# Bell Numbers (Number of ways to Partition a Set)

• Difficulty Level : Medium
• Last Updated : 12 Dec, 2022

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} }. 

Recommended practice

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 using namespace std; int main() {    int n=5;    int s[n+1][n+1];    for(int i=0;ii) 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

## 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 = 5s = [[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 = 0for 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;ii) 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

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#includeusing 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 programint 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 numberimport 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 programfor 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 numberusing 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

 

## Javascript

 

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)

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