无约束最优化方法可变单纯形法(simple)nelder-mead

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本程序是用Ｃ＋＋编写的，从编写的算例来看，应该是没有问题的。所采用的原理和步骤是参考华南理工大学出版社蒋金山等编写的最优化计算方法第８章第三节可变单纯形法。欢迎各位批评指正。

#include<> #include<> #include<> int i,j;

double d[3][100]={{0,1,0,0},{0,0,1,0},{0,0,0,1}},f[100];//d[][]为单纯形的顶点，本算例中未知数个数为3，则顶点个数为4

double g,h,l,q,s=1,t=2,u=,v=,y=0;//s为反射系数，t为扩展系数，u为压缩系数，v为允许误差 int o,F,r,D,e,lj=0,N=4;//N为顶点的个数，o为最大值点的位置，F为最小值点的位置，r为次大值点的位置

void function1(int e)//求解函数f[e] {

f[e]=(d[0][e]-3)*(d[0][e]-3)+2*(d[1][e]+2)*(d[1][e]+2)+(d[2][e]-4)*(d[2][e]-4);//函数为f=(x1-3)^2+2(x2+2)^2+(x3-4)^2,求其最小值 }

void function2() {

while((++lj)<100)//最大迭代次数

{ for(i=0,g=f[i];iif(g1

g=f[i+1];o=i+1; } else

if(i==0) o=i; }

for(i=0,h=f[i];iif(h>f[i+1]) {

h=f[i+1];F=i+1; } else if(i==0) F=i; }

for(i=0,l=f[i];iif(i==o&&i==0) {

l=f[i+1]; r=i+1; continue; }

if(ll=f[i+1];r=i+1;

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} else if(i==0) r=i; }

for(i=0;id[i][N]=0; for(j=0;jif(j!=o) {

d[i][N]=d[i][N]+d[i][j]; } }

d[i][N]=d[i][N]/(N-1);//平均值 }

function1(N);

for(i=0;id[i][N+1]=d[i][N]+(d[i][N]-d[i][o])*s;//反射 }

function1(N+1); if(f[N+1]for(i=0;i3

d[i][N+2]=d[i][N]+(d[i][N+1]-d[i][N])*t;//扩展 }

function1(N+2);

if(f[N+2]for(i=0;id[i][o]=d[i][N+2]; }

f[o]=f[N+2]; y=0;

for(i=0;iy=y+(f[i]-f[N])*(f[i]-f[N]); }

y=sqrt(y/(N*); if(yfor(i=0;iprintf(\"d[%d][%d]=%f\ printf(\"\\n\"); }

printf(\"f[%d]=%f\ break; }

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else {

function2(); break; }

} else {

for(i=0;id[i][o]=d[i][N+1]; }

f[o]=f[N+1]; y=0;

for(i=0;iy=y+(f[i]-f[N])*(f[i]-f[N]); }

y=sqrt(y/(N*); if(yfor(i=0;iprintf(\"d[%d][%d]=%f\ printf(\"\\n\"); }

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printf(\"f[%d]=%f\ break; } else {

function2(); break; } } }

if(f[N+1]>=f[F]&&f[N+1]<=f[r]) {

for(i=0;id[i][o]=d[i][N+1]; }

f[o]=f[N+1]; y=0;

for(i=0;iy=y+(f[i]-f[N])*(f[i]-f[N]); }

y=sqrt(y/(N*); if(yfor(i=0;i6

printf(\"d[%d][%d]=%f\ printf(\"\\n\"); }

printf(\"f[%d]=%f\ break; } else {

function2(); break; } }

if(f[N+1]>f[r]) {

if(f[N+1]>f[o])//压缩 { D=o; } else { D=N+1; } for(i=0;id[i][N+3]=d[i][N]+(d[i][D]-d[i][N])*u; }

function1(N+3);

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if(f[N+3]<=f[D]) {

for(i=0;id[i][o]=d[i][N+3]; }

f[o]=f[N+3]; y=0;

for(i=0;iy=y+(f[i]-f[N])*(f[i]-f[N]); }

y=sqrt(y/(N*); if(yfor(i=0;iprintf(\"d[%d][%d]=%f\ printf(\"\\n\"); }

printf(\"f[%d]=%f\ break; } else {

function2(); break;

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} } else {

for(i=0;ifor(j=0;jd[i][j]=d[i][j]+*(d[i][F]-d[i][j]);//收缩 } }

for(i=0;ifunction1(i); } y=0;

for(i=0;iy=y+(f[i]-f[N])*(f[i]-f[N]); }

y=sqrt(y/(N*); if(yfor(i=0;iprintf(\"d[%d][%d]=%f\ printf(\"\\n\");

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}

printf(\"f[%d]=%f\ break; } else {

function2(); break; } }

} } }

void main() {

for(i=0;ifunction1(i); }

function2();

}

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