遗传算法求函数最值(C语言实现)

好的，我可以为您提供一个遗传算法求函数最值的C语言实现。下面是代码：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>

#define POP_SIZE 100         // 种群数量
#define GENES_LEN 6          // 基因长度
#define CROSSOVER_RATE 0.7   // 交叉概率
#define MUTATION_RATE 0.001  // 变异概率
#define MAX_GENERATION 1000  // 最大迭代次数

// 随机生成一个二进制基因
int *generate_genes() {
    int *genes = (int *)malloc(GENES_LEN * sizeof(int));
    for (int i = 0; i < GENES_LEN; i++) {
        genes[i] = rand() % 2;
    }
    return genes;
}

// 计算二进制基因所代表的十进制数值
double decode(int *genes) {
    double value = 0.0;
    for (int i = 0; i < GENES_LEN; i++) {
        value += genes[i] * pow(2, GENES_LEN - 1 - i);
    }
    return value;
}

// 计算目标函数的值
double fitness(double x) {
    return sin(10 * M_PI * x) / (2 * x) + pow(x - 1, 4);
}

// 计算种群中每个个体的适应度
double *calculate_fitness(int **pop) {
    double *fitness_values = (double *)malloc(POP_SIZE * sizeof(double));
    for (int i = 0; i < POP_SIZE; i++) {
        double x = decode(pop[i]);
        fitness_values[i] = fitness(x);
    }
    return fitness_values;
}

// 从种群中选择两个个体进行交叉
void crossover(int *parent1, int *parent2, int *child1, int *child2) {
    if ((double)rand() / RAND_MAX < CROSSOVER_RATE) {
        int crossover_point = rand() % GENES_LEN;
        for (int i = 0; i < crossover_point; i++) {
            child1[i] = parent1[i];
            child2[i] = parent2[i];
        }
        for (int i = crossover_point; i < GENES_LEN; i++) {
            child1[i] = parent2[i];
            child2[i] = parent1[i];
        }
    } else {
        for (int i = 0; i < GENES_LEN; i++) {
            child1[i] = parent1[i];
            child2[i] = parent2[i];
        }
    }
}

// 对个体进行变异
void mutate(int *genes) {
    for (int i = 0; i < GENES_LEN; i++) {
        if ((double)rand() / RAND_MAX < MUTATION_RATE) {
            genes[i] = !genes[i];
        }
    }
}

// 选择一个个体
int *select(double *fitness_values, int **pop) {
    double sum = 0.0;
    for (int i = 0; i < POP_SIZE; i++) {
        sum += fitness_values[i];
    }
    double r = (double)rand() / RAND_MAX * sum;
    double s = 0.0;
    for (int i = 0; i < POP_SIZE; i++) {
        s += fitness_values[i];
        if (s >= r) {
            return pop[i];
        }
    }
    return pop[POP_SIZE - 1];
}

// 打印最优解
void print_best(double best_x, double best_fitness, int generation) {
    printf("Generation %d: x = %lf, f(x) = %lf\n", generation, best_x, best_fitness);
}

int main() {
    srand(time(NULL));

    // 初始化种群
    int **pop = (int **)malloc(POP_SIZE * sizeof(int *));
    for (int i = 0; i < POP_SIZE; i++) {
        pop[i] = generate_genes();
    }

    // 开始迭代
    double best_fitness = -1e100;
    double best_x = 0.0;
    for (int generation = 1; generation <= MAX_GENERATION; generation++) {
        double *fitness_values = calculate_fitness(pop);

        // 计算种群中最优解
        for (int i = 0; i < POP_SIZE; i++) {
            double x = decode(pop[i]);
            double f = fitness(x);
            if (f > best_fitness) {
                best_fitness = f;
                best_x = x;
            }
        }
        print_best(best_x, best_fitness, generation);

        // 新种群
        int **new_pop = (int **)malloc(POP_SIZE * sizeof(int *));
        for (int i = 0; i < POP_SIZE; i++) {
            // 选择两个个体进行交叉
            int *parent1 = select(fitness_values, pop);
            int *parent2 = select(fitness_values, pop);
            int *child1 = (int *)malloc(GENES_LEN * sizeof(int));
            int *child2 = (int *)malloc(GENES_LEN * sizeof(int));
            crossover(parent1, parent2, child1, child2);

            // 对交叉后的两个个体进行变异
            mutate(child1);
            mutate(child2);

            new_pop[i] = child1;
            new_pop[i + 1] = child2;
        }

        // 释放旧种群
        for (int i = 0; i < POP_SIZE; i++) {
            free(pop[i]);
        }
        free(pop);

        // 更新种群
        pop = new_pop;
        free(fitness_values);
    }

    return 0;
}

这个程序使用遗传算法来求解一个函数的最值。其中，种群数量为100，每个个体的基因长度为6位，交叉概率为0.7，变异概率为0.001，最大迭代次数为1000次。您可以根据需要进行修改。