Unique Factorization of a Given Number in C++

Here is a C++ Program to get all the unique factorization of a given integer such that addition of a partition results an integer. In this program, a positive integer n is given, and we shall generate all possible unique ways to represent n as sum of positive integers.

Algorithm to Perform Unique Factorization

Below is the algorithm to perform the unique factorization of a given number:

• Begin
• function displayAllUniqueParts(int m):
• 1) Set Index of last element k in a partition to 0
• 2) Initialize first partition as number itself, p[k]=m
• 3) Create a while loop which first prints current partition, then generates next partition. The loop stops when the current partition has all 1s.
• 4) Display current partition as displayArray(p, k + 1)
• 5) Generate next partition:
• 6) Initialize val = 0. Find the rightmost non-one value in p[]. Also, update the val so that we know how much value can be accommodated. If k < 0, all the values are 1 so there are no more partitions Decrease the p[k] found above and adjust the val.
• 7) If val is more, then the sorted order is violated. Divide val in different values of size p[k] and copy these values at different positions after p[k]. Copy val to next position and increment position.
• End

C++ Program to Perform Unique Factorization

Below is a complete C++ program where we generate and print all unique number partitions using a loop.

#include<iostream>
using namespace std;
void displayArray(int p[], int m) //to print the array
{
   for (int i = 0; i < m; i++)
   cout << p[i] << " ";
   cout << endl;
}
void displayAllUniqueParts(int m)
{
   int p[m];
   int k = 0;
   p[k] = m;
   while (true)
   {
      displayArray(p, k + 1);
      int val = 0; //initialize val
      while (k >= 0 && p[k] == 1)
      {
         val += p[k]; //update val
         k--;
      }
      if (k < 0)
      return;
      p[k]--;
      val++;
      while (val > p[k]) //if val is more
      {
         p[k + 1] = p[k];
         val = val - p[k];
         k++;
      }
      p[k + 1] = val;
      k++;
   }
}
int main()
{
   cout << "Display All Unique Partitions of 3\n";
   displayAllUniqueParts(3);
   cout << "\nDisplay All Unique Partitions of 4\n";
   displayAllUniqueParts(4);
   cout << "\nDisplay All Unique Partitions of 5\n";
   displayAllUniqueParts(5);
   return 0;
}

The output below shows all unique factorizations (partitions) of the given numbers.

Display All Unique Partitions of 3
3 
2 1 
1 1 1 

Display All Unique Partitions of 4
4 
3 1 
2 2 
2 1 1 
1 1 1 1 

Display All Unique Partitions of 5
5 
4 1 
3 2 
3 1 1 
2 2 1 
2 1 1 1 
1 1 1 1 1