Bell Numbers (Number of ways to Partition a Set)

Hello Everyone,

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} }.

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.
    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.

// 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;

}

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.