人工智能那些事2遗传算法求解TSP问题
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遗传算法求解TSP问题
python程序:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
plt.rcParams['font.sans-serif'] = ["SimHei"]
# 载入数据
def load_data():
df = pd.read_csv('./TSP问题测试数据集/oliver30.tsp', sep=" ", skiprows=6, header=None)
city = np.array(df[0][0:len(df) - 1]) # 最后一行为EOF,不读入
city_name = city.tolist()
city_x = np.array(df[1][0:len(df) - 1])
city_y = np.array(df[2][0:len(df) - 1])
city_location = list(zip(city_x, city_y))
return city_name, city_location
# 计算两个城市的欧式距离
def dist_cal(x, y):
return ((x[0] - y[0]) ** 2 + (x[1] - y[1]) ** 2) ** 0.5
# 求距离矩阵
def matrix_dis(city_name, city_location):
city_num = len(city_name) - 1
res = np.zeros((city_num, city_num))
for i in range(city_num):
for j in range(i + 1, city_num):
res[i, j] = dist_cal(city_location[i], city_location[j]) # 求两点欧式距离
res[j, i] = res[i, j] # 距离矩阵:对角线为0,对称
return res
# 初始化种群
def rand_pop(city_num, pop_num, pop, distance, matrix_distance):
rand_ch = np.array(range(city_num))
for i in range(pop_num):
np.random.shuffle(rand_ch)
pop[i, :] = rand_ch
distance[i] = comp_dis(city_num, matrix_distance, rand_ch) # 这里的适应度其实是距离
# 计算每个个体的总距离
def comp_dis(city_num, matrix_distance, one_path):
res = 0
for i in range(city_num - 1):
res += matrix_distance[one_path[i], one_path[i + 1]]
res += matrix_distance[one_path[-1], one_path[0]] # 最后一个城市和第一个城市的距离,需单独处理
return res
# 打印出当前路径
def print_path(city_num, one_path):
res = str(one_path[0] + 1) + '-->'
for i in range(1, city_num):
res += str(one_path[i] + 1) + '-->'
res += str(one_path[0] + 1)
print("最佳路径为:")
print(res)
# 轮盘赌的方式选择子代
def select_sub(pop_num, pop, distance):
fit = 1. / distance # 适应度函数
p = fit / sum(fit)
q = p.cumsum() # 累积概率
select_id = []
for i in range(pop_num):
r = np.random.rand() # 产生一个[0,1)的随机数
for j in range(pop_num):
if r < q[0]:
select_id.append(0)
break
elif q[j] < r <= q[j + 1]:
select_id.append(j + 1)
break
next_gen = pop[select_id, :]
return next_gen
# 交叉操作-每个个体对的某一位置进行交叉
def cross_sub(city_num, pop_num, next_gen, cross_prob, evbest_path):
for i in range(0, pop_num):
best_gen = evbest_path.copy()
if cross_prob >= np.random.rand():
next_gen[i, :], best_gen = intercross(city_num, next_gen[i, :], best_gen)
# 具体的交叉方式:部分映射交叉(Partial-Mapped Crossover)
def intercross(city_num, ind_a, ind_b):
r1 = np.random.randint(city_num)
r2 = np.random.randint(city_num)
while r2 == r1:
r2 = np.random.randint(city_num)
left, right = min(r1, r2), max(r1, r2)
ind_a1 = ind_a.copy()
ind_b1 = ind_b.copy()
for i in range(left, right + 1):
ind_a2 = ind_a.copy()
ind_b2 = ind_b.copy()
ind_a[i] = ind_b1[i]
ind_b[i] = ind_a1[i]
# 每个个体包含的城市序号是唯一的,因此交叉时若两个不相同,就会产生冲突
x = np.argwhere(ind_a == ind_a[i])
y = np.argwhere(ind_b == ind_b[i])
# 产生冲突,将不是交叉区间的数据换成换出去的原数值,保证城市序号唯一
if len(x) == 2:
ind_a[x[x != i]] = ind_a2[i]
if len(y) == 2:
ind_b[y[y != i]] = ind_b2[i]
return ind_a, ind_b
# 变异方式:翻转变异
def mutation_sub(city_num, pop_num, next_gen, mut_prob):
for i in range(pop_num):
if mut_prob >= np.random.rand():
r1 = np.random.randint(city_num)
r2 = np.random.randint(city_num)
while r2 == r1:
r2 = np.random.randint(city_num)
if r1 > r2:
temp = r1
r1 = r2
r2 = temp
next_gen[i, r1:r2] = next_gen[i, r1:r2][::-1]
# 绘制路径图
def draw_path(city_num, city_location, pop, distance):
fig, ax = plt.subplots()
x, y = zip(*city_location)
ax.scatter(x, y, linewidths=0.1)
for i, txt in enumerate(range(1, len(city_location) + 1)):
ax.annotate(txt, (x[i], y[i]))
res0 = pop
x0 = [x[i] for i in res0]
y0 = [y[i] for i in res0]
ax.annotate("起点", (x0[0], y[0]))
ax.annotate("终点", (x0[-1], y[-1]))
# 绘制箭图
for i in range(city_num - 1):
plt.quiver(x0[i], y0[i], x0[i + 1] - x0[i], y0[i + 1] - y0[i], color='b', width=0.005, angles='xy', scale=1,
scale_units='xy')
plt.quiver(x0[-1], y0[-1], x0[0] - x0[-1], y0[0] - y0[-1], color='b', width=0.005, angles='xy', scale=1,
scale_units='xy')
plt.title("遗传算法优化路径-最短距离:" + str(distance))
plt.xlabel("城市位置横坐标")
plt.xlabel("城市位置纵坐标")
plt.savefig("map.png")
plt.show()
# 绘制最优解随迭代次数的关系
def draw_iter(iteration, best_distance_list):
iteration = np.linspace(1, iteration, iteration)
plt.plot(iteration, best_distance_list)
plt.xlabel("迭代次数")
plt.ylabel("最短路径长度")
plt.savefig("figure.png")
plt.show()
def main():
city_name, city_location = load_data()
matrix_distance = matrix_dis(city_name, city_location)
city_num = len(city_name) - 1 # 城市数量
pop_num = 300 # 群体个数
cross_prob = 0.95 # 交叉概率
mut_prob = 0.5 # 变异概率
iteration = 100 # 迭代代数
# 初始化初代种群和距离,个体为整数,距离为浮点数
pop = np.array([0] * pop_num * city_num).reshape(pop_num, city_num)
distance = np.zeros(pop_num)
# 初始化种群
rand_pop(city_num, pop_num, pop, distance, matrix_distance)
# draw_path(city_num, city_location, pop[0], distance) # 绘制初代图像
evbest_path = pop[0]
evbest_distance = float("inf")
best_path_list = []
best_distance_list = []
# 循环迭代遗传过程
for i in range(iteration):
# 选择
next_gen = select_sub(pop_num, pop, distance)
# 交叉
cross_sub(city_num, pop_num, next_gen, cross_prob, evbest_path)
# 变异
mutation_sub(city_num, pop_num, next_gen, mut_prob)
# 更新每个个体距离值(1/适应度)
for j in range(pop_num):
distance[j] = comp_dis(city_num, matrix_distance, next_gen[j, :])
index = distance.argmin() # index 记录最小总路程
# 为了防止曲线波动,每次记录最优值,如迭代后出现退化,则将当前最好的个体回退替换为历史最佳
if distance[index] <= evbest_distance:
evbest_distance = distance[index]
evbest_path = next_gen[index, :]
else:
distance[index] = evbest_distance
next_gen[index, :] = evbest_path
# 存储每一步的最优路径(个体)及距离
best_path_list.append(evbest_path)
best_distance_list.append(evbest_distance)
# 绘制迭代次数和最优解的关系曲线
draw_iter(iteration, best_distance_list)
best_path = evbest_path
best_distance = evbest_distance
# 迭代完成,打印出最佳路径
print_path(city_num, best_path)
print("当前最佳距离为:", best_distance)
# 绘制路径图
draw_path(city_num, city_location, best_path, best_distance)
if __name__ == '__main__':
main()
pdf版本和数据集可以加Q群 710413249 获取
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