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I was given two weeks and one other programmer in Biz Stone to write the software. And we did it.
我和比兹·斯通手下的另一个程序设计师,得到了两周时间来编写软件。
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And I think one of the key messages from that play I took away was that
我觉得我从那个戏剧里得到的一个核心信息是，
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Actually, what I did was used 107kJ/mol a different source so I have 107.
事实上，我使用了另一个路径,计算得到了。
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If it were just a little bit different I could say, all right, I have a different approximation. But when it's this different, something is wrong. Right?
但是如果仅仅有点差异的话,我可以认为没什么大问题,我得到了一个不太一样的估算，但这确实有差异，出问题了对不对？
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I could have done this in another way but I am going to just be extra deliberate so I make sure that this copy ends with back slash zero and voila, now, I have a copy.
我可以用另外一种方式来处理,我确保这个拷贝由反斜杠0结尾,那样，现在我得到了一个拷贝。
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I'm going to say that an electron's a wave no matter how much my father says differently, and I'm going to get a Nobel Prize for that, and he does.
我要说电子是一个波,无法这与我父亲的观点如何对立，我也要为此得到一个诺贝尔奖，他确实做到了。
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I just pulled out a common key.
我能得到一个通解
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And if I did this, and again, don't scribble too much in your notes but if we just make it clear what's going on here, I'm actually going to delete these strategies since they're never going to be played I end up with a little box again.
如果我再进行一次,别在笔记上乱画,我们只是想知道最后会怎样,因为这些策略不会被人采用,所以我剔除掉它们,最后我得到了一个更小的方格
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How do we get from there to the conclusion that I can't believe that I'm going to die, I'm going to cease to exist as a person?
怎样从中得到一个结论,说明我不相信我会死，我会停止作为人而存在？
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I guess like, you see on a globe, you see different time zones marked on,
我猜就像，你观察一个地球仪，你看得到不同的时区标志在上面，
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And I imagine the first person who articulated the question aloud probably met with the response saying, "What a stupid question. Of course things fall down."
我想象第一个人,大声提出这个问题,可能得到这样的回答，多么愚蠢的问题，物体当然下降“
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You know, I was pretty fortunate to get a staff position with a company.
我很幸运，能够在一家公司里得到一个职位。
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OK. Having done this now, I can simply go ahead and run this, and in fact if I go up here to run, you'll see I've got both an option to check the module, though in this case I'm just going to run it.
好，讲完了这些，我可以去执行程序了，实际上如果我在这里运行，你们会看到我同样得到了,一个可以同时检查模块的选项，虽然在这个例子中我就是直接去运行程序了。
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All I have done is said I have a fixed system here.
我所做的说明我得到一个固定的系统。
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If I wanted to get full, real division, I should make one of them a float.
如果我想要得到真正的除法答案，我需要把他们之中的一个,变为浮点数类型。
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Compare after 30 or 60 seconds with the person next to you and see if before I reveal the answer we can't get everyone on to the exact same page.
和你身边的人比较30到60秒钟,看看是否在我揭露答案之前，我们不能使每个人,得到一个完全相同的页面。
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And that's a wonderful thing to have because it gives you that modularity, that encapsulation that basically says, when I create a point, the only way I can get at the values, is by using one of the defined methods, in this case it could be Cartesian, Cartesian and get all the pieces of that.
这是很棒的一件事情,因为它让你有了，模块性以及封装性，这基本上也就是说，当我创建了一个点，我能够得到它的值的唯一的方式，就是用一个定义好的方法，在这个例子中也就是。
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All right. So let me give you an example.
我得到了一个是。
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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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If I do this, try and get the 10th element of a list that's only eight long. I get what looks like an error, but it's actually throwing an exception.
如果我来这么做,试着去取长度为8的列表的第十个元素,我貌似能得到一个错误。
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And, in fact, instead of having just an electrode here, what I could do is I could have an electrode that is.
实际上，我不是要得到一个电极，我所做的是,有一个电极。
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Just take a look: if this is 9, 5 divided by 9 is always going to be 0 point something, and if you thus have two integers and you're rounding down, which is what happens when you do integral math we're using this operator, I'm going to get zero times whatever.
稍微看一看：如果这是9，5除以9会得到,0点几，如果你用两个整型数,你舍去小数，这就是当你们,用整型数使用这个操作的所发生的事情，我将得到数值0乘以任何一个数字。
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Let me cluster them because really what I have, sorry, separate them out. I've gone from one problem size eight down to eight problems of size one.
让我把它们聚集起来，因为已经得到我想要的，抱歉，把它们分开来,现在我从一个长度为8的问题,得到了八个长度为1的问题。
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But I know when I multiply a vector by a number, I get a vector in the same direction.
但是我所知道的是当矢量乘以一个常数,我会得到一个同方向的矢量
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OK, so all I do is I take into account that I've got all the positive charge, whatever it is, it's a nucleus.
我做的全部是考虑,我得到所有的正电荷，无论是什么，都是一个核。
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So, I'm getting a string and so what really am I doing in this line of code?
那么，我得到一个字符串,这行代码实际上是干什么的呢？
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So I can concatenate techs plus ivys and assign that result to univs, and then when I print it you'll notice I just get a list of five strings.
和ivys数组用加号串联运算,并把结果赋值给univs数组，接下来我显示下univs的结果,你注意到我得到了一个。
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I expect that if I add -A to A, I should get the guy who plays the role of 0 in this world, which is the vector of no length.
我希望如果用 -A 加上 A,应该得到一个零矢量,也就是一个没有模长的矢量
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I've got a list, walk you through it an element at a time, do I look at each element of the list more than once?
你一次只能得到他的一个元素,我是不是把数组里面的每个元素，都过了大于等于一次？,你不这样认为么？大家有什么建议？
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I can't rely on the user. I want to make sure I get a float in it, so how do I do that?
我想确保这儿我得到的是一个浮点数,我该怎么做呢？
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