0.618法(golden ratio search)

 最优化方法课程里面的，照书上的方法编了一个程序。

 

#include < iostream >
#include < cmath >
using   namespace  std;
const   double  delta = 0.00001 ;
// the function
double  func( double  x)
... {
                   return exp(x)+exp(-x);
}
int  main()
... {
    //the interval [a,b]
    double a=-1,b=1;
    double s=a+0.382*(b-a);
    double t=a+0.618*(b-a);
    double f1=func(s);
    double f2=func(t);
    int k=0;
    while(true)...{
        k++;
        printf("f(a=%6.2lf, s=%6.2lf, t=%6.2lf, b=%6.2lf) = (%6.4lf, %6.4lf, %6.4lf, %6.4lf) ",a,s,t,b,func(a),func(s),func(t),func(b));
        if(f1>f2)...{
            if(b-s<=delta)...{
                printf("******************************************************* ");
                printf("after %d iterators ",k);
                printf("when x= %.2lf, f(x)= %.2lf ",t,func(t));
                break;
            }
            else...{
                a=s;s=t;
                f1=func(t);
                t=a+0.618*(b-a);
                f2=func(t);
            }
        }
        else...{
            if(t-a<=delta)...{
                printf("******************************************************* ");
                printf("after %d iterators ",k);
                printf("when x= %.2lf, f(x)= %.2lf ",s,func(s));
                break;
            }
            else...{
                b=t;t=s;
                f2=func(s);
                s=a+0.382*(b-a);
                f1=func(s);
            }
        }
    }
    return 0;
}