12 - 3 - Mathematics Behind Large Margin Classification (Optional) (20 min).srt

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In this video, I'd like to
在本节课中 我将
(字幕整理：中国海洋大学 黄海广,haiguang2000@qq.com )

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tell you a bit about the
介绍一些

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math behind large margin classification.
大间隔分类背后的数学原理

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This video is optional, so please feel free to skip it.
本节为选学部分 你完全可以跳过它

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It may also give you better
但是听听这节课可能让你对

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intuition about how the
支持向量机中的优化问题

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optimization problem of the
以及如何得到

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support vex machine, how that
大间距分类器

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leads to large margin classifiers.
产生更好的直观理解

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In order to get started, let
首先

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me first remind you of a
让我来给大家复习一下

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couple of properties of what vector inner products look like.
关于向量内积的知识

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Let's say I have two vectors
假设我有两个向量

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U and V, that look like this.
u 和 v 我将它们写在这里

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So both two dimensional vectors.
两个都是二维向量

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Then let's see what U
我们看一下

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transpose V looks like.
u 转置乘以 v 的结果

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And U transpose V is
u 转置乘以 v

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also called the inner products
也叫做向量 u 和 v

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between the vectors U and V.
之间的内积

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Use a two dimensional vector, so
由于是二维向量 我可以

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I can on plot it on this figure.
将它们画在这个图上

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So let's say
我们说

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that's the vector U. And
这就是向量 u

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what I mean by that is
即

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if on the horizontal axis that
在横轴上

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value takes whatever value
取值为某个u1

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U1 is and on the
而在纵轴上

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vertical axis the height
高度是

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of that is whatever U2
某个 u2 作为U的

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is the second component
第二个分量

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of the vector U. Now, one
现在

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quantity that will be nice
很容易计算的

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to have is the norm
一个量就是向量 u 的

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of the vector U. So, these
范数

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are, you know, double bars on
这是双竖线

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the left and right that denotes
左边一个 右边一个

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the norm or length of
表示 u 的范数

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U. So this just means; really the
即 u 的长度

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euclidean length of the
即向量 u 的欧几里得长度

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vector U. And this
根据

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is Pythagoras theorem is just
毕达哥拉斯定理 等于

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equal to U1
它等于 u1 平方

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squared plus U2
加上 u2 平方

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squared square root, right?
开根号

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And this is the length of the vector U. That's a real number.
这是向量 u 的长度 它是一个实数

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Just say you know, what is the length
现在你知道了

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of this, what is the
这个的长度是多少

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length of this vector down here.
这个向量的长度写在这里了

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What is the length of this
我刚刚画的这个

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arrow that I just drew, is the normal view?
向量的长度就知道了

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Now let's go back and
现在让我们回头来看

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look at the vector V because we want to compute the inner product.
向量v 因为我们想计算内积

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So V will be some other
v 是另一个向量

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vector with, you know,
它的两个分量 v1 和 v2

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some value V1, V2.
是已知的

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And so, the vector
向量 v

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V will look like that, towards V like so.
可以画在这里

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Now let's go back
现在让我们

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and look at how to compute
来看看如何计算

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the inner product between U
u 和 v 之间的内积

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and V. Here's how you can do it.
这就是具体做法

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Let me take the vector V and
我们将向量 v

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project it down onto the
投影到

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vector U. So I'm going
向量 u 上

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to take a orthogonal projection or
我们做一个直角投影

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a 90 degree projection, and project
或者说一个90度投影

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it down onto U like so.
将其投影到 u 上

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And what I'm going to do
接下来我度量

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measure length of this
这条红线的

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red line that I just drew here.
长度

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So, I'm going to call the length of
我称这条红线的

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that red line P. So, P
长度为 p 因此 p

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is the length or is
就是长度 或者说是

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the magnitude of the projection
向量 v 投影到

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of the vector V onto the
向量 u 上的量

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vector U. Let me just write that down.
我将它写下来

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So, P is the length
p 是 v

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of the projection of the
投影到

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vector V onto the
向量 u 上的

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vector U. And it is
长度

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possible to show that unit
因此可以

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product U transpose V, that
将 u 转置乘以 v

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this is going to be equal
写作

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to P  times the
p 乘以

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norm or the length of
u 的范数或者说

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the vector U. So, this
u的长度

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is one way to compute the inner product.
这是计算内积的一种方法

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And if you actually do
如果你从几何上

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the geometry figure out what
画出 p 的值

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P is and figure out what the norm of U is.
同时画出 u 的范数

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This should give you the same
你也会同样地

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way, the same answer as
计算出内积

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the other way of computing unit product.
答案是一样的

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Right.
对吧

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Which is if you take U
另一个计算公式是

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transpose V then U transposes
u 转置乘以 v 就是

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this U1 U2, its a
[u1 u2] 这个一行两列的矩阵

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one by two matrix, 1
乘以

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times V. And so
v 因此

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this should actually give you
可以得到

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U1, V1 plus U2, V2.
u1×v1 加上 u2×v2

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And so the theorem of
根据线性代数的知识

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linear algebra that these two
这两个公式

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formulas give you the same answer.
会给出同样的结果

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And by the way, U transpose
顺便说一句

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V is also equal to
u 转置乘以 v

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V transpose U. So if
等于 v 转置乘以 u

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you were to do the same process
因此如果你将 u 和 v 交换位置

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in reverse, instead of projecting
将 u 投影到 v 上

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V onto U, you could project
而不是将 v 投影到 u 上

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U onto V. Then, you know, do
然后做同样地计算

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the same process, but with the rows of U and V reversed.
只是把 u 和 v 的位置交换一下

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And you would actually, you should
你事实上可以

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actually get the same number whatever that number is.
得到同样的结果

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And just to clarify what's
申明一点

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going on in this equation the
在这个等式中

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norm of U is a real
u 的范数是一个实数

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number and P is also a real number.
p也是一个实数

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And so U transpose V is
因此 u 转置乘以 v

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the regular multiplication as two real numbers of
就是两个实数

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the length of P times the normal view.
正常相乘

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Just one last detail, which is
最后一点

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if you look at the norm of
需要注意的就是p值

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P, P is actually signed so to the right.
p事实上是有符号的

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And it can either be positive or negative.
即它可能是正值 也可能是负值

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So let me say what I mean
我的意思是说

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by that, if U
如果 u

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is a vector that looks like
是一个类似这样的向量

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this and V is a vector that looks like this.
v 是一个类似这样的向量

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So if the angle between U
u 和 v 之间的

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and V is greater than ninety degrees.
夹角大于90度

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Then if I project V onto
则如果将 v

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U, what I get
投影到 u 上 会得到

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is a projection it looks like
这样的一个投影

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this and so that length
这是 p 的长度

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P. And in this
在这个情形下

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case, I will still have
我们仍然有

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that U transpose V is
u 转置乘以 v

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equal to P times the
是等于 p 乘以

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norm of U. Except in
u 的范数

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this example P will be negative.
唯一一点不同的是 p 在这里是负的

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So, you know, in inner products if the angle
在内积计算中 如果 u 和 v 之间的夹角

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between U and V is less
小于90度

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than ninety degrees, then P
那么那条红线的长度

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is the positive length for that red line
p 是正值

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whereas if the angle of this
然而如果

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angle of here is greater
这个夹角

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than 90 degrees then P
大于90度 则p

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here will be negative of
将会是负的

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the length of the super line of that little line segment right over there.
就是这个小线段的长度是负的

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So the inner product between two
因此两个向量之间的内积

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vectors can also be negative
也是负的

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if the angle between them is greater than 90 degrees.
如果它们之间的夹角大于90度

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So that's how vector inner
这就是关于向量内积的知识

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products work. We're going to
我们接下来将会

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use these properties of vector
使用这些关于向量内积的

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inner product to try
性质 试图来

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to understand the support
理解支持向量机

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vector machine optimization objective over there. Here
中的目标函数

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is the optimization objective for the
这就是我们先前给出的

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support vector machine that we worked out earlier. Just for
支持向量机模型中的目标函数

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the purpose of this slide I
为了讲解方便

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am going to make one simplification or
我做一点简化

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once just to make the objective easy
仅仅是为了让目标函数

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to analyze and what I'm going to do is
更容易被分析 我接下来忽略掉截距

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ignore the indeceptrums. So, we'll just ignore theta 0 and set that to be equal to 0. To
令 θ0 等于 0

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make things easier to plot, I'm also going to set N the number of features to be equal to 2. So, we have only 2 features,
这样更容易画示意图 我将特征数 n 置为2 因此我们仅有

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X1 and X2.
两个特征 x1 和 x2

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Now, let's look at the objective function.
现在 我们来看一下目标函数

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The optimization objective of the
支持向量机的优化目标函数

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SVM. What we have only two features.
当我们仅有两个特征

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When N is equal to 2.
即 n=2 时

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This can be written,
这个式子可以写作

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one half of
二分之一

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theta one squared plus theta two squared.
θ1 平方加上 θ2 平方

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Because we only have two parameters, theta one and thetaa two.
我们只有两个参数 θ1 和θ2

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What I'm going to do is rewrite this a bit.
接下来我重写一下

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I'm going to write this as one
我将其重写成

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half of theta one
二分之一 θ1 平方

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squared plus theta two squared and
加上 θ2 平方

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the square root squared.
开平方根后再平方

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And the reason I can do that,
我这么做的根据是

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is because for any number, you know, W, right, the
对于任何数 w

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square roots of W and
w的平方根 再取平方

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then squared, that's just equal
得到的就是

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to W. So square roots
w 本身 因此平方根 然后平方

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and squared should give you the same thing.
并不会改变值的大小

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What you may notice is that
你可能注意到

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this term inside is that's
括号里面的这一项

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equal to the norm
是向量 θ

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or the length of the
的范数

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vector theta and what
或者说是向量 θ 的长度

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I mean by that is that
我的意思是

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00:07:20,200 --> 00:07:21,640
if we write out the
如果我们将

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vector theta like this, as
向量 θ 写出来

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00:07:23,080 --> 00:07:24,320
you know theta one, theta two.
θ1 θ2

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Then this term that I've just
那么我刚刚画红线的这一项

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underlined in red, that's exactly
就是向量 θ

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the length, or the norm, of the vector theta.
的长度或范数

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We are calling the definition
这里我们用的是之前

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of the norm of the vector that we have on the previous line.
学过的向量范数的定义

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And in fact this is actually
事实上这就

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00:07:37,400 --> 00:07:38,320
equal to the length of the
等于向量 θ

205
00:07:38,370 --> 00:07:39,760
vector theta, whether you write
的长度

206
00:07:40,020 --> 00:07:41,620
it as theta zero, theta 1, theta 2.
当然你可以将其写作  θ0 θ1 θ2

207
00:07:42,280 --> 00:07:45,230
That is, if theta zero is equal to zero, as I assume here.
如果 θ0等于0

208
00:07:45,860 --> 00:07:46,770
Or just the length of theta
那就是

209
00:07:46,900 --> 00:07:48,680
1, theta 2; but for
θ1 θ2 的长度

210
00:07:48,830 --> 00:07:50,450
this line I am going to ignore theta 0.
在这里我将忽略 θ0

211
00:07:50,940 --> 00:07:52,710
So let me just, you know, treat theta
将 θ 仅仅写作这样

212
00:07:53,150 --> 00:07:54,730
as this, let me just
这样来写 θ

213
00:07:54,960 --> 00:07:56,360
write theta, the normal
θ 的范数

214
00:07:56,720 --> 00:07:58,480
theta as this theta 1,
仅仅和 θ1 θ2 有关

215
00:07:58,620 --> 00:08:00,180
theta 2 only, but the
但是

216
00:08:00,260 --> 00:08:01,220
math works out either way,
数学上不管你是否包含 θ0

217
00:08:01,460 --> 00:08:03,790
whether we include theta zero here or not.
其实并没有差别

218
00:08:03,970 --> 00:08:05,870
So it's not going to matter for the rest of our derivation.
因此在我们接下来的推导中去掉θ0不会有影响

219
00:08:07,630 --> 00:08:09,120
And so finally this means
这意味着

220
00:08:09,390 --> 00:08:11,440
that my optimization objective is equal
我们的目标函数是

221
00:08:11,750 --> 00:08:13,100
to one half of the
等于二分之一

222
00:08:13,190 --> 00:08:14,610
norm of theta squared.
θ范数的平方

223
00:08:16,190 --> 00:08:17,230
So all the support vector machine
因此支持向量机

224
00:08:17,530 --> 00:08:19,010
is doing in the optimization
做的全部事情就是

225
00:08:19,910 --> 00:08:21,500
objective is it's minimizing the
极小化参数向量 θ

226
00:08:21,590 --> 00:08:23,100
squared norm of the square
范数的平方 或者说

227
00:08:23,470 --> 00:08:24,840
length of the parameter vector theta.
长度的平方

228
00:08:28,330 --> 00:08:29,160
Now what I'd like to do
现在我将要

229
00:08:29,370 --> 00:08:30,790
is look at these terms, theta
看看这些项

230
00:08:31,090 --> 00:08:33,670
transpose X and understand better what they're doing.
θ 转置乘以x 更深入地理解它们的含义

231
00:08:34,230 --> 00:08:36,600
So given the parameter vector theta and given
给定参数向量θ 给定一个样本 x

232
00:08:36,930 --> 00:08:39,880
and example x, what is this is equal to?
这等于什么呢?

233
00:08:40,820 --> 00:08:42,120
And on the previous slide, we
在前一页幻灯片上

234
00:08:42,230 --> 00:08:44,070
figured out what U transpose
我们画出了

235
00:08:44,870 --> 00:08:45,850
V looks like, with different
在不同情形下

236
00:08:46,110 --> 00:08:47,880
vectors U and V. And so we're
u转置乘以v的示意图

237
00:08:48,130 --> 00:08:50,340
going to take those definitions, you know, with theta
我们将会使用这些概念

238
00:08:50,980 --> 00:08:52,300
and X(i) playing the
θ 和 x(i) 就

239
00:08:52,410 --> 00:08:53,310
roles of U and V.
类似于 u 和 v

240
00:08:54,400 --> 00:08:57,430
And let's see what that picture looks like.
让我们看一下示意图

241
00:08:57,860 --> 00:08:59,160
So, let's say I plot. Let's say I look at
我们考察一个

242
00:08:59,430 --> 00:09:01,130
just a single training example. Let's say I
单一的训练样本

243
00:09:01,230 --> 00:09:03,360
have a positive example the drawing
我有一个正样本在这里

244
00:09:03,720 --> 00:09:05,050
was across there and let's say that is
用一个叉来表示这个样本

245
00:09:05,800 --> 00:09:09,310
my example X(i), what
x(i)

246
00:09:09,500 --> 00:09:10,970
that really means is
意思是 在

247
00:09:12,100 --> 00:09:13,510
plotted on the horizontal axis
水平轴上

248
00:09:14,450 --> 00:09:16,210
some value X(i) 1
取值为 x(i)1

249
00:09:17,140 --> 00:09:19,620
and on the vertical axis
在竖直轴上

250
00:09:21,240 --> 00:09:22,290
X(i) 2.
取值为 x(i)2

251
00:09:22,650 --> 00:09:24,070
That's how I plot my training examples.
这就是我画出的训练样本

252
00:09:25,400 --> 00:09:27,160
And although we haven't been really
尽管我没有将其

253
00:09:27,320 --> 00:09:28,310
thinking of this as a vector, what
真的看做向量

254
00:09:28,570 --> 00:09:29,600
this really is, this is a
它事实上

255
00:09:29,650 --> 00:09:30,910
vector from the origin
就是一个

256
00:09:31,610 --> 00:09:33,520
from 0, 0 out to
始于原点

257
00:09:34,560 --> 00:09:36,210
the location of this training example.
终点位置在这个训练样本点的向量

258
00:09:37,830 --> 00:09:39,460
And now let's say we have
现在 我们有

259
00:09:39,980 --> 00:09:41,850
a parameter vector and
一个参数向量

260
00:09:42,080 --> 00:09:43,620
I'm going to plot
我会将它也

261
00:09:43,800 --> 00:09:45,720
that as vector, as well.
画成向量

262
00:09:46,390 --> 00:09:48,410
What I mean by that is if I plot theta 1
我将 θ1 画在这里

263
00:09:49,100 --> 00:09:53,530
here and theta 2 there
将 θ2 画在这里

264
00:09:56,230 --> 00:09:57,050
so what is the inner
那么内积

265
00:09:57,290 --> 00:09:58,940
product theta transpose X(i).
θ 转置乘以 x(i) 将会是什么呢

266
00:09:59,220 --> 00:10:01,240
While using our earlier method,
使用我们之前的方法

267
00:10:01,990 --> 00:10:03,360
the way we compute that is we
我们计算的方式就是

268
00:10:04,310 --> 00:10:06,170
take my example and
我将训练样本

269
00:10:06,320 --> 00:10:08,710
project it onto my parameter vector theta.
投影到参数向量 θ

270
00:10:09,830 --> 00:10:10,700
And then I'm going to look
然后我来看一看

271
00:10:10,950 --> 00:10:13,070
at the length of this segment
这个线段的长度

272
00:10:13,680 --> 00:10:14,660
that I'm coloring in, in red.
我将它画成红色

273
00:10:15,090 --> 00:10:16,500
And I'm going to
我将它称为

274
00:10:16,670 --> 00:10:19,480
call that P superscript I
p 上标 (i)

275
00:10:20,330 --> 00:10:21,330
to denote that this is a
用来表示这是

276
00:10:21,610 --> 00:10:22,920
projection of the i-th training example
第 i 个训练样本

277
00:10:24,860 --> 00:10:25,540
onto the parameter vector theta.
在参数向量 θ 上的投影

278
00:10:26,900 --> 00:10:28,140
And so what we have is
根据我们之前幻灯片的内容

279
00:10:28,350 --> 00:10:30,790
that theta transpose X(i) is
我们知道的是

280
00:10:30,920 --> 00:10:32,830
equal to following what
θ 转置乘以 x(i)

281
00:10:32,960 --> 00:10:34,210
we have on the previous slide, this
等于

282
00:10:34,430 --> 00:10:35,350
is going to be equal to
将会等于

283
00:10:36,560 --> 00:10:40,000
P times the
p 乘以

284
00:10:40,090 --> 00:10:42,090
length of the norm of the vector theta.
向量 θ 的长度 或 范数

285
00:10:43,410 --> 00:10:44,690
And this is of course also equal to
这就等于

286
00:10:44,750 --> 00:10:46,660
theta 1 x1
θ1 乘以 x1

287
00:10:47,920 --> 00:10:50,610
plus theta 2 x2. So each
加上 θ2 x2

288
00:10:50,810 --> 00:10:52,360
of these is, you know, an equally
这两种方式是等价的

289
00:10:52,680 --> 00:10:54,080
valid way of computing the
都可以用来计算

290
00:10:54,180 --> 00:10:56,160
inner product between theta and X(i).
θ 和 x(i) 之间的内积

291
00:10:57,780 --> 00:10:57,780
Okay.
好

292
00:10:58,140 --> 00:10:59,040
So where does this leave us?
这告诉了我们什么呢

293
00:10:59,280 --> 00:11:00,770
What this means is that, this
这里表达的意思是

294
00:11:01,020 --> 00:11:02,890
constrains that theta transpose X(i)
这个 θ 转置乘以 x(i)

295
00:11:03,130 --> 00:11:05,330
be greater than or equal to one or less than minus one.
大于等于1 或者小于-1的

296
00:11:06,110 --> 00:11:06,860
What this means is that it
约束是可以被

297
00:11:06,970 --> 00:11:07,830
can replace the use of constraints
p(i)乘以x大于等于1

298
00:11:08,610 --> 00:11:12,000
that P(i) times X
这个约束

299
00:11:12,320 --> 00:11:13,500
be greater than or equal to one.
所代替的

300
00:11:13,680 --> 00:11:16,280
Because theta transpose X(i) is
因为 θ 转置乘以  x(i)

301
00:11:16,400 --> 00:11:19,470
equal to P(i) times the norm of theta.
等于 p(i) 乘以 θ 的范数

302
00:11:21,250 --> 00:11:23,080
So writing that into our optimization objective.
将其写入我们的优化目标

303
00:11:23,910 --> 00:11:24,870
This is what we get
我们将会得到

304
00:11:25,130 --> 00:11:26,290
where I have, instead of
没有了约束

305
00:11:27,090 --> 00:11:28,400
theta transpose X(i), I now
θ 转置乘以x(i)

306
00:11:28,620 --> 00:11:30,920
have this P(i) times the norm of theta.
而变成了 p(i) 乘以 θ 的范数

307
00:11:31,970 --> 00:11:32,970
And just to remind you we
需要提醒一点

308
00:11:33,090 --> 00:11:34,240
worked out earlier too that
我们之前曾讲过

309
00:11:34,460 --> 00:11:36,310
this optimization objective can be
这个优化目标函数可以

310
00:11:36,510 --> 00:11:38,130
written as one half times
被写成二分之一乘以

311
00:11:38,500 --> 00:11:39,910
the norm of theta squared.
θ 平方的范数

312
00:11:41,730 --> 00:11:43,490
So, now let's consider
现在让我们考虑

313
00:11:44,210 --> 00:11:45,550
the training example that we
下面这里的

314
00:11:45,700 --> 00:11:47,100
have at the bottom and
训练样本

315
00:11:47,450 --> 00:11:49,620
for now, continuing to use the simplification that
现在 继续使用之前的简化 即

316
00:11:50,180 --> 00:11:51,340
theta 0 is equal to 0.
θ0 等于0

317
00:11:52,030 --> 00:11:54,810
Let's see what decision boundary the support vector machine will choose.
我们来看一下支持向量机会选择什么样的决策界

318
00:11:55,860 --> 00:11:57,710
Here's one option, let's say
这是一种选择

319
00:11:57,870 --> 00:11:59,190
the support vector machine were to
我们假设支持向量机会

320
00:11:59,340 --> 00:12:01,750
choose this decision boundary.
选择这个决策边界

321
00:12:02,690 --> 00:12:05,110
This is not a very good choice because it has very small margins.
这不是一个非常好的选择 因为它的间距很小

322
00:12:05,530 --> 00:12:08,210
This decision boundary comes very close to the training examples.
这个决策界离训练样本的距离很近

323
00:12:09,810 --> 00:12:12,360
Let's see why the support vector machine will not do this.
我们来看一下为什么支持向量机不会选择它

324
00:12:14,130 --> 00:12:15,420
For this choice of parameters
对于这样选择的参数 θ

325
00:12:16,410 --> 00:12:18,280
it's possible to show that the
可以看到

326
00:12:19,030 --> 00:12:21,250
parameter vector theta is actually
参数向量 θ 事实上

327
00:12:21,760 --> 00:12:23,350
at 90 degrees to the decision boundary.
是和决策界是90度正交的

328
00:12:24,060 --> 00:12:25,440
And so, that green decision boundary
因此这个绿色的决策界

329
00:12:26,250 --> 00:12:27,550
corresponds to a parameter vector
对应着一个参数向量 θ

330
00:12:27,920 --> 00:12:29,650
theta that points in that direction.
指向这个方向

331
00:12:30,730 --> 00:12:32,280
And by the way, the simplification that
顺便提一句 θ0 等于0

332
00:12:32,510 --> 00:12:34,120
theta 0 equals 0 that
的简化仅仅意味着

333
00:12:34,300 --> 00:12:35,410
just means that the decision boundary
决策界必须

334
00:12:35,910 --> 00:12:37,960
must pass through the origin, (0,0) over there.
通过原点 (0,0)

335
00:12:38,330 --> 00:12:40,350
So now, let's
现在让我们看一下

336
00:12:40,690 --> 00:12:41,780
look at what this implies
这对于优化目标函数

337
00:12:41,840 --> 00:12:43,590
for the optimization objective.
意味着什么

338
00:12:45,260 --> 00:12:46,420
Let's say that this example here.
比如这个样本

339
00:12:47,460 --> 00:12:48,560
Let's say that's my first example, you know,
我们假设它是我的第一个样本

340
00:12:50,480 --> 00:12:50,650
X1.
x(1)

341
00:12:51,690 --> 00:12:52,630
If we look at the projection
如果我考察这个样本

342
00:12:53,320 --> 00:12:54,870
of this example onto my parameters theta.
到参数 θ 的投影

343
00:12:56,180 --> 00:12:56,520
That's the projection.
这就是投影

344
00:12:57,660 --> 00:12:59,230
And so that little red line segment.
这个短的红线段

345
00:13:00,450 --> 00:13:01,720
That is equal to P1.
就等于p(1)

346
00:13:02,380 --> 00:13:04,650
And that is going to be pretty small, right.
它非常短 对么

347
00:13:05,860 --> 00:13:08,590
And similarly, if this
类似地 这个样本

348
00:13:09,610 --> 00:13:10,710
example here, if this happens
如果它恰好是

349
00:13:11,170 --> 00:13:12,620
to be X2, that's my second example.
x(2) 是我的第二个训练样本

350
00:13:13,880 --> 00:13:16,620
Then, if I look at the projection of this this example onto theta.
则它到 θ 的投影在这里

351
00:13:18,080 --> 00:13:18,170
You know.
你知道的

352
00:13:18,440 --> 00:13:20,460
Then, let me draw this one in magenta.
我将它画成粉色

353
00:13:21,600 --> 00:13:23,690
This little magenta line segment, that's
这个短的粉色线段

354
00:13:24,000 --> 00:13:25,820
going to be P2. That's
它是 p(2)

355
00:13:26,070 --> 00:13:27,370
the projection of the second example
第二个样本到

356
00:13:27,770 --> 00:13:28,870
onto my, onto the direction
我的参数向量 θ

357
00:13:30,100 --> 00:13:32,650
of my parameter vector theta which goes like this.
的投影

358
00:13:33,870 --> 00:13:34,250
And so, this little
因此

359
00:13:35,270 --> 00:13:35,270
projection line segment is getting pretty small.
这个投影非常短

360
00:13:36,850 --> 00:13:38,420
P2 will actually be a negative number, right so P2 is
p(2) 事实上是一个负值

361
00:13:38,560 --> 00:13:42,490
in the opposite direction.
p(2) 是在相反的方向

362
00:13:43,710 --> 00:13:45,250
This vector has greater
这个向量

363
00:13:45,560 --> 00:13:47,130
than 90 degree angle with my
和参数向量 θ 的夹角

364
00:13:47,270 --> 00:13:48,980
parameter vector theta, it's going to be less than 0.
大于90度 p(2) 的值小于0

365
00:13:50,280 --> 00:13:51,580
And so what we're finding is that
我们会发现

366
00:13:51,850 --> 00:13:54,880
these terms P(i) are
这些 p(i)

367
00:13:55,200 --> 00:13:57,230
going to be pretty small numbers.
将会是非常小的数

368
00:13:58,210 --> 00:13:59,080
So if we look at
因此当我们考察

369
00:13:59,110 --> 00:14:01,650
the optimization objective and see, well, for positive examples
优化目标函数的时候 对于正样本而言

370
00:14:02,490 --> 00:14:04,860
we need P(i) times
我们需要 p(i) 乘以

371
00:14:05,220 --> 00:14:07,590
the norm of theta to be bigger than either one.
θ 的范数大于等于1

372
00:14:08,670 --> 00:14:10,640
But if P(i) over
但是如果 p(i) 在这里

373
00:14:10,860 --> 00:14:12,140
here, if P1 over here
如果 p(1) 在这里

374
00:14:12,770 --> 00:14:14,160
is pretty small, that means
非常小 那就意味着

375
00:14:14,410 --> 00:14:15,580
that we need the norm of
我们需要 θ 的范数

376
00:14:15,650 --> 00:14:18,420
theta to be pretty large, right? If
非常大 对么

377
00:14:19,830 --> 00:14:20,840
P1 of theta is small
因为如果 p(1) 很小

378
00:14:21,790 --> 00:14:23,110
and we want P1 you
而我们希望 p(1)

379
00:14:23,410 --> 00:14:24,600
know times in all of theta
乘以 θ

380
00:14:24,920 --> 00:14:25,890
to be bigger than either one, well
大于等于1

381
00:14:26,400 --> 00:14:27,300
the only way for that
令其实现的

382
00:14:27,510 --> 00:14:28,440
to be true for the profit that
唯一的办法就是

383
00:14:28,650 --> 00:14:29,750
these two numbers to be large
这两个数较大

384
00:14:30,020 --> 00:14:31,120
if P1 is small, as we
如果 p(1) 小

385
00:14:31,240 --> 00:14:32,980
said we want the norm of theta to be large.
我们就希望 θ 的范数大

386
00:14:34,150 --> 00:14:36,450
And similarly for our
类似地

387
00:14:36,650 --> 00:14:38,560
negative example, we need P2
对于负样本而言 我们需要 p(2)

388
00:14:39,750 --> 00:14:41,070
times the norm of
乘以

389
00:14:41,350 --> 00:14:44,990
theta to be
θ 的范数

390
00:14:45,160 --> 00:14:46,910
less than or equal to minus one.
小于等于-1

391
00:14:47,760 --> 00:14:48,540
And we saw in this
我们已经在这个样本中

392
00:14:48,710 --> 00:14:50,200
example already that P2
看到 p(2)

393
00:14:50,260 --> 00:14:51,520
is going pretty small negative number,
会是一个非常小的数

394
00:14:52,040 --> 00:14:53,290
and so the only way for
因此唯一的办法

395
00:14:53,420 --> 00:14:54,490
that to happen as well is
就是

396
00:14:54,530 --> 00:14:56,730
for the norm of theta to be
θ 的范数变大

397
00:14:57,010 --> 00:14:59,630
large, but what
但是我们

398
00:14:59,790 --> 00:15:00,900
we are doing in the optimization
的目标函数是

399
00:15:01,290 --> 00:15:02,400
objective is we are
希望

400
00:15:02,540 --> 00:15:03,840
trying to find a setting
找到一个参数 θ

401
00:15:04,170 --> 00:15:05,320
of parameters where the norm
它的范数

402
00:15:05,550 --> 00:15:07,100
of theta is small, and so
是小的

403
00:15:07,830 --> 00:15:09,040
you know, so this doesn't
因此 这看起来

404
00:15:09,330 --> 00:15:10,070
seem like such a good direction
不像是一个好的

405
00:15:10,610 --> 00:15:14,160
for the parameter vector and theta.
参数向量 θ 的选择

406
00:15:14,450 --> 00:15:15,510
In contrast, just look at a different decision boundary.
相反的 来看一个不同的决策边界

407
00:15:17,040 --> 00:15:19,500
Here, let's say, this SVM chooses
比如说 支持向量机选择了

408
00:15:20,510 --> 00:15:21,280
that decision boundary.
这个决策界

409
00:15:22,870 --> 00:15:23,980
Now the is going to be very different.
现在状况会有很大不同

410
00:15:24,420 --> 00:15:25,890
If that is the decision boundary,
如果这是决策界

411
00:15:26,190 --> 00:15:27,380
here is the
这就是

412
00:15:27,450 --> 00:15:28,770
corresponding direction for theta.
相对应的参数 θ 的方向

413
00:15:29,210 --> 00:15:30,920
So, with the direction
因此 在这个

414
00:15:31,000 --> 00:15:32,110
boundary you know, that
决策界之下

415
00:15:32,300 --> 00:15:33,560
vertical line that corresponds
垂直线是决策界

416
00:15:34,470 --> 00:15:35,960
to it is possible to
使用线性代数的知识

417
00:15:36,190 --> 00:15:37,880
show using linear algebra that
可以说明

418
00:15:38,070 --> 00:15:39,140
the way to get that green decision
这个绿色的决策界

419
00:15:39,460 --> 00:15:41,190
boundary is have the vector of theta be
有一个垂直于它的

420
00:15:41,390 --> 00:15:42,620
at 90 degrees to it,
向量 θ

421
00:15:43,610 --> 00:15:44,470
and now if you look
现在如果你考察

422
00:15:44,560 --> 00:15:45,570
at the projection of your
你的数据在横轴 x

423
00:15:45,710 --> 00:15:47,540
data onto the vector
上的投影

424
00:15:48,800 --> 00:15:50,010
x, lets say its before
比如 这个我之前提到的样本

425
00:15:50,010 --> 00:15:52,620
this example is my example of x1. So when
我的样本 x(1)

426
00:15:52,890 --> 00:15:54,600
I project this on to x,
当我将它投影到横轴x上

427
00:15:55,410 --> 00:15:59,110
or onto theta, what I find is that this is P1.
或说投影到θ上 就会得到这样的p(1)

428
00:16:00,650 --> 00:16:02,410
That length there is P1.
它的长度是 p(1)

429
00:16:03,750 --> 00:16:05,820
The other example, that
另一个样本

430
00:16:06,260 --> 00:16:08,620
example is and I
那个样本是x(2)

431
00:16:08,840 --> 00:16:11,300
do the same projection and
我做同样的投影

432
00:16:11,410 --> 00:16:12,580
what I find is that this
我会发现

433
00:16:12,780 --> 00:16:14,680
length here is a
这是 p(2) 的长度

434
00:16:15,610 --> 00:16:17,880
P2 really that is going to be less than 0.
它是负值

435
00:16:18,830 --> 00:16:19,940
And you notice that now
你会注意到

436
00:16:20,480 --> 00:16:22,490
P1 and P2, these lengths
现在 p(1) 和 p(2)

437
00:16:23,810 --> 00:16:24,740
of the projections are going to
这些投影长度

438
00:16:24,780 --> 00:16:26,800
be much bigger, and so
是长多了

439
00:16:27,440 --> 00:16:28,460
if we still need to enforce
如果我们仍然要满足这些约束

440
00:16:28,890 --> 00:16:30,700
these constraints that P1 of
p(1) 乘以 θ 的范数

441
00:16:30,800 --> 00:16:33,040
the norm of theta is phase
是比1大的

442
00:16:33,230 --> 00:16:35,670
number one because P1 is so much bigger now.
则因为 p(1) 变大了

443
00:16:36,580 --> 00:16:39,110
The normal can be smaller.
θ 的范数就可以变小了

444
00:16:41,960 --> 00:16:43,090
And so, what this means is
因此这意味着

445
00:16:43,210 --> 00:16:44,320
that by choosing the decision
通过选择

446
00:16:44,730 --> 00:16:45,760
boundary shown on the right
右边的决策界

447
00:16:46,010 --> 00:16:47,610
instead of on the left, the
而不是左边的那个

448
00:16:47,850 --> 00:16:49,000
SVM can make the
支持向量机可以

449
00:16:49,080 --> 00:16:50,560
norm of the parameters theta
使参数 θ 的范数

450
00:16:50,840 --> 00:16:52,420
much smaller. So, if we can
变小很多 因此如果我们想

451
00:16:52,550 --> 00:16:54,080
make the norm of theta smaller and
令 θ 的范数变小

452
00:16:54,260 --> 00:16:55,140
therefore make the squared norm of
从而令 θ 范数的平方

453
00:16:55,590 --> 00:16:57,080
theta smaller, which is
变小 就能让

454
00:16:57,210 --> 00:16:58,130
why the SVM
支持向量机

455
00:16:58,710 --> 00:17:00,920
would choose this hypothesis on the right instead.
选择右边的决策界

456
00:17:02,800 --> 00:17:04,260
And this is how
这就是

457
00:17:05,580 --> 00:17:07,160
the SVM gives rise
支持向量机如何能

458
00:17:07,500 --> 00:17:09,550
to this large margin certification effect.
有效地产生大间距分类的原因

459
00:17:10,700 --> 00:17:11,620
Mainly, if you look
看这条绿线

460
00:17:11,820 --> 00:17:13,250
at this green line, if you look at this green
这个绿色的决策界

461
00:17:13,490 --> 00:17:14,990
hypothesis we want the
我们希望

462
00:17:15,070 --> 00:17:16,250
projections of my positive
正样本和负样本投影到

463
00:17:17,190 --> 00:17:18,780
and negative examples onto theta to be large, and
θ 的值大

464
00:17:19,200 --> 00:17:20,360
the only way for that to
要做到这一点

465
00:17:20,710 --> 00:17:23,490
hold true this is if surrounding the green line.
的唯一方式就是选择这条绿线做决策界

466
00:17:24,950 --> 00:17:27,710
There's this large margin, there's
这是大间距决策界

467
00:17:27,880 --> 00:17:31,460
this large gap that separates
来区分开

468
00:17:33,970 --> 00:17:37,240
positive and negative examples is
正样本和负样本

469
00:17:38,020 --> 00:17:40,740
really the magnitude of this gap.
这个间距的值

470
00:17:41,080 --> 00:17:42,050
The magnitude of this margin
这个间距的值

471
00:17:43,040 --> 00:17:44,900
is exactly the values of
就是p(1) p(2) p(3)

472
00:17:45,060 --> 00:17:47,730
P1, P2, P3 and so on.
等等的值

473
00:17:47,890 --> 00:17:48,970
And so by making the margin
通过让间距变大

474
00:17:49,480 --> 00:17:51,270
large, by these tyros P1,
通过这些p(1) p(2) p(3) 等等

475
00:17:51,470 --> 00:17:53,650
P2, P3 and so on that's
的值 支持向量机

476
00:17:53,830 --> 00:17:55,520
the SVM can end up with
最终可以找到

477
00:17:55,670 --> 00:17:56,860
a smaller value for the
一个较小的 θ 范数

478
00:17:56,960 --> 00:17:59,450
norm of theta which is what it is trying to do in the objective.
这正是支持向量机中最小化目标函数的目的

479
00:18:00,250 --> 00:18:01,290
And this is why this
以上就是为什么

480
00:18:01,960 --> 00:18:03,300
machine ends up with enlarge
支持向量机最终会找到

481
00:18:03,790 --> 00:18:05,510
margin classifiers because itss
大间距分类器的原因

482
00:18:05,640 --> 00:18:07,570
trying to maximize the norm
因为它试图极大化这些

483
00:18:07,720 --> 00:18:08,910
of these P1 which is the distance from
p(i) 的范数 它们是

484
00:18:09,060 --> 00:18:10,450
the training examples to the decision boundary.
训练样本到决策边界的距离

485
00:18:14,360 --> 00:18:16,450
Finally, we did this whole derivation
最后一点 我们的推导

486
00:18:17,200 --> 00:18:18,590
using this simplification that the
自始至终使用了这个简化假设

487
00:18:18,750 --> 00:18:21,150
parameter theta 0 must be equal to 0.
就是参数 θ0 等于0

488
00:18:21,860 --> 00:18:23,440
The effect of that as
就像我之前提到的

489
00:18:23,560 --> 00:18:25,380
I mentioned briefly, is that if
这个的作用是

490
00:18:25,540 --> 00:18:26,560
theta 0 is equal to
θ0 等于 0

491
00:18:26,830 --> 00:18:28,280
0 what that means
的意思是

492
00:18:28,460 --> 00:18:29,770
is that we are entertaining decision
我们让决策界

493
00:18:30,200 --> 00:18:31,510
boundaries that pass through the
通过原点

494
00:18:31,750 --> 00:18:33,640
origins of decision boundaries pass through
让决策界通过原点

495
00:18:33,800 --> 00:18:35,510
the origin like that, if you
就像这样

496
00:18:35,710 --> 00:18:37,980
allow theta zero to
如果你令

497
00:18:38,080 --> 00:18:39,540
be non 0 then what
θ0 不是0的话

498
00:18:39,870 --> 00:18:41,190
that means is that you entertain the
含义就是你希望

499
00:18:41,380 --> 00:18:43,120
decision boundaries that did not
决策界不通过原点

500
00:18:43,390 --> 00:18:45,730
cross through the origin, like that one I just drew.
比如这样

501
00:18:46,380 --> 00:18:47,940
And I'm not going to do
我将不会做全部的推导

502
00:18:48,010 --> 00:18:49,540
the full derivation that. It
实际上

503
00:18:49,650 --> 00:18:50,600
turns out that this same
支持向量机产生大间距分类器的结论

504
00:18:51,060 --> 00:18:52,720
large margin proof works in
会被证明同样成立

505
00:18:52,780 --> 00:18:54,240
pretty much in exactly the same way.
证明方式是非常类似的

506
00:18:54,390 --> 00:18:56,100
And there's a generalization of this
是我们刚刚做的

507
00:18:56,850 --> 00:18:57,830
argument that we just went through
证明的推广

508
00:18:58,030 --> 00:18:59,400
them long ago through that shows
之前视频中说过

509
00:18:59,870 --> 00:19:01,540
that even when theta
即便 θ0

510
00:19:01,840 --> 00:19:03,690
0 is non 0, what
不等于0

511
00:19:03,960 --> 00:19:06,940
the SVM is trying to do when you have this optimization objective.
支持向量机要做的事情都是优化这个目标函数

512
00:19:08,200 --> 00:19:09,620
Which again corresponds to the
对应着

513
00:19:09,720 --> 00:19:11,570
case of when C is very large.
C值非常大的情况

514
00:19:14,010 --> 00:19:15,110
But it is possible to
但是可以说明的是

515
00:19:15,290 --> 00:19:16,510
show that, you know, when theta
即便 θ0 不等于 0

516
00:19:16,810 --> 00:19:18,420
is not equal to 0 this
支持向量机

517
00:19:18,620 --> 00:19:19,750
support vector machine is still
仍然会

518
00:19:20,100 --> 00:19:21,360
finding is really trying
找到

519
00:19:21,640 --> 00:19:22,650
to find the large margin
正样本和负样本之间的

520
00:19:23,040 --> 00:19:24,060
separator that between the positive and negative
大间距分隔

521
00:19:24,630 --> 00:19:28,200
examples. So that
总之

522
00:19:28,420 --> 00:19:31,060
explains how this support vector machine is a large margin classifier.
我们解释了为什么支持向量机是一个大间距分类器

523
00:19:32,920 --> 00:19:34,020
In the next video we
在下一节我们

524
00:19:34,190 --> 00:19:35,080
will start to talk about how
将开始讨论

525
00:19:35,400 --> 00:19:36,480
to take some of these
如何利用支持向量机的原理

526
00:19:36,710 --> 00:19:37,980
SVM ideas and start to
应用它们

527
00:19:38,190 --> 00:19:39,200
apply them to build a complex
建立一个复杂的

528
00:19:39,900 --> 00:19:41,370
nonlinear classifiers.
非线性分类器