极简单cnn编程测试

我们前面已经推导了极简的cnn数学流程，这里编程测试一下，从极简bpnet改过来，c#版本！

我们用九宫格"X"，来训练1万次，然后测试，看看效果：

测个：工

然后，全1

测个，0做的x，不错，识别到了！有意思！

我们再测试一个：0（1做的圈）

再测：x

效果还不错！

完整代码如下：

using System;

using System.Collections.Generic;

using System.ComponentModel;

using System.Data;

using System.Drawing;

using System.Linq;

using System.Text;

using System.Windows.Forms;

namespace 识别x第二版//及其简单的bpnet

{

public partial class Form1 : Form

{

//九宫格x，即

//1 0 1

//0 1 0

//1 0 1

double 常数a = 0.02;

float a1 = 1;

float a2 = 0;

float a3 = 1;

float a4 = 0;

float a5 = 1;

float a6 = 0;

float a7 = 1;

float a8 = 0;

float a9 = 1;

public double sigmoid(double x)

{

return 1 / (1.0 + Math.Exp(-x));

}

public double dsigmoid(double x)

{

return x * (1 - x);

}

public Form1()

{

InitializeComponent();

for (int xx = 0; xx < 2; xx++)

wcnn $xx$ = new double $4$ ;

for (int xx = 0; xx < 2; xx++)

wcnn的偏差 $xx$ = new double $4$ ;

a0 $0$ = a1;

a0 $1$ = a2;

a0 $2$ = a3;

a0 $3$ = a4;

a0 $4$ = a5;

a0 $5$ = a6;

a0 $6$ = a7;

a0 $7$ = a8;

a0 $8$ = a9;

}

double\[\] a0 = new double $9$ ;

double\[\] aI = new double $4$ ;

double aII = 0;

double\[\] z1 = new double $4$ ;

double z2 = 0;

double\[\]\[\] wcnn = new double $2$ \[\];// wcnn $0$ 第一层卷积核, wcnn $1$ 第二层卷积核

double\[\]\[\] wcnn的偏差 = new double $2$ \[\];

double o = 0;

public void forward()

{

//1,第一层卷积

z1 $0$ = a0 $0$ * wcnn $0$ $0$ + a0 $1$ * wcnn $0$ $1$ + a0 $3$ * wcnn $0$ $2$ + a0 $4$ * wcnn $0$ $3$ ;//1,2,4,5

z1 $1$ = a0 $1$ * wcnn $0$ $0$ + a0 $2$ * wcnn $0$ $1$ + a0 $4$ * wcnn $0$ $2$ + a0 $5$ * wcnn $0$ $3$ ;//2,3,5,6

z1 $2$ = a0 $3$ * wcnn $0$ $0$ + a0 $4$ * wcnn $0$ $1$ + a0 $6$ * wcnn $0$ $2$ + a0 $7$ * wcnn $0$ $3$ ;//4,5,7,8

z1 $3$ = a0 $4$ * wcnn $0$ $0$ + a0 $5$ * wcnn $0$ $1$ + a0 $7$ * wcnn $0$ $2$ + a0 $8$ * wcnn $0$ $3$ ;//5,6,8,9

//2, z1->aI,第二层输入

aI $0$ = sigmoid(z1 $0$ );

aI $1$ = sigmoid(z1 $1$ );

aI $2$ = sigmoid(z1 $2$ );

aI $3$ = sigmoid(z1 $3$ );

//3,第三层

z2 = aI $0$ * wcnn $1$ $0$ + aI $1$ * wcnn $1$ $1$ + aI $2$ * wcnn $1$ $2$ + aI $3$ * wcnn $1$ $3$ ;

aII = sigmoid(z2);

o = aII;

//4

double 偏差 = (1 - o) * (1 - o) * 1 / 2.0;

}

double E偏差zII = 0;

public void back()

{

//1，计算偏差,e/a2的偏导，a/z2偏导

E偏差zII = -(1 - o) * dsigmoid(o);//delta

//2，计算w【1】的偏导，第二层卷积核

wcnn的偏差 $1$ $0$ = E偏差zII * aI $0$ ;//e偏差z2*zII偏导w2【11】

wcnn的偏差 $1$ $1$ = E偏差zII * aI $1$ ;//e偏差z2*zII偏导w2【12】

wcnn的偏差 $1$ $2$ = E偏差zII * aI $2$ ;//e偏差z2*zII偏导w2【21】

wcnn的偏差 $1$ $3$ = E偏差zII * aI $3$ ;//e偏差z2*zII偏导w2【22】

//e对z1的偏导=e对a1的偏导*a1对z1的偏导

double delta11 = E偏差zII * wcnn $1$ $0$ *dsigmoid(aI $0$ );

double delta12 = E偏差zII * wcnn $1$ $1$ *dsigmoid(aI $1$ );

double delta21 = E偏差zII * wcnn $1$ $2$ *dsigmoid(aI $2$ );

double delta22 = E偏差zII * wcnn $1$ $3$ * dsigmoid(aI $3$ );

//3，计算w【0】的偏导，第一层卷积核

wcnn的偏差 $0$ $0$ = delta11*a0 $0$ + delta12*a0 $1$ + delta21*a0 $3$ + delta22*a0 $4$ ;//w1的偏差

wcnn的偏差 $0$ $1$ = delta11*a0 $1$ + delta12*a0 $2$ + delta21*a0 $4$ + delta22*a0 $5$ ;//w2的偏差

wcnn的偏差 $0$ $2$ = delta11*a0 $3$ + delta12*a0 $4$ + delta21*a0 $6$ + delta22*a0 $7$ ;//w3的偏差

wcnn的偏差 $0$ $3$ = delta11*a0 $4$ + delta12*a0 $5$ + delta21*a0 $7$ + delta22*a0 $8$ ;//w4的偏差

//更新

//w1,第二层的一个卷积核

for (int i = 0; i < 4; i++)

{

wcnn $1$ $i$ = wcnn $1$ $i$ - 常数a * wcnn的偏差 $1$ $i$ ;

}

//w0，第一层的一个卷积核

for (int i = 0; i < 4; i++)

{

wcnn $0$ $i$ = wcnn $0$ $i$ - 常数a * wcnn的偏差 $0$ $i$ ;

}

private void button1_Click(object sender, EventArgs e)//训练

{

for (int i = 0; i < 10000; i++)

{

forward();

back();

}

double last偏差 = (1 - o) * (1 - o) * 1 / 2.0;

textBox10.Text = last偏差.ToString();

button1.Enabled = false;

MessageBox.Show("请换图测试！");

}

Random ran = new Random();

public double randomNormalDistribution()

{

double u = 0.0, v = 0.0, w = 0.0, c = 0.0;

{

//获得两个（-1,1）的独立随机变量

u = ran.NextDouble() * 2 - 1.0;

v = ran.NextDouble() * 2 - 1.0;

w = u * u + v * v;

} while (w == 0.0 || w >= 1.0);

//这里就是 Box-Muller转换

c = Math.Sqrt((-2 * Math.Log(w)) / w);

//返回2个标准正态分布的随机数，封装进一个数组返回

//当然，因为这个函数运行较快，也可以扔掉一个

//return $u\*c,v\*c$ ;

//double u1 = ran.NextDouble();

//double u2 = ran.NextDouble();

//double z = Math.Sqrt(-2 * Math.Log(u1)) * Math.Sin(2 * Math.PI * u2);

//return z;

return u * c;

}

private void button2init_Click(object sender, EventArgs e)

{

a1 =(float)Convert.ToDouble(textBox1.Text);

a2 = (float)Convert.ToDouble(textBox2.Text);

a3 = (float)Convert.ToDouble(textBox3.Text);

a4 = (float)Convert.ToDouble(textBox4.Text);

a5 = (float)Convert.ToDouble(textBox5.Text);

a6 = (float)Convert.ToDouble(textBox6.Text);

a7 = (float)Convert.ToDouble(textBox7.Text);

a8 = (float)Convert.ToDouble(textBox8.Text);

a9 = (float)Convert.ToDouble(textBox9.Text);

a0 $0$ = a1;

a0 $1$ = a2;

a0 $2$ = a3;

a0 $3$ = a4;

a0 $4$ = a5;

a0 $5$ = a6;

a0 $6$ = a7;

a0 $7$ = a8;

a0 $8$ = a9;

// layer.map $i$ .bias = randomNormalDistribution() / Math.Sqrt(9 / 2f);

for (int xx = 0; xx < 4; xx++)

{

wcnn的偏差 $0$ $xx$ = randomNormalDistribution() / Math.Sqrt(9 / 2f);

wcnn的偏差 $1$ $xx$ = randomNormalDistribution() / Math.Sqrt(4 / 2f);

}

button2init.Enabled = false;

}

private void button2Test_Click(object sender, EventArgs e)

{

a1 = (float)Convert.ToDouble(textBox1.Text);

a2 = (float)Convert.ToDouble(textBox2.Text);

a3 = (float)Convert.ToDouble(textBox3.Text);

a4 = (float)Convert.ToDouble(textBox4.Text);

a5 = (float)Convert.ToDouble(textBox5.Text);

a6 = (float)Convert.ToDouble(textBox6.Text);

a7 = (float)Convert.ToDouble(textBox7.Text);

a8 = (float)Convert.ToDouble(textBox8.Text);

a9 = (float)Convert.ToDouble(textBox9.Text);

a0 $0$ = a1;

a0 $1$ = a2;

a0 $2$ = a3;

a0 $3$ = a4;

a0 $4$ = a5;

a0 $5$ = a6;

a0 $6$ = a7;

a0 $7$ = a8;

a0 $8$ = a9;

forward();

double last偏差 = (1 - o) * (1 - o) * 1 / 2.0;

MessageBox.Show(last偏差.ToString());

}