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标题:哪位大神能给看看 帮忙注释下
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bob421559385
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哪位大神能给看看 帮忙注释下
#include <stdio.h>
#include <stdlib.h>
#include <math.h>


static void  initialization(void);
static void  evaluation(int gen);
static void  selection(void);
static void  crossover(void);
static void  mutation(void);
static void  objective_function(void);
static int   constraint_check(double x[]);

#define N 3  // number of variables
#define M 1  // number of objectives
#define TYPE 1 // 1=max;-1=min
#define GEN 400 // maximum generation number
#define POP_SIZE 30
#define P_MUTATION 0.2
#define P_CROSSOVER 0.3

double  CHROMOSOME[POP_SIZE+1][N+1];
double  OBJECTIVE[POP_SIZE+1][M+1];
double  q[POP_SIZE+1];

double myu(double a, double b) // Uniform Distribution
{
  double y;
  if(a>b) {
    printf("\nThe first parameter should be less than the second!");
    exit(1);
  }
  y = (double)rand()/(RAND_MAX);
  return (a+(b-a)*y);
}

static void objective_function(void)
{
    double x1,x2,x3;
    int i;
    for(i = 1; i <= POP_SIZE; i++) {
        x1 = CHROMOSOME[i][1];
        x2 = CHROMOSOME[i][2];
        x3 = CHROMOSOME[i][3];
        OBJECTIVE[i][1] = sqrt(x1)+sqrt(x2)+sqrt(x3);
    }
    for(i=1;i<=POP_SIZE;i++)
      OBJECTIVE[i][0]= OBJECTIVE[i][1];
}

static int constraint_check(double x[])
{
    double a;
    int n;
    for(n=1;n<=N;n++) if(x[n]<0) return 0;
    a = x[1]*x[1]+2*x[2]*x[2]+3*x[3]*x[3];
    if(a>1) return 0;
    return 1;
}

static void initialization(void)
{
  double x[N+1]; // N is the number of variables
  int i,j;
  for(i=1; i<=POP_SIZE; i++){
      mark:
      for(j=1; j<=N; j++) x[j]=myu(0,1);
      if(constraint_check(x)==0) goto mark;
      for(j=1; j<=N; j++) CHROMOSOME[i][j]=x[j];
  }
}

main()
{
  int i, j;
  double a;

  q[0]=0.05; a=0.05;
  for(i=1; i<=POP_SIZE; i++) {a=a*0.95; q[i]=q[i-1]+a;}
  initialization();
  evaluation(0);
  for(i=1; i<=GEN; i++) {
      selection();
      crossover();
      mutation();
      evaluation(i);
      printf("\nGeneration NO.%d\n", i);
      printf("x=(");
      for(j=1; j<=N; j++) {
          if(j<N) printf("%3.4f,",CHROMOSOME[0][j]);
          else printf("%3.4f",CHROMOSOME[0][j]);
      }
      if(M==1) printf(")\nf=%3.4f\n", OBJECTIVE[0][1]);
      else {
          printf(")\nf=(");
          for(j=1; j<=M; j++) {
             if(j<M) printf("%3.4f,", OBJECTIVE[0][j]);
             else printf("%3.4f", OBJECTIVE[0][j]);
          }
          printf(")  Aggregating Value=%3.4f\n",OBJECTIVE[0][0]);
      }
  }
  printf("\n");
  return 1;
}

static void evaluation(int gen)
{
  double a;
  int   i, j, k, label;
  objective_function();
  if(gen==0){
     for(k=0; k<=M; k++) OBJECTIVE[0][k]=OBJECTIVE[1][k];
     for(j = 1; j <= N; j++) CHROMOSOME[0][j]=CHROMOSOME[1][j];
  }
  for(i=1; i<POP_SIZE; i++){        // i is staring from 1
      label=0;  a=OBJECTIVE[i][0];
      for(j=i+1; j<=POP_SIZE; j++)
         if((TYPE*a)<(TYPE*OBJECTIVE[j][0])) {
             a=OBJECTIVE[j][0];
             label=j;
         }
      if(label!=0) {
         for(k=0; k<=M; k++) {
             a=OBJECTIVE[i][k];
             OBJECTIVE[i][k]=OBJECTIVE[label][k];
             OBJECTIVE[label][k]=a;
         }
         for(j=1; j<=N; j++) {
             a=CHROMOSOME[i][j];
             CHROMOSOME[i][j]=CHROMOSOME[label][j];
             CHROMOSOME[label][j]=a;
         }
      }
  }
  //add the following 2 line, add i=0
       for(k=0; k<=M; k++) OBJECTIVE[0][k]=OBJECTIVE[1][k];
     for(j = 1; j <= N; j++) CHROMOSOME[0][j]=CHROMOSOME[1][j];
}

static void selection()
{
  double r, temp[POP_SIZE+1][N+1];
  int   i, j, k;
  for(i=1; i<=POP_SIZE; i++) {
      r=myu(0, q[POP_SIZE]);
      for(j=0; j<=POP_SIZE; j++) {
          if(r<=q[j]) {
              for(k=1; k<=N; k++) temp[i][k]=CHROMOSOME[j][k];
              break;
          }
      }
  }
  for(i=1; i<=POP_SIZE; i++)
     for(k=1; k<=N; k++)
         CHROMOSOME[i][k]=temp[i][k];
}

static void crossover()
{
  int   i, j, jj, k, pop;
  double r, x[N+1], y[N+1];
  pop=POP_SIZE/2;
  for(i=1; i<=pop; i++) {
     if(myu(0,1)>P_CROSSOVER) continue;
     j=(int)myu(1,POP_SIZE);
     jj=(int)myu(1,POP_SIZE);
     r=myu(0,1);
     for(k=1; k<=N; k++) {
         x[k]=r*CHROMOSOME[j][k]+(1-r)*CHROMOSOME[jj][k];
         y[k]=r*CHROMOSOME[jj][k]+(1-r)*CHROMOSOME[j][k];
     }
     if(constraint_check(x)==1)
         for(k=1; k<=N; k++) CHROMOSOME[j][k]=x[k];
     if(constraint_check(y)==1)
         for(k=1; k<=N; k++) CHROMOSOME[jj][k]=y[k];
  }
}

static void mutation(void)
{
  int i, j, k;
  double x[N+1], y[N+1], infty, direction[N+1];
  double INFTY=10, precision=0.0001;
  for(i=1; i<=POP_SIZE; i++) {
      if(myu(0,1)>P_MUTATION) continue;
      for(k=1; k<=N; k++) x[k] = CHROMOSOME[i][k];
      for(k=1; k<=N; k++)
          if(myu(0,1)<0.5) direction[k]=myu(-1,1);
          else direction[k]=0;
      infty=myu(0,INFTY);
      while(infty>precision) {
          for(j=1; j<=N; j++) y[j]=x[j]+infty*direction[j];
          if(constraint_check(y)==1) {
             for(k=1; k<=N; k++) CHROMOSOME[i][k]=y[k];
             break;
          }
          infty=myu(0,infty);
      }
  }
}
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2014-02-21 10:27
bob421559385
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等 级:新手上路
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回复 楼主 bob421559385
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2014-02-21 10:28
快速回复:哪位大神能给看看 帮忙注释下
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