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And realize too, as we say on this home page here, by default, we anonomize you when you log into this bulletin board, whereby, you're all logged in as quote unquote students.
并且认识到，像我们说的这个主页，默认情况下，当你登陆这个电子公告牌，我们禁止你们，都匿名学生登陆。
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And with this data, where students were able ; to implement last year their own E-trade-like website; whereby you have accounts and you log in your hand age of your users 10,000 virtual dollars and with them can they get stock quotes, by stocks, sell stocks and the like, all of this accomplished just after a few weeks time.
通过这些数据,学生们就可以做出类似电子商务的网站;,只要你有账号就能进入你的账户,里面有1万的虚拟美元,这样你就能去查询股票报价,进行买卖股票之类的事,这一切仅仅用了几个星期的时间就完成。
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Now hopefully you're keeping a sheet of paper with you and you're writing down what you eat so you don't forget everything, and then you sit down at your computer at the end of the day and you log it all in.
希望你们可以随时带张纸,以便随时记下,吃的东西以防忘记,然后坐在电脑前,在一天结束时输入电脑
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A log algorithm typically is one where you cut the size of the problem down by some multiplicative factor.
对数级复杂度的算法就是指,通过一系列常量级步数的操作,可以将问题的规模。
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And if you look at this series, the series log of one, what is the series expansion?
如果你们看这个式子，它的对数分之一，它的指数是多少？
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For example, this is x=t, x=t^, you're going to have x=sin t and cos t and log t.
例如,这是x=t,x=t^,你还会遇到正弦,余弦,对数函数
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So you can log into this, not only chat with this interactively, but on the right-hand side here, which is just a wallpaper right now, you can actually grant us view or control of your own screen.
所以你可以登录进去，不仅仅是,相互间的聊天，在这里的右手边，这只是一张壁纸，你可以允许我们查看或者控制你的屏幕。
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OK. So this is, in fact, log. Now, having said that I actually snuck something by you.
就是对数次,我已经教给你们一些。
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Boy, there's a dumb question, because I've been telling you n log n for the last two lectures the complexity is n log n, but let's see if it really is.
孩子们，这是一个愚蠢的问题，因为前两节课的时候我就已经告诉你们了,复杂度是,但是让我们来看一下是不是真的是这样。
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Again, if you didn't hear it, the answer was it's log.
因为每次我可以把这个问题的规模，缩小一半，你答对了。
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But mathematically, we've mentioned this before, log N or really to be precise, log base 2 of N, is the way you express this mathematically.
但从数学上说，之前我们已经提到过了，准确地说是log，N，以2为底N的对数，这就是它在数学上的表示。
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Yeah. Log. It's a good think, but why do you think it's log? Ah-ha. It's not a bad instinct, the length is getting shorter each time, but what's one of the characteristics of a log algorithm? It drops in half each time.
对了，对数，这是个好想法,但是你们为什么认为是对数呢？,啊哈，这样的本能不错，每次长度都会缩小些,但是对数算法的特性是什么。
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If we can sort things, you know, we get this n log n behavior, and we got a n log n behavior overall. But can we actually do better in terms of searching.
如果我们可以排序，如你所知，我们有n，log，n级别的算法，并且我有一个整体的n，log，n级别的算法,但是我们在搜索方面可以做的更好吗？
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Simply by clicking that link you can then log into your account; and then on the left hand side what appears is a little menu of all of your friends who are also using this tool and you can select any one of them and actually see what that person is shopping.
只需轻点鼠标便可进入自己的账户;,页面的左边有一个菜单栏,其中是你所有也在使用这个工具的朋友,你可以选中其中一个,查看ta所选择的课程。
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If you get them backwards, logr you will integrate one over r and will get log r.
如果你逆推的话，对1/r积分得到。
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It'll say is this really what you ate, and if you ate it, then you say yes I ate this, and then it adds to the electronic log.
系统会问这是否确实是你吃的,如果确实是 就选择是,你的选择会被记入电子日志
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And if you ask the TAs in recitation tomorrow, they'll tell you that you see a lot of n log n algorithms in computer science.
如果你明天在复习课上问助教的话，他们会告诉你在计算机科学中,存在着非常多的n，log，n规模的算法。
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With this, if I can assume that accessing the i'th element of a list is constant, then you can't see that the rest of that analysis looks just like the log analysis I did before, and each step, no matter which branch I'm taking, I'm cutting the problem down in half.
读取数组中的第i个元素,是个常量时间的操作的话，我也就能像以前那样得到,这个算法是对数级复杂度的分析,并且每一步不管我选择哪个区间，我都可以把问题的规模缩小一半。
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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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If you go and work that out, log2 log of two and you have two log two, by math you can show that for the line r series m 2*log2 is equal to two natural log of two, which is 1 1.386 which is greater than one.
如果你把它计算出来，这二者的对数，你会得到2个，通过数学，你会得到第r列，第m个值,等于,也就是1。386，这大于。
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