So this guy here gets rid of this, and I have my answer as V1 over V2 to =p2/p1 the gamma power is equal to p2 over p1.
这两项抵消掉了,于是等式,变成了^γ
Instead of taking that one vase, the thief could take two radios. And get more value. So the greedy thief, in some sense, gets the wrong answer. But maybe isn't so dumb.
这个贼可以带走两个收音机,这样总价值更大,因此这个贪婪的贼,在某种意义上没有得到正确的答案,但是可能它也不笨。
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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