香蕉函数的各类最优解法

g(1,1)=-4*(x2-x1^2)*x1;
g(2,1)=4*x2-2*x1^2-2;
q =[12*x1^2-4*x2,-4*x1;-4*x1,4];
%梯度的模
g_mo(k)=sqrt(g(1,1)^2+g(2,1)^2);
while(g_mo(k)>eps)
l=l+1;
%确定搜索方向
p(:,k)=-inv(q)*g(:,k);
f2=t2*t2+2*t2;
}
}
t=(t1+t2)/2.0;
printf("f(x)=x*x+2*x minmum point is:");
printf("%f\n",t);
printf("f(x) minmum is:%f\n",t*t+2*t);
getch();
}
二分法
#include<stdio.h>
%用牛顿法求香蕉函数minf=(x2-x1^2)^2+(x2-1)^2
X(:,1)=[2,2]';
eps=0.00001;
k=1;
l=0%迭代的次数
x1=X(1,1);x2=X(2,1);
%f=(x2-x1^2)^2+(x2-1)^2
f_val(k)=(x2-x1^2)^2+(x2-1)^2;
%求梯度和hesse矩阵
if(g_mo(k)<eps)
X(:,k);
end
p(:,1)=-g(:,1);
l=0;%迭代的次数
while(1)
l=l+1;
%一维搜索
t=sym('t');
x1=X(1,k)+t*p(1,k);
x2=X(2,k)+t*p(2,k);
f=2*x1^2+x2^2-x1*x2;
s=diff(f,t);
#include<math.h>
main()
{float a,b,c,fc,t,eps;
printf("please input a,b,eps\n ");
scanf("%f,%f,%f",&a,&b,&eps);
while(b-a>eps)
{c=(a+b)/2.0;
fc=2*c+2;
if(fc<0) a=c;
g_mo(k+1)=sqrt(g(1,k+1)^2+g(2,k+1)^2);
if(g_mo(k+1)<eps)
break;
end
if(k==n)
%从Xn+1重新迭代
p(:,1)=-g(:,k+1);
g(:,1)=g(:,k+1);
X(:,1)=X(:,k+1);
H=eye(2,2);
k=1;
else
else if(fine(ti)>fine(t0)) t1=ti;
else { t1=ti;t2=t0,t0=(t1+t2)/2;}
}
else if(ti>t0)
{
if(fine(ti)<fine(t0)) { t1=t0;t0=ti;}
else if(fine(ti)>fine(t0)) t2=ti;
x0=[1,1]';
Q=[2,0;0,8];
g=[2,8]';
g0=g.*x0;
x1=x0-(g0'*g0)/(g0'*Q*g0)*g0;
g1=g.*x1;
h0=eye(2,2);
g2=g1;
eps=10^(-6);
k=1;
while g2'*g2>eps
g_mo(k)=sqrt(g(1,1)^2+g(2,1)^2);
if(g_mo(k)<eps)
X(:,k);
end
p(:,1)=-g(:,1);
l=0;
while(1)
l=l+1;
%一维搜索
t=sym('t');
x1=X(1,k)+t*p(1,k);
x2=X(2,k)+t*p(2,k);
f=(x2-x1^2)^2+(x2-1)^2
}
main()
{
float x[2]={2,2},e;
printf("\nInput e\n");
scanf("%f",&e);
printf("\n\tx[2]\t\t\t f(x)\t\t\t g\n");
newton(x,e);
getch();
}
抛物线法
#include <stdio.h>
#include <math.h>
minf(X)=(x2-x1^2)^2+(x2-1)^2
clear
X(:,1)=[2,2]';
eps=0.00001;
k=1;
n=2;
H=eye(2,2);
x1=X(1,1);x2=X(2,1);
%f(X)=(x2-x1^2)^2+(x2-1)^2
g(1,1)=-4*(x2-x1^2)*x1;
g(2,1)=4*x2-2*x1^2-2;
float g0[2],b[2]={0,0},q;
int i,j;
g0[0]=2*x[0]-10-x[1];
g0[1]=2*x[1]-4-x[0];
printf("(%f\t%f)\t%f\t(%f\t%f)\n",x[0],x[1],f(x),g0[0],g0[1]);
q=sqrt(pow(g0[0],2)+pow(g0[1],2));
%用最速下降法求香蕉函数minf=(x2-x1^2)^2+(x2-1)^2
X(:,1)=[2,2]';
eps=0.00001;
k=1;
l=0;
x1=X(1,1);x2=X(2,1);
%f=(x2-x1^2)^2+(x2-1)^2
f_val(k)=(x2-x1^2)^2+(x2-1)^2;
%求梯度和hesse矩阵
scanf("%f",&e);
printf("\nInput t0 in [0,2]\n") ;
scanf("%f",&t0);
printf("t1\t\t t2\t\t t0\t\t ti\n");
jixiao(t1,t2,t0,e);
getch();
}
%min f(x)=x1^2+4*x2^2 用变尺度法求解
if(g_mo(k+1)<eps)
X(:,k+1);
break;
end
if(k==n)
%从Xn+1重新迭代
p(:,1)=-g(:,k+1);
X(:,1)=X(:,k+1);
k=1;
else
%构造新的迭代方向
tt=(g_mo(k+1)/g_mo(k))^2;
p(:,k+1)=-g(:,k+1)+tt*p(:,k);
%一维搜索
t=sym('t');
x1=X(1,k)+t*p(1,k);
x2=X(2,k)+t*p(2,k);
f=(x2-x1^2)^2+(x2-1)^2;
s=diff(f,t);
s1=finverse(s);
t=subs(s1,'0',t);
s1=0;
X(:,k+1)=X(:,k)+t*p(:,k);
s1=diff(f,t);
s2=finverse(s1);
t=subs(s2,'0',t);
S1=0;
s2=0;
%迭代
X(:,k+1)=X(:,k)+t*p(:,k);
x1=X(1,k+1);x2=X(2,k+1);
g(1,k+1)=-4*(x2-x1^2)*x1;
g(2,k+1)=4*x2-2*x1^2-2;
#include <math.h>
float f(float *x)
{
float fi;
fi=60-10*x[0]-4*x[1]+pow(x[0],2)+pow(x[1],2)-x[0]*x[1];
return (fi);
}
newton(float *x,float e)
{
float H[2][2]={{2,-1},{-1,2}},H0[2][2]={{2.0/3.0,1.0/3.0},{1.0/3,2.0/3.0}};
x1=X(1,k+1);x2=X(2,k+1);
f_val(k+1)=(x2-x1^2)^2+(x2-1)^2;
g(1,1)=-4*(x2-x1^2)*x1;
g(2,1)=4*x2-2*x1^2-2;
g_mo=sqrt(g(1,1)^2+g(2,1)^2);
k=k+1;
end
X(:,k)
l
%用DFP法求香蕉函数
if(q>e)
{
for(i=0;i<2;i++)