编写并运行一个AI搜索程序,从一开始就运行搜索,直到找到结果或结果。但是,当我运行它时,我没有得到搜索结果,而是失败了,没有。如果你知道这个问题的原因是什么,我们将不胜感激。
grid = [[0, 0, 1, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 1, 0],
[0, 0, 1, 1, 1, 0],
[0, 0, 0, 0, 1, 0]]
init = [0, 0]
goal = [len(grid)-1, len(grid[0])-1]
cost = 1
delta = [[-1, 0], # go up
[ 0,-1], # go left
[ 1, 0], # go down
[ 0, 1]] # go right
delta_name = ['^', '<', 'v', '>']
def search():
closed = [[0 for row in range(len(grid[0]))] for col in range(len(grid))]
closed[init[0]][init[1]] = 1
x = init[0]
y =init[1]
g = 0
open = [[g, x, y]]
found = False
resign = False
while found is False and resign is False:
if len(open) == 0:
resign = True
print 'fail'
else:
open.sort()
open.reverse()
next = open.pop()
x = next[3]
y = next[4]
g = next[1]
if x == goal[0] and y == goal[1]:
found = next
print next
else:
for i in range(len(delta)):
x2 = x + delta[i][0]
y2 = y + delta[i][1]
if x2 >= 0 and x2 < len(grid) and y2 >= 0 and y2 < len(grid):
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g + cost
open.append([g2, x2, y2])
closed[x2][y2] = 1
print search()
发布于 2018-05-27 09:31:17
第一个问题是代码的这一部分:
x = next[3]
y = next[4]
g = next[1]
open
列表中的每个元素只有三个条目,因此3
和4
是无效的索引。这可能应改为:
x = next[1]
y = next[2]
g = next[0]
第二个问题在本部分的第一行:
if x2 >= 0 and x2 < len(grid) and y2 >= 0 and y2 < len(grid):
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g + cost
x2
和y2
都与len(grid)
进行了比较,但是您似乎没有方格,因此其中一个检查将是不正确的。它可能应该改为:
if x2 >= 0 and x2 < len(grid) and y2 >= 0 and y2 < len(grid[0]):
if closed[x2][y2] == 0 and grid[x2][y2] == 0:
g2 = g + cost
潜在的第三个问题是,search()
函数的意图似乎是返回一些东西,但它没有任何return
语句。然后它将自动返回None
,这意味着底部的print search()
语句总是简单地打印None
。从您的问题中还不清楚您希望函数返回什么,因此我无法确定如何修复它。
还不妨指出,本部分的任何一项评论都令人困惑:
delta = [[-1, 0], # go up
[ 0,-1], # go left
[ 1, 0], # go down
[ 0, 1]] # go right
或者将变量名(如x
和y
)用于坐标是令人困惑的。在技术意义上没有问题,但是在这个实现中,x
坐标是由条目修改的,注释上写着“向上”和“向下”,而y
坐标被修改为“向左”和“向右”。
https://stackoverflow.com/questions/50546040
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