最优化-求解多元函数的最优解源码C

#include <stdio.h> 
#include <stdlib.h> 
#include <math.h> 
double det=0.0001;         
double ak=0.003;              
double f(double *x)                  
{ 
 double y; 
 y=x[0]*x[0]+2*x[1]*x[1]-4*x[0]-2*x[0]*x[1]; 
 return y; 
 } 
void fun(double *x_k1,double *x_k,double *s_k,double a,int n)      
{ 
    int i; 
 for(i=0;i<n;i++) 
  x_k1[i]=x_k[i]+a*s_k[i]; 
} 
 void Get_Search_area(int n,double *x_k,double *s_k,double *a1,double *a3) 
{                                                                    
 double a=ak,s,f0,a2=0; 
 double f1=0,f2=0,f3=0; 
 double *x_k1; 
 x_k1=(double*)malloc(n*sizeof(double)); 
 f2=f(x_k); 
 while(1) 
 { 
  *a1=a2+a; 
  fun(x_k1,x_k,s_k,*a1,n); 
     f1=f(x_k1); 
     if(f1>f2) 
  { 
   if(a==ak) 
   { 
    *a3=*a1; 
    f3=f1; 
    a=-a; 
    continue; 
   } 
   else if(*a1>*a3) 
    { 
     s=*a1; 
     *a1=*a3; 
     *a3=s; 
     f0=f1; 
     f1=f3; 
     f3=f0; 
     break; 
    } 
   else break; 
  } 
  else  
  { 
   *a3=a2; 
   a2=*a1; 
   f3=f2; 
   f2=f1; 
   a=2*a; 
   continue; 
  } 
 } 
}  
void Gold_division(int n,double a1,double a4,double *x_k,double *s_k) 
{ 
 double a2,a3,f2,f3,a_star; 
 double *x_k2,*x_k3; int i; 
 x_k2=(double*)malloc(n*sizeof(double)); 
 x_k3=(double*)malloc(n*sizeof(double));  
 for(i=0;i<n;i++) 
 { 
  x_k2[i]=0; 
  x_k3[i]=0; 
 } 
 a2=a1+0.382*(a4-a1); 
 fun(x_k2,x_k,s_k,a2,n); 
 f2=f(x_k2); 
 a3=a1+0.618*(a4-a1); 
 fun(x_k3,x_k,s_k,a3,n); 
 f3=f(x_k3);  
 do 
 { 
  if(f2<=f3) 
  { 
   a4=a3; 
   a3=a2; 
   f3=f2; 
   a2=a1+0.382*(a4-a1); 
   fun(x_k2,x_k,s_k,a2,n); 
   f2=f(x_k2);
  } 
  else 
  { 
   a1=a2; 
   a2=a3; 
   f2=f3; 
   a3=a1+0.618*(a4-a1); 
         fun(x_k3,x_k,s_k,a3,n); 
   f3=f(x_k3); 
  } 
 } while(fabs(a4-a1)>det);  
  a_star=(a1+a4)*0.5; 
  fun(x_k,x_k,s_k,a_star,n); 
 free(x_k2); 
 free(x_k3); 
} 
void lunhuan(int n,double *x_k,double *f0) 
{ 
 double sum; 
 double *a_star=0,*f_star=0,*x0; 
 double **s_k; 
    double a1=0,a3=0; 
 double *a_1,*a_3; 
 int i,j; 
 a_1=&a1; 
 a_3=&a3; 
 x0=(double*)malloc(n* sizeof(double)); 
 s_k=(double**)malloc(n*sizeof(double*)); 
 for(i=0;i<n;i++) 
 { 
  s_k[i]=(double*)malloc(n* sizeof(double)); 
  for(j=0;j<n;j++) 
   if(i==j) 
    s_k[i][j]=1; 
   else 
    s_k[i][j]=0; 
 } 
 while(1) 
 { 
  for(i=0;i<n;i++) 
   x0[i]=x_k[i];  
  for(i=0;i<n;i++)  
  { 
   Get_Search_area(n,x_k,s_k[i],a_1,a_3);  
      Gold_division(n,*a_1,*a_3,x_k,s_k[i]); 
  } 
  sum=0; 
  for(i=0;i<n;i++) 
   sum+=(x0[i]-x_k[i])*(x0[i]-x_k[i]); 
  if(sqrt(sum)<=det) 
   break; 
 } 
 *f0=f(x_k); 
 free(x0); 
 for(i=0;i<n;i++) 
 { 
  free(s_k[i]); 
 } 
 free(s_k); 
}  
int main() 
{ 
 double *x_k,*f; 
 double f0=0;
 int n=2,i; 
 f=&f0; 
 puts("原函数为f=(x[0]-2)*(x[0]-2)+(x[1]-3)*(x[1]-3);\n"); 
 x_k=(double*)malloc(n*sizeof(double)); 
 printf("输入起始坐标x_k:\n"); 
 for(i=0;i<n;i++) 
  scanf("%lf",&x_k[i]); 
  lunhuan(n,x_k,f); 
 puts("函数极小点坐标:\n"); 
 for(i=0;i<n;i++) 
  printf("%lf;",x_k[i]); 
 printf("\n"); 
 puts("函数最小值:\n"); 
 printf("%lf\n",*f); 
 free(x_k); 
 return 0; 
}