无限循环 - Rubiks立方体扰乱器

时间:2014-11-24 19:54:55

标签: python loops rubiks-cube

我在Python中遇到Rubiks Cube扰码器的问题 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)

所以,当我正在运行我的程序时,他将会

  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])

他正在循环......我试过&#34;我&#34;而不是&#34;长度 - 1&#34;但它没有工作(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了&#34; i&#34;长篇大论,当我的节目满足功能时,他告诉我们指数超出范围&#34;我不知道为什么,如果i> = 2,它就不会超出范围,因为&#34; lenght&#34;(在函数中)== 1,2,3等等和& #34;长度 - 1&#34; == 0,1,2。 任何想法如何解决这个问题?

顺便说一句。例如正确争夺Rubiks Cube:

R2 U2 R2 B' U2 B2 R2 F' U2 L' B2 F2 U' F2 R' B D R B R'

1 个答案:

答案 0 :(得分:1)

    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。永远不会在此处输入while循环,因为当您到达good_move行时,print肯定会为假。也许你的意思是每次都有相同的条件。

    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]。它在列表中搜索x的第一个实例并将其删除,无论它在何处。如果要删除列表的最后一个元素,可以将其分割。

            scramble = scramble[:-1]

或删除它。

            del scramble[-1]

现在你的程序应该正常结束。样本结果:

["F'", 'D', 'B', 'D', 'B2', "U'", 'R2', 'L2', "D'", 'B2', 'F', "R'", 'B2', 'R', "F'", "R'", "B'", 'U2', 'F', 'L2']