Merge pull request #175 from abhisheks008/main

Bell Numbers Problem using Dynamic Programming

Author: prathimacode-hub

Dynamic Programming/Bell Numbers/Images/bell1.png

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+# Bell Numbers Problem using Dynamic Programming
+Language used : **Python 3**
+
+## 🎯 Aim
+The aim of this script is to find out the nth Bell numbers using Dynamic Programming method..
+
+## 👉 Purpose
+The main purpose of this script is to show the implementation of Dynamic Programming to find out nth Bell Numbers.
+
+## 📄 Description
+- **What is 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)`.
+- 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} }
+```
+## 📈 Workflow of the script
+- `bellNumber` -- This is the working function which will check and return the nth bell numbers one after another using the Dynamic Programming method.
+- `main` - This is the driver program for this script.
+
+## 📃 Explanation
+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
+```
+
+## 🧮 Algorithm
+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}. 
+```
+## 💻 Input and Output 
+- **Test Case 1 :**
+
+![](https://github.com/abhisheks008/PyAlgo-Tree/blob/main/Dynamic%20Programming/Bell%20Numbers/Images/bell1.png)
+
+- **Test Case 2 :**
+
+![](https://github.com/abhisheks008/PyAlgo-Tree/blob/main/Dynamic%20Programming/Bell%20Numbers/Images/bell2.png)
+
+- **Test Case 3 :**
+
+![](https://github.com/abhisheks008/PyAlgo-Tree/blob/main/Dynamic%20Programming/Bell%20Numbers/Images/bell3.png)
+
+
+## ⏰ Time and Space complexity
+- **Time Complexity:** `O(n^2)`.
+
+---------------------------------------------------------------
+## 🖋️ Author
+**Code contributed by, _Abhishek Sharma_, 2021 [@abhisheks008](github.com/abhisheks008)**
+
+[![forthebadge made-with-python](http://ForTheBadge.com/images/badges/made-with-python.svg)](https://www.python.org/)

Dynamic Programming/Bell Numbers/Images/bell2.png

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+# A Python program to find n'th Bell number
+# Approach : Dynamic Programming
+# Abhishek S, 2021
+
+# ----------------------------------------------------------
+
+# Problem statement : Given a set of n elements, find the 
+#                     number of ways of partitioning it.
+
+# ----------------------------------------------------------
+
+# Solution : Used the Dynamic Approach to solve the problem.
+
+# This is the working function which will check and return the
+# nth bell numbers one after another using the Dynamic Programming
+# method
+
+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
+print ("- Bell Numbers using Dynamic Programming Approach -")
+print ("---------------------------------------------------")
+print ()
+z = int(input("Enter the number of numbers you wanna print : "))
+print ()
+print ("---------------------------------------------------")
+print ()
+print ("Output : ")
+print ()
+for n in range(z):
+    print('Bell Number', n, 'is', bellNumber(n))
+    
+    
+
+# ----------------------------------------------------------
+
+# Enter the number of numbers you wanna print : 6
+
+# ----------------------------------------------------------
+
+# 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
+
+# ----------------------------------------------------------
+
+# Code contributed by, Abhishek Sharma, 2021