Bell Numbers (Number of ways to Partition a Set) - GeeksforGeeks

Bell Numbers (Number of ways to Partition a Set) - GeeksforGeeks
Given a set of n elements, find number of ways of partitioning it.

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.

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}. 

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];

}

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.
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