# Multiset Partitioning Problem

In mathematics, a partition of a set is a grouping of its elements into non-empty subsets, in such a way that the number of elements in the original set always equals the sum of the number of elements in each partition.

Equivalently, a family of sets P is a partition of X if and only if all of the following conditions hold:

• The family P does not contain the empty set
• The union of the sets in P is equal to X. The sets in P are said to cover X.
• The intersection of any two distinct sets in P is empty i.e. the elements of P are said to be pairwise disjoint.

The number of possible partitions of a set of n elements is B(n) known as the Bell number. As we know, this problem is NP-Complete i.e. it has a non-polynomial time solution.

###### Solving for a smaller set in Polynomial time

But for a smaller set, we could try to find a way to generate all partitions.
Let’s take an example, given a collection of numbers that may contain duplicates, find all partitions of it. (all possible ways of dividing the collection.) For a set i.e. when each element in a multiset has multiplicity 1, there’s a nice solution here. But it doesn’t solve the problem when there are duplicate elements present in the set.

Let the multiset be {1, 1, 2}, it has 4 partitions:
partition 1 = { {1}, {1}, {2} }
partition 2 = { {1}, {1, 2} }
partition 3 = { {1, 1}, {2} }
partition 4 = { {1, 1, 2} }

Here’s a brute-force solution for a set with/without duplicates using backtracking and recursion

Code Implementation

//
//  main.cpp
//  Multiset Partitioning Problem
//
//  Created by Himanshu on 11/01/23.
//

#include <iostream>
#include <set>
#include <vector>
#include <algorithm>
using namespace std;

void solve (set<vector<vector<int>>>& solution, vector<int> inputSet,
vector<vector<int>>& partitions, vector<int> partition,
int n, int i) {

int numberOfElements = 0;
for (int i=0; i<partitions.size(); i++) {
numberOfElements += partitions[i].size();
}

if (numberOfElements == n) {
vector<vector<int>> newPartitions = partitions;
for (int i=0; i<newPartitions.size(); i++) {
sort (newPartitions[i].begin(), newPartitions[i].end());
}
sort(newPartitions.begin(), newPartitions.end());
solution.insert(newPartitions);
return;
}

for (int j=i; j<n; j++) {
partition.push_back(inputSet[j]);
partitions.push_back(partition);
vector<int> partitionNew;
solve(solution, inputSet, partitions, partitionNew, n, j+1);
partitions.pop_back();
}
}

void permute (set<vector<vector<int>>>& solution, vector<int>& inputSet,
int i, int n) {

if (i == n) {
vector<int> partition;
vector<vector<int>> partitions;
solve(solution, inputSet, partitions, partition, (int) inputSet.size(), 0);
return;
}

for (int j=i; j<=n; j++) {
swap(inputSet[i], inputSet[j]);
permute(solution, inputSet, i+1, n);
swap(inputSet[i], inputSet[j]);
}
}

int main() {
vector<int> inputSet {1, 1, 2, 3}, partition;
set<vector<vector<int>>> partitions;
set<vector<vector<int>>>::iterator it;

permute(partitions, inputSet, 0, (int) inputSet.size()-1);

int counter = 1;
for (it = partitions.begin(); it!=partitions.end(); it++) {
cout<<"Partitions set "<<counter++<<":"<<endl;
for (int i = 0; i<(*it).size(); i++) {
for(int j = 0; j<(*it)[i].size(); j++) {
cout<<(*it)[i][j]<<" ";
}
cout<<endl;
}
}

return 0;
}


Output

Partitions set 1:
1
1
2
3
Partitions set 2:
1
1
2 3
Partitions set 3:
1
1 2
3
Partitions set 4:
1
1 2 3
Partitions set 5:
1
1 3
2
Partitions set 6:
1 1
2
3
Partitions set 7:
1 1
2 3
Partitions set 8:
1 1 2
3
Partitions set 9:
1 1 2 3
Partitions set 10:
1 1 3
2
Partitions set 11:
1 2
1 3 

Here’s the working example: Generate all Partitions