How Can We Recursively Generate All Partitions of a Set?

Partitioning a Set: A Recursive Approach

The task of finding all partitions of a set arises frequently in computer science and mathematics. A partition divides a set into disjoint subsets, collectively containing all elements of the original set.

To begin, let's consider a simpler problem: dividing a set into two subsets. For an n-element set, we can represent a partition using a binary bitmask. Each bit corresponds to an element of the set, where a 0 indicates placement in the first subset and a 1 indicates the second. This ensures that each element is assigned to exactly one subset.

To handle the presence of the first element in the first subset, we only consider bitmasks where the first bit is 0. This reduces the number of bitmasks to 2^(n-1).

To generalize this approach, we can partition a set into multiple subsets recursively. We start with all two-part partitions, then divide the second subset into two parts, then the third subset, and so on. This recursive process produces all possible partitions.

Here is an example implementation in C# that generates all partitions for a given array:

using System;
using System.Collections.Generic;

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

        private static IEnumerable<T[][]> GetAllPartitions<T>(T[][] fixedParts, T[] suffixElements)
        {
            // Trivial partition: fixed parts followed by remaining elements as one block
            yield return fixedParts.Concat(new[] { suffixElements }).ToArray();

            // Get all two-part partitions of suffix elements and subdivide recursively
            var suffixPartitions = GetTuplePartitions(suffixElements);

            foreach (Tuple<T[], T[]> suffixPartition in suffixPartitions)
            {
                var recursivePartitions = GetAllPartitions(fixedParts.Concat(new[] { suffixPartition.Item1 }).ToArray(), suffixPartition.Item2);
                foreach (T[][] partition in recursivePartitions)
                {
                    yield return partition;
                }
            }
        }

        private static IEnumerable<Tuple<T[], T[]>> GetTuplePartitions<T>(T[] elements)
        {
            if (elements.Length < 2)
                yield break;

            for (int pattern = 1; pattern < 1 << (elements.Length - 1); pattern++)
            {
                List<T>[] resultSets = { new List<T> { elements[0] }, new List<T>() };

                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());
            }
        }
    }
}

Calling GetAllPartitions with an array of elements generates all possible partitions for that set.

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