archieve: homework5

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2025-05-19 10:23:14 +08:00
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import numpy as np
import random
import matplotlib.pyplot as plt
import copy
# 各个城市的坐标
# City_Map = [[106.54,29.59]
# ,[91.11,29.97]
# ,[87.68,43.77]
# ,[106.27,38.47]
# ,[111.65,40.82]
# ,[108.33,22.84]
# ,[126.63,45.75]
# ,[125.35,43.88]
# ,[123.38,41.8]
# ,[114.48,38.03]
# ,[112.53,37.87]
# ,[101.74,36.56]
# ,[117,36.65]
# ,[113.6,34.76]
# ,[118.78,32.04]
# ,[117.27,31.86]]
City_Map = 100 * np.random.rand(20, 2) # 随机产生20个城市
DNA_SIZE = len(City_Map) # 编码长度
POP_SIZE = 200 # 种群大小
CROSS_RATE = 0.6 # 交叉率
MUTA_RATE = 0.2 # 变异率
Iterations = 1000 # 迭代次数
def distance(DNA): # 根据DNA的路线计算距离
dis = 0
temp = City_Map[DNA[0]]
for i in DNA[1:]:
dis = dis + ((City_Map[i][0] - temp[0]) ** 2 + (City_Map[i][1] - temp[1]) ** 2) ** 0.5
temp = City_Map[i]
return dis + ((temp[0] - City_Map[DNA[0]][0]) ** 2 + (temp[1] - City_Map[DNA[0]][1]) ** 2) ** 0.5
def getfitness(pop): # 计算种群适应度,这里适应度用距离的倒数表示
temp = []
for i in range(len(pop)):
temp.append(1 / (distance(pop[i])))
return temp - np.min(temp)
def select(pop, fitness): # 根据适应度选择,以赌轮盘的形式,适应度越大的个体被选中的概率越大
s = fitness.sum()
temp = np.random.choice(np.arange(len(pop)), size=POP_SIZE, replace=True, p=(fitness / s))
p = []
for i in temp:
p.append(pop[i])
return p
def mutation(DNA, MUTA_RATE): # 进行变异
if np.random.rand() < MUTA_RATE: # 以MUTA_RATE的概率进行变异
mutate_point1 = np.random.randint(0, DNA_SIZE) # 随机产生一个实数,代表要变异基因的位置
mutate_point2 = np.random.randint(0, DNA_SIZE) # 随机产生一个实数,代表要变异基因的位置
while (mutate_point1 == mutate_point2): # 保证2个所选位置不相等
mutate_point2 = np.random.randint(0, DNA_SIZE)
DNA[mutate_point1], DNA[mutate_point2] = DNA[mutate_point2], DNA[mutate_point1] # 2个所选位置进行互换
def crossmuta(pop, CROSS_RATE): # 交叉变异
new_pop = []
for i in range(len(pop)): # 遍历种群中的每一个个体,将该个体作为父代
n = np.random.rand()
if n >= CROSS_RATE: # 大于交叉概率时不发生变异,该子代直接进入下一代
temp = pop[i].copy()
new_pop.append(temp)
if n < CROSS_RATE: # 小于交叉概率时发生变异
list1 = pop[i].copy()
list2 = pop[np.random.randint(POP_SIZE)].copy() # 选取种群中另一个个体进行交叉
status = True
while status: # 产生2个不相等的节点,中间部分作为交叉段,采用部分匹配交叉
k1 = random.randint(0, len(list1) - 1)
k2 = random.randint(0, len(list2) - 1)
if k1 < k2:
status = False
k11 = k1
fragment1 = list1[k1: k2]
fragment2 = list2[k1: k2]
list1[k1: k2] = fragment2
list2[k1: k2] = fragment1
del list1[k1: k2]
left1 = list1
offspring1 = []
for pos in left1:
if pos in fragment2:
pos = fragment1[fragment2.index(pos)]
while pos in fragment2:
pos = fragment1[fragment2.index(pos)]
offspring1.append(pos)
continue
offspring1.append(pos)
for i in range(0, len(fragment2)):
offspring1.insert(k11, fragment2[i])
k11 += 1
temp = offspring1.copy()
mutation(temp, MUTA_RATE)
new_pop.append(temp) # 把部分匹配交叉后形成的合法个体加入到下一代种群
return new_pop
def print_info(pop): # 用于输出结果
fitness = getfitness(pop)
maxfitness = np.argmax(fitness) # 得到种群中最大适应度个体的索引
# 打印结果
print("最优的基因型:", pop[maxfitness])
print("最短距离:", distance(pop[maxfitness]))
# 按最优结果顺序把地图上的点加入到best_map列表中
best_map = []
for i in pop[maxfitness]:
best_map.append(City_Map[i])
best_map.append(City_Map[pop[maxfitness][0]])
X = np.array((best_map))[:, 0]
Y = np.array((best_map))[:, 1]
# 绘制地图以及路线
plt.figure()
plt.rcParams['font.sans-serif'] = ['SimHei']
plt.scatter(X, Y)
for dot in range(len(X) - 1):
plt.annotate(pop[maxfitness][dot], xy=(X[dot], Y[dot]), xytext=(X[dot], Y[dot]))
plt.annotate('start', xy=(X[0], Y[0]), xytext=(X[0] + 1, Y[0]))
plt.plot(X, Y)
if __name__ == "__main__": # 主循环
# 生成初代种群pop
pop = []
list = list(range(DNA_SIZE))
for i in range(POP_SIZE):
random.shuffle(list)
l = list.copy()
pop.append(l)
best_dis = []
# 进行选择,交叉,变异,并把每代的最优个体保存在best_dis中
for i in range(Iterations): # 迭代N代
pop = crossmuta(pop, CROSS_RATE)
fitness = getfitness(pop)
maxfitness = np.argmax(fitness)
best_dis.append(distance(pop[maxfitness]))
pop = select(pop, fitness) # 选择生成新的种群
print_info(pop) # 打印信息
print('逐代的最小距离:', best_dis)
# 画图
plt.figure()
plt.plot(range(Iterations), best_dis)
plt.show()
plt.close()