非线性方程与方程组的求解

原创
2010/03/26 16:01
阅读数 819

#include "stdafx.h"
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
#include "NonLinearEquation.h"
#include "MatrixAlgo.h"
#include "LinearEquation.h"
#include "MatrixEigenvalue.h"

//求非线性方程一个实根的连分式解法
int dpqrt(double *x, double eps)
{
 int i,j,m,it,l;
 double a[10],y[10],z,h,x0,q;
 l=10; q=1.0e+35; x0=*x; h=0.0;
 while (l!=0)
 {
  l=l-1;
  j=0;
  it=l;
  while (j<=7)
  {
   if (j<=2)
    z=x0+0.1*j;
   else
    z=h;
   y[j]=dpqrtf(z);
   h=z;
   if (j==0)
    a[0]=z;
   else
   {
    m=0;
    i=0;
    while ((m==0)&&(i<=j-1))
    {
     if (fabs(h-a[i])+1.0==1.0)
      m=1;
     else
      h=(y[j]-y[i])/(h-a[i]);
     i=i+1;
    }
    a[j]=h;
    if (m!=0)
     a[j]=q;
    h=0.0;
    for (i=j-1; i>=0; i--)
    {
     if (fabs(a[i+1]+h)+1.0==1.0)
      h=q;
     else
      h=-y[i]/(a[i+1]+h);
    }
    h=h+a[0];
   }
   if (fabs(y[j])>=eps)
    j=j+1;
   else
   {
    j=10;
    l=0;
   }
  }
  x0=h;
 }
 *x=h;
 return(10-it);
}


//求非线性方程一个实根的牛顿法
int dnewt(double *x, double eps, int js)
{
 int k,l;
 double y[2],d,p,x0,x1;
 l=js;
 k=1;
 x0=*x;
 dnewtf(x0,y);
 d=eps+1.0;
 while ((d>=eps)&&(l!=0))
 {
  if (fabs(y[1])+1.0==1.0)
  {
   printf("err\n");
   return(-1);
  }
  x1=x0-y[0]/y[1];
  dnewtf(x1,y);
  d=fabs(x1-x0);
  p=fabs(y[0]);
  if (p>d)
   d=p;
  x0=x1;
  l=l-1;
 }
 *x=x1;
 k=js-l;
 return(k);
}


//求复系数代数方程全部根的牛顿-下山法
int dcsrt(double ar[], double ai[], int n, double xr[], double xi[])
{
 int m,i,jt,k,is,it;
 double t,x,y,x1,y1,dx,dy,p,q,w,dd,dc,c;
 double g,u,v,pq,g1,u1,v1;
 m=n;
 p=sqrt(ar[m]*ar[m]+ai[m]*ai[m]);
 while ((m>0)&&(p+1.0==1.0))
 {
  m=m-1;
  p=sqrt(ar[m]*ar[m]+ai[m]*ai[m]);
 }
 if (m<=0)
 {
  printf("fail\n");
  return(-1);
 }
 for (i=0; i<=m; i++)
 {
  ar[i]=ar[i]/p;
  ai[i]=ai[i]/p;
 }
 for (i=0; i<=m/2; i++)
 {
  w=ar[i];
  ar[i]=ar[m-i];
  ar[m-i]=w;
  w=ai[i];
  ai[i]=ai[m-i];
  ai[m-i]=w;
 }
 k=m; is=0; w=1.0;
 jt=1;
 while (jt==1)
 {
  pq=sqrt(ar[k]*ar[k]+ai[k]*ai[k]);
  while (pq<1.0e-12)
  {
   xr[k-1]=0.0;
   xi[k-1]=0.0;
   k=k-1;
   if (k==1)
   {
    p=ar[0]*ar[0]+ai[0]*ai[0];
    xr[0]=-w*(ar[0]*ar[1]+ai[0]*ai[1])/p;
    xi[0]=w*(ar[1]*ai[0]-ar[0]*ai[1])/p;
    return(1);
   }
   pq=sqrt(ar[k]*ar[k]+ai[k]*ai[k]);
  }
  q=log(pq);
  q=q/(1.0*k);
  q=exp(q);
  p=q;
  w=w*p;
  for (i=1; i<=k; i++)
  {
   ar[i]=ar[i]/q;
   ai[i]=ai[i]/q;
   q=q*p;
  }
  x=0.0001;
  x1=x;
  y=0.2;
  y1=y;
  dx=1.0;
  g=1.0e+37;
l40:
  u=ar[0];
  v=ai[0];
  for (i=1; i<=k; i++)
  {
   p=u*x1;
   q=v*y1;
   pq=(u+v)*(x1+y1);
   u=p-q+ar[i];
   v=pq-p-q+ai[i];
  }
  g1=u*u+v*v;
  if (g1>=g)
  {
   if (is!=0)
   {
    it=1;
    g65(&x,&y,&x1,&y1,&dx,&dy,&dd,&dc,&c,&k,&is,&it);
    if (it==0)
     goto l40;
   }
   else
   {
    g60(&t,&x,&y,&x1,&y1,&dx,&dy,&p,&q,&k,&it);
    if (t>=1.0e-03)
     goto l40;
    if (g>1.0e-18)
    {
     it=0;
     g65(&x,&y,&x1,&y1,&dx,&dy,&dd,&dc,&c,&k,&is,&it);
     if (it==0)
      goto l40;
    }
   }
   g90(xr,xi,ar,ai,&x,&y,&p,&w,&k);
  }
  else
  {
   g=g1;
   x=x1;
   y=y1;
   is=0;
   if (g<=1.0e-22)
    g90(xr,xi,ar,ai,&x,&y,&p,&w,&k);
   else
   {
    u1=k*ar[0];
    v1=ai[0];
    for (i=2; i<=k; i++)
    {
     p=u1*x;
     q=v1*y;
     pq=(u1+v1)*(x+y);
     u1=p-q+(k-i+1)*ar[i-1];
     v1=pq-p-q+(k-i+1)*ai[i-1];
    }
    p=u1*u1+v1*v1;
    if (p<=1.0e-20)
    {
     it=0;
     g65(&x,&y,&x1,&y1,&dx,&dy,&dd,&dc,
     &c,&k,&is,&it);
     if (it==0)
      goto l40;
     g90(xr,xi,ar,ai,&x,&y,&p,&w,&k);
    }
    else
    {
     dx=(u*u1+v*v1)/p;
     dy=(u1*v-v1*u)/p;
     t=1.0+4.0/k;
     g60(&t,&x,&y,&x1,&y1,&dx,&dy,&p,&q,&k,&it);
     if (t>=1.0e-03)
      goto l40;
     if (g>1.0e-18)
     {
      it=0;
      g65(&x,&y,&x1,&y1,&dx,&dy,&dd,&dc,&c,&k,&is,&it);
      if (it==0)
       goto l40;
     }
     g90(xr,xi,ar,ai,&x,&y,&p,&w,&k);
    }
   }
  }
  if (k==1)
   jt=0;
  else
   jt=1;
 }
 return(1);
}

void g60(double *t, double *x, double *y, double *x1, double *y1, double *dx, double *dy, double *p, double *q, int *k, int *it)
{
 *it=1;
 while (*it==1)
 {
  *t=*t/1.67;
  *it=0;
  *x1=*x-*t*(*dx);
  *y1=*y-*t*(*dy);
  if (*k>=30)
  {
   *p=sqrt(*x1*(*x1)+*y1*(*y1));
   *q=exp(75.0/(*k));
   if (*p>=*q)
    *it=1;
  }
 }
 return;
}

void g90(double xr[], double xi[], double ar[], double ai[], double *x, double *y, double *p, double *w, int *k)
{
 int i;
 for (i=1; i<=*k; i++)
 {
  ar[i]=ar[i]+ar[i-1]*(*x)-ai[i-1]*(*y);
  ai[i]=ai[i]+ar[i-1]*(*y)+ai[i-1]*(*x);
 }
 xr[*k-1]=*x*(*w);
 xi[*k-1]=*y*(*w);
 *k=*k-1;
 if (*k==1)
 {
  *p=ar[0]*ar[0]+ai[0]*ai[0];
  xr[0]=-*w*(ar[0]*ar[1]+ai[0]*ai[1])/(*p);
  xi[0]=*w*(ar[1]*ai[0]-ar[0]*ai[1])/(*p);
 }
 return;
}

void g65(double *x, double *y, double *x1, double *y1, double *dx, double *dy, double *dd, double *dc, double *c, int *k, int *is, int *it)
{
 if (*it==0)
 {
  *is=1;
  *dd=sqrt(*dx*(*dx)+*dy*(*dy));
  if (*dd>1.0)
   *dd=1.0;
  *dc=6.28/(4.5*(*k));
  *c=0.0;
 }
 while(1==1)
 {
  *c=*c+*dc;
  *dx=*dd*cos(*c);
  *dy=*dd*sin(*c);
  *x1=*x+*dx;
  *y1=*y+*dy;
  if (*c<=6.29)
  {
   *it=0;
   return;
  }
  *dd=*dd/1.67;
  if (*dd<=1.0e-07)
  {
   *it=1;
   return;
  }
  *c=0.0;
 }
}

//求实系数代数方程全部报的牛顿-下山法
int dsrrt(double a[], int n, double xr[], double xi[])
{
 int m,i,jt,k,is,it;
 double t,x,y,x1,y1,dx,dy,p,q,w,dd,dc,c;
 double g,u,v,pq,g1,u1,v1;
 m=n;
 while ((m>0)&&(fabs(a[m])+1.0==1.0))
  m=m-1;
 if (m<=0)
 {
  printf("fail\n");
  return(-1);
 }
 for (i=0; i<=m; i++)
  a[i]=a[i]/a[m];
 for (i=0; i<=m/2; i++)
 {
  w=a[i];
  a[i]=a[m-i];
  a[m-i]=w;
 }
 k=m;
 is=0;
 w=1.0;
 jt=1;
 while (jt==1)
 {
  pq=fabs(a[k]);
  while (pq<1.0e-12)
  {
   xr[k-1]=0.0;
   xi[k-1]=0.0;
   k=k-1;
   if (k==1)
   {
    xr[0]=-a[1]*w/a[0];
    xi[0]=0.0;
    return(1);
   }
   pq=fabs(a[k]);
  }
  q=log(pq);
  q=q/(1.0*k);
  q=exp(q);
  p=q; w=w*p;
  for (i=1; i<=k; i++)
  {
   a[i]=a[i]/q;
   q=q*p;
  }
  x=0.0001;
  x1=x;
  y=0.2;
  y1=y;
  dx=1.0;
  g=1.0e+37;
l40:
  u=a[0];
  v=0.0;
  for (i=1; i<=k; i++)
  {
   p=u*x1; q=v*y1;
   pq=(u+v)*(x1+y1);
   u=p-q+a[i];
   v=pq-p-q;
  }
  g1=u*u+v*v;
  if (g1>=g)
  {
   if (is!=0)
   {
    it=1;
    g65(&x,&y,&x1,&y1,&dx,&dy,&dd,&dc,&c,&k,&is,&it);
    if (it==0)
     goto l40;
   }
   else
   {
    g60(&t,&x,&y,&x1,&y1,&dx,&dy,&p,&q,&k,&it);
    if (t>=1.0e-03)
     goto l40;
    if (g>1.0e-18)
    {
     it=0;
     g65(&x,&y,&x1,&y1,&dx,&dy,&dd,&dc,&c,&k,&is,&it);
     if (it==0)
      goto l40;
    }
   }
   g90(xr,xi,a,&x,&y,&p,&q,&w,&k);
  }
  else
  {
   g=g1;
   x=x1;
   y=y1;
   is=0;
   if (g<=1.0e-22)
    g90(xr,xi,a,&x,&y,&p,&q,&w,&k);
   else
   {
    u1=k*a[0];
    v1=0.0;
    for (i=2; i<=k; i++)
    {
     p=u1*x;
     q=v1*y;
     pq=(u1+v1)*(x+y);
     u1=p-q+(k-i+1)*a[i-1];
     v1=pq-p-q;
    }
    p=u1*u1+v1*v1;
    if (p<=1.0e-20)
    {
     it=0;
     g65(&x,&y,&x1,&y1,&dx,&dy,&dd,&dc,&c,&k,&is,&it);
     if (it==0)
      goto l40;
     g90(xr,xi,a,&x,&y,&p,&q,&w,&k);
    }
    else
    {
     dx=(u*u1+v*v1)/p;
     dy=(u1*v-v1*u)/p;
     t=1.0+4.0/k;
     g60(&t,&x,&y,&x1,&y1,&dx,&dy,&p,&q,&k,&it);
     if (t>=1.0e-03)
      goto l40;
     if (g>1.0e-18)
     {
      it=0;
      g65(&x,&y,&x1,&y1,&dx,&dy,&dd,&dc,&c,&k,&is,&it);
      if (it==0)
       goto l40;
     }
     g90(xr,xi,a,&x,&y,&p,&q,&w,&k);
    }
   }
  }
  if (k==1)
   jt=0;
  else
   jt=1;
 }
 return(1);
}

//求非线性方程组一组实根的蒙特卡络法
void dnmtc(double x[], int n, double b, int m, double eps)
{
 int k,i;
 double a,r,*y,z,z1;
 y=(double *)malloc(n*sizeof(double));
 a=b;
 k=1;
 r=1.0;
 z=dnmtcf(x,n);
 while (a>=eps)
 {
  for (i=0; i<=n-1; i++)
   y[i]=-a+2.0*a*mrnd1(&r)+x[i];
  z1=dnmtcf(y,n);
  k=k+1;
  if (z1>=z)
  {
   if (k>m)
   {
    k=1;
    a=a/2.0;
   }
  }
  else
  {
   k=1;
   for (i=0; i<=n-1; i++)
    x[i]=y[i];
   z=z1;
   if (z<eps)
    return;
  }
 }
 return;
}

//求非线性方程组一组实根的拟牛顿法
int dnetn(int n, double eps, double t, double h, double x[], int k)
{
 int i,j,l;
 double am,z,beta,d,*y,*a,*b;
 y=(double *)malloc(n*sizeof(double));
 a=(double *)malloc(n*n*sizeof(double));
 b=(double *)malloc(n*sizeof(double));
 l=k; am=1.0+eps;
 while (am>=eps)
 {
  dnetnf(x,b,n);
  am=0.0;
  for (i=0; i<=n-1; i++)
  {
   z=fabs(b[i]);
  if (z>am) am=z;
  }
  if (am>=eps)
  {
   l=l-1;
   if (l==0)
   {
    free(y);
    free(b);
    free(a);
    printf("fail\n");
    return(0);
   }
   for (j=0; j<=n-1; j++)
   {
    z=x[j];
    x[j]=x[j]+h;
    dnetnf(x,y,n);
    for (i=0; i<=n-1; i++)
     a[i*n+j]=y[i];
    x[j]=z;
   }
   if (agaus(a,b,n)==0)
   {
    free(y);
    free(a);
    free(b);
    return(-1);
   }
   beta=1.0;
   for (i=0; i<=n-1; i++)
    beta=beta-b[i];
   if (fabs(beta)+1.0==1.0)
   {
    free(y);
    free(a);
    free(b);
    printf("fail\n");
    return(-2);
   }
   d=h/beta;
   for (i=0; i<=n-1; i++)
    x[i]=x[i]-d*b[i];
   h=t*h;
  }
 }
 free(y);
 free(a);
 free(b);
 return(k-l);
}

//求非线性方程组一组实根的梯度法
int dsnse(int n, double eps, double x[], int js)
{
 int l,j;
 double f,d,s,*y;
 y=(double *)malloc(n*sizeof(double));
 l=js;
 f=dsnsef(x,y,n);
 while (f>=eps)
 {
  l=l-1;
  if (l==0)
  {
   free(y);
   return(js);
  }
  d=0.0;
  for (j=0; j<=n-1; j++)
   d=d+y[j]*y[j];
  if (d+1.0==1.0)
  {
   free(y);
   return(-1);
  }
  s=f/d;
  for (j=0; j<=n-1; j++)
   x[j]=x[j]-s*y[j];
  f=dsnsef(x,y,n);
 }
 free(y);
 return(js-l);
}

//求非线性方程一个实根的蒙特卡洛法
void dmtcl(double *x, double b, int m, double eps)
{
 int k;
 double xx,a,r,y,x1,y1;
 a=b;
 k=1;
 r=1.0;
 xx=*x;
 y=dmtclf(xx);
 while (a>=eps)
 {
  x1=mrnd1(&r);
  x1=-a+2.0*a*x1;
  x1=xx+x1;
  y1=dmtclf(x1);
  k=k+1;
  if (fabs(y1)>=fabs(y))
  {
   if (k>m)
   {
    k=1;
    a=a/2.0;
   }
  }
  else
  {
   k=1;
   xx=x1;
   y=y1;
   if (fabs(y)<eps)
   {
    *x=xx;
    return;
   }
  }
 }
 *x=xx;
 return;
}

//求实系数代数方程全部根的QR方法
int dqrrt(double a[], int n, double xr[], double xi[], double eps, int jt)
{
 int i,j;
 double *q;
 q=(double *)malloc(n*n*sizeof(double));
 for (j=0; j<=n-1; j++)
  q[j]=-a[n-j-1]/a[n];
 for (j=n; j<=n*n-1; j++)
  q[j]=0.0;
 for (i=0; i<=n-2; i++)
  q[(i+1)*n+i]=1.0;
 i=chhqr(q,n,xr,xi,eps,jt);
 free(q);
 return(i);
}

//求非线性方程组最小二乘解的广义逆法
int dngin(int m, int n, double eps1, double eps2, double x[], int ka)
{
 int i,j,k,l,kk,jt;
 double y[10],b[10],alpha,z,h2,y1,y2,y3,y0,h1;
 double *p,*d,*pp,*dx,*u,*v,*w;
 p=(double *)malloc(m*n*sizeof(double));
 d=(double *)malloc(m*sizeof(double));
 pp=(double *)malloc(n*m*sizeof(double));
 dx=(double *)malloc(n*sizeof(double));
 u=(double *)malloc(m*m*sizeof(double));
 v=(double *)malloc(n*n*sizeof(double));
 w=(double *)malloc(ka*sizeof(double));
 l=60; alpha=1.0;
 while (l>0)
 {
  dnginf(m,n,x,d);
  dngins(m,n,x,p);
  jt=agmiv(p,m,n,d,dx,pp,eps2,u,v,ka);
  if (jt<0)
  {
   free(p);
   free(d);
   free(pp);
   free(w);
   free(dx);
   free(u);
   free(v);
   return(jt);
  }
  j=0;
  jt=1;
  h2=0.0;
  while (jt==1)
  {
   jt=0;
   if (j<=2)
    z=alpha+0.01*j;
   else
    z=h2;
   for (i=0; i<=n-1; i++)
    w[i]=x[i]-z*dx[i];
   dnginf(m,n,w,d);
   y1=0.0;
   for (i=0; i<=m-1; i++)
    y1=y1+d[i]*d[i];
   for (i=0; i<=n-1; i++)
    w[i]=x[i]-(z+0.00001)*dx[i];
   dnginf(m,n,w,d);
   y2=0.0;
   for (i=0; i<=m-1; i++)
    y2=y2+d[i]*d[i];
   y0=(y2-y1)/0.00001;
   if (fabs(y0)>1.0e-10)
   {
    h1=y0;
    h2=z;
    if (j==0)
    {
     y[0]=h1;
     b[0]=h2;
    }
    else
    {
     y[j]=h1;
     kk=0;
     k=0;
     while ((kk==0)&&(k<=j-1))
     {
      y3=h2-b[k];
      if (fabs(y3)+1.0==1.0)
       kk=1;
      else
       h2=(h1-y[k])/y3;
      k=k+1;
     }
     b[j]=h2;
     if (kk!=0)
      b[j]=1.0e+35;
     h2=0.0;
     for (k=j-1; k>=0; k--)
      h2=-y[k]/(b[k+1]+h2);
     h2=h2+b[0];
    }
    j=j+1;
    if (j<=7)
     jt=1;
    else
     z=h2;
   }
  }
  alpha=z;
  y1=0.0;
  y2=0.0;
  for (i=0; i<=n-1; i++)
  {
   dx[i]=-alpha*dx[i];
   x[i]=x[i]+dx[i];
   y1=y1+fabs(dx[i]);
   y2=y2+fabs(x[i]);
  }
  if (y1<eps1*y2)
  {
   free(p);
   free(pp);
   free(d);
   free(w);
   free(dx);
   free(u);
   free(v);
   return(1);
  }
  l=l-1;
 }
 free(p);
 free(pp);
 free(d);
 free(dx);
 free(u);
 free(v);
 free(w);
 return(0);
}


//求非线性方程实根的对分法
int ddhrt(double a, double b, double h, double eps, double x[], int m)
{
 int n,js;
 double z,y,z1,y1,z0,y0;
 n=0;
 z=a;
 y=ddhrtf(z);
 while ((z<=b+h/2.0)&&(n!=m))
 {
  if (fabs(y)<eps)
  {
   n=n+1;
   x[n-1]=z;
   z=z+h/2.0;
   y=ddhrtf(z);
  }
  else
  {
   z1=z+h;
   y1=ddhrtf(z1);
   if (fabs(y1)<eps)
   {
    n=n+1;
    x[n-1]=z1;
    z=z1+h/2.0;
    y=ddhrtf(z);
   }
   else if (y*y1>0.0)
   {
    y=y1;
    z=z1;
   }
   else
   {
    js=0;
    while (js==0)
    {
     if (fabs(z1-z)<eps)
     {
      n=n+1;
      x[n-1]=(z1+z)/2.0;
      z=z1+h/2.0;
      y=ddhrtf(z);
      js=1;
     }
     else
     {
      z0=(z1+z)/2.0;
      y0=ddhrtf(z0);
      if (fabs(y0)<eps)
      {
       x[n]=z0;
       n=n+1;
       js=1;
       z=z0+h/2.0;
       y=ddhrtf(z);
      }
      else if ((y*y0)<0.0)
      {
       z1=z0;
       y1=y0;
      }
      else
      {
       z=z0;
       y=y0;
      }
     }
    }
   }
  }
 }
 return(n);
}


//求实函数或复函数方程一个复根的蒙特卡洛法
void dcmtc(double *x, double *y, double b, int m, double eps)
{
 int k;
 double xx,yy,a,r,z,x1,y1,z1;
 a=b;
 k=1;
 r=1.0;
 xx=*x;
 yy=*y;
 z=dcmtcf(xx,yy);
 while (a>=eps)
 {
  x1=-a+2.0*a*mrnd1(&r);
  x1=xx+x1;
  y1=-a+2.0*a*mrnd1(&r);
  y1=yy+y1;
  z1=dcmtcf(x1,y1);
  k=k+1;
  if (z1>=z)
  {
   if (k>m)
   {
    k=1;
    a=a/2.0;
   }
  }
  else
  {
   k=1;
   xx=x1;
   yy=y1;
   z=z1;
   if (z<eps)
   {
    *x=xx;
    *y=yy;
    return;
   }
  }
 }
 *x=xx;
 *y=yy;
 return;
}


//求非线性方程一个实根的埃待金达代法
int datkn(double *x, double eps, int js)
{
 int flag,l;
 double u,v,x0;
 l=0;
 x0=*x;
 flag=0;
 while ((flag==0)&&(l!=js))
 {
  l=l+1;
  u=datknf(x0);
  v=datknf(u);
  if (fabs(u-v)<eps)
  {
   x0=v;
   flag=1;
  }
  else
   x0=v-(v-u)*(v-u)/(v-2.0*u+x0);
 }
 *x=x0;
 l=js-l;
 return(l);
}

double ddhrtf(double x)
{
 double z;
 z = (((((x-5.0)*x + 3.0)*x + 1.0) * x - 7.0) * x + 7.0) * x - 20.0;;
 return z;
}

void dnewtf(double x, double y[2])
{
 y[0] = x * x * (x - 1.0) - 1.0;
 y[1] = 3.0 * x * x - 2.0 * x;
}

double datknf(double x)
{
 double z;
 z = 6.0 - x * x;
 return z;
}

double dpqrtf(double x)
{
 double z;
 z = x * x * (x - 1.0) - 1.0;
 return z;
}

/*
f[0] = x[0] - 5 * x[1]^2 + 7 * x[2]^2 + 12 = 0
f[1] = 3 * x[0] * x[1] + x[0] * x[2] - 11 * x[0] = 0
f[2] = 2 * x[1] * x[2] + 40 * x[0] = 0
*/
double dsnsef(double x[], double y[], int n)
{
 double z, f1, f2, f3, df1, df2, df3;
 n = n;
 f1 = x[0] - 5.0 * x[1] * x[1] + 7.0 * x[2] + 12.0;
 f2 = 3.0 * x[0] * x[1] + x[0] * x[2] - 11.0 * x[0];
 f3 = 2.0 * x[1] * x[2] + 40.0 * x[0];
 z = f1 * f1 + f2 * f2 + f3 * f3;
 df1 = 1.0;
 df2 = 3.0 * x[1] + x[2] - 11.0;
 df3 = 40.0;
 y[0] = 2.0 * (f1 * df1 + f2 * df2 + f3 * df3);
 df1 = 10.0 * x[1];
 df2 = 3.0 * x[0];
 df3 = 2.0 * x[2];
 y[1] = 2.0 * (f1 * df1 + f2 * df2 + f3 * df3);
 df1 = 14.0 * x[2];
 df2 = x[0];
 df3 = 2.0 * x[1];
 y[2] = 2.0 * (f1 * df1 + f2 * df2 + f3 * df3);
 return z;
}

/*
f[0] = x[0] ^ 2 + x[1] ^ 2 + x[2] ^ 2 - 1 = 0
f[1] = 2 * x[0] ^ 2 + x[1] ^ 2 - 4 * x[2] = 0
f[3] = 3 * x[0] ^ 2 - 4 * x[1] + x[2] ^ 2 = 0
*/
void dnetnf(double x[], double y[], int n)
{
 y[0] = x[0] * x[0] + x[1] * x[1] + x[2] * x[2] - 1;
 y[1] = 2 * x[0] * x[0] + x[1] * x[1] - 4 * x[2];
 y[3] = 3 * x[0] * x[0] - 4 * x[1] + x[2] * x[2];
}

/*
f[0] = x[0] ^ 2 + 10 * x[0] * x[1] + 4 * x[1] ^ 2  + 0.7401006 = 0
f[1] = x[0] ^ 2 - 3 * x[0] * x[1] + 2 * x[1] - 1.0201228 = 0
其中m = 2, n = 2。取初值为(0.5,-1.0),eps1 = 0.000001,eps2 = 0.000001。
*/
void dnginf(int m, int n, double x[], double d[])
{
 m = m;
 n = n;
 d[0] = x[0] * x[0] + 10 * x[0] * x[1] + 4 * x[1] * x[1]  + 0.7401006;
 d[1] = x[0] * x[0] - 3 * x[0] * x[1] + 2 * x[1] - 1.0201228;
}

void dngins(int m, int n, double x[], double p[])
{
 m = m;
 n = n;
 p[0] = 2.0 * x[0] + 10.0 * x[1];
 p[1] = 10.0 * x[0] + 8.0 * x[1];
 p[2] = 2.0 * x[0] - 3.0 * x[1];
 p[3] = -3.0 * x[0] + 4.0 * x[1];
}

/*
f(x) = e ^ (-x ^ 3) - sin(x) / cos(x) + 800 = 0
*/
double dmtclf(double x)
{
 double z;
 z = exp(-x * x * x) - sin(x)/cos(x) + 800.0;
 return z;
}

/*
f(z) =z ^ 2 - 6z + 13 = 0
取初值z = 0.5 + 0.5j, b = 1.0, m = 10, ε= 0.000001。其中
f(z) = (x ^ 2 - y ^ 2 - 6x + 13) + j(2xy - 6y)
*/
double dcmtcf(double x, double y)
{
 double u, v, z;
 u = x * x - y * y - 6.0 * x + 13.0;
 v = 2.0 * x * y - 6.0 * y;
 z = sqrt(u * u + v * v);
 return z;
}

/*
f[0] = 3x[1] + x[2] + 2x[3] ^ 2 - 3 = 0;
f[1] = -3x[1] + 5x[2] ^ 2 + 2x[1]x[3] - 1 = 0;
f[2] = 25x[1]x[2] + 20x[3] + 12 = 0;
取初值X = (0, 0, 0)^T, b = 2.0, m = 10, ε = 0.00001。
*/
double dnmtcf(double x[], int n)
{
 double f, f1, f2, f3;
 n = n;
 f1 = 3.0 * x[0] + x[1] + 2.0 * x[2] * x[2] - 3.0;
 f2 = -3.0 * x[0] + 5.0 * x[1] * x[1] + 2.0 * x[0] * x[2] - 1.0;
 f3 = 25.0 * x[0] * x[1] + 20.0 * x[2] + 12.0;
 f = sqrt(f1 * f1 + f2 * f2 + f3 * f3);
 return f;
}
                              ----根据《C语言常用算法程序集》整理

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