北京等六城市的距离表如下,请据此表求解TSP问题。即给定N个城市,从某一城市开始走遍所有城市但不许重复,最后回到出发点,要求总路径最短。 上海 天津 石家庄 太原 呼和浩特 北京 1078 119 263 398 401 上海 0 963 989 1096 1381 天津 0 262 426 504 石家庄 0 171 394 太原 0 341 要求使用遗传算法求解并给我代码
时间: 2024-02-18 20:02:26 浏览: 60
好的,我可以帮您解答这个问题,并提供遗传算法的Python代码。
遗传算法是一种启发式算法,能够在大规模的搜索空间中找到较优的解。在TSP问题中,可以将每个城市看作是一个基因,将一条路径看作是一个个体,通过交叉、变异等操作来生成新的个体,最终找到一条总路径最短的路径。
以下是基于遗传算法的Python代码,用于求解上述TSP问题:
```python
import numpy as np
# 城市距离矩阵
dist_matrix = np.array([[0, 1078, 119, 263, 398, 401],
[963, 0, 989, 1096, 1381, 1078],
[262, 0, 0, 426, 504, 119],
[171, 262, 0, 0, 394, 263],
[0, 0, 171, 341, 0, 398],
[0, 0, 0, 0, 0, 0]])
# 遗传算法参数
POP_SIZE = 50 # 种群大小
CROSS_RATE = 0.8 # 交叉概率
MUT_RATE = 0.02 # 变异概率
N_GENERATIONS = 200 # 迭代次数
# 生成初始种群
def init_pop(n_cities, pop_size):
pop = np.zeros((pop_size, n_cities), dtype=int)
for i in range(pop_size):
pop[i] = np.random.permutation(n_cities)
return pop
# 计算总路程
def get_dist(path):
dist = 0
for i in range(len(path)-1):
dist += dist_matrix[path[i]][path[i+1]]
dist += dist_matrix[path[-1]][path[0]]
return dist
# 计算适应度
def get_fitness(pop):
fitness = np.zeros((len(pop),))
for i, path in enumerate(pop):
fitness[i] = 1 / get_dist(path)
return fitness
# 选择操作
def select(pop, fitness):
idx = np.random.choice(np.arange(len(pop)), size=len(pop), replace=True, p=fitness/fitness.sum())
return pop[idx]
# 交叉操作
def crossover(parent, pop):
if np.random.rand() < CROSS_RATE:
i_ = np.random.randint(0, len(pop), size=1)
cross_points = np.random.randint(0, 2, len(parent)).astype(np.bool)
keep_city = parent[~cross_points]
swap_city = pop[i_, np.isin(pop[i_].ravel(), keep_city, invert=True)]
parent[:] = np.concatenate((keep_city, swap_city))
return parent
# 变异操作
def mutate(child):
for point in range(len(child)):
if np.random.rand() < MUT_RATE:
swap_point = np.random.randint(0, len(child))
child[point], child[swap_point] = child[swap_point], child[point]
return child
# 进化操作
def evolve(pop):
fitness = get_fitness(pop)
pop = select(pop, fitness)
pop_copy = pop.copy()
for parent in pop:
child = crossover(parent, pop_copy)
child = mutate(child)
parent[:] = child
return pop
# 主函数
def main(n_cities, pop_size, n_generations):
pop = init_pop(n_cities, pop_size)
for i in range(n_generations):
pop = evolve(pop)
best_path = pop[np.argmax(get_fitness(pop))]
best_dist = get_dist(best_path)
print('Generation:', i, '| Best distance:', best_dist)
return best_path, best_dist
if __name__ == '__main__':
best_path, best_dist = main(n_cities=6, pop_size=POP_SIZE, n_generations=N_GENERATIONS)
print('Best path:', best_path, '| Best distance:', best_dist)
```
代码中的注释已经比较详细了,大致思路是通过初始化种群、计算适应度、选择操作、交叉操作和变异操作等步骤来进行进化,最终找到一条总路径最短的路径。
希望能帮助到您!
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