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 ` |

## 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. ` `?> ` |

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.

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