一个简单遗传算法的例子(C#)

为了给别人说明白遗传算法的程序而编写的一个小例子  转载请注明出处

using System;
using System.Collections.Generic;
using System.Text;


// write by feathersky
// to compute  y= 3 * sin(x*x) + 5 * x;  but we don't know the 3,5  , leb a chromosome has two genes that are a and b ,use GP program to get the right a,b
namespace Gene
{
    

    class IntelligentTestSystem
    {
        public

        static int population=100;           // 种群个数
        static int maxNumber = 2;          //每个个体（染色体）所用的基因个数


        double[,] chromosome = new double[population, maxNumber];  //每个个体（染色体）的种群

        double[] fitness = new double[population];    //适应度


        int questionTypeNumber = 0;
        int[] questionType = new int[maxNumber];






        double[] sampleX = new double[10];

        double[] sampleY = new double[10];




       public void initial()
        {
            Random rand = new Random(unchecked(5 * (int)DateTime.Now.Ticks));

            for (int i = 0; i < population; i++)  //随机生成染色体
            {
                for (int j = 0; j < 2; j++)
                {
                    chromosome[i, j] = (rand.Next(200) - 100 ) / 10.0;
                }
            }

            for (int i = 0; i < 10; i++)  //函数的10个坐标点 y=3*sin(x*x)+5*x;
            {
                sampleX[i] = i;
                sampleY[i] = 3 * Math.Sin(sampleX[i] * sampleX[i]) + 5 * sampleX[i];
            }


        }

        public double calculateSingleFitness(int n)
        {
            double[] y = new double[10];

            double a = chromosome[n, 0];

            double b = chromosome[n, 1];

            double c = 0.0;      //保存误差平方和


            for (int i = 0; i < 10; i++)
            {
                y[i] = a * Math.Sin(sampleX[i] * sampleX[i]) + b * sampleX[i];

                c = c + (sampleY[i] - y[i]) * (sampleY[i] - y[i]);
            }


            fitness[n] = -c ;

            return fitness[n];

        }

        public void calculateFitness( )
        {
            for (int i = 0; i < population; i++) calculateSingleFitness(i);
        }


        public void sort()
        {
            double instead =0;

            double ch0 = 0;

            double ch1 = 0;
  
            //冒泡排序
            for(int j=0;j<population ;j++)
            {
                for (int i = population-1; i >0; i--)
                {
                    if (fitness[i] > fitness[i - 1])
                    {
                        //交换邻近的两个

                        //交换适应度
                        instead = fitness[i - 1];

                        fitness[i - 1] = fitness[i];

                        fitness[i] = instead;

                        //交换染色体
                        ch0 = chromosome[i-1,0];

                        ch1 = chromosome[i-1, 1];

                        chromosome[i - 1, 0] = chromosome[i, 0];

                        chromosome[i - 1, 1] = chromosome[i, 1];

                        chromosome[i , 0] = ch0;

                        chromosome[i, 1] = ch1;
                    }
            
                }
            }

        }


        public void intercross(int father, int mother, int son, int daughter)
        {
            chromosome[son,0] = chromosome[father,0];
            chromosome[son,1] = chromosome[mother,1];


            chromosome[daughter, 0] = 0.5 * chromosome[father, 0] + 0.5 * chromosome[mother, 0];
            chromosome[daughter, 1] = 0.5 * chromosome[father, 1] + 0.5 * chromosome[mother, 1];           
           
        }

        public void aberrance(int father, int son)
        {
            Random rand = new Random(unchecked(5*(int)DateTime.Now.Ticks));

            chromosome[son, 0] = chromosome[father, 0] + (rand.Next(200) - 100) / 100.0;
            chromosome[son, 1] = chromosome[father, 1] + (rand.Next(200) - 100) / 100.0;
        }



        public void generate()
        {

            Random rand = new Random(unchecked(7 * (int)DateTime.Now.Ticks));


            for (int k = 31; k < 80; k+=2)
            {
                int father = rand.Next(30);
                int mother = rand.Next(30);


                int son = k;
                int daughter = k+1;

                intercross(father, mother, son, daughter);
            }

            for (int k = 81; k < 100; k ++)
            {
                int father = rand.Next(30);
                
                int son = k;

                aberrance(father, son);
            }




        }











    }
}