人工智能实战_第二次做业_陈泽寅

第二次做业

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题目说明:

• 详细说明

项目	内容
课程	人工智能实战2019
做业要求	做业
我在这个课程的目标是	了解人工智能理论，提高coding能力
这个做业在哪一个具体方面帮助我实现目标	经过学习神经网络反向传播算法更好地巩固本身的AI基础

做业要求、

• 实现双变量的反向线性传播
• 经过梯度降低求解参数\(\omega\)和\(b\)的最终值

解题思路

\[ \frac{\partial{z}}{\partial{b}}=\frac{\partial{z}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{b}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{b}}=y \cdot 3+x \cdot 2=3y+2x \]\[ \frac{\partial{z}}{\partial{w}}=\frac{\partial{z}}{\partial{x}} \cdot \frac{\partial{x}}{\partial{w}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{w}}=y \cdot 2+x \cdot 0 = 2y \]
因此，在每次迭代中，要从新计算下面两个值：
\[ \Delta b=\frac{\Delta z}{3y+2x} \]\[ \Delta w=\frac{\Delta z}{2y} \]
可是因为$
\Delta b\(和\)\Delta\omega\(都对\)\Delta z$形成一半的影响，所以咱们这里的公式应该改为
\[ \Delta b=\frac{\Delta z /2}{3y+2x} \]\[ \Delta w=\frac{\Delta z /2}{2y} \]

而后每次更新\(b\)和\(\omega,z\)的值,循环迭代，直到\(z -150\leq 10^-5\)

代码实现(每次均更新\(\Delta b\)和\(\Delta \omega\))

# -*-coding:utf-8-*-
from numpy import *

target = 150.0
w = 3.0
b = 4.0
z = 162
min = 1e-5;
print(z)
iter_count = 0
while ((abs(150 - z)) > min):
    iter_count +=1
    x = 2 * w + 3 * b
    y = 2 * b + 1
    z = x * y
    delta_b = ((z - 150) / ((2 * x + 3 * y)*2))
    delta_w = ((z - 150) / ((2 * y)*2))
    w = w - delta_w
    b = b - delta_b
    print("loop: %d, w = %.8f, b = %.8f, z = %.8f, error = %.8f"%(iter_count,w,b,z,abs(150-z)));
print("final result:loop = %d, w = %.8f, b = %.8f,z = %.8f,error = %.8f"%(iter_count,w,b,z,abs(z-150)));

运行结果

loop: 1, w = 2.66666667, b = 3.90476190, z = 162.00000000, error = 12.00000000
loop: 2, w = 2.66151866, b = 3.90326327, z = 150.18140590, error = 0.18140590
loop: 3, w = 2.66151740, b = 3.90326291, z = 150.00004434, error = 0.00004434
loop: 4, w = 2.66151740, b = 3.90326291, z = 150.00000000, error = 0.00000000
final result:loop = 4, w = 2.66151740, b = 3.90326291,z = 150.00000000,error = 0.00000000

当咱们在反向传播过程当中认为\(\Delta b\)和\(\Delta \omega\)的值保持不变

delta_b = ((z - 150) / ((63)*2))
 delta_w = ((z - 150) / ((18)*2))

运行结果

loop: 1, w = 2.66666667, b = 3.90476190, z = 162.00000000, error = 12.00000000
loop: 2, w = 2.66162761, b = 3.90332218, z = 150.18140590, error = 0.18140590
loop: 3, w = 2.66147411, b = 3.90327832, z = 150.00552614, error = 0.00552614
loop: 4, w = 2.66146940, b = 3.90327697, z = 150.00016964, error = 0.00016964
loop: 5, w = 2.66146925, b = 3.90327693, z = 150.00000521, error = 0.00000521
final result:loop = 5, w = 2.66146925, b = 3.90327693,z = 150.00000521,error = 0.00000521

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比较与分析

• 咱们能够看到，上面两种方法都可以在有限次迭代中达到想要的结果，可是第一种明显效果更好，速度也更快，由于神经网络的反馈是实时的，也就是说参数的影响会直接做用于下一轮迭代，这也能使神经网络调整参数的速度更加快。所以咱们须要在每次迭代时先更新参数的值，后计算下一轮的数据。