需要帮助查明遗传算法中的问题 Single-Point Java 中的交叉机制

Question

我一直在使用 Java 实现一个简单的遗传算法 (GA)。我的 GA 的步骤基本上是二进制编码、锦标赛选择、single-point 交叉和 bit-wise 变异。种群中的每个个体都由一个 class 表示，该 class 由二元基因和一个适应度值组成。

public class Individual {
    int gene[];
    int fitness;

    public Individual(int n){
        this.gene = new int[n];
    }
}

下面的代码不包括 bit-wise 突变部分，因为我在 GA 的 single-point 交叉部分遇到了问题。我实现 single-point 交叉算法的方法是随机为两个连续的单个数组元素找到一个点，然后交换它们的尾巴。然后对每对个体重复尾巴交换。我还创建了 printGenome() 方法来打印出所有要比较的数组，交叉过程后的结果数组没有正确交换。我单独测试了我的 single-point 交叉算法，它有效。但是，当我在下面的代码中尝试运行时，交叉根本不起作用。我可以知道这是因为锦标赛选择算法有问题吗？还是其他原因（愚蠢的错误）？我一直在修改它，但仍然无法查明错误。

如能提供任何帮助和信息，我将不胜感激！ :)

public class GeneticAlgorithm {

    public static void main(String[] args) {
        int p = 10;
        int n = 10;
        Individual population[];

        //create new population
        population = new Individual[p];

        for (int i = 0; i < p; i++) {
            population[i] = new Individual(n);
        }

        //fills individual's gene with binary randomly
        for (int i = 0; i < p; i++) {
            for (int j = 0; j < n; j++) {
                population[i].gene[j] = (Math.random() < 0.5) ? 0 : 1;
            }
            population[i].fitness = 0;
        }

        //evaluate each individual
        for (int i = 0; i < p; i++) {
            for (int j = 0; j < n; j++) {
                if (population[i].gene[j] == 1) {
                    population[i].fitness++;
                }
            }
        }

        //total fitness check
        System.out.println("Total fitness check #1 before tournament selection: " + getTotalFitness(population, p));
        System.out.println("Mean fitness check #1 before tournament selection: " + getMeanFitness(population, p));
        System.out.println("");

        //tournament selection
        Individual offspring[] = new Individual[p];

        for (int i = 0; i < p; i++) {
            offspring[i] = new Individual(n);
        }

        int parent1, parent2;
        Random rand = new Random();
        for (int i = 0; i < p; i++) {
            parent1 = rand.nextInt(p); //randomly choose parent
            parent2 = rand.nextInt(p); //randomly choose parent

            if (population[parent1].fitness >= population[parent2].fitness) {
                offspring[i] = population[parent1];
            } else {
                offspring[i] = population[parent2];
            }
        }

        //total fitness check
        System.out.println("Total fitness check #2 after tournament selection: " + getTotalFitness(offspring, p));
        System.out.println("Mean fitness check #2 after tournament selection: " + getMeanFitness(offspring, p));
        System.out.println("");

        //genome check
        System.out.println("Before Crossover: ");
        printGenome(offspring, p, n);

        //crossover
        for (int i = 0; i < p; i = i + 2) {
            int splitPoint = rand.nextInt(n);
            for (int j = splitPoint; j < n; j++) {
                int temp = offspring[i].gene[j];
                offspring[i].gene[j] = offspring[i + 1].gene[j];
                offspring[i + 1].gene[j] = temp;
            }
        }

        //genome check
        System.out.println("After Crossover:");
        printGenome(offspring, p, n);

        //evaluate each individual by counting the number of 1s after crossover
        for (int i = 0; i < p; i++) {
            offspring[i].fitness = 0;
            for (int j = 0; j < n; j++) {
                if (offspring[i].gene[j] == 1) {
                    offspring[i].fitness++;
                }
            }
        }

        //total fitness check
        System.out.println("Total fitness check #3 after crossover: " + getTotalFitness(offspring, p));
        System.out.println("Mean fitness check #3 after crossover: " + getMeanFitness(offspring, p));
    }

    public static void printGenome(Individual pop[], int p, int n) {
        for (int i = 0; i < p; i++) {
            for (int j = 0; j < n; j++) {
                System.out.print(pop[i].gene[j]);
            }
            System.out.println("");
        }
    }

    public static int getTotalFitness(Individual pop[], int p) {
        int totalFitness = 0;
        for (int i = 0; i < p; i++) {
            totalFitness = totalFitness + pop[i].fitness;
        }
        return totalFitness;
    }

    public static double getMeanFitness(Individual pop[], int p) {
        double meanFitness = getTotalFitness(pop, p) / (double) p;
        return meanFitness;
    }

}

Answer 1

问题是，在您的选择中，您（很可能）在复制个体，当您说：

offspring[i] = population[parent1]

您实际上是在 offspring[i] 中存储对 population[parent1] 的引用。因此，您的后代数组可以多次包含相同的引用，因此同一个对象将与多个合作伙伴多次参与交叉。

作为解决方案，您可以存储一个克隆而不是对同一对象的引用。在个人中添加：

    public Individual clone(){
        Individual clone = new Individual(gene.length);
        clone.gene = gene.clone();
        return clone;
    }

在您的选择中（注意添加的 .clone()）：

    for (int i = 0; i < p; i++) {
        parent1 = rand.nextInt(p); //randomly choose parent
        parent2 = rand.nextInt(p); //randomly choose parent

        if (population[parent1].fitness >= population[parent2].fitness) {
            offspring[i] = population[parent1].clone();
        } else {
            offspring[i] = population[parent2].clone();
        }
    }

这样后代中的每个元素都是不同的对象，即使基因组相同。

这解决了 Java 部分。关于 GA 理论，我希望有些东西，比如你的健身指标只是占位符，对吧？

需要帮助查明遗传算法中的问题 Single-Point Java 中的交叉机制

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