统计学习方法——第二章的c++实现

1、东西搞丢了，只写一部分

2、算法那收敛性

证明逻辑：

a、γ是y_i*(w_opt*X_i)最小值

b、R是X_i最小值

c、k<=（R/γ）^2

难打公式，直接说，Wk由Wk-1迭代而来，所以事实上，只需要找到迭代第三项就可以了，迭代第三项有R有eta，Wk的迭代公式有，通过放缩，得到k，eta，R的关系。

与此同时，得到Wk与Wopt的关系，这里，Wopt是最终结果的W，同样，将Wk拆分，与上面a配合，得到eta与R 关系，结合上面这条，得证。

下面是C++代码

感知机原始代码：

#include <iostream>
using namespace std;

int main_2() {
    double x1[] = {,,};
    double x2[] = {,,};
    double y[] = {,,-};
    double w[] = {,};
    double b[] = {  };
    int index = ;
    double eta = ;
    while ( index <= ) {
        double temp = y[index] * (x1[index] * w[] + x2[index] *w[] + b[]);
        if (temp <= ) {
            w[] = w[] + eta * (x1[index])*y[index]; cout << w[] << endl;
            w[] = w[] + eta * (x2[index])*y[index]; cout << w[] << endl;
            b[] += y[index]; cout << b[] << endl;
            index = ;//这一步很重要，实现了每当出现问题，全部重新开始

            //cout << w[0] << "   " << w[1];
            //cout << b[0];

        }
        else
        {
            index = index+;
        }

    };
    cout << w[]<<"   "<<w[]<<endl;
    cout << b[];
    system("pause");
    return ;
}

感知机对偶代码：

 #include<iostream>

 using namespace std;
 int main()
 {
     int x1[] = { ,, };
     int x2[] = { ,, };
     int y[] = { ,,- };

     int b = ;
     int a[] = { ,, };
     int G[][];
     for (int i = ; i < ; i++)
         for (int j = ; j < ; j++)
         {
             G[i][j] = x1[i] * x1[j] + x2[i] * x2[j];
             cout << G[i][j] << "\t";
             if (j == )

         }

     int index = ;
     int eta = ;
     while (index < )
     {
         int temp = ;
         for (int j = ; j < ; j++)
         {
             temp += a[j] * y[j] * G[j][index];
         }
         int all = y[index] * (temp + b);
         if (all <= )
         {
             a[index] += eta;
             b +=  y[index];
             index = ;

         }
         else
         {
             index++;
         }

     }
     cout << "a:" << a[] << "   " << a[] << "   " << a[] << endl;
     cout << "b:" << b;

     int w[] = { , };
     for (int i = ; i < ; i++)
     {
         w[] += a[i] * y[i] * x1[i];
         w[] += a[i] * y[i] * x2[i];
     }
     b = ;
     for (int i = ; i < ; i++)
     {
         b += a[i] * y[i];
     }
     cout << endl << "w:" << w[] << "   " << w[] << endl;
     cout << "b:" << b;
     getchar();
     return ;
 }