Bell Numbers (Number of ways to Partition a Set)

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} }. 

The 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. 

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).
Below is Dynamic Programming based implementation of the above recursive code using the Stirling number-

C++

#include <iostream>
using namespace std;

int main() {
    int n=5;
    int s[n+1][n+1];
    for(int i=0;i<n+1;i++){
        for(int j=0;j<n+1;j++){
            if(j>i) s[i][j]=0;
            else if(i==j) s[i][j]=1;
            else if(i==0 || j==0) s[i][j]=0;
            else{

                s[i][j]= j*s[i-1][j] + s[i-1][j-1];
            }

        }
    }
    int ans=0;
    for(int i=0;i<n+1;i++){
        ans += s[n][i];
    }
    cout<<ans;


    return 0;
}

Java

/*package whatever //do not write package name here */
// Java program to find number of ways of partitioning it.

import java.io.*;
// "static void main" must be defined in a public class.
public class GFG {
    public static void main(String[] args)
    {
        int n = 5;
        int[][] s = new int[n + 1][n + 1];
        for (int i = 0; i < n + 1; i++) {
            for (int j = 0; j < n + 1; j++) {
                if (j > i)
                    s[i][j] = 0;
                else if (i == j)
                    s[i][j] = 1;
                else if (i == 0 || j == 0)
                    s[i][j] = 0;
                else {
                    s[i][j]
                        = j * s[i - 1][j] + s[i - 1][j - 1];
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n + 1; i++) {
            ans += s[n][i];
        }
        System.out.println(ans);
    }
}

// The code is contributed by Gautam goel (gautamgoel962)

Python3

# python program to find number of ways of partitioning it.
n = 5
s = [[0 for _ in range(n+1)] for _ in range(n+1)]
for i in range(n+1):
    for j in range(n+1):
        if j > i:
            continue
        elif(i==j):
            s[i][j] = 1
        elif(i==0 or j==0):
            s[i][j]=0
        else:
            s[i][j] = j*s[i-1][j] + s[i-1][j-1]
ans = 0
for i in range(0,n+1):
    ans+=s[n][i]
print(ans)

C#

// C# Program to find number of ways of partitioning it.
using System;

public class Program {
    static public void Main(string[] args) {

        int n = 5;

        int[, ] s = new int[n + 1, n + 1];

        for (int i = 0; i < n + 1; i++) {

            for (int j = 0; j < n + 1; j++) {

                if (j > i)
                    s[i, j] = 0;

                else if (i == j)
                    s[i, j] = 1;

                else if (i == 0 || j == 0)
                    s[i, j] = 0;

                else
                    s[i, j]
                        = j * s[i - 1, j] + s[i - 1, j - 1];
            }
        }

        int ans = 0;

        for (int i = 0; i < n + 1; i++)
            ans += s[n, i];

        Console.WriteLine(ans);
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Javascript

// JavaScript program to find number of ways of partitioning it.

let n=5;
let s = new Array(n+1);
for(let i=0;i<n+1;i++){
    s[i] = new Array(n+1);
    for(let j=0;j<n+1;j++){
        if(j>i) s[i][j]=0;
        else if(i==j) s[i][j]=1;
        else if(i==0 || j==0) s[i][j]=0;
        else{

            s[i][j]= j*s[i-1][j] + s[i-1][j-1];
        }

    }
}
let ans=0;
for(let i=0;i<n+1;i++){
    ans += s[n][i];
}
console.log(ans)

// The code is contributed by Gautam goel (gautamgoel962)

Output

52

Time complexity: O(N²) 
Auxiliary Space: O(N²) 

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++14

// 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.

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>

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.
?>

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: O(N²) 
Auxiliary Space: O(N²)

Space Optimized DP Approach:

We can use a 1-D list to represent the previous row of the Bell triangle. We initialize dp[0] to 1, since there is only one way to partition an empty set.

To compute the Bell numbers for n > 0, we first set dp[0] = dp[i-1], since the first element in each row is the same as the last element in the previous row. Then, we use the recurrence relation dp[j] = prev + dp[j-1] to compute the Bell number for each partition, where prev is the value of dp[j] in the previous iteration of the inner loop. We update prev to the temporary variable temp before updating dp[j].

Finally, we return dp[0], which is the Bell number for the partition of a set with n elements into non-empty subsets.

C++

#include <iostream>
#include <vector>

// Function to calculate the Bell number for 'n'
int bellNumbers(int n)
{
    // Initialize the previous row of the Bell triangle with
    // dp[0] = 1
    std::vector<int> dp(n + 1, 0);
    dp[0] = 1;

    for (int i = 1; i <= n; i++) {
        // The first element in each row is the same as the
        // last element in the previous row
        int prev = dp[0];
        dp[0] = dp[i - 1];
        for (int j = 1; j <= i; j++) {
            // The Bell number for n is the sum of the Bell
            // numbers for all previous partitions
            int temp = dp[j];
            dp[j] = prev + dp[j - 1];
            prev = temp;
        }
    }

    return dp[0];
}

int main()
{
    int n = 5;
    std::cout << bellNumbers(n) << std::endl;

    return 0;
}

Java

import java.util.Arrays;

public class BellNumbers {

    // Function to calculate the Bell number for 'n'
    static int bellNumbers(int n)
    {
        // Initialize the previous row of the Bell triangle
        // with dp[0] = 1
        int[] dp = new int[n + 1];
        Arrays.fill(dp, 0);
        dp[0] = 1;

        for (int i = 1; i <= n; i++) {
            // The first element in each row is the same as
            // the last element in the previous row
            int prev = dp[0];
            dp[0] = dp[i - 1];

            for (int j = 1; j <= i; j++) {
                // The Bell number for n is the sum of the
                // Bell numbers for all previous partitions
                int temp = dp[j];
                dp[j] = prev + dp[j - 1];
                prev = temp;
            }
        }

        return dp[0];
    }

    public static void main(String[] args)
    {
        int n = 5;
        System.out.println(bellNumbers(n));
    }
}

Python3

def bell_numbers(n):
    # Initialize the previous row of the Bell triangle with dp[0] = 1
    dp = [1] + [0] * n

    for i in range(1, n+1):
        # The first element in each row is the same as the last element in the previous row
        prev = dp[0]
        dp[0] = dp[i-1]
        for j in range(1, i+1):
            # The Bell number for n is the sum of the Bell numbers for all previous partitions
            temp = dp[j]
            dp[j] = prev + dp[j-1]
            prev = temp

    return dp[0]


n = 5
print(bell_numbers(n))

C#

using System;

class Program {
    // Function to calculate the Bell number for 'n'
    static int BellNumbers(int n)
    {
        // Initialize the previous row of the Bell triangle
        // with dp[0] = 1
        int[] dp = new int[n + 1];
        dp[0] = 1;

        for (int i = 1; i <= n; i++) {
            // The first element in each row is the same as
            // the last element in the previous row
            int prev = dp[0];
            dp[0] = dp[i - 1];

            for (int j = 1; j <= i; j++) {
                // The Bell number for n is the sum of the
                // Bell numbers for all previous partitions
                int temp = dp[j];
                dp[j] = prev + dp[j - 1];
                prev = temp;
            }
        }

        return dp[0];
    }

    static void Main()
    {
        int n = 5;
        Console.WriteLine(BellNumbers(n));
    }
}

Output

52

Time Complexity: 
Auxiliary Space:

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.