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The B-M-I of someone overweight is between twenty-five and twenty-nine point five.
VOA: special.2009.07.21
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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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Okay. I'm a producer for a record label called B-Boy Records. One of the original labels
好的，我是B-Boy唱片公司的制作人。B-Boy是最早成立的唱片公司之一。
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And,an obese person has a B-M-I above thirty.
VOA: special.2009.07.21
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Please choose a rhythm for excerpt one you can choose rhythm A which is or you could choose rhythm B So I'm about to play a piece. It's by Chopin.
请选一个和你听到的相符的节奏,你可以选节奏甲,也可以选节奏乙,我现在要弹一段,肖邦的曲子
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Well, A has to get whatever is in B. So now I'm clobbering, as they say, this value here, and at this point in the story, this is not good.
好的，A将获得B中的数据，现在我把这个值删掉，在这个地方,这是不好的。
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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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So prices are determined as follows: prices depend on some parameters I'm just going to call a and b, and let me write the equation and then we'll see what it looks like.
价格由如下因素决定,价格取决于两个参数a和b,我们用方程写一下吧
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I'm doing a test there to say, x if the string x is less than the value of b, and x does not appear before b as strings, then I was going to do, oh, a couple of things, because they're at the same block level.
我正在做一个测试来看看，字符串,是不是小于b的值，而且x不在b之前作为字符串出现，如果是真的话我就要做，一系列的操作，因为他们是在同一个块等级上。
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So we can draw that for 1 s a, we can also draw it for 1 s b, and what I'm saying for the molecular wave function is that we have the interference between the two, and we have a constructive interference, so we end up adding these two wave functions together.
所以我们可以对1sa画出它来，我们也可以对1sb画出它来，对于分子波函数我要说的是,它们两者之间会干涉，这里我们有相长干涉，所以我们得到的是波两个波函数加起来。
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OK? You can take a quick look at it, a even with the wrong glasses on, it says if a 1 sorry, b is equal to 1, a I'm just going to return a.
你们可以快速的过一遍，它是这么写的：如果,对不起，是b等于，那么就直接返回。
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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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So I'm done when b minus k equals 1, or k equals b minus 1.
因此b-k=1或k-b=1的时候，我就可以停下来了。
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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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I see no Mike Smiths because I'm on page 1 A where the A's simply are so I turn to the B's and the C's S and the D's and so forth and finally I get to the S's but in the worst case I've looked through 1,000 or so pages.
第一页没有找到叫Mike，Smith的人1,因为第一页姓名首字母是,于是我又接着找B，C，D……最终到了字母,但最坏的情况是,我翻了几千页才找到。
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But I'm not always going to have b as even.
如果b是奇数。
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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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Okay, so I'm going to draw for the case, I'm going to draw the best response of Player I and I'm going to draw the best response for Player II in a minute, for the case B equals 1/4.
好的,我们要画这个案例的图像,我要在图中画出参与人I的最佳对策,一会还要画出参与人II的最佳对策,在这里B等于1/4
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And I can solve them, and if I solve them out, I'm going to get something like, let me just be careful, I'm going to get something like: 1 minus B S1* is equal to 1, or S1* equals S2* is equal to 1 over .
我们都会解这个方程,如果解完方程,我们会得到,我得仔细点别算错,我们会得到,1-BS1*=1,或S1*=S2*=1/
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Yeah. You're jumping slightly ahead of me, but basically, I'm done when this is equal to 1 right? Because I get down to the base case, so I'm done when b u is over 2 to the k is equal to 1, and you're absolutely right, that's when k is log base 2 of b.
因为这就是最基本的情况了，因此当b/2的k次方等于1的时候就停止了1,你说的太对了，就是k等于b的以2为底的对数的时候,你们都坐的挺靠后的啊，我不知道是不是我讲的不太明白？
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