5.2 用状态空间搜索法求解农夫、狐狸、鸡、小米问题。农夫、狐狸、鸡、小米都在一条河的左岸，现在要把它们全部送到右岸去。农夫有一条船，过河时，除农夫外，船上至多能载狐狸、鸡和小米中的一样。狐狸要吃鸡，鸡要吃小米，除非农夫在那里。试规划出一个确保全部安全的过河计划。(提示:8.用四元组(农夫，狐狸，鸡，米)表示状态，其中每个元素都可为0或1,0表示在左岸，1表示在右岸:b.把每次过河的一种安排作为一个算符，每次过河都必须有农夫，因为只有他可以划船。)

6.2 用遗传算法求解下列函数的最大值,f(x)=0.4+sinc(4x)+1.1sinc(4x+2)+0.8sinc(x-2)+0.7sinc(6x-4),x∈[-2,2]

以下是一个简单的使用遗传算法求解函数最大值的代码（Python）：

import numpy as np

# 定义目标函数def fitness_function(x):

    return 0.4+sinc(4x)+1.1sinc(4x+2)+0.8sinc(x-2)+0.7sinc(6x-4)

# 初始化种群def initialize_population(population_size, chromosome_length):

    return np.random.randint(0, 30, (population_size, chromosome_length))

# 计算适应度def calculate_fitness(population):

    return np.array([fitness_function(chromosome) for chromosome in population])

# 选择操作def selection(population, fitness):

    return population[np.argmax(fitness)]

# 交叉操作def crossover(parent1, parent2):

    crossover_point = np.random.randint(1, len(parent1))

    child1 = np.concatenate((parent1[:crossover_point], parent2[crossover_point:]))

    child2 = np.concatenate((parent2[:crossover_point], parent1[crossover_point:]))

    return child1, child2

# 变异操作def mutation(chromosome):

    mutation_point = np.random.randint(len(chromosome))

    chromosome[mutation_point] = np.random.randint(0, 10)

    return chromosome

# 遗传算法主程序def genetic_algorithm(population_size, chromosome_length, generations):

    population = initialize_population(population_size, chromosome_length)

    for _ in range(generations):

        fitness = calculate_fitness(population)

        parent1 = selection(population, fitness)

        parent2 = selection(population, fitness)

        child1, child2 = crossover(parent1, parent2)

        child1 = mutation(child1)

        child2 = mutation(child2)

        population = np.vstack((population, child1, child2))

    best_chromosome = selection(population, calculate_fitness(population))

    return best_ch