问题描述
假设我有一个从 0 到 9 的 20 个随机整数列表.我想将列表分成 N
个子集,以便子集总和的比率等于给定值,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 N = 2
情况.
Say I have a list of 20 random integers from 0 to 9. I want to divide the list into N
subsets so that the ratio of subset sums equal to given values, and I want to find all possible partitions. I wrote the following code and got it work for the N = 2
case.
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
If you calculate the sum of each subset, you will find the ratio is 44 : 33 = 4 : 3, which are exactly the input values. However, I want the function to work for any number of subsets. For example, I expect
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.有什么建议吗?
I have been thinking about this problem for a month and I found this to be extremely hard. My conclusion is that this problem can only be solved by a recursion. I would like to know if there are any relatively fast algorithm for this. Any suggestions?
推荐答案
是的,需要递归.基本逻辑是将一个部分分为一部分和其余部分,然后以所有可能的方式递归拆分其余部分.我已经按照你的假设假设一切都是可区分的,这会产生很多可能性,可能太多而无法一一列举.尽管如此:
Yes, recursion is called for. The basic logic is to do a bipartition into one part and the rest and then recursively split the rest in all possible ways. I've followed your lead in assuming that everything is distinguishable, which creates a lot of possibilities, possibly too many to enumerate. Nevertheless:
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)
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