生成所有“独特"的集合的子集(不是幂集) [英] Generate all "unique" subsets of a set (not a powerset)

问题描述

假设我们有一个集合 S，其中包含一些子集:

Let's say we have a Set S which contains a few subsets:

- [a,b,c]
- [a,b]
- [c]
- [d,e,f]
- [d,f]
- [e]

假设 S 包含六个唯一元素:a、b、c、d、e 和 f.

Let's also say that S contains six unique elements: a, b, c, d, e and f.

我们如何才能找到所有可能的 S 子集，其中包含 S 的每个唯一元素恰好一次?

How can we find all possible subsets of S that contain each of the unique elements of S exactly once?

函数/方法的结果应该是这样的:

The result of the function/method should be something like that:

1. [[a,b,c], [d,e,f]];
2. [[a,b,c], [d,f], [e]];
3. [[a,b], [c], [d,e,f]];
4. [[a,b], [c], [d,f], [e]].

是否有任何最佳实践或任何标准方法来实现这一目标?

Is there any best practice or any standard way to achieve that?

如果有伪代码、Ruby 或 Erlang 示例，我将不胜感激.

I would be grateful for a Pseudo-code, Ruby or Erlang example.

推荐答案

听起来你正在寻找的是 分区 集合/数组.

It sounds like what you are looking for are the partitions of a set/array.

这样做的一种方法是递归:

One way of doing this is recursively:

• 一个 1 元素数组 [x] 恰好有一个分区
• 如果 x 是 x = [head] + tail 形式的数组，则 x 的分区是通过获取 tail 的每个分区并将 head 添加到每个可能的来生成的.例如，如果我们生成 [3,2,1] 的分区，那么从 [2,1] 的分区 [[2,1]] 我们可以将 3 添加到 [2,1] 或作为单独的元素，这为我们提供了 [1,2,3] 的 5 个分区中的 2 个分区 [[3,2,1] 或 [[2,1], [3]]

ruby 实现有点像

def partitions(x)
  if x.length == 1
   [[x]]
  else
    head, tail = x[0], x[1, x.length-1]
    partitions(tail).inject([]) do |result, tail_partition|
      result + partitions_by_adding_element(tail_partition, head)
    end
  end
end

def partitions_by_adding_element(partition, element)
  (0..partition.length).collect do |index_to_add_at|
    new_partition = partition.dup
    new_partition[index_to_add_at] = (new_partition[index_to_add_at] || []) + [element]
    new_partition
  end
end

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