Python 15个谜题的这个实现的终止条件应该是什么？

我正在尝试实现一个解决方案，用于在Python中输出一个问题的移动序列。这是MOOC可选作业的一部分。问题陈述在中给出

我有一个执行有效转换的程序版本（如下所示）

我首先识别空单元格的邻居（由0表示）并将它们放入列表中。然后，我从列表中随机选择一个邻居来执行与空单元格的交换。所有的掉期都被累积在一个不同的列表中，以记录解谜的动作顺序。然后在程序结束时输出

然而，用空单元格进行交换的随机数字选择只是永远在进行。为了避免循环的“无限”（非常长的周期），我现在将交换的数量限制为30

from random import randint
def find_idx_of_empty_cell(p):
    for i in range(len(p)):
        if p[i] == 0:
            return i

def pick_random_neighbour_idx(neighbours_idx_list):
    rand_i = randint(0, len(neighbours_idx_list)-1)
    return neighbours_idx_list[rand_i]

def perform__neighbour_transposition(p, tar_idx, src_idx):
    temp = p[tar_idx]
    p[tar_idx] = p[src_idx]
    p[src_idx] = temp

def solve_15_puzzle(p):
    standard_perm = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0]
    neighbours_idx_list = []
    moves_sequence = []
    empty_cell_idx = find_idx_of_empty_cell(p)
    previous_empty_cell_idx = empty_cell_idx

    while (not(p == standard_perm) and len(moves_sequence) < 30):
        if not (empty_cell_idx in [0,4,8,12]):
            neighbours_idx_list.append(empty_cell_idx - 1)
        if not (empty_cell_idx in [3,7,11,15]):
            neighbours_idx_list.append(empty_cell_idx + 1)
        if not (empty_cell_idx in [0,1,2,3]):
            neighbours_idx_list.append(empty_cell_idx - 4)
        if not (empty_cell_idx in [12,13,14,15]):
            neighbours_idx_list.append(empty_cell_idx + 4)

        if previous_empty_cell_idx in neighbours_idx_list:
            neighbours_idx_list.remove(previous_empty_cell_idx)

        chosen_neighbour_idx = pick_random_neighbour_idx(neighbours_idx_list)
        moves_sequence.append(p[chosen_neighbour_idx])
        perform__neighbour_transposition(p, empty_cell_idx, chosen_neighbour_idx)
        previous_empty_cell_idx = empty_cell_idx
        empty_cell_idx = chosen_neighbour_idx
        neighbours_idx_list = []

    if (p == standard_perm):
        print("Solution: ", moves_sequence)

---

15块瓷砖的问题更难，因为乍一看似乎更难

计算最佳（最短）解是一个困难的问题，已经证明，随着N的增加寻找最优解是NP困难的

找到（非最优）解决方案要容易得多。例如，一个非常简单的算法是：

• 将当前位置的“距离”定义为每个磁贴距所需位置的距离之和
• 从给定的位置开始，做一些随机移动
• 如果移动后的距离提高或保持不变，则保留更改，否则撤消更改并返回起点

这种算法可以描述为一种多步骤随机爬山方法，能够解决15个难题（只需确保允许足够多的随机移动，以避开局部最小值）

Python可能不是解决这个问题的最佳语言，但是如果您使用PyPy实现，您可以在合理的时间内获得解决方案

我的实现为一个在几秒钟内被1000个随机移动混合在一起的谜题找到了解决方案，例如：

(1, 5, 43, [9, [4, 10, 14, 11, 15, 3, 8, 1, 13, None, 9, 7, 12, 2, 5, 6]])
(4, 17, 41, [9, [4, 10, 14, 11, 15, 3, 8, 1, 12, None, 6, 2, 5, 13, 9, 7]])
(7, 19, 39, [11, [4, 10, 14, 11, 15, 3, 1, 2, 12, 6, 8, None, 5, 13, 9, 7]])
(9, 54, 36, [5, [4, 14, 3, 11, 15, None, 10, 2, 12, 6, 1, 8, 5, 13, 9, 7]])
(11, 60, 34, [10, [4, 14, 3, 11, 15, 10, 1, 2, 12, 6, None, 8, 5, 13, 9, 7]])
(12, 93, 33, [14, [4, 14, 11, 2, 15, 10, 3, 8, 12, 6, 1, 7, 5, 13, None, 9]])
(38, 123, 31, [11, [4, 14, 11, 2, 6, 10, 3, 8, 15, 12, 1, None, 5, 13, 9, 7]])
(40, 126, 30, [13, [15, 6, 4, 2, 12, 10, 11, 3, 5, 14, 1, 8, 13, None, 9, 7]])
(44, 172, 28, [10, [15, 4, 2, 3, 12, 6, 11, 8, 5, 10, None, 14, 13, 9, 1, 7]])
(48, 199, 23, [11, [15, 6, 4, 3, 5, 12, 2, 8, 13, 10, 11, None, 9, 1, 7, 14]])
(61, 232, 22, [0, [None, 15, 4, 3, 5, 6, 2, 8, 1, 12, 10, 14, 13, 9, 11, 7]])
(80, 276, 20, [10, [5, 15, 4, 3, 1, 6, 2, 8, 13, 10, None, 7, 9, 12, 14, 11]])
(105, 291, 19, [4, [9, 1, 2, 4, None, 6, 8, 7, 5, 15, 3, 11, 13, 12, 14, 10]])
(112, 313, 17, [9, [1, 6, 2, 4, 9, 8, 3, 7, 5, None, 14, 11, 13, 15, 12, 10]])
(113, 328, 16, [15, [1, 6, 2, 4, 9, 8, 3, 7, 5, 15, 11, 10, 13, 12, 14, None]])
(136, 359, 15, [4, [1, 6, 2, 4, None, 8, 3, 7, 9, 5, 11, 10, 13, 15, 12, 14]])
(141, 374, 12, [15, [1, 2, 3, 4, 8, 6, 7, 10, 9, 5, 12, 11, 13, 15, 14, None]])
(1311, 385, 11, [14, [1, 2, 3, 4, 8, 5, 7, 10, 9, 6, 11, 12, 13, 15, None, 14]])
(1329, 400, 10, [13, [1, 2, 3, 4, 6, 8, 7, 10, 9, 5, 11, 12, 13, None, 15, 14]])
(1602, 431, 9, [4, [1, 2, 3, 4, None, 6, 8, 7, 9, 5, 11, 10, 13, 15, 14, 12]])
(1707, 446, 8, [5, [1, 2, 3, 4, 6, None, 7, 8, 9, 5, 15, 12, 13, 10, 14, 11]])
(1711, 475, 7, [12, [1, 2, 3, 4, 6, 5, 7, 8, 9, 10, 15, 12, None, 13, 14, 11]])
(1747, 502, 6, [8, [1, 2, 3, 4, 6, 5, 7, 8, None, 9, 10, 12, 13, 14, 15, 11]])
(1824, 519, 5, [14, [1, 2, 3, 4, 9, 6, 7, 8, 5, 10, 15, 12, 13, 14, None, 11]])
(1871, 540, 4, [10, [1, 2, 3, 4, 9, 6, 7, 8, 5, 10, None, 12, 13, 14, 15, 11]])
(28203, 555, 3, [9, [1, 2, 3, 4, 5, 6, 7, 8, 9, None, 10, 12, 13, 14, 11, 15]])
(28399, 560, 2, [10, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, None, 12, 13, 14, 11, 15]])
(28425, 581, 1, [11, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, None, 13, 14, 15, 12]])
(28483, 582, 0, [15, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, None]])

最后一行表示在24483次实验后，它在582次移动后找到了目标位置。请注意，582肯定远远不是最佳的，因为众所周知，在经典版的15个拼图中，没有位置需要超过80个动作

移动次数后的数字是“曼哈顿距离”，例如，最后第四行是位置：

---

其中，与解决方案的曼哈顿距离之和为3。

请为此问题创建一个标签“15拼图”。我认为这是必需的，但我没有这样做的必要特权。15个拼图有

（4*4）/2

（即10461394944000）有效配置；只是随机行走，等待在解决方案中绊倒，这将需要一段时间：-）。。。还请注意，“最困难”的配置至少需要80次单次交换移动到solve@6502这似乎是大海捞针。我该怎么做呢？补充了一个解释；不发布代码，因为这是一个有趣的实验；-）我理解你解决这个难题的策略的本质。不过，我不知道曼哈顿距离是多少，也不知道如何计算。在稍后的某个时候再次尝试这个谜题之前，我会对它做一些研究。为什么说Python不适合解决这个问题？我听说它是当今使用的最强大的语言之一。@flamingo_stark:Manhattan距离就是

abs（x1-x2）+abs（y1-y2）

。Python是一个非常非常好的语言（我喜欢它），但是缺省实现cpython并不炫耀执行速度，你应该考虑，例如，对于数字和数组操作的一般低级算法来说，它是C的100倍慢。PyPy实现将其提升到Javascript级别，虽然仍然比C慢，但速度要快得多。

(1, 5, 43, [9, [4, 10, 14, 11, 15, 3, 8, 1, 13, None, 9, 7, 12, 2, 5, 6]])
(4, 17, 41, [9, [4, 10, 14, 11, 15, 3, 8, 1, 12, None, 6, 2, 5, 13, 9, 7]])
(7, 19, 39, [11, [4, 10, 14, 11, 15, 3, 1, 2, 12, 6, 8, None, 5, 13, 9, 7]])
(9, 54, 36, [5, [4, 14, 3, 11, 15, None, 10, 2, 12, 6, 1, 8, 5, 13, 9, 7]])
(11, 60, 34, [10, [4, 14, 3, 11, 15, 10, 1, 2, 12, 6, None, 8, 5, 13, 9, 7]])
(12, 93, 33, [14, [4, 14, 11, 2, 15, 10, 3, 8, 12, 6, 1, 7, 5, 13, None, 9]])
(38, 123, 31, [11, [4, 14, 11, 2, 6, 10, 3, 8, 15, 12, 1, None, 5, 13, 9, 7]])
(40, 126, 30, [13, [15, 6, 4, 2, 12, 10, 11, 3, 5, 14, 1, 8, 13, None, 9, 7]])
(44, 172, 28, [10, [15, 4, 2, 3, 12, 6, 11, 8, 5, 10, None, 14, 13, 9, 1, 7]])
(48, 199, 23, [11, [15, 6, 4, 3, 5, 12, 2, 8, 13, 10, 11, None, 9, 1, 7, 14]])
(61, 232, 22, [0, [None, 15, 4, 3, 5, 6, 2, 8, 1, 12, 10, 14, 13, 9, 11, 7]])
(80, 276, 20, [10, [5, 15, 4, 3, 1, 6, 2, 8, 13, 10, None, 7, 9, 12, 14, 11]])
(105, 291, 19, [4, [9, 1, 2, 4, None, 6, 8, 7, 5, 15, 3, 11, 13, 12, 14, 10]])
(112, 313, 17, [9, [1, 6, 2, 4, 9, 8, 3, 7, 5, None, 14, 11, 13, 15, 12, 10]])
(113, 328, 16, [15, [1, 6, 2, 4, 9, 8, 3, 7, 5, 15, 11, 10, 13, 12, 14, None]])
(136, 359, 15, [4, [1, 6, 2, 4, None, 8, 3, 7, 9, 5, 11, 10, 13, 15, 12, 14]])
(141, 374, 12, [15, [1, 2, 3, 4, 8, 6, 7, 10, 9, 5, 12, 11, 13, 15, 14, None]])
(1311, 385, 11, [14, [1, 2, 3, 4, 8, 5, 7, 10, 9, 6, 11, 12, 13, 15, None, 14]])
(1329, 400, 10, [13, [1, 2, 3, 4, 6, 8, 7, 10, 9, 5, 11, 12, 13, None, 15, 14]])
(1602, 431, 9, [4, [1, 2, 3, 4, None, 6, 8, 7, 9, 5, 11, 10, 13, 15, 14, 12]])
(1707, 446, 8, [5, [1, 2, 3, 4, 6, None, 7, 8, 9, 5, 15, 12, 13, 10, 14, 11]])
(1711, 475, 7, [12, [1, 2, 3, 4, 6, 5, 7, 8, 9, 10, 15, 12, None, 13, 14, 11]])
(1747, 502, 6, [8, [1, 2, 3, 4, 6, 5, 7, 8, None, 9, 10, 12, 13, 14, 15, 11]])
(1824, 519, 5, [14, [1, 2, 3, 4, 9, 6, 7, 8, 5, 10, 15, 12, 13, 14, None, 11]])
(1871, 540, 4, [10, [1, 2, 3, 4, 9, 6, 7, 8, 5, 10, None, 12, 13, 14, 15, 11]])
(28203, 555, 3, [9, [1, 2, 3, 4, 5, 6, 7, 8, 9, None, 10, 12, 13, 14, 11, 15]])
(28399, 560, 2, [10, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, None, 12, 13, 14, 11, 15]])
(28425, 581, 1, [11, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, None, 13, 14, 15, 12]])
(28483, 582, 0, [15, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, None]])