Question

是否有一种有效的方法来计算所有小于给定无符号整数的无符号整数，并且设置位数相同？

到目前为止我尝试了什么

给定一个无符号整数，表示为按降序排列的设置位列表，我可以使用colex函数来计算具有相同设置位数的较小无符号整数。我在Python 3中实现了递归过程（显示逻辑）和迭代过程。

想法来自this paper。

我宁愿找到一个不依赖于排序位的解决方案。

from math import factorial


def n_choose_k(n, k):
    return 0 if n < k else factorial(n) // (factorial(k) * factorial(n - k))


def indexset_recursive(bitset, lowest_bit=0):
    """Return number of bitsets with same number of set bits but less than
    given bitset.

    Args:
      bitset (sequence) - Sequence of set bits in descending order.
      lowest_bit (int) - Name of the lowest bit. Default = 0.

    >>> indexset_recursive([51, 50, 49, 48, 47, 46, 45])
    133784559
    >>> indexset_recursive([52, 51, 50, 49, 48, 47, 46], lowest_bit=1)
    133784559
    >>> indexset_recursive([6, 5, 4, 3, 2, 1, 0])
    0
    >>> indexset_recursive([7, 6, 5, 4, 3, 2, 1], lowest_bit=1)
    0

    """
    m = len(bitset)
    first = bitset[0] - lowest_bit
    if m == 1:
        return first
    else:
        t = n_choose_k(first, m)
        return t + indexset_recursive(bitset[1:], lowest_bit)


def indexset(bitset, lowest_bit=0):
    """Return number of bitsets with same number of set bits but less than
    given bitset.

    Args:
      bitset (sequence) - Sequence of set bits in descending order.
      lowest_bit (int) - Name of the lowest bit. Default = 0.

   >>> indexset([51, 50, 49, 48, 47, 46, 45])
    133784559
    >>> indexset([52, 51, 50, 49, 48, 47, 46], lowest_bit=1)
    133784559
    >>> indexset([6, 5, 4, 3, 2, 1, 0])
    0
    >>> indexset([7, 6, 5, 4, 3, 2, 1], lowest_bit=1)
    0

    """
    m = len(bitset)
    g = enumerate(bitset)
    return sum(n_choose_k(bit - lowest_bit, m - i) for i, bit in g)

计算小于具有相同设置位数的给定无符号整数的无符号整数

0 个答案: