python - 如何检查排列是否具有相等的奇偶性？

Question

我正在寻找一种方法来检查 2个排列（由列表表示）是否具有相同的奇偶性。请注意，我对它们是偶校验还是奇校验不感兴趣，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 大师向我展示一些很酷的技巧，以在更少、更优雅的 Python 代码中实现相同的目标。

score 7 · Accepted Answer

如果我们结合两个排列，当每个排列具有相同的奇偶性时，结果将具有偶校验，如果它们具有不同的奇偶性，则结果将具有奇校验。因此，如果我们解决奇偶校验问题，比较两个不同的排列是微不足道的。

奇偶校验可以如下确定：选择一个任意元素，找到排列移动到的位置，重复直到回到你开始的那个位置。您现在找到了一个循环：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量少一个交换来撤消它。现在选择另一个你还没有处理过的元素并重复，直到你看到每个元素。请注意，您总共需要每个元素一次交换减去每个循环一次交换。

时间复杂度在排列的大小上是 O(N)。请注意，尽管我们在循环中有一个循环，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])

此外，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

score 6 · Accepted Answer

这是我对您的代码的调整

使用 list() 复制 perm1 - 这意味着您可以将元组等传递给函数，使其更通用
在 for 循环中使用 enumerate() 而不是 len(..) 和数组查找以获得更简洁的代码
制作一个 perm1_map 并保持最新，以停止对 perm1 的昂贵 O(N) 搜索和一个昂贵的列表切片
直接返回布尔值而不是 if ... return True else return False

就这个

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

score 5 · Accepted Answer

上一个答案的一个小变种 - 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

score 4 · Accepted Answer

我天真的解决方案：

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

score 2 · Accepted Answer

这是稍微重构的 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

score 2 · Accepted Answer

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。

score 0 · Accepted Answer

我的直觉告诉我，仅计算两个排列之间的差异就会比从一个排列到另一个排列所需的交换次数多一个。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，因此您拥有偶数奇偶校验。

其他：

ABCD, CDBA.

有四个差异，因此有三个互换，因此是奇校验。

score 0 · Accepted Answer

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

python - 如何检查排列是否具有相等的奇偶性？

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