快速 python 算法，用於從子集總和等於給定比率的數字列表中找到所有可能的分區

Question

假設我有一個從 0 到 9 的 20 個隨機整數的列表。我想將該列表划分為N個子集，以便子集和的比率等於給定值，並且我想找到所有可能的分區。 我編寫了以下代碼並使其適用於N = 2的情況。

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])

Output：

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

如果你計算每個子集的總和，你會發現比例是 44:33 = 4:3，這正是輸入值。 但是，我希望 function 適用於任意數量的子集。 例如，我期望

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

返回類似的東西

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

我已經考慮這個問題一個月了，我發現這非常困難。 我的結論是，這個問題只能通過遞歸來解決。 我想知道是否有任何相對快速的算法。 有什么建議么？

Answer 1

是的，需要遞歸。 基本邏輯是將 rest 分成一個部分，然后以所有可能的方式遞歸拆分 rest。 我按照你的說法假設一切都是可區分的，這創造了很多可能性，可能太多而無法列舉。 盡管如此：

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)