How Can We Generate All Set Partitions of a Given Set?

Generating All Set Partitions

One of the fundamental problems in combinatorial mathematics is finding all partitions of a given set. A set partition divides the set into non-empty disjoint subsets, referred to as blocks or parts.

In this problem, we seek a method to enumerate all partitions of a set with distinct elements. Consider the set {1, 2, 3}. Its partitions are:

• {{1}, {2}, {3}}
• {{1, 2}, {3}}
• {{1, 3}, {2}}
• {{1}, {2, 3}}
• {{1, 2, 3}}

Partitioning Algorithm

The task can be broken down into two subproblems: partitioning into two parts and partitioning a part into multiple parts.

Two-Part Partitioning

For an n-element set, all two-part partitions can be generated by representing each element as a bit in an n-bit pattern. A 0 bit indicates placement in the first part, and a 1 bit indicates placement in the second part. To avoid duplicate results when swapping parts, we always assign the first element to the first part. This leaves (2^(n-1))-1 unique two-part patterns.

Recursive Partitioning

With the two-part partitioning technique in place, we can recursively construct all partitions.

1. Start with an empty fixed part and the original set as the suffix.
2. Generate two-part partitions of the suffix.
3. For each suffix partition, recursively partition its second part into multiple parts.
4. Combine the fixed parts with the recursive partitions to obtain all partitions containing the fixed parts.
5. Repeat step 4 until all elements are partitioned.

C# Implementation

The following C# implementation employs the recursive partitioning algorithm:

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) {
            // Trivial partition: fixed parts followed by all suffix elements as a single block
            yield return fixedParts.Concat(new[] { suffixElements }).ToArray();

            // Two-group-partitions of suffix elements and their recursive sub-partitions
            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>> GetTuplePartitions<t>(T[] elements) {
            if (elements.Length [] resultSets = {
                    new List<t> { elements[0] },
                    new List<t>() 
                };
                
                for (int index = 1; index > (index - 1)) & 1].Add(elements[index]);
                }

                yield return Tuple.Create(resultSets[0].ToArray(), resultSets[1].ToArray());
            }
        }
    }
}</t></t></t></tuple></t></t></t></t></t>

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