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But when you solve the Schrodingerequation, you don't get just a set of solutions that are dependent upon one number.
但当你解薛定谔的方程式时，你没得到有一个统一答案的,一系列解决方法。
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They're easy to spot, you can't run the program with them there, so you're not going to get weird answers.
它们很容易被捕捉到，你没法带着这些错误去运行程序，因此你不会得到奇怪的答案。
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And, that's a good thing to know because if you come out with an answer that's somewhere near the diameter of the universe and it's supposed to be the diameter of an atom, then you will know that you probably made a mistake.
这是我们需要知道的,因为当你做一道题得到的答案,是宇宙的直径,而正确答案是原子直径时，你能很快发现你的错误。
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You'll run into errors. Yeah?
你将得到一个错误答案，请说？
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bug This is an easy bug to come up with. But imagine, if you don't do the test, you're going to get answers that don't make any sense.
这是一个容易发现的，但是想象下，如果你没做测试的话,你会得到没有意义的答案，实际上。
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If you ask a dumb question there is only a dumb answer.
如果你问一个愚蠢的问题,只能得到愚蠢的答案。
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Second thing we've got to worry about is, what's a basic step? All right, if I bury a whole lot of computation inside of something, I can say, wow, this program, you know, runs in one step. Unfortunately, that one step calls the Oracle at Delphi and gets an answer back. Maybe not quite what you want.
我们需要担心的第二件事情就是，什么该作为一个基本的步骤呢？,如果我把一大堆的计算过程放到里面，我可以说，噢，这个程序你知道的，一步就完成了，不幸的是,这一步可能要靠预言家才能得到答案,这可能跟你想要的结果不太相同吧。
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And, that gives you a number for the fluorine, minus 0.29. Now, if you do the same for the hydrogen, you'll come up with the same number.
然后给出F的数字，负的0。29，如果同样的用H来计算，你会得到一样的答案。
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Sorry? OK. You're doing the whole thing for me What's the complexity just of this inner loop m How many times do I go through that loop?
就这一块代码的,我执行了多少次这个循环？,次对不对？，here？，Just，this，piece。,我一会儿就会去得到你的答案的？
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It could be the pair of none, none, it could be the pair of, you know, whatever the answer was that we put up there. That value comes back out and is now available inside the scope of Barnyard. OK. And Barnyard then uses that. Question?
也可以是，你知道，任何我们在这里得到的答案,这个返回的值在Barnyard范围内,就可以使用了,然后Barnyard就使用了这个值,有什么问题吗？
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