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.
- Start with an empty fixed part and the original set as the suffix.
- Generate two-part partitions of the suffix.
- For each suffix partition, recursively partition its second part into multiple parts.
- Combine the fixed parts with the recursive partitions to obtain all partitions containing the fixed parts.
- 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[], 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()); } } } }
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