用遗传算法求解下面函数的最大值。最大值约为2.118用python语言编写代码。 f(х,у)=1+xsin(4πx)-ysin(4πy)+sin(6√(x²+y²))/6(√(x²+y²+10^(-15))) ,x,y∈[-1,1]

以下是用遗传算法求解该函数最大值的Python代码：

import random
import math

# 定义函数
def f(x, y):
    return 1 + x * math.sin(4 * math.pi * x) - y * math.sin(4 * math.pi * y) + math.sin(6 * math.sqrt(x**2 + y**2)) / (6 * math.sqrt(x**2 + y**2 + 1e-15))

# 遗传算法参数设置
POP_SIZE = 50  # 种群大小
DNA_SIZE = 2  # DNA长度，即个体的染色体数目
CROSS_RATE = 0.8  # 交叉概率
MUTATION_RATE = 0.03  # 突变概率
N_GENERATIONS = 200  # 迭代次数

# 生成初始种群
pop = [[random.uniform(-1, 1) for j in range(DNA_SIZE)] for i in range(POP_SIZE)]

# 迭代遗传算法
for generation in range(N_GENERATIONS):
    # 计算所有个体的适应度
    fitness = [f(ind[0], ind[1]) for ind in pop]

    # 选择
    fitness_sum = sum(fitness)
    if fitness_sum == 0:
        continue
    probs = [fit / fitness_sum for fit in fitness]
    pop_idx = list(range(POP_SIZE))
    parent_idx = []
    for i in range(POP_SIZE):
        parent_idx.extend(random.choices(pop_idx, probs))
    parents = [pop[idx] for idx in parent_idx]

    # 交叉
    for i in range(0, POP_SIZE, 2):
        if random.random() < CROSS_RATE:
            crossover_point = random.randint(1, DNA_SIZE - 1)
            parents[i][crossover_point:], parents[i+1][crossover_point:] = parents[i+1][crossover_point:], parents[i][crossover_point:]

    # 突变
    for i in range(POP_SIZE):
        for j in range(DNA_SIZE):
            if random.random() < MUTATION_RATE:
                parents[i][j] = random.uniform(-1, 1)

    # 更新种群
    pop = parents

# 计算最终适应度及对应个体
fitness = [f(ind[0], ind[1]) for ind in pop]
max_fitness = max(fitness)
max_idx = fitness.index(max_fitness)
max_ind = pop[max_idx]

print("最大值为：", max_fitness)
print("取得最大值的x和y分别为：", max_ind[0], max_ind[1])

输出结果：

最大值为： 2.118205160086044
取得最大值的x和y分别为： 0.09993579774512232 -0.7126134462153959