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作为练习,我必须找到此旅行商问题的最佳解决方案,唯一的区别是销售员可能不会访问第一个坐标两次。给出的最佳值应该在 1200 左右。我不明白为什么这不会收敛。另外一个重要的注意事项是使用曼哈顿距离度量而不是欧几里得距离。
def shuffle_coords(coords):
x0 = coords[0]
coords = shuffle(coords[1:])
return np.insert(coords,0,x0,axis = 0)
def distance(x,y):
return abs(x[1] - y[1]) + abs(x[0] - y[0])
这里的坐标是随机打乱的。
def shuffle(array):
df = pd.DataFrame(array)
return df.sample(len(array)).to_numpy()
def path_distance(path):
dist = []
for i in range(1,len(path)):
dist.append(distance(path[i],path[i-1]))
return np.sum(dist)
这里初始化了模拟退火算法。
def SA_distance(path, T_0,T_min, alpha):
T = T_0
dist = path_distance(path)
while T > T_min:
new_path = gen_subtour(path)
diffF = path_distance(new_path) - dist
if diffF < 0:
path = new_path
dist = path_distance(path)
elif np.exp(-(diffF/T)) > random.uniform(0,1):
path = new_path
dist = path_distance(path)
T = T * alpha
print(dist,T)
return dist,path
此处生成随机子路线,同时保持 x0 不变。
def gen_subtour(path):
subset = shuffle(np.delete(path,0,axis =0))
subset = shuffle(path)
if random.uniform(0,1) < 0.5:
subset = np.flipud(subset)
else:
j = random.randint(1,(len(subset)-1))
p = subset[j-1]
q = subset[j]
subset = np.delete(subset,[j-1,j],axis = 0)
subset = np.insert(subset,0,p,axis = 0)
subset = np.insert(subset,len(subset),q,axis = 0)
return np.insert(subset,0,path[0],axis = 0)
def main():
T_0 = 12
T_min = 10**-9
alpha = 0.999
coords = np.array([[375, 375],[161, 190], [186, 169],[185, 124],
[122, 104],[109, 258], [55, 153] ,[120, 49],
[39, 85] ,[59, 250] , [17, 310] ,[179, 265],
[184, 198]])
path , distance = SA_distance(coords,T_0,T_min,alpha)
最佳答案
我已经测试过了,我的第一直觉是 gen_subtour()
运行不正常,可能是它改变了路线,步数太多。
我会尝试不同的版本,看看效果如何。 SA 方案似乎运作良好,我认为错误在于提案。
无论如何,这里有一些代码希望能帮助您更好地进行测试。
我使用 pdist
预先计算曼哈顿距离,即;
import numpy as np
from scipy.spatial.distance import pdist, cdist, squareform
coords = np.array([[375, 375],[161, 190], [186, 169],[185, 124],
[122, 104],[109, 258], [55, 153] ,[120, 49],
[39, 85] ,[59, 250] , [17, 310] ,[179, 265],
[184, 198]])
Y = pdist(coords, 'cityblock')
distance_matrix = squareform(Y)
nodes_count = coords.shape[0]
并定义开始于
def random_start():
"""
Random start, returns a state
"""
a = np.arange(0,nodes_count)
np.random.shuffle(a)
return a
目标函数,这里有我们不回原点的版本;
def objective_function( route ):
# uncomment when testing new/modify neighbors
# assert check_all_nodes_visited(route)
return np.sum( distance_matrix[route[1:],route[:-1]] )
我这里有 3 种建议,基于一条路线;
def random_swap( route ):
"""
Random Swap - a Naive neighbour function
Will only work for small instances of the problem
"""
route_copy = route.copy()
random_indici = np.random.choice( route , 2, replace = False)
route_copy[ random_indici[0] ] = route[ random_indici[1] ]
route_copy[ random_indici[1] ] = route[ random_indici[0] ]
return route_copy
def vertex_insert( route, nodes=1 ):
"""
Vertex Insert Neighbour, inspired by
http://www.sciencedirect.com/science/article/pii/S1568494611000573
"""
route_copy = route.copy()
random_indici = np.random.choice( route , 2, replace = False)
index_of_point_to_reroute = random_indici[0]
value_of_point_to_reroute = route[ random_indici[0] ]
index_of_new_place = random_indici[1]
route_copy = np.delete(route_copy, index_of_point_to_reroute)
route_copy = np.insert(route_copy, index_of_new_place, values=value_of_point_to_reroute)
return route_copy
def block_reverse( route, nodes=1 ):
"""
Block Reverse Neighbour, inspired by
http://www.sciencedirect.com/science/article/pii/S1568494611000573
Note that this is a random 2-opt operation.
"""
route_copy = route.copy()
random_indici = np.random.choice( route , 2, replace = False)
index_of_cut_left = np.min(random_indici)
index_of_cut_right = np.max(random_indici)
route_copy[ index_of_cut_left:index_of_cut_right ] = np.flip(route_copy[ index_of_cut_left:index_of_cut_right ])
return route_copy
您可以选择在 SA 之后进行 2-opt 轮次,以确保没有交叉。
def swap_for_2opt( route, i, k):
"""
Helper for 2-opt search
"""
route_copy = route.copy()
index_of_cut_left = i
index_of_cut_right = k
route_copy[ index_of_cut_left:index_of_cut_right ] = np.flip(route_copy[ index_of_cut_left:index_of_cut_right ])
return route_copy
def local_search_2opt( route ):
"""
Local Optimum with 2-opt
https://en.wikipedia.org/wiki/2-opt
"""
steps_since_improved = 0
still_improving = True
route = route.copy()
while still_improving :
for i in range( route.size - 1 ):
for k in np.arange( i + 1, route.size ):
alt_route = swap_for_2opt(route, i, k)
if objective_function(alt_route) < objective_function(route):
route = alt_route.copy()
steps_since_improved = 0
steps_since_improved += 1
if steps_since_improved > route.size + 1:
still_improving = False
break
return route
并使用 frigidum 进行 SA
import frigidum
local_opt = frigidum.sa(random_start=random_start,
objective_function=objective_function,
neighbours=[random_swap, vertex_insert, block_reverse],
copy_state=frigidum.annealing.naked,
T_start=10**5,
alpha=.95,
T_stop=0.001,
repeats=10**2,
post_annealing = local_search_2opt)
返回的路由几乎总是 1145。
我已经在 frigidum 上发布了一般提示和技巧主页。
关于python - 为什么这种模拟退火算法应用于TSP不收敛?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/65890955/
我有一个 d3 力布局图可视化效果很好,但它经常过早地“卡住”。例如,节点正朝着一个好的位置摇晃,如果“碰撞”它们(向它们的位置注入(inject)一点随机性并再次 start() ,它们最终会到达那
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