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[] = {3,4,1};
    double x2[] = {3,3,1};
    double y[] = {1,1,-1};
    double w[] = {0,0};
    double b[] = { 0 };
    int index = 0;
    double eta = 1;
    while ( index <= 2) {
        double temp = y[index] * (x1[index] * w[0] + x2[index] *w[1] + b[0]);
        if (temp <= 0) {
            w[0] = w[0] + eta * (x1[index])*y[index]; cout << w[0] << endl;
            w[1] = w[1] + eta * (x2[index])*y[index]; cout << w[1] << endl;
            b[0] += y[index]; cout << b[0] << endl;
            index = 0;//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+1;
        }
        
    };
    cout << w[0]<<"   "<<w[1]<<endl;
    cout << b[0];
    system("pause");
    return 0;
}

感知机对偶代码：

#include<iostream>

using namespace std;
int main()
{
    int x1[] = { 3,4,1 };
    int x2[] = { 3,3,1 };
    int y[] = { 1,1,-1 };

    int b = 0;
    int a[] = { 0,0,0 };
    int G[3][3];
    for (int i = 0; i < 3; i++)
        for (int j = 0; j < 3; j++)
        {
            G[i][j] = x1[i] * x1[j] + x2[i] * x2[j];
            cout << G[i][j] << "	";
            if (j == 2)

        }

    int index = 0;
    int eta = 1;
    while (index < 3)
    {
        int temp = 0;
        for (int j = 0; j < 3; j++)
        {
            temp += a[j] * y[j] * G[j][index];
        }
        int all = y[index] * (temp + b);
        if (all <= 0)
        {
            a[index] += eta;
            b +=  y[index];
            index = 0;

        }
        else
        {
            index++;
        }

    }
    cout << "a:" << a[0] << "   " << a[1] << "   " << a[2] << endl;
    cout << "b:" << b;


    int w[] = { 0,0 };
    for (int i = 0; i < 3; i++)
    {
        w[0] += a[i] * y[i] * x1[i];
        w[1] += a[i] * y[i] * x2[i];
    }
    b = 0;
    for (int i = 0; i < 3; i++)
    {
        b += a[i] * y[i];
    }
    cout << endl << "w:" << w[0] << "   " << w[1] << endl;
    cout << "b:" << b;
    getchar();
    return 0;
}