快速python算法，从子集总和等于给定比率的数字列表中找到所有可能的分区

Question

假设我有一个从 0 到 9 的 20 个随机整数列表。我想将列表划分为N子集，以便子集总和的比率等于给定值，并且我想找到所有可能的分区。我编写了以下代码并使其适用于该N = 2案例。

import random
import itertools

#lst = [random.randrange(10) for _ in range(20)]
lst = [2, 0, 1, 7, 2, 4, 9, 7, 6, 0, 5, 4, 7, 4, 5, 0, 4, 5, 2, 3]

def partition_sum_with_ratio(numbers, ratios):
    target1 = round(int(sum(numbers) * ratios[0] / (ratios[0] + ratios[1])))
    target2 = sum(numbers) - target1
    p1 = [seq for i in range(len(numbers), 0, -1) for seq in
          itertools.combinations(numbers, i) if sum(seq) == target1
          and sum([s for s in numbers if s not in seq]) == target2]

    p2 = [tuple(n for n in numbers if n not in seq) for seq in p1]

    return list(zip(p1, p2))

partitions = partition_sum_with_ratios(lst, ratios=[4, 3])
print(partitions[0])

输出：

((2, 0, 1, 2, 4, 6, 0, 5, 4, 4, 5, 0, 4, 5, 2), (7, 9, 7, 7, 3))

如果计算每个子集的总和，您会发现比率为 44 : 33 = 4 : 3，这正是输入值。但是，我希望该函数适用于任意数量的子集。例如，我期望

partition_sum_with_ratio(lst, ratios=[4, 3, 3])

返回类似的东西

((2, 0, 1, 2, 4, 6, 0, 5, 4, 4, 3), (5, 0, 4, 5, 2, 7), (9, 7, 7))

我已经考虑这个问题一个月了，我发现这非常困难。我的结论是，这个问题只能通过递归来解决。我想知道是否有任何相对较快的算法。有什么建议？

Answer 1

是的，需要递归。基本逻辑是将一部分分为一部分和其余部分，然后以所有可能的方式递归地分割其余部分。我跟随你的思路，假设一切都是可区分的，这创造了很多可能性，可能多得无法一一列举。尽管如此：

import itertools


def totals_from_ratios(sum_numbers, ratios):
    sum_ratios = sum(ratios)
    totals = [(sum_numbers * ratio) // sum_ratios for ratio in ratios]
    residues = [(sum_numbers * ratio) % sum_ratios for ratio in ratios]
    for i in sorted(
        range(len(ratios)), key=lambda i: residues[i] * ratios[i], reverse=True
    )[: sum_numbers - sum(totals)]:
        totals[i] += 1
    return totals


def bipartitions(numbers, total):
    n = len(numbers)
    for k in range(n + 1):
        for combo in itertools.combinations(range(n), k):
            if sum(numbers[i] for i in combo) == total:
                set_combo = set(combo)
                yield sorted(numbers[i] for i in combo), sorted(
                    numbers[i] for i in range(n) if i not in set_combo
                )


def partitions_into_totals(numbers, totals):
    assert totals
    if len(totals) == 1:
        yield [numbers]
    else:
        for first, remaining_numbers in bipartitions(numbers, totals[0]):
            for rest in partitions_into_totals(remaining_numbers, totals[1:]):
                yield [first] + rest


def partitions_into_ratios(numbers, ratios):
    totals = totals_from_ratios(sum(numbers), ratios)
    yield from partitions_into_totals(numbers, totals)


lst = [2, 0, 1, 7, 2, 4, 9, 7, 6, 0, 5, 4, 7, 4, 5, 0, 4, 5, 2, 3]
for part in partitions_into_ratios(lst, [4, 3, 3]):
    print(part)