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} }.
The 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).
Below is Dynamic Programming based implementation of the above recursive code using the Stirling number-
C++
#include <iostream> using namespace std; int main() { int n=5; int s[n+1][n+1]; for ( int i=0;i<n+1;i++){ for ( int j=0;j<n+1;j++){ if (j>i) s[i][j]=0; else if (i==j) s[i][j]=1; else if (i==0 || j==0) s[i][j]=0; else { s[i][j]= j*s[i-1][j] + s[i-1][j-1]; } } } int ans=0; for ( int i=0;i<n+1;i++){ ans += s[n][i]; } cout<<ans; return 0; } |
Java
/*package whatever //do not write package name here */ // Java program to find number of ways of partitioning it. import java.io.*; // "static void main" must be defined in a public class. public class GFG { public static void main(String[] args) { int n = 5 ; int [][] s = new int [n + 1 ][n + 1 ]; for ( int i = 0 ; i < n + 1 ; i++) { for ( int j = 0 ; j < n + 1 ; j++) { if (j > i) s[i][j] = 0 ; else if (i == j) s[i][j] = 1 ; else if (i == 0 || j == 0 ) s[i][j] = 0 ; else { s[i][j] = j * s[i - 1 ][j] + s[i - 1 ][j - 1 ]; } } } int ans = 0 ; for ( int i = 0 ; i < n + 1 ; i++) { ans += s[n][i]; } System.out.println(ans); } } // The code is contributed by Gautam goel (gautamgoel962) |
Python3
# python program to find number of ways of partitioning it. n = 5 s = [[ 0 for _ in range (n + 1 )] for _ in range (n + 1 )] for i in range (n + 1 ): for j in range (n + 1 ): if j > i: continue elif (i = = j): s[i][j] = 1 elif (i = = 0 or j = = 0 ): s[i][j] = 0 else : s[i][j] = j * s[i - 1 ][j] + s[i - 1 ][j - 1 ] ans = 0 for i in range ( 0 ,n + 1 ): ans + = s[n][i] print (ans) |
C#
// C# Program to find number of ways of partitioning it. using System; public class Program { static public void Main( string [] args) { int n = 5; int [, ] s = new int [n + 1, n + 1]; for ( int i = 0; i < n + 1; i++) { for ( int j = 0; j < n + 1; j++) { if (j > i) s[i, j] = 0; else if (i == j) s[i, j] = 1; else if (i == 0 || j == 0) s[i, j] = 0; else s[i, j] = j * s[i - 1, j] + s[i - 1, j - 1]; } } int ans = 0; for ( int i = 0; i < n + 1; i++) ans += s[n, i]; Console.WriteLine(ans); } } // This code is contributed by Tapesh(tapeshdua420) |
Javascript
// JavaScript program to find number of ways of partitioning it. let n=5; let s = new Array(n+1); for (let i=0;i<n+1;i++){ s[i] = new Array(n+1); for (let j=0;j<n+1;j++){ if (j>i) s[i][j]=0; else if (i==j) s[i][j]=1; else if (i==0 || j==0) s[i][j]=0; else { s[i][j]= j*s[i-1][j] + s[i-1][j-1]; } } } let ans=0; for (let i=0;i<n+1;i++){ ans += s[n][i]; } console.log(ans) // The code is contributed by Gautam goel (gautamgoel962) |
52
Time complexity: O(N2)
Auxiliary Space: O(N2)
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++14
// 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. ?> |
Javascript
<script> // Javascript program to find n'th Bell number // Function to find n'th Bell Number function bellNumber(n) { let bell = new Array(n+1); for (let i = 0; i < n + 1; i++) { bell[i] = new Array(n + 1); } bell[0][0] = 1; for (let 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 (let j=1; j<=i; j++) bell[i][j] = bell[i-1][j-1] + bell[i][j-1]; } return bell[n][0]; } for (let n=0; n<=5; n++) document.write( "Bell Number " + n + " is " +bellNumber(n) + "</br>" ); </script> |
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: O(N2)
Auxiliary Space: 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.
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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