集合中加权元素的组合，其中加权和等于固定整数（python中）

我想在一个集合中找到所有可能的加权元素组合，其中它们的权重之和正好等于给定的权重W

假设我想从集合

{A'，B'，C'，D'，E'}

中选择k个元素，其中

权重={A'：2'，B'：1'，C'：3'，D'：2'，E'：1}

和

W=4

那么这将产生：

（'A'、'B'、'E'）
（‘A’、‘D’）
（‘B’，‘C’）
（‘B’、‘D’、‘E’）
（'C'，'E'）

我意识到蛮力方法是找到给定集合的所有置换（使用

itertools.permutations

）并用加权和W拼接出前k个元素。但我处理的是每个集合至少20个元素，这在计算上很昂贵

我认为使用背包的一种变体会有所帮助，其中只考虑重量（而不是价值），并且重量之和必须等于到W（而不是更低）

我想用python实现这一点，但任何cs理论提示都会有所帮助。优雅的奖励积分

在所有n中循环！排列太贵了。而是生成所有2^n子集

屈服

[('A', 'D'), ('C', 'B'), ('C', 'E'), ('A', 'B', 'E'), ('B', 'E', 'D')]

通过将列表理解的返回部分更改为

tuple（sorted（x））

，或者将

list

调用

powerset

替换为一个to

sorted

，可以将其转换为排序元组有效的方法是使用前k项创建具有相同权重的元素集

从k=0的空集开始，然后使用k-1中的组合为k创建组合。除非可以使用负权重，否则可以修剪总权重大于W的组合

下面是使用您的示例的结果：

comb[k，w]是使用前k个元素具有总权重w的元素集。
大括号用于集合。
S+e是通过将元素e添加到S的每个成员而创建的集合集

comb[0,0]={}
comb[1,0]={comb[0,0]}
comb[1,2]={comb[0,0]+'A'}
comb[2,0]={comb[1,0]}
comb[2,1]={comb[1,0]+'B'}
comb[2,2]={comb[1,2]}
comb[2,3]={comb[1,2]'B'}
comb[3,0]={comb[2,0]}
comb[3,1]={comb[2,1]}
comb[3,2]={comb[2,2]}
comb[3,3]={comb[2,3],comb[2,0]+'C'}
comb[3,4]={comb[2,3]+'C'}
comb[4,0]={comb[3,0]}
comb[4,1]={comb[3,1]}
comb[4,2]={comb[3,2],comb[3,0]+'D'}
comb[4,3]={comb[3,3],comb[3,1]+'D'}
comb[4,4]={comb[3,4],comb[3,2]+'D'}
comb[5,0]={comb[4,0]}
comb[5,1]={comb[4,1],comb[4,0]+'E'}
comb[5,2]={comb[4,2],comb[4,1]+'E'}
comb[5,3]={comb[4,3],comb[4,2]+'E'}
comb[5,4]={comb[4,4],comb[4,3]+'E'}

答案是comb[5,4]，它简化为：

{
  {{'B'}+'C'},
  {{'A'}+'D'},
  {
    {{'A'}+'B'},
    {'C'},
    {'B'}+'D'             
  }+'E'
}

---

给出所有的组合。

这类集合中的项目数有上限吗？如果您这样做了，并且它最多约为40，那么中所述的计算可能非常简单，并且比蛮力计算的复杂度要低得多

注意：使用比Python dict更节省内存的数据结构，这也适用于更大的集合。一个高效的实现应该能够轻松地处理大小为60的集合

下面是一个示例实现：

from collections import defaultdict
from itertools import chain, combinations, product

# taken from the docs of the itertools module
def powerset(iterable):
     "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
     s = list(iterable)
     return chain.from_iterable(combinations(s, r) for r in xrange(len(s) + 1))

def gen_sums(weights):
    """Given a set of weights, generate a sum --> subsets mapping.

    For each posible sum, this will create a list of subsets of weights
    with that sum.

    >>> gen_sums({'A':1, 'B':1})
    {0: [()], 1: [('A',), ('B',)], 2: [('A', 'B')]}
    """
    sums = defaultdict(list)
    for weight_items in powerset(weights.items()):
        if not weight_items:
            sums[0].append(())
        else:
            keys, weights = zip(*weight_items)
            sums[sum(weights)].append(keys)
    return dict(sums)

def meet_in_the_middle(weights, target_sum):
    """Find subsets of the given weights with the desired sum.

    This uses a simplified meet-in-the-middle algorithm.

    >>> weights = {'A':2, 'B':1, 'C':3, 'D':2, 'E':1}
    >>> list(meet_in_the_middle(weights, 4))
    [('B', 'E', 'D'), ('A', 'D'), ('A', 'B', 'E'), ('C', 'B'), ('C', 'E')]
    """
    # split weights into two groups
    weights_list = weights.items()
    weights_set1 = dict(weights_list[:len(weights)//2])
    weights_set2 = dict(weights_list[len(weights_set1):])

    # generate sum --> subsets mapping for each group of weights,
    # and sort the groups in descending order
    set1_sums = sorted(gen_sums(set1).items())
    set2_sums = sorted(gen_sums(set2).items(), reverse=True)

    # run over the first sorted list, meanwhile going through the
    # second list and looking for exact matches
    set2_sums = iter(set2_sums)
    try:
        set2_sum, subsets2 = set2_sums.next()
        for set1_sum, subsets1 in set1_sums:
            set2_target_sum = target_sum - set1_sum
            while set2_sum > set2_target_sum:
                set2_sum, subsets2 = set2_sums.next()
            if set2_sum == set2_target_sum:
                for subset1, subset2 in product(subsets1, subsets2):
                    yield subset1 + subset2
    except StopIteration: # done iterating over set2_sums
        pass

---

itertools.permutation

返回一个迭代器，而不是所有置换的列表。因此，您可以逐步遍历结果并在第k个匹配处停止。@TimPietzcker-仍然无法提高计算复杂度，除非迭代器返回按某个特定值排序的结果（在本例中为权重之和），我猜它不会。您的评论很好@fylow，但别忘了检查答案左侧的复选标记，用一些冷酷、刻薄的分数来量化你的感激之情！明白了@PaulMcGuire，我对堆栈溢出不是很熟悉。如果集合的大小至少为20，那么上面的操作将需要相当长的时间。对于一组N个权重，复杂度为O（2^N）。对于30码左右的套装，这永远不会结束。。。我发布了一个简单的解决方案，它使用空间/时间权衡来处理更大的权重集。我在示例中犯了一个错误，导致它错过了（'B'，'C'）——现在已修复。

from collections import defaultdict
from itertools import chain, combinations, product

# taken from the docs of the itertools module
def powerset(iterable):
     "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
     s = list(iterable)
     return chain.from_iterable(combinations(s, r) for r in xrange(len(s) + 1))

def gen_sums(weights):
    """Given a set of weights, generate a sum --> subsets mapping.

    For each posible sum, this will create a list of subsets of weights
    with that sum.

    >>> gen_sums({'A':1, 'B':1})
    {0: [()], 1: [('A',), ('B',)], 2: [('A', 'B')]}
    """
    sums = defaultdict(list)
    for weight_items in powerset(weights.items()):
        if not weight_items:
            sums[0].append(())
        else:
            keys, weights = zip(*weight_items)
            sums[sum(weights)].append(keys)
    return dict(sums)

def meet_in_the_middle(weights, target_sum):
    """Find subsets of the given weights with the desired sum.

    This uses a simplified meet-in-the-middle algorithm.

    >>> weights = {'A':2, 'B':1, 'C':3, 'D':2, 'E':1}
    >>> list(meet_in_the_middle(weights, 4))
    [('B', 'E', 'D'), ('A', 'D'), ('A', 'B', 'E'), ('C', 'B'), ('C', 'E')]
    """
    # split weights into two groups
    weights_list = weights.items()
    weights_set1 = dict(weights_list[:len(weights)//2])
    weights_set2 = dict(weights_list[len(weights_set1):])

    # generate sum --> subsets mapping for each group of weights,
    # and sort the groups in descending order
    set1_sums = sorted(gen_sums(set1).items())
    set2_sums = sorted(gen_sums(set2).items(), reverse=True)

    # run over the first sorted list, meanwhile going through the
    # second list and looking for exact matches
    set2_sums = iter(set2_sums)
    try:
        set2_sum, subsets2 = set2_sums.next()
        for set1_sum, subsets1 in set1_sums:
            set2_target_sum = target_sum - set1_sum
            while set2_sum > set2_target_sum:
                set2_sum, subsets2 = set2_sums.next()
            if set2_sum == set2_target_sum:
                for subset1, subset2 in product(subsets1, subsets2):
                    yield subset1 + subset2
    except StopIteration: # done iterating over set2_sums
        pass