machine learning實驗7 矩陣求逆

#include"stdio.h"#include <math.h>double function(double matrix[3][6],double theta[3],int sample_i){        double ret=0.0;        for(int i=0;i<3;++i)        {                ret+=matrix[sample_i][i]*theta[i];        }        return ret;}double theta[3]={1,1,1};int main(void){        double matrix[3][6]={                        {1,2,3,1,0,0},                        {2,1,1,0,1,0},                        {3,1,2,0,0,1},                    };        double alfa = 0.1;        double c = 0;        double d = 1;        for(int z=0;z<3;++z)        {                double loss = 0.0;                for(int j = 0;j<3;++j)                {                        double sum=function(matrix,theta,j);                        loss += pow((pow(sum+c,d)-matrix[j][z+3]),2);                }//printf("loss : %lf\n",loss);                  for(int i=0;i<200;++i)                {                        for(int sample_i = 0; sample_i<3;sample_i++)                        {                                double result = function(matrix,theta,sample_i)+c;                                for(int j=0;j<3;++j)                                {                                        theta[j] = theta[j] - alfa*(pow(result,d)-matrix[sample_i][z+3])*d*pow((result),d-1)*matrix[sample_i][j];                                }                        }                        double loss = 0.0;                        for(int j = 0;j<3;++j)                        {                                double sum=function(matrix,theta,j);                                loss += pow((pow(sum+c,d)-matrix[j][z+3]),2);                        }                        //printf("%d,loss  now: %lf,%lf,%lf,%lf\n",i,loss,theta[0],theta[1],theta[2]);                }                printf("%lf,%lf,%lf\n",theta[0],theta[1],theta[2]);        }        return 0;}                             

以上代碼從實驗6稍微修改而來

求一個矩陣的逆陣，如下例：

 1 2 3  x1 y1 z1   1 0 0
 2 1 1  x2 y2 z2   0 1 0
 3 1 2  x3 y3 z3   0 0 1

精確結果應為：
        -1/2   1/2     1/2
        1/2    7/4     5/4 
        1/2   -3/2     3/4

迭代的結果為

-0.249835,0.250747,0.249447
0.247973,1.740834,-1.243221
0.253544,-1.233976,0.738149