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OK. What would you guess the order of growth here is? Yeah. Why? Good. Exactly. Right?
如果没听到答案的同学,答案是对数级的？
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Cut the problem in half. Cut the problem in half again. And that's a typical characterization of a log algorithm.
是每次除以特定的量,将问题减一半，再减一半,如此，这就是对数算法的典型特性。
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You might remember vaguely logarithms from high school math and such but what this suggests for us , the computer scientists, is that this is certainly a smarter, a faster algorithm.
你可能还会依稀记得,高中数学里的对数,这就给了我们这些计算机科学家们，一些启示,即，这种算法更智能，更迅速。
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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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The integral of one over a quantity is the natural log.
这种形式的积分结果,是自然对数。
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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, this is several months and this logarithmic scale on the Y-axis represents antibody concentration.
这是在几个月内,这里使用的是对数刻度,Y轴代表的是抗体的浓度
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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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It's an example of a very common tool that's going to be really useful to us, not just for doing search, but for doing a whole range of problems. That is, in essence, the template the describes a log style algorithm.
不仅仅是做搜索,还可以解决一整类问题,本质上，这个模板就描述了,对数形的算法,我们一会再回来。
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So in fact, what does that suggest about the order of growth here? What is the complexity of this? Yeah. Logarithmic. Why?
复杂度是多少？,对的，对数级的么？，为什么呢？,学生：对数级的？
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It's a good sign that this is logarithmic, and I'm going to come back in a second to why logs are a great thing.
为什么对数级复杂度是个好事情，让我们再来看一个算法,噢，抱歉是让我们再来看两个算法。
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Wrev So minus w reversible, 1 p2 is less than p1, so p2 over p1 is less than one, log of something less than 1 is negative times negative.
那么，而p2小于p1，p2/p1小于，取对数后是负数,负负得正。
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N 6 Sixteen, so that's 16 times log base 2 of 16 and though I'm writing small here, log base 2 of 16, 16 this gives me 4 'cause 2 to the 4 equals 16.
是多少呢？，Well，，N，is，what？,16，那就是16乘以以2为底16的对数6,在这儿我将2写小一些，以2为底16的对数是4，因为2^4等于。
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It's nR log of p2 over p1 for the process where there's a pressure change.
结果是dS等于nR乘以p2除以p1的对数,这是对压强变化的结果。
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Thank you. I mean, I know I sort of said it to you, but you're right. It's logarithmic, right?
但你们是正确的，是对数级的,它有这样的效率是因为他每次能把问题减半？
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and we like log algorithms, because they're really fast. A typical characteristic of a log algorithm is a pro-- or sorry, an algorithm where it reduces the size of the problem by a constant factor.
并且我们也很喜欢对数算法,因为它很快,对数算法的典型特性是高速，哦，抱歉，是他能以常数因子的速度,降低问题的大小，很明显。
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Let's suppose n is 1000, and we're running at nanosecond speed.
假入我们一秒钟运算十亿次，我们已经看过了对数级，线性增长的。
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OK. There's some constants in there, but this is order log b.
对数级的，这太重要了，接下来我要给你们看个例子。
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This case, I reduced the size of the problem in half.
这很好的表明了这是，对数级复杂度的问题，我马上就要解释。
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It's got that property of, it cuts things in half.
换个角度来思考下为什么是对数级的。
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Log? Linear? Exponential? Quadratic?
对数？线性？指数？平方？
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We've seen log, we've seen linear, we've seen quadratic, we've seen exponential.
我们看过了对数级的，线性的，二次平方的，指数级的算法。
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What does a logarithmic running time look like?
这个对数函数是怎样的？
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In the log case, it's divide by an amount.
而在对数算法中。
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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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And as a consequence, it is log.
因此这是对数级复杂度的算法。
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We also saw a logarithmic algorithm.
我们也看过对数算法。
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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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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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So it's nR log V2 over V1.
结果是nR乘以V2除以V1的对数。
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