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I'm going to let t of b be the number of steps it takes to solve the problem of size b.
我会设立一个t作为,计算指数为b的时候解决问题需要的步骤数。
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B "Ok, so I don't care if I don't get a B in the course.
我不在乎成绩是不是。
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OK. So if I look at this code, first of all I'm calling search, it just has one call, so looks like search is constant, except I don't know what happens inside of b search. So I've got to look at b search.
首先调用一下搜索，就一步调用,看起来搜索是固定的，除非我不知道二分搜索的原理，那我们来看看二分搜索吧,所以让我们看看，第一行打印出来的内容。
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And you know, A) I don't have the energy to do this, and B) it would probably violate four hundred different privacy laws or whatever But if I took all those faces and morphed them together, I would get a very pretty face ? And how do we know this?
你们知道，第一，我没有精力这么做，第二，这可能会违反400条不同的隐私条例,但如果我把你们的样子,混合在一起，会得到一张很漂亮的脸,我们怎么知道？
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And a bit, B-I-T, just a shorthand for binary digit, 2 so a digit of zero over a one, bi means two, you only have two digits.
比特，B-I-T，是binary，digit的简写，从0到1的数字，而bi表示，所以你就只有两个数字。
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So, let's just arbitrarily put it between these two in this case here, but actually there's no reason we couldn't also put it between oxygen b and c, so I'm going to draw another structure where we have it here.
那么，让我们任意地将它们放在这两个之间，但实际上我们没有理由,不能把它们放在氧原子，B，与，C，之间，因此我将把另外一个结构画在这里。
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If you think of two towns down here, Brive,b-r-i-v-e,famous for its rugby, and Tulle,t-u-l-l-e,which is a capital.
如果你可以想起来这儿的两个城镇,布瑞福,b-r-i-v-e,以英式橄榄球运动而负有盛名,图勒,t-u-l-l-e,是一个省的首府
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If I can tell a story in which A exists and B doesn't exist, it's got to follow that A and B are not the same thing.
如果我跟你说A存在但B不存在,就可以得出结论A和B不是一个东西
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I personally don't like swings that much and it's the B-/B+ range, so I'd much rather prefer that to a swing from A to C, and that's my reason.
我不喜欢成绩波动很大的,比如B-/B+这个范围,所以我还是喜欢像A到C这样小点的,这就是我的原因
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If I reduce that it would be 3 plus t of b minus 3, so in general 3*k+t this is 3 k plus t of b minus k. OK.
把b减去一个,在外面加个3就可以了,因此也就是。
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It says, in either case in general, t of b-- and this is where I'm going to abuse notation a little bit but I can basically bound it by t, 12 steps plus t of b over 2.
我可以用一个,比12+t的数代表,这里有点不准确的地方,具体的步数依赖于奇数偶数，但是你们可以看到在两个case中。
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t In the b even case, again I'm going to let t of b be the number of steps I want to go through.
如果b为偶数，那么我还是要用,来代表解决这个问题需要的步骤数。
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In particular, how would I write an expression for t of b minus 1?
尤其是,我是怎么写出来t的表达式的？
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And if I pull it out one more level, 12k it's 12 plus 12 plus t of b over 8, 12 k because I'll have 12 of those to add up, plus t of b over 2 to the k.
总结一下也就是说，在k步以后，总步数应该是，那这种情况什么时候才能停止呢？,才能到达最基本的情况呢？
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On the next step though, this, we get substituted by that. Right, on the next step, I'm back in the even case, it's going to take six more steps, plus t of b minus 1. Oops, sorry about that, over 2.
这一步就是偶数了，这一步会让我们得到，6+t这样的结果，因为b-1现在是偶数了,别忽略这里的细节。
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This is their profit. And I won't tell you what B is for now, but let's just.. I mean I won't tell you exactly what it is, but I'll explain it.
这就是利润表达式,先不讨论B,我是说我们先不讨论B到底多大,我们照常计算
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