我在python中使用Rubiks Cube扰码器遇到了一些问题。
有我的代码：
from random import randint
moves = ["F", "F'", "R", "R'", "L", "L'", "U", "U'", "D", "D'", "B", "B'", "F2", "R2", "L2", "U2", "D2", "B2"]
scramble = []
lenght = len(scramble)
lenght_moves = len(moves) - 1
def good_move(scramble, lenght):
if scramble[lenght] == "R" or scramble[lenght] == "R'" or scramble[lenght] == "R2":
if scramble[lenght - 1] == "R" or scramble[lenght - 1] == "R'" or scramble[lenght - 1] == "R2":
return False
if scramble[lenght] == "L" or scramble[lenght] == "L'" or scramble[lenght] == "L2":
if scramble[lenght - 1] == "L" or scramble[lenght - 1] == "L'" or scramble[lenght - 1] == "L2":
return False
if scramble[lenght] == "F" or scramble[lenght] == "F'" or scramble[lenght] == "F2":
if scramble[lenght - 1] == "F" or scramble[lenght - 1] == "F'" or scramble[lenght - 1] == "F2":
return False
if scramble[lenght] == "U" or scramble[lenght] == "U'" or scramble[lenght] == "U2":
if scramble[lenght - 1] == "U" or scramble[lenght - 1] == "U'" or scramble[lenght - 1] == "U2":
return False
if scramble[lenght] == "D" or scramble[lenght] == "D'" or scramble[lenght] == "D2":
if scramble[lenght - 1] == "D" or scramble[lenght - 1] == "D'" or scramble[lenght - 1] == "D2":
return False
if scramble[lenght] == "B" or scramble[lenght] == "B'" or scramble[lenght] == "B2":
if scramble[lenght - 1] == "B" or scramble[lenght - 1] == "B'" or scramble[lenght - 1] == "B2":
return False
return True
while (lenght < 20):
print (lenght)
print (scramble)
random = randint(0, lenght_moves)
if lenght - 1 >= 1:
if good_move(scramble, lenght - 1) == False:
print ("I'm here")
while (good_move(scramble, lenght - 1)) != False:
random = randint(0, lenght_moves)
print (random)
scramble.remove(lenght - 1)
scramble.append(moves[random])
else:
scramble.append(moves[random])
else:
scramble.append(moves[random])
lenght = len(scramble)
print (scramble)
因此，当我运行程序时，他将
if lenght - 1 >= 1:
if good_move(scramble, lenght - 1) == False:
print ("I'm here")
while (good_move(scramble, lenght - 1)) != False:
random = randint(0, lenght_moves)
print (random)
scramble.remove(lenght - 1)
scramble.append(moves[random])
而且他正在循环播放...我尝试使用“ i”而不是“ length-1”，但是没有用（索引超出范围等）。
moves = ["F", "F'", "R", "R'", "L", "L'", "U", "U'", "D", "D'", "B", "B'", "F2", "R2", "L2", "U2", "D2", "B2"]
scramble = []
length = len(scramble)
length_moves = len(moves) - 1
def good_move(scramble, length):
if scramble[length] == "R" or scramble[length] == "R'" or scramble[length] == "R2":
if scramble[length - 1] == "R" or scramble[length - 1] == "R'" or scramble[length - 1] == "R2":
return False
if scramble[length] == "L" or scramble[length] == "L'" or scramble[length] == "L2":
if scramble[length - 1] == "L" or scramble[length - 1] == "L'" or scramble[length - 1] == "L2":
return False
if scramble[length] == "F" or scramble[length] == "F'" or scramble[length] == "F2":
if scramble[length - 1] == "F" or scramble[length - 1] == "F'" or scramble[length - 1] == "F2":
return False
if scramble[length] == "U" or scramble[length] == "U'" or scramble[length] == "U2":
if scramble[length - 1] == "U" or scramble[length - 1] == "U'" or scramble[length - 1] == "U2":
return False
if scramble[length] == "D" or scramble[length] == "D'" or scramble[length] == "D2":
if scramble[length - 1] == "D" or scramble[length - 1] == "D'" or scramble[length - 1] == "D2":
return False
if scramble[length] == "B" or scramble[length] == "B'" or scramble[length] == "B2":
if scramble[length - 1] == "B" or scramble[length - 1] == "B'" or scramble[length - 1] == "B2":
return False
return True
i = 0
while (i < 20):
print (length)
print (scramble)
random = randint(0, length_moves)
if i >= 2:
if good_move(scramble, i) == False:
print ("I'm here")
while (good_move(scramble, i)) != False:
random = randint(0, length_moves)
print (random)
scramble.remove(i)
scramble.append(moves[random])
else:
scramble.append(moves[random])
else:
scramble.append(moves[random])
i += 1
print (scramble)
例如，在第二个代码中，我放入了“ i”长度，当我的程序正在执行功能时，他告诉我“索引超出范围”，我不知道为什么，如果i> = 2则不能超出范围，因为“长度”（在函数中）== 1,2,3，依此类推，而“长度-1” == 0,1,2。
任何想法如何解决这个问题？
顺便说一句。例如，正确地争夺Rubiks Cube：
R2 U2 R2 B' U2 B2 R2 F' U2 L' B2 F2 U' F2 R' B D R B R'
最佳答案
if good_move(scramble, lenght - 1) == False:
print ("I'm here")
while (good_move(scramble, lenght - 1)) != False:
这是第一个问题。在这里永远不会输入
while循环，因为当您到达good_move行时，print肯定为假。也许您的意思是每次都具有相同的条件。 if good_move(scramble, lenght - 1) == False:
print ("I'm here")
while (good_move(scramble, lenght - 1)) == False:
scramble.remove(lenght - 1)
这是第二个问题。
list.remove(x)不会从列表中删除list[x]。无论在哪里，它都会在列表中搜索x的第一个实例并将其删除。如果要删除列表的最后一个元素，可以将其切掉。 scramble = scramble[:-1]
或删除它。
del scramble[-1]
现在，您的程序应正确结束。样本结果：
["F'", 'D', 'B', 'D', 'B2', "U'", 'R2', 'L2', "D'", 'B2', 'F', "R'", 'B2', 'R', "F'", "R'", "B'", 'U2', 'F', 'L2']
关于python - 无限循环-Rubiks立方体加密器，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/27113060/