Question

这是Rubik的Cube加扰生成器的尝试。有时我会一次又一次地转过同一张脸，例如（R，R'）。我尝试使用for和while循环修复此问题，但没有成功。

import ast
import _ast

def get_variables(node):
    variables = set()
    if hasattr(node, 'body'):
        for subnode in node.body:
            variables |= get_variables(subnode)
    elif isinstance(node, _ast.Assign):
        for name in node.targets:
            if isinstance(name, _ast.Name):
                variables.add(name.id)
    return variables

print(get_variables(ast.parse(open('file.py').read())))

Answer 1

我为您提供此解决方案：

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)

Answer 2

我已将您的加扰功能重写为更加有效，并且可以完全按照您的期望工作。

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)

Answer 3

如果添加的最后一个值相同，则无法添加列表。

这是一个更好的while循环，可以满足您的需求：

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))

此循环确保列表的长度固定。您可以将修饰符添加到此列表中。

但是。.如果您不习惯使用修饰符的概念，那么出于随机性和简单性的考虑，最好的做法是将上面的循环与带有所有可能排列的长列表一起使用。

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()

以上代码的输出为：

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

Python比较列表中的顺序项，如果相同则将其删除

3 个答案: