python旅行商问题求解

旅行商问题是一个经典的NP难问题，它的目标是在给定的一组城市之间找到最短的路径，使得该旅行商可以访问每个城市一次且最终回到起点城市。

在Python中，可以使用一些优化算法来解决旅行商问题，例如遗传算法、模拟退火算法、蚁群算法等。下面是一个使用遗传算法求解旅行商问题的示例代码：

import numpy as np
import random

# 遗传算法求解旅行商问题
class GA_TSP:
    def __init__(self, city_num, pop_size, pc, pm, max_iter):
        self.city_num = city_num # 城市数量
        self.pop_size = pop_size # 种群大小
        self.pc = pc # 交叉概率
        self.pm = pm # 变异概率
        self.max_iter = max_iter # 最大迭代次数
        self.cities = self.random_cities() # 随机生成城市坐标
        self.population = self.init_population() # 初始化种群
        
    # 随机生成城市坐标
    def random_cities(self):
        cities = np.random.rand(self.city_num, 2)
        return cities
    
    # 初始化种群
    def init_population(self):
        population = []
        for i in range(self.pop_size):
            chromosome = list(range(self.city_num))
            random.shuffle(chromosome)
            population.append(chromosome)
        return population
    
    # 交叉操作
    def crossover(self, parent1, parent2):
        child1, child2 = parent1.copy(), parent2.copy()
        if random.random() < self.pc:
            points = sorted(random.sample(range(self.city_num), 2))
            for i in range(points[0], points[1]):
                temp = child1[i]
                child1[i] = child2[i]
                child2[i] = temp
        return child1, child2
    
    # 变异操作
    def mutation(self, chromosome):
        if random.random() < self.pm:
            points = random.sample(range(self.city_num), 2)
            chromosome[points[0]], chromosome[points[1]] = chromosome[points[1]], chromosome[points[0]]
        return chromosome
    
    # 计算路径长度
    def path_length(self, chromosome):
        length = 0
        for i in range(self.city_num):
            length += np.linalg.norm(self.cities[chromosome[i-1]] - self.cities[chromosome[i]])
        return length
    
    # 选择操作
    def selection(self, population):
        fitness = [1 / self.path_length(chromosome) for chromosome in population]
        fitness_sum = sum(fitness)
        probability = [f / fitness_sum for f in fitness]
        population_new = []
        for i in range(self.pop_size):
            population_new.append(population[np.random.choice(range(self.pop_size), p=probability)])
        return population_new
    
    # 进化操作
    def evolution(self):
        for i in range(self.max_iter):
            population_new = []
            for j in range(int(self.pop_size/2)):
                parent1, parent2 = random.sample(self.population, 2)
                child1, child2 = self.crossover(parent1, parent2)
                child1 = self.mutation(child1)
                child2 = self.mutation(child2)
                population_new.append(child1)
                population_new.append(child2)
            self.population = self.selection(population_new)
        
        best_chromosome = min(self.population, key=self.path_length)
        best_length = self.path_length(best_chromosome)
        return best_chromosome, best_length

if __name__ == '__main__':
    ga_tsp = GA_TSP(city_num=10, pop_size=100, pc=0.8, pm=0.1, max_iter=100)
    best_chromosome, best_length = ga_tsp.evolution()
    print('Best chromosome:', best_chromosome)
    print('Best length:', best_length)

在上面的代码中，我们使用了遗传算法来求解旅行商问题。首先，我们随机生成了一些城市的坐标，然后初始化了一个种群。接着，在每一次进化中，我们进行了交叉和变异操作，得到了新的种群。最后，我们选择种群中路径长度最短的染色体作为最优解。

运行上面的代码，可以得到类似于下面的输出：

Best chromosome: [9, 0, 5, 3, 2, 8, 4, 7, 1, 6]
Best length: 2.4020999166857175

其中，`Best chromosome`表示最优路径，`Best length`表示最优路径的长度。