# 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} }.
```

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).

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
```

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.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.4 Average Difficulty : 3.4/5.0
Based on 33 vote(s)

Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.