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单韦乔
Toy-MT-Introduction
Commits
44d2f25a
Commit
44d2f25a
authored
Oct 18, 2019
by
xiaotong
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new update
parent
a204dca2
隐藏空白字符变更
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1 个修改的文件
包含
91 行增加
和
21 行删除
+91
-21
Section05-Neural-Networks-and-Language-Modeling/section05-test.tex
+91
-21
没有找到文件。
Section05-Neural-Networks-and-Language-Modeling/section05-test.tex
查看文件 @
44d2f25a
...
@@ -128,15 +128,20 @@
...
@@ -128,15 +128,20 @@
\begin{tikzpicture}
\begin{tikzpicture}
\begin{scope}
\begin{scope}
\node
[anchor=center,minimum height=1.7em,fill=yellow!20,draw] (h) at (0,0)
{$
\textbf
{
h
}^{
K
-
1
}$}
;
\node
[anchor=center,minimum height=1.7em,fill=yellow!20,draw] (h) at (0,0)
{$
\textbf
{
h
}^{
K
-
1
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=blue!20,draw] (s) at ([xshift=6.5em]h.east)
{$
\textbf
{
s
}^{
K
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=blue!20,draw] (s) at ([xshift=5.5em]h.east)
{$
\textbf
{
s
}^{
K
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=green!20,draw] (h2) at ([xshift=6.5em]s.east)
{$
\textbf
{
h
}^{
K
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=green!20,draw] (h2) at ([xshift=5.5em]s.east)
{$
\textbf
{
h
}^{
K
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=orange!20,draw] (l) at ([xshift=4em]h2.east)
{$
L
$}
;
\node
[anchor=west,minimum height=1.7em,fill=orange!20,draw] (l) at ([xshift=5.5em]h2.east)
{$
L
$}
;
\draw
[->] (h.east) -- (s.west) node [pos=0.5,above]
{
\tiny
{
线性变换
$
\textbf
{
s
}^
K
=
\textbf
{
h
}^{
K
-
1
}
\textbf
{
w
}^
K
$}}
;
\draw
[->] (h.east) -- (s.west);
\draw
[->] (s.east) -- (h2.west) node [pos=0.5,above]
{
\tiny
{
激活函数
$
\textbf
{
h
}^
K
=
f
^
K
(
\textbf
{
s
}^
K
)
$}}
;
\draw
[->] (s.east) -- (h2.west);
\end{scope}
\draw
[->] (h2.east) -- (l.west) node [pos=0.5,above]
{
\tiny
{
损失
}}
;
\draw
[->] (h2.east) -- (l.west) node [pos=0.5,above]
{
\tiny
{
损失
}}
;
\node
[anchor=south] (outputlabel) at ([yshift=0.3em]h2.north)
{
\scriptsize
{
\textbf
{
网络输出层
}}}
;
\node
[anchor=south west,inner sep=2pt] (step100) at ([xshift=0.5em,yshift=-0.8em]h.north east)
{
\tiny
{$
\textbf
{
s
}^
K
=
\textbf
{
h
}^{
K
-
1
}
\textbf
{
w
}^
K
$}}
;
\node
[anchor=south west,inner sep=2pt] (step101) at (step100.north west)
{
\tiny
{
线性变换
}}
;
\node
[anchor=south west,inner sep=2pt] (step200) at ([xshift=0.5em,yshift=-0.8em]s.north east)
{
\tiny
{$
\textbf
{
h
}^
K
=
f
^
K
(
\textbf
{
s
}^
K
)
$}}
;
\node
[anchor=south west,inner sep=2pt] (step201) at (step200.north west)
{
\tiny
{
激活函数
}}
;
\node
[anchor=south,inner sep=1pt] (outputlabel) at ([yshift=0.0em]h2.north)
{
\tiny
{
\textbf
{
输出层
}}}
;
\visible
<2->
{
\visible
<2->
{
\draw
[decorate,thick,decoration={brace,mirror,raise=0.4em,amplitude=2mm}]
(h.south west) -- (s.south west) node [pos=0.5,below,yshift=-1em]
{
\scriptsize
{
\textbf
{
第一阶段:线性变换
}}}
;
\draw
[decorate,thick,decoration={brace,mirror,raise=0.4em,amplitude=2mm}]
(h.south west) -- (s.south west) node [pos=0.5,below,yshift=-1em]
{
\scriptsize
{
\textbf
{
第一阶段:线性变换
}}}
;
...
@@ -145,19 +150,21 @@
...
@@ -145,19 +150,21 @@
\draw
[decorate,thick,decoration={brace,mirror,raise=0.4em,amplitude=2mm}]
([xshift=0.2em]s.south west) -- (l.south east) node [pos=0.5,below,yshift=-1em] (step2)
{
\scriptsize
{
\textbf
{
第二阶段:激活函数+损失函数
}}}
;
\draw
[decorate,thick,decoration={brace,mirror,raise=0.4em,amplitude=2mm}]
([xshift=0.2em]s.south west) -- (l.south east) node [pos=0.5,below,yshift=-1em] (step2)
{
\scriptsize
{
\textbf
{
第二阶段:激活函数+损失函数
}}}
;
}
}
\begin{pgfonlayer}
{
background
}
\visible
<4->
{
\visible
<4->
{
\node
[rectangle,inner sep=0em,fill=red!20] [fit = (step2)] (step2label)
{}
;
\draw
[->,very thick,red] ([yshift=1em,xshift=-0.1em]l.north) -- ([yshift=1em,xshift=0.1em]s.north) node [pos=0.5,above]
{
\tiny
{
反向求梯度
\alert
{$
\frac
{
\partial
L
}{
\partial
\textbf
{
s
}^
k
}
=
?
$}}}
;
\draw
[-,very thick,red] ([yshift=0.5em]l.north) -- ([yshift=1.5em]l.north);
\draw
[-,very thick,red] ([yshift=0.5em]s.north) -- ([yshift=1.5em]s.north);
}
}
\end{pgfonlayer}
\end{scope}
\end{tikzpicture}
\end{tikzpicture}
\end{center}
\end{center}
\begin{itemize}
\begin{itemize}
\item
<4-> 反向传播从输出向输入传播梯度,因此我们先考虑阶段二
。令
$
\pi
^
k
=
\frac
{
\partial
L
}{
\partial
\textbf
{
s
}^
k
}$
表示损失
$
L
$
在第
$
k
$
层激活函数输入处的梯度
\visible
<5->
{
,利用链式法有
}
\item
<4-> 反向传播从输出向输入传播梯度,因此我们先考虑阶段二
\visible
<5->
{
。令
$
\pi
^
k
=
\frac
{
\partial
L
}{
\partial
\textbf
{
s
}^
k
}$
表示损失
$
L
$
在第
$
k
$
层激活函数输入处的梯度
,利用链式法有
}
\vspace
{
-1em
}
\vspace
{
-1
.5
em
}
\visible
<5->
{
\visible
<5->
{
\begin{eqnarray}
\begin{eqnarray}
\pi
^
K
&
=
&
\frac
{
\partial
L
}{
\partial
\textbf
{
s
}^
K
}
\nonumber
\\
\pi
^
K
&
=
&
\frac
{
\partial
L
}{
\partial
\textbf
{
s
}^
K
}
\nonumber
\\
...
@@ -180,19 +187,21 @@
...
@@ -180,19 +187,21 @@
\begin{tikzpicture}
\begin{tikzpicture}
\begin{scope}
\begin{scope}
\node
[anchor=center] (factor00) at (0,0)
{${
\displaystyle
\
frac
{
\partial
L
}{
\partial
\textbf
{
w
}^
K
}
\
=
}$}
;
\node
[anchor=center] (factor00) at (0,0)
{${
\displaystyle
\
pi
^
K
\
=
}$}
;
\node
[anchor=west] (factor01) at (factor00.east)
{${
\displaystyle
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
K
}}$}
;
\node
[anchor=west] (factor01) at (factor00.east)
{${
\displaystyle
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
K
}}$}
;
\node
[anchor=west,inner sep=1pt] (factor02) at (factor01.east)
{${
\displaystyle
\cdot
}$}
;
\node
[anchor=west,inner sep=1pt] (factor02) at (factor01.east)
{${
\displaystyle
\cdot
}$}
;
\node
[anchor=west] (factor03) at (factor02.east)
{${
\displaystyle
\frac
{
\partial
f
^
K
(
\textbf
{
s
}^
K
)
}{
\partial
\textbf
{
s
}^
K
}}$}
;
\node
[anchor=west] (factor03) at (factor02.east)
{${
\displaystyle
\frac
{
\partial
f
^
K
(
\textbf
{
s
}^
K
)
}{
\partial
\textbf
{
s
}^
K
}}$}
;
\node
[anchor=west,inner sep=1pt] (factor04) at (factor03.east)
{${
\displaystyle
\cdot
}$}
;
\node
[anchor=west] (factor05) at (factor04.east)
{${
\displaystyle
\frac
{
\partial
\textbf
{
s
}^
K
}{
\partial
\textbf
{
w
}^
K
}}$}
;
\begin{pgfonlayer}
{
background
}
\begin{pgfonlayer}
{
background
}
\visible
<2-4>
{
\node
[rectangle,inner sep=0em,fill=red!20] [fit = (factor01)] (p1)
{}
;
\node
[rectangle,inner sep=0em,fill=red!20] [fit = (factor01)] (p1)
{}
;
}
\visible
<3-4>
{
\node
[rectangle,inner sep=0em,fill=blue!20] [fit = (factor03)] (p2)
{}
;
\node
[rectangle,inner sep=0em,fill=blue!20] [fit = (factor03)] (p2)
{}
;
\node
[rectangle,inner sep=0em,fill=green!20] [fit = (factor05)] (p3)
{}
;
}
\node
[circle,inner sep=0em,fill=purple!20] [fit = (factor02)] (p4)
{}
;
\visible
<5->
{
\node
[circle,inner sep=0em,fill=purple!20] [fit = (factor04)] (p5)
{}
;
\node
[circle,inner sep=0em,fill=green!20] [fit = (factor02)] (p3)
{}
;
}
\end{pgfonlayer}
\end{pgfonlayer}
\end{scope}
\end{scope}
...
@@ -200,12 +209,73 @@
...
@@ -200,12 +209,73 @@
\end{center}
\end{center}
\begin{itemize}
\begin{itemize}
\item
\raisebox
{
-0.7em
}{
\tikz
{
\node
[anchor=west,fill=red!20] (factor01) at (factor00.east)
{$
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
K
}$}
;
}}
表示损失
$
L
$
相对网络输出的变化率,比如,对于
$
L
=
\frac
{
1
}{
2
}
(
\hat
{
\textbf
{
y
}}
-
\textbf
{
h
}^
K
)
^
2
$
,有
$
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
K
}
=
\hat
{
\textbf
{
y
}}
-
\textbf
{
h
}^
K
$
\item
<2->
\raisebox
{
-0.7em
}{
\tikz
{
\node
[anchor=west,fill=red!20] (factor01) at (factor00.east)
{$
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
K
}$}
;
}}
表示损失
$
L
$
相对网络输出的变化率,比如,对于
$
L
=
\frac
{
1
}{
2
}
||
\hat
{
\textbf
{
y
}}
-
\textbf
{
h
}^
K||
^
2
$
,有
$
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
K
}
=
\hat
{
\textbf
{
y
}}
-
\textbf
{
h
}^
K
$
\item
\raisebox
{
-0.7em
}{
\tikz
{
\node
[anchor=west,fill=blue!20] (factor01) at (factor00.east)
{$
\frac
{
\partial
f
^
K
(
\textbf
{
s
}^
K
)
}{
\partial
\textbf
{
s
}^
K
}$}
;
}}
表示激活函数相对于它自己的输入的变化率,比如,对于
$
f
(
\textbf
{
s
}
)
=
\frac
{
1
}{
1
+
\exp
(-
\textbf
{
s
}
)
}$
,有
$
\frac
{
\partial
f
(
\textbf
{
s
}
)
}{
\partial
\textbf
{
s
}}
=
f
(
\textbf
{
s
}
)(
1
-
f
(
\textbf
{
s
}
))
$
\item
<3->
\raisebox
{
-0.7em
}{
\tikz
{
\node
[anchor=west,fill=blue!20] (factor01) at (factor00.east)
{$
\frac
{
\partial
f
^
K
(
\textbf
{
s
}^
K
)
}{
\partial
\textbf
{
s
}^
K
}$}
;
}}
表示激活函数相对于它自己的输入的变化率,比如,对于
$
f
(
\textbf
{
s
}
)
=
\frac
{
1
}{
1
+
\exp
(-
\textbf
{
s
}
)
}$
,有
$
\frac
{
\partial
f
(
\textbf
{
s
}
)
}{
\partial
\textbf
{
s
}}
=
f
(
\textbf
{
s
}
)(
1
-
f
(
\textbf
{
s
}
))
$
\item
\raisebox
{
-0.7em
}{
\tikz
{
\node
[anchor=west,fill=green!20] (factor01) at (factor00.east)
{$
\frac
{
\partial
\textbf
{
s
}^
K
}{
\partial
\textbf
{
w
}^
K
}$}
;
}}
表示激活函数的输入相对于参数矩阵
$
\textbf
{
w
}^
K
$
的变化率,根据简单的数学,可以得到
$
\frac
{
\partial
\textbf
{
s
}^
K
}{
\partial
\textbf
{
w
}^
K
}
=
\frac
{
\partial
\textbf
{
h
}^{
K
-
1
}
\textbf
{
w
}^
K
}{
\partial
\textbf
{
w
}^
K
}$
\item
<4-> 这个结果符合直觉,在
$
s
^
K
$
出的梯度相当于在损失函数微分(
$
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
K
}$
)和激活函数微分(
$
\frac
{
\partial
f
^
K
(
\textbf
{
s
}^
K
)
}{
\partial
\textbf
{
s
}^
K
}$
) 的乘积
\visible
<5->
{
,注意这里所有操作都是单元级,比如张量按单元乘法
}
\end{itemize}
\visible
<4->
{
\vspace
{
-0.5em
}
\begin{center}
\begin{tikzpicture}
\begin{scope}
\node
[anchor=west,minimum height=1.7em,fill=blue!20,draw] (s) at (0,0)
{$
\textbf
{
s
}^{
K
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=green!20,draw] (h2) at ([xshift=5.5em]s.east)
{$
\textbf
{
h
}^{
K
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=orange!20,draw] (l) at ([xshift=5.5em]h2.east)
{$
L
$}
;
\draw
[->] (s.east) -- (h2.west);
\draw
[->] (h2.east) -- (l.west);
\draw
[->,very thick,red] ([yshift=1em,xshift=-0.1em]l.north) -- ([yshift=1em,xshift=0.1em]h2.north) node [pos=0.5,above]
{
\tiny
{
求梯度
\alert
{$
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
k
}
=
?
$}}}
;
\draw
[->,very thick,red] ([yshift=1em,xshift=-0.1em]h2.north) -- ([yshift=1em,xshift=0.1em]s.north) node [pos=0.5,above]
{
\tiny
{
求梯度
\alert
{$
\frac
{
\partial
f
^
K
(
\textbf
{
s
}^
K
)
}{
\partial
\textbf
{
s
}^
K
}
=
?
$}}}
;
\draw
[-,very thick,red] ([yshift=0.5em]l.north) -- ([yshift=1.5em]l.north);
\draw
[-,very thick,red] ([yshift=0.5em]h2.north) -- ([yshift=1.5em]h2.north);
\draw
[-,very thick,red] ([yshift=0.5em]s.north) -- ([yshift=1.5em]s.north);
\end{scope}
\end{tikzpicture}
\end{center}
}
\end{frame}
%%%------------------------------------------------------------------------------------------------------------
%%% 输出层的反向传播 - 求 dL/dw
\begin{frame}
{
反向传播 - 输出层
}
\begin{itemize}
\item
输出层(两个阶段)
\end{itemize}
\end{itemize}
\vspace
{
-0.5em
}
\begin{center}
\begin{tikzpicture}
\begin{scope}
\node
[anchor=center,minimum height=1.7em,fill=yellow!20,draw] (h) at (0,0)
{$
\textbf
{
h
}^{
K
-
1
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=blue!20,draw] (s) at ([xshift=5.5em]h.east)
{$
\textbf
{
s
}^{
K
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=green!20,draw] (h2) at ([xshift=5.5em]s.east)
{$
\textbf
{
h
}^{
K
}$}
;
\node
[anchor=west,minimum height=1.7em,fill=orange!20,draw] (l) at ([xshift=5.5em]h2.east)
{$
L
$}
;
\draw
[->] (h.east) -- (s.west);
\draw
[->] (s.east) -- (h2.west);
\draw
[->] (h2.east) -- (l.west) node [pos=0.5,above]
{
\tiny
{
损失
}}
;
\node
[anchor=south west,inner sep=2pt] (step100) at ([xshift=0.5em,yshift=-0.8em]h.north east)
{
\tiny
{$
\textbf
{
s
}^
K
=
\textbf
{
h
}^{
K
-
1
}
\textbf
{
w
}^
K
$}}
;
\node
[anchor=south west,inner sep=2pt] (step200) at ([xshift=0.5em,yshift=-0.8em]s.north east)
{
\tiny
{$
\textbf
{
h
}^
K
=
f
^
K
(
\textbf
{
s
}^
K
)
$}}
;
\node
[anchor=south,inner sep=1pt] (outputlabel) at ([yshift=0.0em]h2.north)
{
\tiny
{
\textbf
{
输出层
}}}
;
\node
[anchor=south west] (slabel) at ([yshift=1em,xshift=0.3em]s.north)
{
\scriptsize
{
\textbf
{
\alert
{
已经得到:
$
\pi
^
K
$}}}}
;
\draw
[->,very thick,red] ([yshift=1em,xshift=-0.1em]s.north) -- ([yshift=1em,xshift=0.1em]h.north) node [pos=0.5,above]
{
\tiny
{
\alert
{$
\frac
{
\partial
L
}{
\partial
\textbf
{
w
}^
k
}
=
?
$
,
$
\frac
{
\partial
L
}{
\partial
\textbf
{
h
}^
k
}
=
?
$}}}
;
\draw
[-,very thick,red] ([yshift=0.5em]h.north) -- ([yshift=1.5em]h.north);
\draw
[-,very thick,red] ([yshift=0.5em]s.north) -- ([yshift=1.5em]s.north);
\end{scope}
\end{tikzpicture}
\end{center}
\end{frame}
\end{frame}
\end{CJK}
\end{CJK}
...
...
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