• 一个简单的遗传算法应用(java版)


    计算智能课的遗传算法上完了,老师布置了一个简单的应用,用java实现了下。

    首先是题目要求:

    A given function is as follows:

    Use genetic algorithm to find a near-maximal value in

    f==xsin(10*pi*x)+2

    [-1,2]. In addition, the required precision is six places after the decimal point.

    代码实现如下:

    import java.util.Random;
    public class GA {
    public static final double A=-1;//下界
    public static final double B=2;//
    public static final int POP_SIZE=30;//种群数目
    public static final int M=22; //编码位数
    public static String[]pop=new String[POP_SIZE];//种群编码
    public static double[]result=new double[POP_SIZE];//种群代表的结果
    public static final int LENGTH=22;//编码长度,因为要精确到小数点后六位,所以编为22位长,有一公式可参考

    public static final int MJ2=4194304;//2^22
    public static double[]fitness=new double[POP_SIZE];//存放种群适应度
    public static final double PC=0.95;//交叉率
    public static final double PM=0.05;//变异率
    public static double[]p=new double[POP_SIZE];//轮盘赌方法个体适应度概率
    public static double[]q=new double[POP_SIZE];//q[i]是前n项p之和
    public static Random random=new Random();//用于产生随机数的工具
    public static Best best=new Best();//记录最佳答案的对象

    /*
    * 构造函数,初始化种群
    */
    public GA(double d[])
    {
       for (int i = 0; i < d.length; i++) {
        result[i]=d[i];
       }
    }
    /*
    * 编码方法,将解值表示为二进制字节码形式
    */
    public void encoding()

    {
      
       for (int i = 0; i < result.length; i++) {
        double d1=((result[i]-A)/(B-A))*(MJ2-1);
       
        pop[i]=Integer.toBinaryString((int)d1);
       }

       for (int i = 0; i < pop.length; i++) {
        if (pop[i].length()<LENGTH) {
         int k=LENGTH-pop[i].length();
         for (int j = 0; j <k ; j++) {
          pop[i]="0"+pop[i];
         }
        }
       }
       for (int i = 0; i < pop.length; i++) {
        //System.out.println(pop[i]+" "+pop[i].length());
       }
      

    }
    /*
    * 解码方法,讲二进制字节码还原为解
    */
    public void decoding()
    {
       for (int i = 0; i < pop.length; i++) {
        int k=Integer.parseInt(pop[i], 2);
        result[i]=A+k*(B-A)/(MJ2-1);
       
       }
       for (int i = 0; i < result.length; i++) {
       // System.out.println(result[i]);
       }
    }
    /*
    * 适应度函数
    */
    public void fitness()
    {
       for (int i = 0; i < result.length; i++) {
        fitness[i]=result[i]*(Math.sin(10*Math.PI*result[i]))+2;
        //System.out.println(fitness[i]);
       }
    }
    /*
    * 交叉操作
    */
    public void crossover()
    {
      
      
       for (int i = 0; i < POP_SIZE/2; i++) {
        double d=random.nextDouble();
        if(d<PC)
        {
         int k1=random.nextInt(POP_SIZE);
         int k2=random.nextInt(POP_SIZE);
         do {
          k1=(int)random.nextInt(POP_SIZE);
          k2=(int)random.nextInt(POP_SIZE);
         } while (k1==k2);
         int position=random.nextInt(LENGTH);
         //System.out.println("crossover position="+position+" "+k1+ " "+k2);
         String s11=null,s12=null,s21=null,s22=null;
         //System.out.println(pop[k1]+"   "+pop[k1].length());
         s11=pop[k1].substring(0, position);    
         s12=pop[k1].substring(position,LENGTH);
        
         //System.out.println(pop[k2]+"   "+pop[k2].length());
         s21=pop[k2].substring(0, position);
         s22=pop[k2].substring(position,LENGTH);
        
         pop[k1]=s11+s22;
         pop[k2]=s21+s12;
        
        
        }
       
       }
    }
    /*
    * 变异操作
    */
    public void mutation()
    {
       for (int i = 0; i < pop.length; i++) {
        for (int j = 0; j < LENGTH; j++) {
         double k=random.nextDouble();
         if(PM>k)
         {
          mutation(i,j);
         }
        }
       }
    }
    public void mutation(int i,int j)
    {
       String s=pop[i];
       StringBuffer sb=new StringBuffer(s);
       if(sb.charAt(j)=='0')
        sb.setCharAt(j, '1');
       else
        sb.setCharAt(j, '0');
       pop[i]=sb.toString();
      
    }
    /*
    * 轮盘赌方法
    */
    public void roulettewheel()
    {
       decoding();
       fitness();
      
       double sum=0;
       for (int i = 0; i <POP_SIZE; i++) {
        sum=fitness[i]+sum;
       }
       for (int i = 0; i < POP_SIZE; i++) {
        p[i]=fitness[i]/sum;
       }
       for (int i = 0; i < POP_SIZE; i++) {
        for (int j = 0; j < i+1; j++) {
         q[i]=p[j];
        }
       }
       double []ran=new double[POP_SIZE];
       String[] tempPop=new String[POP_SIZE];
       for (int i = 0; i < ran.length; i++) {
        ran[i]=random.nextDouble();
       }
       for (int i = 0; i < ran.length; i++) {
        int k = 0;
        for (int j = 0; j < q.length; j++) {
         if(ran[i]<q[j])
         {
          k=j;
          break;
         }
         else continue;
        }
        tempPop[i]=pop[k];
       }
       for (int i = 0; i < tempPop.length; i++) {
        pop[i]=tempPop[i];
       }
    }

    /*
    * 一次进化
    */
    public void evolution()
    {
       encoding();
       crossover();
       mutation();
       decoding();
       fitness();
       roulettewheel();
       findResult();
      
    }
    /*
    *整个进化过程,n 表示进化多少代
    */
    public void dispose(int n)
    {
       for (int i = 0; i < n; i++) {
        evolution();
       }
    }
    /*
    * 取得结果
    */
    public double findResult()
    {
       if(best==null)
        best=new Best();
       double max=best.fitness;
       for (int i = 0; i < fitness.length; i++) {
        if(fitness[i]>max)
        {
         best.fitness=fitness[i];
         best.x=result[i];
         best.str=pop[i];
         //System.out.println(best.fitness);
        }
       }
       return max;
    }
    /*
    * 取得x值
    *
    */
    public double findx()
    {
       fitness();
       double max=0;
       int index=0;
       for (int i = 0; i < fitness.length; i++) {
        System.out.println(result[i]);
        if(fitness[i]>max)
        {
         max=fitness[i];
         index=i;
        }
       }
       return result[index];
    }
    public static void main(String[] args) {
       //d为初试种群
       double d[]={-0.953181,-0.851234,-0.749723,-0.645386,-0.551234,-0.451644,-0.351534,-0.239566,-0.151234,0.145445,
            0.245445,0.285174,0.345445,0.445445,0.542445,0.645445,0.786445,0.845445,0.923238,1.245445,
            1.383453,1.454245,1.584566,1.644345,1.741545,1.845445,1.981254,-0.012853,0.083413,1.801231};
       //初始化其它参数
       GA ga=new GA(d);
       System.out.println("种群进化中....");
       //进化,这里进化10000次
       ga.dispose(10000);
       System.out.println("+++++++++++++++++++++++++++结果为:");
       System.out.println("x="+best.x);
       System.out.println("f="+best.fitness);
      
    }

    }
    class Best { // 存储最佳的
    public int generations;
    public String str;
    public double fitness;
    public double x;
    }

  • 相关阅读:
    [转]jQuery 1.4的十五大新功能实例精讲
    数据库优化使用索引优化存储过程
    SQL语句练习实例之一——找出最近的两次晋升日期与工资额
    Sql Server中三种字符串合并方法的性能比较
    在IIS7下配置ASP.NET v1.1(Visual Studio .NET 2003)环境
    数据库优化方法 (一)
    WEB版一次选择多个文件进行批量上传(swfupload)的解决方案
    [转]根据性能监视器,分析性能瓶颈
    数据库优化方法(三)
    英文SilverLight 4 tools for vs 2010 安装于vs 2010中文版,无法拖拽数据源问题解决方法
  • 原文地址:https://www.cnblogs.com/macula7/p/1960673.html
Copyright © 2020-2023  润新知