C++实现：双输入单输出系统：建立一个 感知器网络，实现上述样本的分类。计算出相应的网络权值矩阵w。

练习题：

双输入单输出系统：

x1(1)=1 x2(1)=1 d(1)=1

x1(2)=-0.5 x2(2)=-1 d(2)=-1

x1(3)=3 x2(3)=1 d(3)=1

x1(4)=-2 x2(4)=-1 d(4)=-1

建立一个 感知器网络，实现上述样本的分类。

计算出相应的网络权值矩阵w。

说明：

用了opencv画图

代码

#include<iostream>
#include"math.h"
#include <opencv2/core/core.hpp>
//#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
//
//#include "linefinder.h"
//#include "edgedetector.h"

using namespace cv;
using namespace std;
int main()
{
double W[4][3];

for(int i=0;i<4;i++)
{
W[i][1]=0.5;
W[i][2]=0.5;
}

double X[4][3]={-1,1,1,
-1,-0.5,-1,
-1,3,1,
-1,-2,-1};
double d[4]={1,-1,1,-1};
double e[4];
double E[10]={0,0,0,0,0,0,0,0,0,0};
double b=5;
//W[][0]=b;
for(int i=0;i<4;i++)
W[i][0]=b;
double l=0.1;
/*W[0][1]=rand()%10*0.1;
W[0][2]=rand()%10*0.1;
W[1][1]=rand()%10*0.1;
W[1][2]=rand()%10*0.1;
W[2][1]=rand()%10*0.1;
W[2][2]=rand()%10*0.1;
W[3][1]=rand()%10*0.1;
W[3][2]=rand()%10*0.1;*/
cout<<W[0][1]<<endl<<W[0][2]<<endl;
cout<<W[1][1]<<endl<<W[1][2]<<endl;
cout<<W[2][1]<<endl<<W[2][2]<<endl;
cout<<W[3][1]<<endl<<W[3][2]<<endl;
double Y[4]={0};
for(int i=0;i<4;i++)
{
Y[i]=W[i][0]*X[i][0]+W[i][1]*X[i][1]+W[i][2]*X[i][2];
//cout<<"Y["<<i<<"]="<<Y[i]<<endl;
if(Y[i]>=0)
Y[i]=1;
else
Y[i]=-1;

e[i]=abs(d[i]-Y[i]);
}
E[0]=e[0]+e[1]+e[2]+e[3];
cout<<"E{0]="<<E[0]<<endl;
//cout<<"e1"<<e[i]<<endl;
int n=0;
while(E
>0)
{
n++;
for(int i=0;i<4;i++)
{

W[i][1]=W[i][1]+l*(d[i]-Y[i])*X[i][1];
W[i][2]=W[i][2]+l*(d[i]-Y[i])*X[i][2];
//cout<<W[i][1]<<endl<<W[i][2]<<endl;
Y[i]=W[i][0]*X[i][0]+W[i][1]*X[i][1]+W[i][2]*X[i][2];
if(Y[i]>=0)
Y[i]=1;
else
Y[i]=-1;

e[i]=abs(d[i]-Y[i]);
//cout<<"e"<<e[i]<<endl;
}

E
=e[0]+e[1]+e[2]+e[3];
/*cout<<"Yi"<<Y[i]<<endl;
cout<<"e["<<i<<"]="<<e[i]<<endl;*/
cout<<"E["<<n<<"]="<<E
<<endl;

}
cout<<"W[]="<<endl;
for(int i=0;i<4;i++)
{
for(int j=0;j<3;j++)
cout<<W[i][j]<<" ";
cout<<endl;
}
//E=e[0]+e[1]+e[2]+e[3];
cout<<"ok"<<endl;
/*IplImage *test=cvCreateImage(cvSize(400,400),IPL_DEPTH_16U,0);
Mat mtx(test);*/
Mat test(400,400,CV_8UC3,Scalar(130,240,205));//3通道图像
line(test,Point(20,400-20),Point(20,0),Scalar(0));//因为mat型图像的原点在右上角，所以比较麻烦，要通过计算转换
line(test,Point(20,400-20),Point(400,400-20),Scalar(0));
for(int i=0;i<10;i++)
{
line(test,Point(20*(i+1),380-20*E[i]),Point(20*(i+2),380-20*E[i+1]),Scalar(220,0,150));
}

/*for(int i=0;i<10;i++)
{
line(test,Point(20*(i+1),20*E[i]),Point(20*(i+2),20*E[i+1]),Scalar(255));
}*/

// Display the test image
namedWindow("E Image");
imshow("E Image",test);
imwrite("test.bmp",test);
waitKey();
return 0;
}