如何检查排列是否具有相同的奇偶性？

Question

我正在寻找一种方法来检查 2 个排列（由列表表示）是否相同平价。请注意，我对它们是偶还是奇奇偶校验不感兴趣，只是相等。

我是 Python 新手，下面给出了我的幼稚解决方案作为答复。我期待 Python 专家向我展示一些很酷的技巧，以在更小、更优雅的 Python 代码中实现相同的目标。

Answer 1

我的天真的解决方案：

def arePermsEqualParity(perm0, perm1):
    """Check if 2 permutations are of equal parity.

    Assume that both permutation lists are of equal length
    and have the same elements. No need to check for these
    conditions.
    """

    transCount = 0
    for loc in range(len(perm0) - 1):                         # Do (len - 1) transpositions
        if perm0[loc] != perm1[loc]:
            sloc = perm1.index(perm0[loc])                    # Find position in perm1
            perm1[loc], perm1[sloc] = perm1[sloc], perm1[loc] # Swap in perm1
            transCount += 1

    # Even number of transpositions means equal parity
    if (transCount % 2) == 0:
        return True
    else:
        return False

Answer 2

以下是稍微重构的 Weeble 的答案：

def arePermsEqualParity(perm0, perm1):
    """Whether permutations are of equal parity."""
    return parity(combine(perm0, perm1))

def combine(perm0, perm1):
    """Combine two permutations into one."""
    return map(perm0.__getitem__, perm1)

def parity(permutation):
    """Return even parity for the `permutation`."""
    return (len(permutation) - ncycles(permutation)) % 2 == 0

def ncycles(permutation):
    """Return number of cycles in the `permutation`."""
    ncycles = 0
    seen = [False] * len(permutation)
    for i, already_seen in enumerate(seen):
        if not already_seen:
            ncycles += 1
            # mark indices that belongs to the cycle
            j = i 
            while not seen[j]: 
                seen[j] = True
                j = permutation[j]
    return ncycles

Answer 3

字典的解决方案有问题。
这是调试版本：

def arePermsEqualParity(perm0, perm1):
    """Check if 2 permutations are of equal parity.

    Assume that both permutation lists are of equal length
    and have the same elements. No need to check for these
    conditions.
    """
    perm1 = list(perm1) ## copy this into a list so we don't mutate the original
    perm1_map = dict((v, i) for i,v in enumerate(perm1))
    transCount = 0
    for loc, p0 in enumerate(perm0):
        p1 = perm1[loc]
        if p0 != p1:
            sloc = perm1_map[p0]                       # Find position in perm1
            perm1[loc], perm1[sloc] = p0, p1           # Swap in perm1
            perm1_map[p0], perm1_map[p1] = loc, sloc   # Swap the map
            transCount += 1
    # Even number of transpositions means equal parity
    return (transCount % 2) == 0

唯一的区别是字典中的交换没有正确进行。

Answer 4

我的直觉告诉我，仅计算两种排列之间的差异就会比从一种排列到另一种排列所需的交换次数多出一个。这反过来会给你平价。

这意味着您根本不需要在代码中进行交换。

例如：

ABCD, BDCA.

存在三个差异，因此需要两次交换才能将一个置换为另一个，因此您具有偶数奇偶性。

另：

ABCD, CDBA.

有四个差异，因此有三个交换，因此奇偶校验。

Answer 5

def equalparity(p,q):
    return sum([p[q[i]] > p[q[j]] for i in range(len(p)) for j in range(i)]) % 2 == 0

Answer 6

如果我们将两个排列组合起来，当每个排列具有相同的奇偶性时，结果将具有偶奇偶性，如果它们具有不同的奇偶性，则结果将具有奇奇偶性。因此，如果我们解决奇偶校验问题，那么比较两种不同的排列就很简单了。

奇偶性可以通过如下方式确定：选择一个任意元素，找到排列将其移动到的位置，重复直到回到开始的位置。您现在已经找到了一个循环：排列将所有这些元素旋转一个位置。您需要比循环中的元素数少一次交换才能撤消它。现在选择另一个您尚未处理过的元素并重复，直到您看到每个元素。请注意，每个元素总共需要一次交换减去每个周期一次交换。

就排列的大小而言，时间复杂度为 O(N)。请注意，虽然循环中有循环，但内部循环只能对排列中的任何元素迭代一次。

def parity(permutation):
    permutation = list(permutation)
    length = len(permutation)
    elements_seen = [False] * length
    cycles = 0
    for index, already_seen in enumerate(elements_seen):
        if already_seen:
            continue
        cycles += 1
        current = index
        while not elements_seen[current]:
            elements_seen[current] = True
            current = permutation[current]
    return (length-cycles) % 2 == 0

def arePermsEqualParity(perm0, perm1):
    perm0 = list(perm0)
    return parity([perm0[i] for i in perm1])

另外，只是为了好玩，这里有一个基于维基百科定义的奇偶校验函数的效率低得多但更短的实现（对于偶数返回 True，对于奇数返回 False）：

def parity(p):
    return sum(
        1 for (x,px) in enumerate(p)
          for (y,py) in enumerate(p)
          if x<y and px>py
        )%2==0

Answer 7

这是我对代码的调整

• 使用 list() 复制 perm1 - 这意味着您可以将元组等传递到函数中，使其更通用
• 在 for 循环中使用 enumerate() 而不是 len(..) 和数组查找更简洁的代码
• 制作一个 perm1_map 并保持最新，以停止对 perm1 进行昂贵的 O(N) 搜索，并且昂贵的列表切片
• 直接返回布尔值，而不是 if ... return True else return False

这是它

def arePermsEqualParity(perm0, perm1):
    """Check if 2 permutations are of equal parity.

    Assume that both permutation lists are of equal length
    and have the same elements. No need to check for these
    conditions.
    """
    perm1 = list(perm1) ## copy this into a list so we don't mutate the original
    perm1_map = dict((v, i) for i,v in enumerate(perm1))
    transCount = 0
    for loc, p0 in enumerate(perm0):
        p1 = perm1[loc]
        if p0 != p1:
            sloc = perm1_map[p0]                       # Find position in perm1
            perm1[loc], perm1[sloc] = p0, p1           # Swap in perm1
            perm1_map[p0], perm1_map[p1] = sloc, loc   # Swap the map
            transCount += 1
    # Even number of transpositions means equal parity
    return (transCount % 2) == 0

Answer 8

上一个答案的一个小变体 - 复制 perm1，并保存数组查找。

def arePermsEqualParity(perm0, perm1):
    """Check if 2 permutations are of equal parity.

    Assume that both permutation lists are of equal length
    and have the same elements. No need to check for these
    conditions.
    """
    perm1 = perm1[:] ## copy this list so we don't mutate the original

    transCount = 0
    for loc in range(len(perm0) - 1):                         # Do (len - 1) transpositions
        p0 = perm0[loc]
        p1 = perm1[loc]
        if p0 != p1:
            sloc = perm1[loc:].index(p0)+loc          # Find position in perm1
            perm1[loc], perm1[sloc] = p0, p1          # Swap in perm1
            transCount += 1

    # Even number of transpositions means equal parity
    if (transCount % 2) == 0:
        return True
    else:
        return False