如何找到集合的所有分区 [英] How to find all partitions of a set

问题描述

我有一组不同的价值观.我正在寻找一种方法来生成该集合的所有分区，即将集合划分为子集的所有可能方式.

I have a set of distinct values. I am looking for a way to generate all partitions of this set, i.e. all possible ways of dividing the set into subsets.

例如，集合 {1, 2, 3} 具有以下分区:

For instance, the set {1, 2, 3} has the following partitions:

{ {1}, {2}, {3} },
{ {1, 2}, {3} },
{ {1, 3}, {2} },
{ {1}, {2, 3} },
{ {1, 2, 3} }.

由于这些是数学意义上的集合，因此顺序无关紧要.例如，{1, 2}, {3} 与 {3}, {2, 1} 相同，不应是单独的结果.

As these are sets in the mathematical sense, order is irrelevant. For instance, {1, 2}, {3} is the same as {3}, {2, 1} and should not be a separate result.

可以在 Wikipedia 上找到集分区的详细定义.

A thorough definition of set partitions can be found on Wikipedia.

推荐答案

我找到了一个简单的递归解决方案.

I've found a straightforward recursive solution.

首先，让我们解决一个更简单的问题:如何找到恰好由两部分组成的所有分区.对于一个 n 元素集，我们可以从 0 到 (2^n)-1 计算一个 int.这将创建每个 n 位模式，每个位对应于一个输入元素.如果该位为 0，我们将元素放在第一部分；如果为 1，则元素放置在第二部分.这留下了一个问题:对于每个分区，我们将得到一个重复的结果，其中两个部分被交换.为了解决这个问题，我们总是将第一个元素放入第一部分.然后我们只通过从 0 到 (2^(n-1))-1 的计数来分配剩余的 n-1 个元素.

First, let's solve a simpler problem: how to find all partitions consisting of exactly two parts. For an n-element set, we can count an int from 0 to (2^n)-1. This creates every n-bit pattern, with each bit corresponding to one input element. If the bit is 0, we place the element in the first part; if it is 1, the element is placed in the second part. This leaves one problem: For each partition, we'll get a duplicate result where the two parts are swapped. To remedy this, we'll always place the first element into the first part. We then only distribute the remaining n-1 elements by counting from 0 to (2^(n-1))-1.

现在我们可以将一个集合分成两部分，我们可以编写一个递归函数来解决剩下的问题.该函数从原始集合开始并找到所有两部分分区.对于这些分区中的每一个，它递归地找到将第二部分分成两部分的所有方法，从而产生所有三部分分区.然后它划分每个分区的最后一部分以生成所有四部分分区，依此类推.

Now that we can partition a set into two parts, we can write a recursive function that solves the rest of the problem. The function starts off with the original set and finds all two-part-partitions. For each of these partitions, it recursively finds all ways to partition the second part into two parts, yielding all three-part partitions. It then divides the last part of each of these partitions to generate all four-part partitions, and so on.

以下是 C# 中的实现.调用

The following is an implementation in C#. Calling

Partitioning.GetAllPartitions(new[] { 1, 2, 3, 4 })

产量

{ {1, 2, 3, 4} },
{ {1, 3, 4}, {2} },
{ {1, 2, 4}, {3} },
{ {1, 4}, {2, 3} },
{ {1, 4}, {2}, {3} },
{ {1, 2, 3}, {4} },
{ {1, 3}, {2, 4} },
{ {1, 3}, {2}, {4} },
{ {1, 2}, {3, 4} },
{ {1, 2}, {3}, {4} },
{ {1}, {2, 3, 4} },
{ {1}, {2, 4}, {3} },
{ {1}, {2, 3}, {4} },
{ {1}, {2}, {3, 4} },
{ {1}, {2}, {3}, {4} }.

using System;
using System.Collections.Generic;
using System.Linq;

namespace PartitionTest {
    public static class Partitioning {
        public static IEnumerable<T[][]> GetAllPartitions<T>(T[] elements) {
            return GetAllPartitions(new T[][]{}, elements);
        }

        private static IEnumerable<T[][]> GetAllPartitions<T>(
            T[][] fixedParts, T[] suffixElements)
        {
            // A trivial partition consists of the fixed parts
            // followed by all suffix elements as one block
            yield return fixedParts.Concat(new[] { suffixElements }).ToArray();

            // Get all two-group-partitions of the suffix elements
            // and sub-divide them recursively
            var suffixPartitions = GetTuplePartitions(suffixElements);
            foreach (Tuple<T[], T[]> suffixPartition in suffixPartitions) {
                var subPartitions = GetAllPartitions(
                    fixedParts.Concat(new[] { suffixPartition.Item1 }).ToArray(),
                    suffixPartition.Item2);
                foreach (var subPartition in subPartitions) {
                    yield return subPartition;
                }
            }
        }

        private static IEnumerable<Tuple<T[], T[]>> GetTuplePartitions<T>(
            T[] elements)
        {
            // No result if less than 2 elements
            if (elements.Length < 2) yield break;

            // Generate all 2-part partitions
            for (int pattern = 1; pattern < 1 << (elements.Length - 1); pattern++) {
                // Create the two result sets and
                // assign the first element to the first set
                List<T>[] resultSets = {
                    new List<T> { elements[0] }, new List<T>() };
                // Distribute the remaining elements
                for (int index = 1; index < elements.Length; index++) {
                    resultSets[(pattern >> (index - 1)) & 1].Add(elements[index]);
                }

                yield return Tuple.Create(
                    resultSets[0].ToArray(), resultSets[1].ToArray());
            }
        }
    }
}

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