再一次回到一位，二位，四位，八位，该死的，32，等等，所以我们可以找到32位2,再64，然后，那么我们就有了。
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Allow me, it's about 2, 3 minutes here to introduce you to someone called Binky.
请给我2，3分钟时间来为你们介绍,一个叫做Binky的人。
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2 Let me go ahead and put the number 42.
我放置数字。
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It means that choosing position 2 always gives me a higher share of the vote than choosing position 1, no matter where the other candidate positions herself.
意味着选择立场2,会比选择立场1,获得更多的选票,无论另一个候选人如何选择
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I going to come back in a second to how it actually does that, but it basically says, get me x value for p 1, get me the x value for p 2, compare them, just as you would normally.
是一个类的实例,我要去取的这个实例，所关联的x值，我稍后会讲讲实际上，这里是怎么实现的，但是基本上它的意思就是，给我p1的x值。
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It'll give me q1* is equal to a - c over 2b - q1* over 2.
会得到,q1*=/2b-q1*/2
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For each q2 that you give me or that Player 2 chooses, I want to find out and draw what is Player 1's best response.
对与参与人2的每个策略q2,我想知道参与人1的最佳对策是什么
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But you'll see that very clearly that Steven's right: that choosing 2 will always get me 5% more of the votes than choosing 1 from here on down.
你会发现斯蒂文说的很对,从这里开始,选立场2总会比,选立场1多获得5%选票
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Exactly, so ignoring these first two positions which were a bit weird, choosing 2 always gave me 5% more votes than choosing 1, regardless of what the other person chooses.
完全正确,所以忽略前两个立场,因为前两个立场有点古怪,我们选立场2总比1多获得5%的选票,无论对手如何选择
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From here on down, if I bothered to do it, we'd find that choosing 3 always gets me 5% more of the vote than I would have had from choosing 2, against any of these higher numbers.
如果我不怕麻烦 我能一直这么写下去,我们发现选择立场3总是比选择立场2,使我多获得5%的选票,在对手选更靠后的立场的情况下
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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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