printf("T1 requires 900 binary genes. You have %d genes and they are %d-ary.\nIt is a 2 objective problem. You have %d objectives. Exiting.\n", genes, alleles, objectives);      exit(-1);    }    //set paramaters to zero  for (i = 0; i < 30; i++)    var[i] = 0;  //convert from binary into integer for the 30 params  for (i = 0; i < 30; i++)    {      mul = 1;      for (j = 29; j >= 0; j--) 	{	  var[i] += mul * s->chrom[i*30+j];	  mul *= 2;	}     }  // normalize the params between 0 and 1  for (i = 0; i < 30; i++)    var[i] /= pow(2,30);  f1 = var[0];  for (i = 1; i < 30; i++)	sum += var[i];    g = 1 + (9*sum)/29.0;    h = 1 - sqrt(f1/g);  f2 = g*h;    s->obj[0]=f1;  s->obj[1]=f2;}void F5(sol *s) // two-objective minimization problem{  int i;  if ((objectives!=2)||(genes!=alleles))    {      printf("You've given invalid command parameters for function F5. Check paes.cc for details. Exiting.\n");      exit(-1);    }  s->obj[0] = genes-1;  // worst score for each objective is genes-1  s->obj[1] = genes-1;    for (i = 0; i < genes-1; i++)    {      if(s->chrom[i+1]== s->chrom[i]+1)   // reduce score of objective 1 if there are adjacent genes having consecutive values	s->obj[0]--;            if(s->chrom[i+1]==s->chrom[i]-1)   // as above but reading in reverse for objective 2	s->obj[1]--;    }}int compare_min(double *first, double *second, int n){  // compares two n-dimensional vectors of objective values for minimization problems  // returns 1 if first dominates second,     返回1：1支配2 -1:2支配1 0：其他  // -1 if second dominates first, and 0 otherwise  int obj = 0;     //比較依據 ：若first的每一分量都〉=second的每一分量，且至少有一個〉則first優于second  int deflt = 0;  int current;    do     
  {      if(*first < *second)		  current = 1;      else if(*second < *first)		  current = -1;      else		  current = 0;      
	  if((current)&&(current==-deflt))	  {		  return(0);	  }      if(current!=0)	  {		  deflt = current;	  }      obj++;      *first++;      *second++;	}while(obj < n);    return(deflt);}int compare_max(double *first, double *second, int n){  // as for compare_min() but for maximization problems   int obj = 0;  int deflt = 0;  int current;    do   {      if(*first > *second)		  current = 1;	  else if(*second > *first)		  current = -1;      else		  current = 0;
      if((current)&&(current==-deflt))	  {		  return(0);	  }	  if(current!=0)	  {		  deflt = current;	  }      obj++;      *first++;      *second++;  }while(obj < n);  
  return(deflt);}int equal(double *first, double *second, int n){  // checks to n-dimensional vectors of objectives to see if they are identical  // returns 1 if they are, 0 otherwise  int obj = 0;    do  {	  if(*first!=*second)		  return(0);      *first++;      *second++;      obj++;      // printf("%d\n",obj);  }  while(obj < n);  
  return(1);}void archive_soln(sol *s)  //將s加入archive{  // given a solution s, add it to the archive if  // a) the archive is empty   // b) the archive is not full and s is not dominated or equal to anything currently in the archive  // c) s dominates anything in the archive                          // d) the archive is full but s is nondominated and is in a no more crowded square than at least one solution  // in addition, maintain the archive such that all solutions are nondominated.    int i;  int repl;  int yes = 0;  int most;  int result;  int join = 0;  int old_arclength;  double *evs;  double *evli;   int set = 0;  int tag[MAX_ARC];  sol *tmp;  if (!(tmp = (sol *)malloc(MAX_ARC*sizeof(sol))))    
  {	  printf("Out of memory\n");      exit(-1);  }  for (i = 0; i < archive; i++)    
  {	  tag[i]=0;  }    if (arclength == 0)  {      add_to_archive(s);      return;  }    i = 0;  result = 0;  while((i < arclength)&&(result!=-1))  {      result = equal(s->obj, (&arc[i])->obj, objectives);      if (result == 1)		  break;      //MINIMIZE MAXIMIZE      if (minmax==0)		  result = compare_min(s->obj, (&arc[i])->obj, objectives);      else		  result = compare_max(s->obj, (&arc[i])->obj, objectives);      //  printf("%d\n", result);          if ((result == 1)&&(join == 0))	  {		  arc[i] = *s;             //將第一個受支配的個體替換掉		  join = 1;	  }	  else if (result == 1)        //后面若還有受支配的。。	  {		  tag[i]=1;		  set = 1;	  }	    	          i++;  }  old_arclength = arclength;  if (set==1)  {	  for (i = 0; i < arclength; i++)	  {		  tmp[i] = arc[i];	  }	  
	  arclength = 0;      for (i = 0; i < old_arclength; i++)	  {		  if (tag[i]!=1)           //把其他受支配的刪除		  {			  arc[arclength]=tmp[i];			  arclength++;		  }	  }  }    if ((join==0)&&(result==0))  // ie solution is non-dominated by the list  {                            //也就是說archive中沒有受s支配的個體	  if (arclength == archive)  //archive已滿	  {	  		  most = grid_pop[s->grid_loc];            //.......根據個體密度來替換		  for (i = 0; i < arclength; i++)		  {			  if (grid_pop[(&arc[i])->grid_loc] > most)			  {				  most = grid_pop[(&arc[i])->grid_loc];				  repl = i;				  yes = 1;				  //   printf("i = %d\n", i);			  }		  }	  
		  if (yes)		  {			  arc[repl] = *s;	     		  }	  }      else	  {		  add_to_archive(s);	  }  }  free(tmp);}int find_loc(double *eval)     //這里eval指向一個個體的目標函數值向量{  // find the grid location of a solution given a vector of its objective values  int loc = 0;  int d;  int n = 1;    int i;    int inc[MAX_OBJ];  double width[MAX_OBJ];    //printf("obj = %d, depth = %d\n", objectives, depth);    //if the solution is out of range on any objective, return 1 more than the maximum possible grid location number  for (i = 0; i < objectives; i++)  {	  if ((eval[i] < gl_offset[i])||(eval[i] > gl_offset[i] + gl_range[i]))		  return((int)pow(2,(objectives*depth)));  }  for (i = 0; i < objectives; i++)  {	  inc[i] = n;      n *=2;      width[i] = gl_range[i];  }	      for (d = 1; d <= depth; d++)  {      for (i = 0; i < objectives; i++)	  {		  if(eval[i] < width[i]/2+gl_offset[i])			  loc += inc[i];		  else			  gl_offset[i] += width[i]/2;	  }      
	  for (i = 0; i < objectives; i++)	  {		  inc[i] *= (objectives *2);		  width[i] /= 2;	  }  }  return(loc);} void update_grid(sol *s){  // recalculate ranges for grid in the light of a new solution s  static int change = 0;  int a, b;  int square;  double offset[MAX_OBJ];  double largest[MAX_OBJ];  double sse;  double product;  struct solution **start;    for (a = 0; a < objectives; a++)    {      offset[a] = LARGE;      largest[a] = -LARGE;    }    for (b = 0; b < objectives; b++)    {      for (a = 0; a < arclength; a++)		{	  if ((&arc[a])->obj[b] < offset[b])	    offset[b] = (&arc[a])->obj[b];	  if ((&arc[a])->obj[b] > largest[b])	    largest[b] = (&arc[a])->obj[b];		   	}    }  //printf("oldCURENT:largest = %f, offset = %f\n", largest[0], offset[0]);   //printf("oldCURENT:largest = %f, offset = %f\n", largest[1], offset[1]);   for (b = 0; b < objectives; b++)    {      if (s->obj[b] < offset[b])	offset[b] = s->obj[b];      if (s->obj[b] > largest[b])	largest[b] = s->obj[b];		       }  sse = 0;  product = 1;      for (a = 0; a < objectives; a++)    {       sse += ((gl_offset[a] - offset[a])*(gl_offset[a] - offset[a]));      sse += ((gl_largest[a] - largest[a])*(gl_largest[a] - largest[a]));      product *= gl_range[a];     }  // printf("sse = %f\n", sse);  if (sse > (0.1 * product * product))	//if the summed squared error (difference) between old and new                                        //minima and maxima in each of the objectives    {                                   //is bigger than 10 percent of the square of the size of the space      change++;                         // then renormalise the space and recalculte grid locations      for (a = 0; a < objectives; a++)	{	  gl_largest[a] = largest[a]+0.2*largest[a];	  gl_offset[a] = offset[a]+0.2*offset[a];	  gl_range[a] = gl_largest[a] - gl_offset[a];	}        for (a = 0; a < pow(2, (objectives*depth)); a++)	{	  grid_pop[a] = 0;	}        for (a = 0; a < arclength; a++)	{	  square = find_loc((&arc[a])->obj);	  (&arc[a])->grid_loc = square;	  grid_pop[square]++;	  	}    }  square = find_loc(s->obj);  s->grid_loc = square;  grid_pop[(int)pow(2,(objectives*depth))] = -5;  grid_pop[square]++;     }