Пример #1
0
void newton( double *u, int n, double *f, double *v)
{
  /* linearized newtons method for nonlinear pde  
   * see p. 149 trottenberg for example */ 

  int i,j,nmem;
  double h,h2,d2;
  double nonlin,denom,*v2;
int aux;
double predenom;

  h=(xmax-xmin)/(double) n; 
  //this is h : h= (xmax-xmin)/n, assuming xmax/min=ymax/min
  h2=h*h;     //  if you change this, you must also change d2 in defect !!!
  d2=1./h2;

	// printf("Loop from 0 0 to i:%d j:%d\n", i, j );
predenom = -4.*d2 +invtau;

  for(i=0;i<n;i++){
	 
	
    for(j=0;j<n;j++){
		aux =index2d(i,j,n);
      denom=  predenom + df(u[aux]);
		// printf("-4.* %f + %f\n", d2, invtau );
      v[aux]=
	(f[aux]-(v[index2d(i-1,j,n)]+v[index2d(i,j-1,n)]
       + v[index2d(i+1,j,n)]+v[index2d(i,j+1,n)] )*d2 )/denom;
    }
  }

}/*newton*/
Пример #2
0
 void get_path_ofs(TPath *path, const TCoords& start_p, const TCoords& finish_p)
 {
     size_t ci = index2d(finish_p.x, finish_p.y);
     size_t si = index2d(start_p.x, start_p.y);
     while (ci != si)
     {
         TCoords p;
         p.x = attrs[ci].ofsx;
         p.y = attrs[ci].ofsy;
         path->push_back(p);
         ci -= W * p.y + p.x;
     }
 }
Пример #3
0
double errordefect( double *d, int n) // calculates one norm of the 
	                             // matrix d
{
  int i,j;
  double diff;

  diff=0.;
  for(i=0; i<n; i++){
    for(j=0; j<n; j++){
      diff += d[index2d(i,j,n)]*d[index2d(i,j,n)];
    }
  }
  return(diff);
}/* errordefect */
Пример #4
0
void interpolate(double *uc,double *uf,int nc,int nf)
{
  int i1,i,j,j1,n;
  int ic,jc;
  /*       first equate the matching points on grid  */

  n=nc;
  for( ic=0;ic<nc;ic++){  /* remember uc,nc is coarse grid, uf nf is fine */
    for( jc=0;jc<nc;jc++){       
      i1=ic*2;
      j1=jc*2;
      uf[index2d(i1,j1,nf)]=uc[index2d(ic,jc,nc)]; /* for uf this is even i, even j */
    }//j1
  }//i1

  /* now match up i pts between defined from previous pts - these are 'odd i', even j */
  n=nf;
  for( i=0;i<n;i+=2){ /* i should increment by 2 */
    for( j=1;j<n;j+=2){       
      uf[index2d(i,j,nf)]=(uf[index2d(i,j-1,nf)] + uf[index2d(i,j+1,nf)])*0.5;
    }//j
  }//i

  for( i=1;i<n;i+=2){
    for( j=0;j<n;j++){
      uf[index2d(i,j,nf)]=(uf[index2d(i-1,j,nf)] + uf[index2d(i+1,j,nf)])*0.5; /** note that i,j has been reversed **/
    }//i
  }//j

}/* interpolate */
Пример #5
0
void mexFunction(int lhsCount, mxArray* lhs[], int rhsCount, const mxArray* rhs[])
{
	mxArray* pi_R;
	unsigned int* pi_R_data;
	unsigned int* pi_R_j_data;
	unsigned int pi_R_j_idx;
	unsigned int* inv_pi = (unsigned int*)mxGetData(rhs[0]);
	size_t max_t = mxGetM(rhs[0]);
	size_t N = mxGetN(rhs[0]);
	unsigned int* t = (unsigned int*)mxGetData(rhs[1]);
	unsigned int n = (unsigned int)mxGetScalar(rhs[2]);
	char* already_ranked = mxMalloc(n * sizeof(char));
	unsigned int* this_inv_pi;

	unsigned int i, j, k;
	unsigned int row;

	mwSize dims[3];
	dims[0] = n - 1;
	dims[1] = max_t;
	dims[2] = N;

	pi_R = mxCreateNumericArray(3, dims, mxUINT32_CLASS, mxREAL);
	pi_R_data = (unsigned int*)mxGetData(pi_R);
	lhs[0] = pi_R;

	for (i = 0; i < N; ++i)
	{
		this_inv_pi = inv_pi + index2d(0, max_t, i);
		memset(already_ranked, 0, n * sizeof(char));
		for (j = 0; j < t[i]; ++j)
		{
			/* Get this sparse vector of R */
			pi_R_j_data = pi_R_data + index3d(0, n - 1, j, max_t, i);
			pi_R_j_idx = 0;

			/* Fill vector with proper entries */
			/* Only row will be inv_pi[j] */
			row = this_inv_pi[j];
			already_ranked[row] = 1;
			for (k = 0; k < n; ++k)
			{
				/* If not already ranked, and not current item being ranked, mark in vector */
				if (already_ranked[k] == 0)
				{
					pi_R_j_data[pi_R_j_idx++] = k * n + row;
				}
			}
			assert(pi_R_j_idx == n - j - 1);
		}
	}

	mxFree(already_ranked);
}
Пример #6
0
void funct(double *f,double *u,int n, int t)
{
  int i,j,k;
  double x,y,z,h,t1;
  double d2,sxsy,sint,sint1,uij,fij,fm1,fm,cost,cost1;
  t1=(t-1)*tau;
  d2=(double)n/(xmax-xmin);
  d2=d2*d2;
  h=(xmax-xmin)/(double) n;
  for(i=0;i<n;i++){
    x=i*h +xmin;
    for(j=0;j<n;j++){ 
      y=j*h +xmin;
      sxsy=sin(x)*sin(y); // definitions to make things easier
      sint=sin(t*tau);
      sint1=sin(t1);
      cost=cos(t*tau);
      cost1=cos(t1);
      uij=u[index2d(i,j,n)];
      // define the functions f^m+1, f^m here to makes things easier
      // eqn is : Nu +f = d_t u  => f=d_t u-Nu
      // need f at this time (t), and previous time (t-1)
      // see comments at begining for more explanation
      fm  = sxsy*cost   + 2*sxsy*sint  - fn(sxsy*sint);
      fm1 = sxsy*cost1 + 2*sxsy*sint1 - fn(sxsy*sint1);
      fij=(u[index2d(i+1,j,n)]+u[index2d(i-1,j,n)]+
	   u[index2d(i,j+1,n)]+u[index2d(i,j-1,n)] -4.*uij)*d2;
      fij+=uij*(2./tau) + fn(uij) + fm1 +fm;
      f[index2d(i,j,n)]=-fij;
  
    }//j
  }//i
} /* funct */ 
Пример #7
0
void defectnl(double *d, double *u,int n,double *f) 
{
  int i,j;
  double h,h2,d2,temp,temp1,temp2,temp3,b1;

  h=(xmax-xmin)/(double) n; //this is h : h= (xmax-xmin)/n
  h2=h*h;
  d2=1./h2;
  for( i=0;i<n;i++){ 
    for( j=0;j<n;j++){  // d = f - Nu
      d[index2d(i,j,n)]=f[index2d(i,j,n)] -
	+ (  (u[index2d(i-1,j,n)]+u[index2d(i,j-1,n)]
	      +u[index2d(i+1,j,n)]+u[index2d(i,j+1,n)] -4*u[index2d(i,j,n)])*d2
	     +invtau*u[index2d(i,j,n)]
	     +fn(u[index2d(i,j,n)]) );

    }
  }
}/* defectnl */
Пример #8
0
double error(double *u,double *uback, int n,int *imx,int *jmx)
{   double diff,diffmx,ave;
 int i,j,ic,k;		/* calculates the difference between u now &
			   u before, as one way to calculate error */
 ave=0;
 diffmx=0;
 ic=0;
 for(i=0;i<n;i++){
   for(j=0;j<n;j++){
     diff=u[index2d(i,j,n)]-uback[index2d(i,j,n)];
     if(diff < 0) diff=-diff;
     ave+=diff;
     ic+=1;
     if(diff > diffmx){
       diffmx=diff; 	
       *imx=i;
       *jmx=j;
     } //if
   }//j
 }//i
 ave=ave/ic;
 return (diffmx);
}/*error*/
Пример #9
0
void ic(double *u,int n,int t) // initial condition
{
  int i,j;
  double x,y,h,timebefore;

  h=(xmax-xmin)/(double) n;

  timebefore=(double)(t-1)*tau;
  for(i=0;i<n;i++){
    x=i*h+xmin; 
    for(j=0;j<n;j++){
      y=j*h+xmin; 
      u[index2d(i,j,n)]=sin(x)*sin(y)*sin(timebefore);
    }//i
  }//j
}/* ic */
Пример #10
0
void restrct2(double *uc, double *uf, int n)
{
  int i, j, i2, j2,nf;
  nf=2*n;
      
  for( i=0;i<n;i++){ /* n is the # of pts on the coarse grid */
    for( j=0;j<n;j++) {
      i2=2*i;
      j2=2*j;
      uc[index2d(i,j,n)]=uf[index2d(i2,j2,nf)]*4.
	+uf[index2d(i2,j2+1,nf)] +uf[index2d(i2+1,j2,nf)]
	+uf[index2d(i2-1,j2,nf)]+uf[index2d(i2,j2-1,nf)];
      uc[index2d(i,j,n)]=uc[index2d(i,j,n)]/8.;

    }/* loop over j */
  }/* loop over i */

} /* end  restrct2 */
Пример #11
0
    bool do_search(const TMap& map, const TCoords& start_p, const TCoords& finish_p)
    {
        static struct { int x, y, d; } dirs[] =
        { { -1, -1, 14 },{ 0, -1, 10 },{ 1, -1, 14 },{ -1, 0, 10 },{ 1, 0, 10 },{ -1, 1, 14 },{ 0, 1, 10 },{ 1, 1, 14 } };
        memset(attrs, 0, sizeof(Attributes) * H * W);
        while (!opened.empty()) opened.pop();

        AttrsPtr current = opened_push(start_p, cost_estimate(start_p, finish_p));
        while (!opened.empty())
        {
            current = opened_pop();
            if (current.pos.x == finish_p.x && current.pos.y == finish_p.y)
                return true;
            for (int i = 0; i < 8; ++i)
            {
                int dx = dirs[i].x;
                int dy = dirs[i].y;
                TCoords npos;
                npos.x = current.pos.x + dx;
                npos.y = current.pos.y + dy;
                if (!inbound(npos.x, npos.y) || map.isobstacle(npos.x, npos.y))
                    continue;
                size_t ni = index2d(npos.x, npos.y);
                if (attrs[ni].state == st_Closed)
                    continue;
                TWeight t_gscore = current.pa->gscore + dirs[i].d;
                if (attrs[ni].state == st_Wild)
                {
                    opened_push(npos, t_gscore + cost_estimate(npos, finish_p));
                }
                else
                {
                    if (t_gscore >= attrs[ni].gscore)
                        continue;
                    rearrange(&attrs[ni], t_gscore + cost_estimate(npos, finish_p));
                }
                attrs[ni].ofsx = dx;
                attrs[ni].ofsy = dy;
                attrs[ni].gscore = t_gscore;
            }
        }
        return false;
    }
Пример #12
0
void newtonnl( double *u, int n, double *f)
{ //newtons method directly for pde
  int i,j,nmem; double h,h2,d2,*u2;
  double nonlin,denom;

  h=(xmax-xmin)/(double) n;
  h2=h*h;     //  if you change this, you must also change d2 in defect !!!
  d2=1./h2;
  for(i=0;i<n;i++){ 
    for(j=0;j<n;j++){ 
      denom= -4.*d2 +invtau+df(u[index2d(i,j,n)]);
      nonlin=u[index2d(i,j,n)]*invtau +fn(u[index2d(i,j,n)]) -f[index2d(i,j,n)];
      u[index2d(i,j,n)]=
	u[index2d(i,j,n)]
	-( (u[index2d(i-1,j,n)]+u[index2d(i,j-1,n)]+ u[index2d(i+1,j,n)]
	   +u[index2d(i,j+1,n)] -4*u[index2d(i,j,n)])*d2 +nonlin )/denom;
    }//j
  }//i

}/*newtonnl*/
Пример #13
0
void restrct(double *uc, double *uf, int n)
{
  int i, j, i2, j2,nf;
  nf=2*n;
      
  /*first equate the matching points on grid */
  /*uc=coarse, uf=fine */
  for( i=0;i<n;i++){ // n is the # of pts on the coarse grid 
    for( j=0;j<n;j++) {
      i2=2*i;
      j2=2*j;
      uc[index2d(i,j,n)]=uf[index2d(i2,j2,nf)] *4.
	+2.*uf[index2d(i2,j2+1,nf)] +2.*uf[index2d(i2+1,j2,nf)]
	+2.*uf[index2d(i2-1,j2,nf)]+2.*uf[index2d(i2,j2-1,nf)]
	+uf[index2d(i2-1,j2+1,nf)] +uf[index2d(i2+1,j2-1,nf)]
	+uf[index2d(i2-1,j2-1,nf)] +uf[index2d(i2+1,j2+1,nf)];
      uc[index2d(i,j,n)]=uc[index2d(i,j,n)]/16.;

      // alternate schemes:
      //           uc[index2d(i,j,n)]=uf[index2d(i2,j2,nf)] ;
      //           or
      //           uc[index2d(i,j,n)]=uf[index2d(i2,j2,nf)] *4.
      //     +uf[index2d(i2,j2+1,nf)] +uf[index2d(i2+1,j2,nf)]
      //     +uf[index2d(i2-1,j2,nf)]+uf[index2d(i2,j2-1,nf)]
      //     uc[index2d(i,j,n)]=uc[index2d(i,j,n)]/8;
		    

    }//j
  }//i

} /* end  restrct */
Пример #14
0
void mgrid(double *u, int n,double *f,int t) 
{
  double *v,*d,*vc,*dc, *uf, *uc,*fc;
  double *uback,*uback2;
  double diffmx,diff,difv,x,y,tol,h;
  double dif,mxd,sumu,sumu2,sum3,sol;
  int i,j,k,imx,jmx,in,nc,irep;
  int ncmem,nsq,nmem;
  char filename[filesize];
  FILE *f1, *f2, *out, *f3, *f8;

  tol=10.e-12; // tol= tolerance. How accurately you want to solve the eqn


  diffmx=tol*5.;
  irep=0;

  nc=n/2;
  nmem=(n+1)*(n+1);  // bigger than [0:n]*[0:n], just to ensure enough space
  ncmem=(nc+1)*(nc+1);// same for the grid that is half-size
  nsq=n*n;  


/* --- requesting space for variables ------------------------------*/

  v=malloc(nmem*sizeof(double));
  if(v==NULL) {printf(" oopsies can't allocate v %d\n",__LINE__);}
  d=malloc(nmem*sizeof(double));
  if(d==NULL) printf(" oopsies can't allocate d %d\n",__LINE__);
  uback=malloc(nmem*sizeof(double));
  if(uback==NULL) printf(" oopsies can't allocate uback %d\n",__LINE__);
  uback2=malloc(nmem*sizeof(double));
  if(uback2==NULL) printf(" oopsies can't allocate uback2  %d\n",__LINE__);
  vc=malloc(ncmem*sizeof(double));
  if(vc==NULL) {printf(" oopsies can't allocate vc  %d\n",__LINE__);}
  uc=malloc(ncmem*sizeof(double));
  if(uc==NULL) {printf(" oopsies can't allocate uc  %d\n",__LINE__);}
  dc=malloc(ncmem*sizeof(double));
  if(dc==NULL) printf(" oopsies can't allocate dc  %d\n",__LINE__);


  for(i=0;i<nmem;i++){  // initialize to 0
    v[i]=0.;
    uback[i]=0.;
    d[i]=0.; }  // end of v,d,uback initialization
  for(i=0;i<ncmem;i++){
    vc[i]=0.;
    dc[i]=0.; }  // end of vc,dc



  h=(xmax-xmin)/(double) n;
  for(i=0;i<n;i++){
    x=xmin+h*i;
    for(j=0;j<n;j++){
      y=xmin+h*j;
      uback2[index2d(i,j,n)]=sin(x)*sin(y)*sin(t*tau);
    }//y     // the actual solution, to compare
  }//x

  diffmx=5*tol; /* tol determines solution tolerance. diffmx will measure 
                   that difference */

  irep=0;  // irep will be the number of MG steps required to solve


  /* +++++++++++++++ start of while loop ++++++++++++++++++*/

  while(diffmx>tol){

    diffmx=0.;   // need to zero this now, will measure later
    irep=irep+1; // counts number of MG steps

		 

    for( in=1;in<=nrep1;in++){  // presmoothing
      newtonnl(u,n,f);
    }//nrep1

    for( i=0;i<nmem;i++){
      uback[i]=u[i]; 
    }//uback=u

     /* defectnl calculates the residual. ie, given Nu=f
      *  d=f-Nu, or a measure of how far away you are from the solution */
        
    defectnl(d,u,n,f);

     /* restrict u, have a separate routine for it, in case you want to 
      * use a different restriction for than for other variables */

     restrct2(uc,u,nc);		 


    /* v is the linearlized version of u
     * given, Nu=f
     * define an operator K such that Kv=d, 
     * then u := u+v  (ie new u = old u+ v, v found from above
     * this can be derived from standard Newton's method:
     *  u := u - (Nu -f)/K
     *  where K=(d/du)N
     *
     *  so, if Nu = (\nabla)^2 u + g(u)
     *  K = (\nabla)^2 + (d/du)g
     *
     *  thus, the eqn we now solve is
     *
     *  Kv = (\nabla)^2v + (d/du)g v =d  (#)
     *
     *  and the next iterate of u is u+v
     *
     *  I apply MG techniques to (#)
     */

    restrct(vc,v,nc); 
    restrct(dc,d,nc); // dc is defect on coarse grid

    pooofle(vc,nc,dc,uc); /* funky multigrid part!
			     this solves for vc    */

    interpolate(vc,v,nc,n);//interpolate vc up to the fine grid

    for(i=1;i<2;i++){
      newton(u,n,d,v);   //   this is 'postsmoothing' for v !!
    }

    for( i=0;i<nmem;i++){
      u[i]+=v[i]; 
      uback[i]=u[i];
    }//u+=v

    for( i=0;i<nrep2;i++){
      newtonnl(u,n,f);   	// postsmoothing on u
    }//nrep2


    /* how to tell when you're done?
     * the classic way is to calculate the defect, and when the norm
     * of the defect is less than some predetermined accuracy (tolerance)
     * you're done.
     *
     * I have found that I get better accuracy - ie the solution remains
     * correct farther into time when I use as my condition the difference
     * between the current u, and the previous value of u. If anyone has
     * any ideas for this, please let me know !
     */

    defectnl(d,u,n,f);
    diff=errordefect(d,n)/n/n;

    diffmx=error(u,uback,n,&imx,&jmx); 
    /* diffmx is the condition to test against. for termination of
     * calculation based on defect, do:
     * diffmx=errordefect(d,n)/n/n;
     * diff=error(u,uback,n,&imx,&jmx);
     */ 


    mxd=-10.;
    sum3=0.;
    for(i=0;i<n;i++){
      for(j=0;j<n;j++){
        dif=uback2[index2d(i,j,n)]-u[index2d(i,j,n)]; 
	sum3+=u[index2d(i,j,n)];
        if(dif < 0.e-10) dif= -dif;  
        if(dif > mxd) {mxd= dif;  // maximum difference between actual &
	imx=i;                    // calculated solution. for fun
	jmx=j;                   // imx & jmx calculate where this happens
	} // end if
      }//j
    }//i

     

    if(irep>2000) break; // some break condition
  } /* while */

  printf("%d %d %1.7e %1.7e %1.7e %d %d \n",t,irep,diffmx,diff,mxd,imx,jmx);
  fflush(stdout);
      
      
  /*
   * this will make a file with the x,y coordinates, the calculated &
   * actual solutions & their difference.
   *
   * uncomment if you want this printed out 
   *
  snprintf(filename,filesize,"solution.%d",t);
  f1=fopen(filename,"w");
  for(i=0;i<n;i++){
    x=xmin+h*i;
    for(j=0;j<n;j++){
      y=xmin+h*j;
      sol=sin(x)*sin(y)*sin(t*tau);
      dif=sol-u[index2d(i,j,n)];
      fprintf(f1,"%lf %lf %lf %lf %lf\n",x,y,u[index2d(i,j,n)],sol,dif,uback[index2d(i,j,n)]);
    }
  }
  fclose(f1);

  */

  free(uback);
  free(uback2);
  free(v);
  free(d);
  free(dc);
  free(uc);
  free(vc);

	     
} /* mg */ 
Пример #15
0
main(int argc, char *argv[])
{
  double *areal; /* areal is the variable I'm solving for. will be called u
		    in the functions that are dependent on this on. I just
		    needed to give it a different name here */
  double *freal; /* freal is the right hand side of (**), ie F */
  double *f,x,y,h,dif,sol;
  int n,nmem,t,i,j;
  FILE *f1 ;
  char filename[filesize];

imin = imax = jmin = jmax =stridemax;
stridemin = 0;

  /* Use command line argument if it exists. Otherwise use default value in
     nmax. */
  n = (argc > 1) ? atoi(argv[1]) : nmax;
  //n=nmax;
  nmem=(n+1)*(n+1);
  t=1;
  f1=fopen("nlmax","w");/* in this file gets printed the maximum of the 
			   calculated and actual solutions. for comparison */
			
//===============================

//Precalculem index2d 270400
// -1.64 -1.64 0.64 [66][66][65]

// int c2d[n+2][n+2][n+1];
// 	calci2d=malloc((n+2)*(n+2)*(n+1)*sizeof(int));
//     calci2d = &c2d[0][0][0];
// 	if(calci2d==NULL)printf(" oopsies can't allocate calci2d %d \n",__LINE__);
// 	int i2d, j2d;
// 	i2d = j2d = 0;
// 	int* p = &c2d[0][0][0];
// 	
	//printf(" %p ______ %p \n ", calci2d, p);
	
			int i2d, j2d;
			
			// printf("malloc calci2d\n");
			calci2d = malloc((n+2)*sizeof(int**));
			if(calci2d==NULL)printf(" oopsies can't allocate calci2d %d \n",__LINE__);
			for(i2d=0; i2d < (n+2); i2d++){
				// printf("malloc calci2d[%d]\n", i2d);
				calci2d[i2d] = malloc((n+2)*sizeof(int*));
				if(calci2d[i2d]==NULL)printf(" oopsies can't allocate calci2d[%d] %d \n", i2d,__LINE__);
				for( j2d = 0; j2d < (n+2); j2d++){
					/* code */
					// printf("malloc calci2d[%d][%d]\n", i2d, j2d);
					calci2d[i2d][j2d] = malloc((n+1)*sizeof(int));
					if(calci2d[i2d][j2d]==NULL)printf(" oopsies can't allocate calci2d[%d][%d] %d \n", i2d,j2d, __LINE__);
				}
			}

	int stride = 0;
	i2d = j2d = 0;
	int *p = &calci2d[0][0][0];
	
	for(i2d=0; i2d < (n+2); i2d ++){
		for(j2d=0; j2d < (n+2); j2d ++){
			for(stride=0; stride < (n+1); stride ++){
				calci2d[i2d][j2d][stride] = (i2d-1+n)%n + stride*((j2d-1+n)%n);
			}
		}
	}
	
	

	// while(i2d<n+2){
	// 	//   if(i<0) i+=stride;
	// 	//   else if(i>=stride) i-=stride;
	// 	//   if(j<0) j+=stride;
	// 	//   else if(j>=stride) j-=stride;
	// 	// int aux = (i+stride*j);
	// 	  // return(i+stride*j);
	// 	
	// 	printf("computing calci2d[%d][%d][%d]", i2d, j2d, stride);
	// 	*p = (i2d-1+n)%n + stride*((j2d-1+n)%n);
	// 	printf(" = %d\n", (i2d-1+n)%n + stride*((j2d-1+n)%n));
	// 	p++;
	// 	
	// 	if(j2d == 1){
	// 		printf("%d == %d", calci2d[i2d][j2d][stride], (i2d-1+n)%n + stride*((j2d-1+n)%n));
	// 		exit(0);
	// 	}
	// 	
	// 	stride ++;
	// 	if(stride == n+1){ 
	// 		stride = 0;
	// 		j2d++;
	// 		if(j2d == n+2){
	// 			j2d = 0;
	// 			i2d++;
	// 		}
	// 	}
	// } //while


//=============================
			



  areal=malloc(nmem*sizeof(double));
  if(areal==NULL)printf(" oopsies can't allocate areal %d \n",__LINE__);
  ic(areal,n,t);//call initial condition if it's the first time through
	 
  freal=malloc(nmem*sizeof(double));
  if(freal==NULL)printf(" oopsies can't allocate freal %d \n",__LINE__);

	h=(xmax-xmin)/(double) n;
	
  while(t < timemax){

    funct(freal,areal,n,t);
    mgrid(areal,n,freal,t);
    //mgrid(areal,nmax,freal,t);
	
    fprintf(f1,"%d %lf %lf \n",t,areal[index2d(n/4,n/4,n)],sin(t*tau));
    fflush(f1);

    t++;
  }/* while */
      
printf( "imin: %d imax: %d\n", imin, imax );
printf( "jmin: %d jmax: %d\n", jmin, jmax );
printf( "stridemin: %d stridemax: %d\n", stridemin, stridemax );

  fclose(f1);
  free(areal);
  free(freal);
  
}/* main */
Пример #16
0
 AttrsPtr(const TCoords& p, Attributes *attrs) { pos = p; pa = &attrs[index2d(pos.x, pos.y)]; }