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} }. Recommended practice
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
import java.io.*;
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);
}
}
|
Python3
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#
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);
}
}
|
Javascript
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)
|
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
#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++)
{
bell[i][0] = bell[i-1][i-1];
for (int j=1; j<=i; j++)
bell[i][j] = bell[i-1][j-1] + bell[i][j-1];
}
return bell[n][0];
}
int main()
{
for (int n=0; n<=5; n++)
cout << "Bell Number " << n << " is "
<< bellNumber(n) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int bellNumber(int n)
{
int[][] bell = new int[n+1][n+1];
bell[0][0] = 1;
for (int i=1; i<=n; i++)
{
bell[i][0] = bell[i-1][i-1];
for (int j=1; j<=i; j++)
bell[i][j] = bell[i-1][j-1] + bell[i][j-1];
}
return bell[n][0];
}
public static void main (String[] args)
{
for (int n=0; n<=5; n++)
System.out.println("Bell Number "+ n +
" is "+bellNumber(n));
}
}
|
Python3
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):
bell[i][0] = bell[i-1][i-1]
for j in range(1, i+1):
bell[i][j] = bell[i-1][j-1] + bell[i][j-1]
return bell[n][0]
for n in range(6):
print('Bell Number', n, 'is', bellNumber(n))
|
C#
using System;
class GFG {
static int bellNumber(int n)
{
int[,] bell = new int[n + 1,
n + 1];
bell[0, 0] = 1;
for (int i = 1; i <= n; i++)
{
bell[i, 0] = bell[i - 1, i - 1];
for (int j = 1; j <= i; j++)
bell[i, j] = bell[i - 1, j - 1] +
bell[i, j - 1];
}
return bell[n, 0];
}
public static void Main ()
{
for (int n = 0; n <= 5; n++)
Console.WriteLine("Bell Number "+ n +
" is "+bellNumber(n));
}
}
|
PHP
<?php
function bellNumber($n)
{
$bell[0][0] = 1;
for ($i = 1; $i <= $n; $i++)
{
$bell[$i][0] = $bell[$i - 1]
[$i - 1];
for ($j = 1; $j <= $i; $j++)
$bell[$i][$j] = $bell[$i - 1][$j - 1] +
$bell[$i][$j - 1];
}
return $bell[$n][0];
}
for ($n = 0; $n <= 5; $n++)
echo("Bell Number " . $n . " is "
. bellNumber($n) . "\n");
?>
|
Javascript
<script>
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++)
{
bell[i][0] = bell[i-1][i-1];
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>
|
OutputBell 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)
Space Optimized DP Approach:
We can use a 1-D list to represent the previous row of the Bell triangle. We initialize dp[0] to 1, since there is only one way to partition an empty set.
To compute the Bell numbers for n > 0, we first set dp[0] = dp[i-1], since the first element in each row is the same as the last element in the previous row. Then, we use the recurrence relation dp[j] = prev + dp[j-1] to compute the Bell number for each partition, where prev is the value of dp[j] in the previous iteration of the inner loop. We update prev to the temporary variable temp before updating dp[j].
Finally, we return dp[0], which is the Bell number for the partition of a set with n elements into non-empty subsets.
Python3
def bell_numbers(n):
dp = [1] + [0] * n
for i in range(1, n+1):
prev = dp[0]
dp[0] = dp[i-1]
for j in range(1, i+1):
temp = dp[j]
dp[j] = prev + dp[j-1]
prev = temp
return dp[0]
n = 5
print(bell_numbers(n))
|
Time Complexity: 
Auxiliary Space: 
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.