Bell Numbers (Number of ways to Partition a Set)

Difficulty Level : Medium
Last Updated : 16 Apr, 2021

Given a set of n elements, find number of ways of partitioning it.
Examples:

Input:  n = 2
Output: Number of ways = 2
Explanation: Let the set be {1, 2}
            { {1}, {2} } 
            { {1, 2} }

Input:  n = 3
Output: Number of ways = 5
Explanation: Let the set be {1, 2, 3}
             { {1}, {2}, {3} }
             { {1}, {2, 3} }
             { {2}, {1, 3} }
             { {3}, {1, 2} }
             { {1, 2, 3} }.

Solution to above questions is Bell Number.
What is a Bell Number?
Let S(n, k) be total number of partitions of n elements into k sets. The value of n’th Bell Number is sum of S(n, k) for k = 1 to n.
$Bell(n) = \sum_{k=0}^{n}S(n,k)$
Value of S(n, k) can be defined recursively as, S(n+1, k) = k*S(n, k) + S(n, k-1)
How does above recursive formula work?
When we add a (n+1)’th element to k partitions, there are two possibilities.
1) It is added as a single element set to existing partitions, i.e, S(n, k-1)
2) It is added to all sets of every partition, i.e., k*S(n, k)
S(n, k) is called Stirling numbers of the second kind
First few Bell numbers are 1, 1, 2, 5, 15, 52, 203, ….
A Simple Method to compute n’th Bell Number is to one by one compute S(n, k) for k = 1 to n and return sum of all computed values. Refer this for computation of S(n, k).
A Better Method is to use Bell Triangle. Below is a sample Bell Triangle for first few Bell Numbers.

1
1 2
2 3 5
5 7 10 15
15 20 27 37 52

The triangle is constructed using below formula.

// If this is first column of current row 'i'
If j == 0
   // Then copy last entry of previous row
   // Note that i'th row has i entries
   Bell(i, j) = Bell(i-1, i-1) 

// If this is not first column of current row
Else 
   // Then this element is sum of previous element 
   // in current row and the element just above the
   // previous element
   Bell(i, j) = Bell(i-1, j-1) + Bell(i, j-1)

Interpretation
Then Bell(n, k) counts the number of partitions of the set {1, 2, …, n + 1} in which the element k + 1 is the largest element that can be alone in its set.
For example, Bell(3, 2) is 3, it is count of number of partitions of {1, 2, 3, 4} in which 3 is the largest singleton element. There are three such partitions:

    {1}, {2, 4}, {3}
    {1, 4}, {2}, {3}
    {1, 2, 4}, {3}.

Below is Dynamic Programming based implementation of above recursive formula.

C++

// A C++ program to find n'th Bell number
#include<iostream>
using namespace std;
 
int bellNumber(int n)
{
   int bell[n+1][n+1];
   bell[0][0] = 1;
   for (int i=1; i<=n; i++)
   {
      // Explicitly fill for j = 0
      bell[i][0] = bell[i-1][i-1];
 
      // Fill for remaining values of j
      for (int j=1; j<=i; j++)
         bell[i][j] = bell[i-1][j-1] + bell[i][j-1];
   }
   return bell[n][0];
}
 
// Driver program
int main()
{
   for (int n=0; n<=5; n++)
      cout << "Bell Number " << n << " is "
           << bellNumber(n) << endl;
   return 0;
}

Java

// Java program to find n'th Bell number
import java.io.*;
 
class GFG 
{
    // Function to find n'th Bell Number
    static int bellNumber(int n)
    {
        int[][] bell = new int[n+1][n+1];
        bell[0][0] = 1;
         
        for (int i=1; i<=n; i++)
        {
            // Explicitly fill for j = 0
            bell[i][0] = bell[i-1][i-1];
  
            // Fill for remaining values of j
            for (int j=1; j<=i; j++)
                bell[i][j] = bell[i-1][j-1] + bell[i][j-1];
        }
         
        return bell[n][0];
    }
     
    // Driver program
    public static void main (String[] args) 
    {
        for (int n=0; n<=5; n++)
            System.out.println("Bell Number "+ n +
                            " is "+bellNumber(n));
    }
}
 
// This code is contributed by Pramod Kumar

Python3

# A Python program to find n'th Bell number
 
def bellNumber(n):
 
    bell = [[0 for i in range(n+1)] for j in range(n+1)]
    bell[0][0] = 1
    for i in range(1, n+1):
 
        # Explicitly fill for j = 0
        bell[i][0] = bell[i-1][i-1]
 
        # Fill for remaining values of j
        for j in range(1, i+1):
            bell[i][j] = bell[i-1][j-1] + bell[i][j-1]
 
    return bell[n][0]
 
# Driver program
for n in range(6):
    print('Bell Number', n, 'is', bellNumber(n))
 
# This code is contributed by Soumen Ghosh

C#

// C# program to find n'th Bell number
using System;
 
class GFG {
     
    // Function to find n'th 
    // Bell Number
    static int bellNumber(int n)
    {
        int[,] bell = new int[n + 1, 
                              n + 1];
        bell[0, 0] = 1;
         
        for (int i = 1; i <= n; i++)
        {
             
            // Explicitly fill for j = 0
            bell[i, 0] = bell[i - 1, i - 1];
 
            // Fill for remaining values of j
            for (int j = 1; j <= i; j++)
                bell[i, j] = bell[i - 1, j - 1] + 
                             bell[i, j - 1];
        }
         
        return bell[n, 0];
    }
     
    // Driver Code
    public static void Main () 
    {
        for (int n = 0; n <= 5; n++)
            Console.WriteLine("Bell Number "+ n +
                              " is "+bellNumber(n));
    }
}
 
// This code is contributed by nitin mittal.

PHP

<?php
// A PHP program to find
// n'th Bell number
 
// function that returns 
// n'th bell number
function bellNumber($n)
{
 
    $bell[0][0] = 1;
    for ($i = 1; $i <= $n; $i++)
    {
         
        // Explicitly fill for j = 0
        $bell[$i][0] = $bell[$i - 1]
                            [$i - 1];
     
        // Fill for remaining 
        // values of j
        for ($j = 1; $j <= $i; $j++)
            $bell[$i][$j] = $bell[$i - 1][$j - 1] + 
                                $bell[$i][$j - 1];
    }
    return $bell[$n][0];
}
 
// Driver Code
for ($n = 0; $n <= 5; $n++)
echo("Bell Number " . $n . " is "
      . bellNumber($n) . "\n");
 
// This code is contributed by Ajit.
?>

Javascript

<script>
 
    // Javascript program to find n'th Bell number
     
    // Function to find n'th Bell Number
    function bellNumber(n)
    {
        let bell = new Array(n+1);
        for(let i = 0; i < n + 1; i++)
        {
            bell[i] = new Array(n + 1);
        }
        bell[0][0] = 1;
           
        for (let i=1; i<=n; i++)
        {
            // Explicitly fill for j = 0
            bell[i][0] = bell[i-1][i-1];
    
            // Fill for remaining values of j
            for (let j=1; j<=i; j++)
                bell[i][j] = bell[i-1][j-1] + bell[i][j-1];
        }
           
        return bell[n][0];
    }
     
    for (let n=0; n<=5; n++)
            document.write("Bell Number "+ n + " is "+bellNumber(n) + "</br>");
             
</script>

Output:

Bell Number 0 is 1
Bell Number 1 is 1
Bell Number 2 is 2
Bell Number 3 is 5
Bell Number 4 is 15
Bell Number 5 is 52

Time Complexity of above solution is O(n²). We will soon be discussing other more efficient methods of computing Bell Numbers.
Another problem that can be solved by Bell Numbers.
A number is squarefree if it is not divisible by a perfect square other than 1. For example, 6 is a square free number but 12 is not as it is divisible by 4.
Given a squarefree number x, find the number of different multiplicative partitions of x. The number of multiplicative partitions is Bell(n) where n is number of prime factors of x. For example x = 30, there are 3 prime factors of 2, 3 and 5. So the answer is Bell(3) which is 5. The 5 partitions are 1 x 30, 2 x15, 3 x 10, 5 x 6 and 2 x 3 x 5.
Exercise:
The above implementation causes arithmetic overflow for slightly larger values of n. Extend the above program so that results are computed under modulo 1000000007 to avoid overflows.
Reference:
https://en.wikipedia.org/wiki/Bell_number
https://en.wikipedia.org/wiki/Bell_triangle
This article is contributed by Rajeev Agrawal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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Table of Contents

Bell Numbers (Number of ways to Partition a Set)

C++

Java

Python3

C#

PHP

Javascript

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