Question

我也试图在java中实现GA来解决方程式（此时不重要）

我有一个实施可比较的染色体课程（按健身排序）和我自己的IChromosome界面。

我在这个课程中提供了交叉，变异等功能：这样的一些事情：

public interface IChromosome
{

    public int getSize();
    public void setSize(int s);

    public IGene[] getGenes();
    public void setGenes(IGene[] genes);

    public IGene getGene(int i);
    public void setGene(int i , IGene val);

    public double getFitness();

    public IChromosome[] crossOver(IChromosome c2);
    public IChromosome mutate();

}

我的GA计划的主要部分是：

public class GeneticAlgorithm
{
    final int populationSize;
    int crossOverRate;// Probability in persent
    int mutationRate;// Probability in persent
    int elitismRate;// in persent
    int TournamentNumber;//number of random Chromosomes to compete in Tournament


    int repeatCount;


    Chromosome [] currentGeneration;
    Chromosome [] nextGeneration;


    public GeneticAlgorithm(int populationSize, int crossOverRate, int mutationRate, int ElitismRate, int TournamentNumber)
    {
        this.populationSize = populationSize;
        this.crossOverRate = crossOverRate;
        this.mutationRate = mutationRate;
        this.elitismRate = ElitismRate;
        this.TournamentNumber = TournamentNumber;

        repeatCount = 0;

        currentGeneration = new Chromosome[populationSize];
        nextGeneration = new Chromosome[populationSize];

        for (int i = 0; i < populationSize; i++)
        {
            currentGeneration[i] =new Chromosome(Chromosome.getPolynomial().getSize());//Also randomize it; //size is from another class
        }

    }


    private Chromosome[] select()
    {
        Chromosome[] temp = new Chromosome[TournamentNumber];

        //choose populationSize random uniqe number
        ArrayList<Integer> nums = new ArrayList<>();
        for (int i = 0; i < populationSize; i++)
        {
            nums.add(i);
        }
        Collections.shuffle(nums);


        for (int i = 0; i < TournamentNumber; i++)
        {
            temp[i]=new Chromosome(currentGeneration[nums.get(i)]);
        }

        Arrays.sort(temp);
        Chromosome[] res = new Chromosome[2];

        //choosing best Chromosomes
        res[0] = new Chromosome(temp[TournamentNumber-1]);
        res[1] = new Chromosome(temp[TournamentNumber-2]);
        return res;
    }


    public void repeat()
    {
        repeatCount++;

        Arrays.sort(currentGeneration,Collections.reverseOrder());
        int elitism = (populationSize*elitismRate)/100;

        for (int i = 0; i < elitism ; i++)
        {
            nextGeneration[i] = new Chromosome(currentGeneration[i]);
        }

        int cop = 0;//Cross Over Probability
        int mp;//mutation Probability
        Random rnd = new Random();
        Chromosome [] sel;
        Chromosome [] temp;
        Chromosome ch1 , ch2;
        for (int i = elitism; i < populationSize; i++)
        {
            cop = rnd.nextInt(100);
            sel = new Chromosome[2];
            sel = select();
            if(cop <= crossOverRate)
            {
                temp = (Chromosome[]) ( sel[0].crossOver(sel[1]) );//tested , And I 'm sure cross over function working well
                ch1 = temp[0];
                ch2 = temp[1];
            }
            else
            {
                ch1 = new Chromosome(sel[0]);
                ch2 = new Chromosome(sel[1]);
            }

            mp = rnd.nextInt(100);
            if(mp< mutationRate)
            {
                ch1.mutate();
            }
            nextGeneration[i]= new Chromosome(ch1);
            if(++i<populationSize)
            {
                nextGeneration[i]= new Chromosome(ch2);
            }
        }//for

        //copiny current generation to nextGeneration
        for (int i = 0; i < populationSize; i++)
        {
            currentGeneration[i] = new Chromosome(nextGeneration[i]);
        }
    }//repeat 

}

每件事似乎都是正常的，但经过几代人（称为repeat()），我生成的所有染色体都将成为一个保存价值。

我多次调试程序，但我找不到任何东西。

问题是什么？

Answer 1

我认为如果其他事情没问题，你的算法会早期收敛。

可能的解决方案是： 1-集精英率为零; 2-降低交叉率; 3-增加突变率。

如果要解决的问题非常简单，GA早期收敛是很自然的。

遗传算法每次都提供相同的数据

1 个答案: