黄金分割法

七星北斗 · 发表于 2008-4-22 08:27:24

#include <stdio.h>
#include <conio.h>
#include <math.h>
#define e 0.001
#define tt 0.01
float function(float x )
{

float y=9*pow(x,3)-7*pow(x,2)-4*x+2;

return (y);

}

void finding(float a[3],float f[3])
{
int i;

float t=tt,a1,f1,ia;

a[0]=0;

f[0]=function(a[0]);

for( i=0;;i++)
{
a[1]=a[0]+t;f[1]=function(a[1]);

if(f[1]<f[0])
break;

if(fabs(f[1]-f[0])>=e)
{
t=-t;
a[0]=a[1]; f[0]=f[1];
}

else {
if(ia==1)

return ;

t=t/2;
ia=1;

}

}
for(i=0;;i++)
{

a[2]=a[1]+t; f[2]=function(a[2]);

if(f[2]>f[1])
break ;

t=2*t;

a[0]=a[1]; f[0]=f[1];

a[1]=a[2]; f[1]=f[2];
}

if(a[0]>a[2])
{

a1=a[0]; f1=f[0];

a[0]=a[2]; f[0]=f[2];

a[2]=a1;f[2]=f1;
}

return ;
}

float gold (float *ff)
{ int i;

float a1[3],f1[3],a[4],f[4];

float aa ;

finding(a1,f1);

a[0]=a1[0]; f[0]=f1[0];

a[3]=a1[2]; f[3]=f1[2];

a[1]=a[0]+0.382*(a[3]-a[0]);

a[2]=a[0]+0.618*(a[3]-a[0]);

f[1]=function(a[1]);

f[2]=function(a[2]);

for( i=0;;i++)

{ if(f[1]>=f[2])

{ a[0]=a[1]; f[0]=f[1];

a[1]=a[2]; f[1]=f[2];

a[2]=a[0]+0.618*(a[3]-a[0]);

f[2]=function(a[2]);

}

else { a[3]=a[2]; f[3]=f[2];

a[2]=a[1]; f[2]=f[1];

a[1]=a[0]+0.382*(a[3]-a[0]); f[1]=function(a[1]);

}

if((a[3]-a[0])<e)

{ aa=(a[1]+a[2])/2; *ff=function(aa);

break ;

}

}

return ;

}

void main()

{
float xx, ff;

xx=gold(&ff);

printf("\nThe Optimal Design Result Is: \n");

printf("\n\tx*=%f\n\tf*=%f",xx,ff);

getch();

}

#include <stdio.h>
#include <math.h>
#define e 0.001
#define tt 0.001
float f(double x)
{
float y=9*pow(x,4)-7*pow(x,3)-4*pow(x,2) +2;
return(y);
}
finding(float *p1,float*p2)
{
float x1=0,x2,x3,t,f1,f2,f3,h=tt;
int n=0;
x2=x1+h;f1=f(x1);f2=f(x2);
if(f2>f1) {h=-h;t=x2;x2=x1;x1=t;}
do

/ Z* K# ~9 p6 x2 F{ x3=x2+h;h=2*h;f3=f(x3);n=n+1;}
while(f3<f2);
if(x1>x3) {t=x1;x1=x3;x3=t;}
*p1=x1;*p2=x3;
return(n);
}
gold(float *p)
{
float a,b,x1,x2,f1,f2;
int n=0;
finding(&a,&b);
do
( h! I' f1 `6 `1 p6 ?1 A# d; N: f
{x1=a+0.382*(b-a);
3 F: X, \7 W F, h; O8 N
x2=a+0.618*(b-a);f1=f(x1);f2=f(x2);n=n+1;

0 o6 X1 J# J. A& I, uif(f1>f2) a=x1;

0 b& r& ^( G2 a: M. P- P, P0 R; Q3 K2 velse b=x2;}
while((b-a)>e);
*p=(x1+x2)/2;return(n);
}

main()
{
float a,b,x,min;int n1,n2;
n1=finding(&a,&b);
n2=gold(&x);
min=f(x);
printf("\n The area is %f to %f.",a,b); printf("\n The number 1 is %d.",n1);
printf("\n The min is %f and the result is %f.",x,min);
printf("\n The number 2 is %d.",n2);}

输出结果

The area is 0.000000 to 1.290000.

The number 1 is 8.

The min is 0.845896 and the result is -0.491089.

The number 2 is 15

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