Question

我正在寻找一种方法来检查2 排列（由列表表示）是否属于相同parity 。请注意，我不感兴趣，如果它们甚至或奇数奇偶校验，只是相等。

我是Python的新手，下面给出了我的天真解决方案作为回复。我期待Python专家向我展示一些很酷的技巧，以便在更小，更优雅的Python代码中实现相同的目标。

Answer 1

以下是我对代码的调整

使用list（）复制perm1 - 这意味着您可以将元组等传递给函数，使其更通用
在for循环中使用enumerate（）而不是len（..）和在neater代码中使用数组查找
创建一个perm1_map并使其保持最新以停止对perm1进行昂贵的O（N）搜索，以及一个昂贵的列表切片
直接返回布尔值，而不是if ...返回True，否则返回False

这是

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 2

如果我们结合两种排列，当每个排列具有相同的奇偶校验时，结果将具有偶校验，如果它们具有不同的奇偶校验，则奇校验。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是微不足道的。

奇偶校验可以确定如下：选择一个任意元素，找到置换移动到的位置，重复直到你回到你开始的位置。您现在已经找到了一个循环：排列将所有这些元素围绕一个位置旋转。您需要一个小于循环中元素数量的交换来撤消它。现在选择你尚未处理的另一个元素并重复，直到你看到每个元素。请注意，总的来说，每个元素需要一次交换，每个周期需要一次交换。

时间复杂度是置换大小的O（N）。请注意，虽然我们在循环中有一个循环，但内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，这是基于Wikipedia中定义的奇偶校验函数的效率低得多但短得多的实现（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 3

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

Answer 4

上一个答案的次要变体 - 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

Answer 5

这是稍微重构的Weeble's answer：

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 6

带有字典的解决方案被窃听。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。

Answer 7

我的直觉告诉我，只计算两种排列之间的差异，就会比交换的数量多一个。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，因此需要两个互换来将一个互换到另一个，因此您甚至可以进行奇偶校验。

另：

ABCD, CDBA.

有四个差异，因此有三个互换，因此奇数奇偶校验。

Answer 8

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

如何检查排列是否具有相等的平价？

8 个答案: