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这是一个魔方打乱发生器的尝试。我有时会一个接一个地转动同一张脸，例如（R，R'）。我试图使用 for 和 while 循环来解决这个问题，但它没有用。

import random
def getscramble():
  moves = ["R","F","U","L","B","D"]
  scramble = []
  finascramble = []

  for x in range(25):
     scramble.append(random.choice(moves))


  for x in range(0,len(scramble)-1):
      if scramble[x] == scramble[x+1]:
        scramble[x] = random.choice(moves)

  for x in range(0,len(scramble)-1):
     if scramble[x] == "R" and scramble[x+1] == "L":
          scramble[x+1] = random.choice(moves)
     if scramble[x] == "U" and scramble[x+1]== "D":
          scramble[x+1] == random.choice(moves)
     if scramble[x] == "F" and scramble[x+1] == "B":
          scramble[x+1] == random.choice(moves)

  modifiers = ["","2","'"]

  for x in range(25):
    randy = random.randint(0,2)
    finascramble.append(scramble[x]+modifiers[randy])
  return " ".join(finascramble)

t = True
while t == True:
  again = input()
  if again == "s":
       print()
       print(getscramble())
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3 回答 3

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如果附加的最后一个值相同，则不能附加列表。

这是一个更好的 while 循环，可以满足您的要求：

i = len(scramble)
while i < 25:
    current_face = random.choice(moves)
    if i == 0:
        scramble.append(current_face)
    elif scramble[i - 1] == current_face:
        continue
    else:
        scramble.append(current_face)
    i += 1
    print(' '.join(scramble))

此循环确保列表具有固定长度。您可以将修饰符添加到此列表中。

但是.. 如果你不接受使用修饰符的概念，为了随机性和简单性，最好的办法是使用上面的循环和长列表以及所有可能的排列。

import random


def getscramble():
    moves = ["R", "F", "U", "L", "B", "D",
             "R'", "F'", "U'", "L'", "B'", "D'",
             "R2", "F2", "U2", "L2", "B2", "D2"]
    scramble = []

    i = len(scramble)
    while i < 25:
        curent_face = random.choice(moves)
        if i == 0:
            scramble.append(curent_face)
        elif (scramble[i - 1])[0] == curent_face[0]:
            continue
        else:
            scramble.append(curent_face)
        i += 1

     print(' '.join(scramble))


 getscramble()

上述代码的输出是：

L' R D2 U L' U' F R2 L' U2 D' R' F' B2 F2 L2 R' D' L2 F' U2 F U2 R D

于 2018-06-30T21:19:43.497 回答
0

我为您提供这个解决方案：

import random

MOVES = ['R', 'L', 'D', 'U', 'B', 'F']
MODFIERS = ['', '2', "'"]


def getScramble(length=25):
    return ''.join([random.choice(MOVES) for _ in range(length)])


def isValid(scramble):
    for seq in ['LR', 'DU', 'BF'] + [move * 2 for move in MOVES]:
        if seq in scramble or seq[::-1] in scramble:
            return False
    return True


while True:
    scramble = getScramble()
    if isValid(scramble):
        break

scramble = ' '.join([move + random.choice(MODFIERS) for move in scramble])
print(scramble)
于 2018-06-30T20:50:02.460 回答
0

我已经根据您的期望重写了您的打乱函数，使其更有效率并且可以充分发挥作用。

import random

moves = ("R", "F", "U", "L", "B", "D")
modifiers = ("", "2", "'")


def getscramble(n=25):

    scramble = random.choices(moves, k=n)

    def is_valid(c, c2):
        together = c + c2
        return (c != c2 and together not in
                ("RL", "UD", "FB", "LR", "DU", "BF"))

    for x in range(1, n):
        before = scramble[x - 1]
        current = scramble[x]
        while not is_valid(before, current):
            current = scramble[x] = random.choice(moves)

    for i, x in enumerate(random.choices(modifiers, k=n)):
        scramble[i] += x

    return " ".join(scramble)
于 2018-06-30T21:03:52.000 回答