反向传播算法推导

反向传播英语：Backpropagation，缩写为 BP ）是“误差反向传播”的简称，是一种与最优化方法（如梯度下降法）结合使用的，用来训练人工神经网络的常见方法。该方法对网络中所有权重计算损失函数的梯度。这个梯度会反馈给最优化方法，用来更新权值以最小化损失函数。

假设，你有这样一个网络层

第一层是输入层，包含两个神经元 $i1$，$i2$，和截距项$b1$；第二层是隐含层，包含两个神经元$h1$,$h2$和截距项$b2$，第三层是输出$o1$,$o2$，每条线上标的$wi$是层与层之间连接的权重，激活函数我们默认为 sigmoid 函数。

现在对他们赋上初值，如下图：

　　 　　 其中， 输入数据 $i1=0.05$，$i2=0.10$;

输出数据 $o1=0.01$，$o2=0.99$;

初始权重
$w1=0.15$，$w2=0.20$, $w3=0.25$，$w4=0.30$; $w5=0.40$，$w6=0.45$, $w7=0.50$，$w8=0.55$;

目标:给出输入数据$i1$，$i2$(0.05 和 0.10)，使输出尽可能与原始输出$o1$,$o2$(0.01 和 0.99)接近。

前向传播过程

1. 输入层---->隐含层：

计算神经元$h1$的输入加权和：

net_{h1} = w_1 * i_1 + w_2 * i_2 + b_1 * 1 net_{h1} = 0.15 * 0.05 + 0.2 * 0.1 + 0.35 * 1 = 0.3775 

计算神经元$h1$的输出$o1$:(此处用到激活函数为 sigmoid 函数)

out_{h1} = \frac{1}{1+e^{-net_{h1}}} = 0.5932 

同理,可计算神经元 $h2$ 的输出 $o2$

out_{h2} = 0.5968 

2. 隐藏层---->输出层:

net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1 out_{o1} = \frac{1}{1+e^{-net_{o1}}} = 0.7514 

同样的,计算神经元 o2 的输出

out_{o2} = 0.7730 

反向传播过程

接下来,就可以进行反向传播的计算了

1. 计算总误差

E_{total} = E_{o1} + E_{o2} 

分别计算$o1$,$o2$的误差

E_{o1} = \frac{1}{2} (target_{o1} - out_{o1})^2 = 0.2748 E_{o2} = \frac{1}{2} (target_{o2} - out_{o2})^2 = 0.0235 

E_{total} = E_{o1} + E_{o2} = 0.2983 

2. 隐含层---->输出层的权值更新：

以权重参数$w5$为例，如果我们想知道$w5$对整体误差产生了多少影响，可以用整体误差对$w5$求偏导求出(链式法则)

\frac {\partial (E_{total} )}{\partial (w_{5})} = \frac {\partial (E_{total} )}{\partial (out_{o1})} + \frac {\partial (out_{o1} )}{\partial (net_{o1})} + \frac {\partial (net_{o1} )}{\partial (w_{5})} 

下面的图可以更直观的看清楚误差是怎样反向传播的

我们分别计算每个式子的值：

计算 $\frac {\partial (E_{total} )}{\partial (out_{o1})}$

 E_{total} = \frac {1}{2}(target_{o1} - out_{o1} )^2 +\frac {1}{2}(target_{o2} - out_{o2} )^2 

\frac {\partial (E_{total} )}{\partial (out_{o1})} = - (target_{o1} - out{o1} ) = 0.7414 

计算 $ \frac {\partial ( out_{o1} )}{\partial (net_{o1})} $

 out_{o1} = \frac{1}{1+e^{-net_{o1}}} \frac {\partial ( out_{o1} )}{\partial (net_{o1})} = out_{o1}(1 - out_{o1} ) = 0.1868 

计算 $ \frac {\partial ( net_{o1} )}{\partial (w_{5})}$

net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1 \frac {\partial ( net_{o1} )}{\partial (w_{5})} = out_{h1} = 0.5932 

最后三者相乘

 \frac {\partial (E_{total} )}{\partial (w_{5})} = \frac {\partial (E_{total} )}{\partial (out_{o1})} * \frac {\partial (out_{o1} )}{\partial (net_{o1})} * \frac {\partial (net_{o1} )}{\partial (w_{5})} = 0.082 

看看上面的公式,我们发现：

\frac {\partial (E_{total} )}{\partial (w_{5})} = -(target_{o1}-out_{o1})*out_{o1}(1-out_{o1})*out_{h1} 

为了表达方便，用$\delta _{o1}$来表示输出层的误差

\delta _{o1} = \frac {\partial (E_{total} )}{\partial (out_{o1})} + \frac {\partial (out_{o1} )}{\partial (net_{o1})} 

 \delta _{o1} = -(target_{o1}-out_{o1})*out_{o1}(1-out_{o1}) 

 \frac {\partial (E_{total} )}{\partial (w_{5})} = \delta _{o1} *out_{h1} 

更新$w_5$的值：

 w_5^+ = w_5 - \eta * \frac {\partial (E_{total} )}{\partial (w_{5})} = 0.3589 

同理,更新 $w_6$,$w_7$,$w_8$

w_6^+ = 0.4086 

w_7^+ = 0.5113 

w_8^+ = 0.5614 

3.隐含层---->隐含层的权值更新：

我们可以依照上述的方法计算 $w_1$, $w_2$, $w_3$, $w_4$，方法其实与上面说的差不多，但是有个地方需要变一下。

在上文计算总误差对 w5 的偏导时，是从：

$out_{o1}$ -> $net_{o1}$ -> $w_5$

但是在隐含层之间的权值更新时，是从：

$out_{h1}$ -> $net_{h1}$ -> $w_1$

计算 $\frac {\partial (E_{total} )}{\partial (out_{h1})}$

\frac {\partial (E_{total} )}{\partial (out_{h1})} = \frac {\partial (E_{o1} )}{\partial (out_{h1})} + \frac {\partial (E_{o2} )}{\partial (out_{h1})} 

先计算$\frac {\partial (E_{o1} )}{\partial (out_{h1})}$

\frac {\partial (E_{o1} )}{\partial (out_{h1})} = \frac {\partial (E_{o1} )}{\partial (net_{o1})} * \frac {\partial (net_{o1} )}{\partial (out_{h1})} 

 \frac {\partial (E_{o1} )}{\partial (net_{o1})} = \frac {\partial (E_{o1} )}{\partial (out_{o1})} * \frac {\partial (out_{o1} )}{\partial (net_{o1})} = 0.1385 

 net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1 

 \frac {\partial (net_{o1} )}{\partial (out_{h1})} = w_5= 0.40 

\frac {\partial (E_{o1} )}{\partial (out_{h1})} = \frac {\partial (E_{o1} )}{\partial (net_{o1})} * \frac {\partial (net_{o1} )}{\partial (out_{h1})} = 0.138 * 0.4 = 0.055 

同理,计算出

 \frac {\partial (E_{o2} )}{\partial (out_{h1})} = -0.019 

两者相加,得到总值

 \frac {\partial (E_{total} )}{\partial (out_{h1})} = \frac {\partial (E_{o1} )}{\partial (out_{h1})} + \frac {\partial (E_{o2} )}{\partial (out_{h1})} = 0.036 

再计算 $\frac {\partial (out_{h1} )}{\partial (net_{h1})}$

 out_{h1} = \frac{1}{1+e^{-net_{h1}}} 

 \frac {\partial (out_{h1} )}{\partial (net_{h1})} = out_{h1} *(1-out_{h1}) = 0.2413 

再计算$ \frac {\partial (net_{h1} )}{\partial (w_{1})} $

 net_{h1} = w_1 * i_1 + w_2 * i_2 + b_1 * 1 \frac {\partial (net_{h1} )}{\partial (w_{1})} = i_1 =0.05 

最后,三者相乘

 \frac {\partial (E_{total} )}{\partial (w_{1})} = \frac {\partial (E_{total} )}{\partial (out_{h1})} * \frac {\partial (out_{h1} )}{\partial (net_{h1})} * \frac {\partial (net_{h1} )}{\partial (w_{1})} 

 \frac {\partial (E_{total} )}{\partial (w_{1})} = 0.036 * 0.2413 * 0.05 = 0.000438 

我们更新$w_1$的值

 w_1^+ = w_1 - \eta * \frac {\partial (E_{total} )}{\partial (w_{1})} = 0.1498 

同理,更新 $w_2$,$w_3$,$w_4$

w_2^+ = 0.1996 

w_3^+ = 0.2498 

w_4^+ = 0.2995 

这样误差反向传播法就完成了，最后我们再把更新的权值重新计算，不停地迭代.

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