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

## 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

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(n^{2}). 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.