# 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 number of elements in the original set always equals to the sum of 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 partition of a set of n elements is B(n) known as Bell number. As we know, this problem is NP-Complete i.e. it has 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 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.

For instance, the multiset {1, 1, 2} has 4 partitions:

partition 1 = { {1}, {1}, {2} } partition 2 = { {1}, {1, 2} } partition 3 = { {1, 1}, {2} } partition 4 = { {1, 1, 2} }

It can be solved using backtracking and recursion. This C++ program might be useful.

//Working Code in C++

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, 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]);
}
} 

Here’s the working example: Generate all Partitions

Thanks