Python 我如何使用欧几里德距离公式来计算这里的启发式？让算法在八个方向上搜索？
我正在研究星型算法，我的启发式设置如下：Python 我如何使用欧几里德距离公式来计算这里的启发式？让算法在八个方向上搜索？,python,path-finding,a-star,heuristics,Python,Path Finding,A Star,Heuristics,我正在研究星型算法，我的启发式设置如下： heuristic = [[9, 8, 7, 6, 5, 4], [8, 7, 6, 5, 4, 3], [7, 6, 5, 4, 3, 2], [6, 5, 4, 3, 2, 1], [5, 4, 3, 2, 1, 0]] 算法在四个方向的移动如下： delta = [[-1, 0 ], # go up [ 0, -1], # g
heuristic = [[9, 8, 7, 6, 5, 4],
[8, 7, 6, 5, 4, 3],
[7, 6, 5, 4, 3, 2],
[6, 5, 4, 3, 2, 1],
[5, 4, 3, 2, 1, 0]]
delta = [[-1, 0 ], # go up
[ 0, -1], # go left
[ 1, 0 ], # go down
[ 0, 1 ]] # go right
delta_name = ['^','<','V','>']
import random
grid = [[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0]]
heuristic = [[9, 8, 7, 6, 5, 4],
[8, 7, 6, 5, 4, 3],
[7, 6, 5, 4, 3, 2],
[6, 5, 4, 3, 2, 1],
[5, 4, 3, 2, 1, 0]]
init = [0,0]
goal = [len(grid)-1,len(grid[0])-1]
#Below the four potential actions to the single field
delta = [[-1 , 0], #up
[ 0 ,-1], #left
[ 1 , 0], #down
[ 0 , 1]] #right
delta_name = ['^','<','V','>'] #The name of above actions
cost = 1
def search():
#open list elements are of the type [g,x,y]
closed = [[0 for row in range(len(grid[0]))] for col in range(len(grid))]
action = [[-1 for row in range(len(grid[0]))] for col in range(len(grid))]
#We initialize the starting location as checked
closed[init[0]][init[1]] = 1
expand=[[-1 for row in range(len(grid[0]))] for col in range(len(grid))]
# we assigned the cordinates and g value
x = init[0]
y = init[1]
g = 0
h = heuristic[x][y]
f = g + h
#our open list will contain our initial value
open = [[f, g, h, x, y]]
found = False #flag that is set when search complete
resign = False #Flag set if we can't find expand
count = 0
#print('initial open list:')
#for i in range(len(open)):
#print(' ', open[i])
#print('----')
while found is False and resign is False:
#Check if we still have elements in the open list
if len(open) == 0: #If our open list is empty, there is nothing to expand.
resign = True
print('Fail')
print('############# Search terminated without success')
print()
else:
#if there is still elements on our list
#remove node from list
open.sort() #sort elements in an increasing order from the smallest g value up
open.reverse() #reverse the list
next = open.pop() #remove the element with the smallest g value from the list
#print('list item')
#print('next')
#Then we assign the three values to x,y and g. Which is our expantion.
x = next[3]
y = next[4]
g = next[1]
expand[x][y] = count
count+=1
#Check if we are done
if x == goal[0] and y == goal[1]:
found = True
print(next) #The three elements above this "if".
print('############## Search is success')
print()
else:
#expand winning element and add to new open list
for i in range(len(delta)): #going through all our actions the four actions
#We apply the actions to x and y with additional delta to construct x2 and y2
x2 = x + delta[i][0]
y2 = y + delta[i][1]
#if x2 and y2 falls into the grid
if x2 >= 0 and x2 < len(grid) and y2 >=0 and y2 <= len(grid[0])-1:
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g + cost
h2 = heuristic[x2][y2]
f2 = g2 + h2
open.append([f2,g2,h2,x2,y2])
#print('append list item')
#print([g2,x2,y2])
#Then we check them to never expand again
closed[x2][y2] = 1
action[x2][y2] = i
for i in range(len(expand)):
print(expand[i])
print()
policy=[[' ' for row in range(len(grid[0]))] for col in range(len(grid))]
x=goal[0]
y=goal[1]
policy[x][y]='*'
while x !=init[0] or y !=init[1]:
x2=x-delta[action[x][y]][0]
y2=y-delta[action[x][y]][1]
policy[x2][y2]= delta_name[action[x][y]]
x=x2
y=y2
for i in range(len(policy)):
print(policy[i])
search()
随机导入
网格=[[0,1,0,0,0,0,0]，
[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0]]
启发式=[[9,8,7,6,5,4]，
[8, 7, 6, 5, 4, 3],
[7, 6, 5, 4, 3, 2],
[6, 5, 4, 3, 2, 1],
[5, 4, 3, 2, 1, 0]]
init=[0,0]
目标=[len（网格）-1，len（网格[0]）-1]
#下面四个潜在的行动，以单一领域
增量=[-1，0]，#向上
[0，-1]，#左
[1,0]，#向下
[0,1]#对
delta_name=['^'，'']#上述操作的名称
成本=1
def search（）：
#开放列表元素的类型为[g，x，y]
闭合=[[0表示范围内的行（len（网格[0]））]表示范围内的列（len（网格））]
action=[-1表示范围内的行（len（网格[0]））]表示范围内的列（len（网格））]
#我们将起始位置初始化为选中状态
已关闭[init[0]][init[1]]=1
expand=[-1表示范围内的行（len（网格[0]）），expand=[-1表示范围内的列（len（网格））]
#我们分配了cordinates和g值
x=init[0]
y=init[1]
g=0
h=启发式[x][y]
f=g+h
#我们的开放列表将包含我们的初始值
开放=[[f，g，h，x，y]]
found=False#搜索完成时设置的标志
如果找不到expand，则设置dedict=False#标志
计数=0
#打印（'初始打开列表：'）
#对于范围内的i（len（open））：
#打印（“”，打开[i]）
#打印（'---'）
当发现错误且辞职错误时：
#检查开放列表中是否仍有元素
如果len（open）=0:#如果我们的open列表为空，则没有可扩展的内容。
辞职=真
打印（'失败'）
打印（“搜索终止但未成功”）
打印（）
其他：
#如果我们的列表中还有元素
#从列表中删除节点
open.sort（）#按从最小g值开始的递增顺序对元素进行排序
open.reverse（）#反转列表
next=open.pop（）#从列表中删除g值最小的元素
#打印（'列表项'）
#打印（‘下一个’）
#然后我们将这三个值赋给x、y和g。这是我们的解释。
x=下一个[3]
y=下一个[4]
g=下一个[1]
展开[x][y]=计数
计数+=1
#检查我们是否完成了
如果x==目标[0]和y==目标[1]：
找到=真
打印（下一页）#此“如果”上方的三个元素。
印刷品（‘搜索就是成功’）
打印（）
其他：
#展开获胜元素并添加到新的开放列表
对于范围内的i（len（delta））：#经历我们所有的动作，四个动作
#我们将动作应用于x和y，并附加delta来构造x2和y2
x2=x+delta[i][0]
y2=y+delta[i][1]
#如果x2和y2落入网格中
如果x2>=0和x2=0和y2，你不应该使用图表：？@Dan说实话，亲爱的，这是我第一次知道，Matplotlib旁边有这样的东西。不，不是绘图。该算法寻找连接图上两个节点的最短路径。所以现在，你的图在上下邻域和左右邻域之间有边。要得到对角线移动，您不需要更改算法，只需在图形中添加连接（边）。@Dan您能帮我解决这个问题吗，先生？我感到困惑。。我怎么办？这有点太开放了，看看这个。也许你不应该用图表：？@Dan说实话，亲爱的，这是我第一次知道，Matplotlib旁边有这样的东西。不，不是情节。该算法寻找连接图上两个节点的最短路径。所以现在，你的图在上下邻域和左右邻域之间有边。要得到对角线移动，您不需要更改算法，只需在图形中添加连接（边）。@Dan您能帮我解决这个问题吗，先生？我感到困惑。。我怎么办？这有点太开放了，看看这个。也可能
import random
grid = [[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0]]
heuristic = [[9, 8, 7, 6, 5, 4],
[8, 7, 6, 5, 4, 3],
[7, 6, 5, 4, 3, 2],
[6, 5, 4, 3, 2, 1],
[5, 4, 3, 2, 1, 0]]
init = [0,0]
goal = [len(grid)-1,len(grid[0])-1]
#Below the four potential actions to the single field
delta = [[-1 , 0], #up
[ 0 ,-1], #left
[ 1 , 0], #down
[ 0 , 1]] #right
delta_name = ['^','<','V','>'] #The name of above actions
cost = 1
def search():
#open list elements are of the type [g,x,y]
closed = [[0 for row in range(len(grid[0]))] for col in range(len(grid))]
action = [[-1 for row in range(len(grid[0]))] for col in range(len(grid))]
#We initialize the starting location as checked
closed[init[0]][init[1]] = 1
expand=[[-1 for row in range(len(grid[0]))] for col in range(len(grid))]
# we assigned the cordinates and g value
x = init[0]
y = init[1]
g = 0
h = heuristic[x][y]
f = g + h
#our open list will contain our initial value
open = [[f, g, h, x, y]]
found = False #flag that is set when search complete
resign = False #Flag set if we can't find expand
count = 0
#print('initial open list:')
#for i in range(len(open)):
#print(' ', open[i])
#print('----')
while found is False and resign is False:
#Check if we still have elements in the open list
if len(open) == 0: #If our open list is empty, there is nothing to expand.
resign = True
print('Fail')
print('############# Search terminated without success')
print()
else:
#if there is still elements on our list
#remove node from list
open.sort() #sort elements in an increasing order from the smallest g value up
open.reverse() #reverse the list
next = open.pop() #remove the element with the smallest g value from the list
#print('list item')
#print('next')
#Then we assign the three values to x,y and g. Which is our expantion.
x = next[3]
y = next[4]
g = next[1]
expand[x][y] = count
count+=1
#Check if we are done
if x == goal[0] and y == goal[1]:
found = True
print(next) #The three elements above this "if".
print('############## Search is success')
print()
else:
#expand winning element and add to new open list
for i in range(len(delta)): #going through all our actions the four actions
#We apply the actions to x and y with additional delta to construct x2 and y2
x2 = x + delta[i][0]
y2 = y + delta[i][1]
#if x2 and y2 falls into the grid
if x2 >= 0 and x2 < len(grid) and y2 >=0 and y2 <= len(grid[0])-1:
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g + cost
h2 = heuristic[x2][y2]
f2 = g2 + h2
open.append([f2,g2,h2,x2,y2])
#print('append list item')
#print([g2,x2,y2])
#Then we check them to never expand again
closed[x2][y2] = 1
action[x2][y2] = i
for i in range(len(expand)):
print(expand[i])
print()
policy=[[' ' for row in range(len(grid[0]))] for col in range(len(grid))]
x=goal[0]
y=goal[1]
policy[x][y]='*'
while x !=init[0] or y !=init[1]:
x2=x-delta[action[x][y]][0]
y2=y-delta[action[x][y]][1]
policy[x2][y2]= delta_name[action[x][y]]
x=x2
y=y2
for i in range(len(policy)):
print(policy[i])
search()