# Number of loops of size k starting from a specific node

Given two positive integer n, k. Consider an undirected complete connected graph of n nodes in a complete connected graph. The task is to calculate the number of ways in which one can start from any node and return to it by visiting K nodes.

Examples:

Input : n = 3, k = 3
Output : 2 Input : n = 4, k = 2
Output : 3


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Lets f(n, k) be a function which return number of ways in which one can start from any node and return to it by visiting K nodes.
If we start and end from one node, then we have K – 1 choices to make for the intermediate nodes since we have already chosen one node in the beginning. For each intermediate choice, you have n – 1 options. So, this will yield (n – 1)k – 1 but then we have to remove all the choices cause smaller loops, so we subtract f(n, k – 1).
So, recurrence relation becomes,
f(n, k) = (n – 1)k – 1 – f(n, k – 1) with base case f(n, 2) = n – 1.
On expanding,
f(n, k) = (n – 1)k – 1 – (n – 1)k – 2 + (n – 1)k – 3 ….. (n – 1) (say eqn 1)

Dividing f(n, k) by (n – 1),
f(n, k)/(n – 1) = (n – 1)k – 2 – (n – 1)k – 3 + (n – 1)k – 4 ….. 1 (say eqn 2)

On adding eqn 1 and eqn 2,
f(n, k) + f(n, k)/(n – 1) = (n – 1)k – 1 + (-1)k
f(n, k) * ( (n -1) + 1 )/(n – 1) = (n – 1)k – 1 + (-1)k Below is the implementation of this approach:

## C++

 // C++ Program to find number of cycles of length  // k in a graph with n nodes.  #include  using namespace std;     // Return the Number of ways from a  // node to make a loop of size K in undirected  // complete connected graph of N nodes  int numOfways(int n, int k)  {      int p = 1;         if (k % 2)          p = -1;         return (pow(n - 1, k) + p * (n - 1)) / n;  }     // Driven Program  int main()  {      int n = 4, k = 2;      cout << numOfways(n, k) << endl;      return 0;  }

## Java

 // Java Program to find number of  // cycles of length k in a graph  // with n nodes.  public class GFG {             // Return the Number of ways      // from a node to make a loop      // of size K in undirected      // complete connected graph of      // N nodes      static int numOfways(int n, int k)      {          int p = 1;                 if (k % 2 != 0)              p = -1;                 return (int)(Math.pow(n - 1, k)                      + p * (n - 1)) / n;      }             // Driver code      public static void main(String args[])      {          int n = 4, k = 2;                 System.out.println(numOfways(n, k));      }  }     // This code is contributed by Sam007.

## Python3

 # python Program to find number of   # cycles of length k in a graph   # with n nodes.     # Return the Number of ways from a  # node to make a loop of size K in  # undirected complete connected   # graph of N nodes  def numOfways(n,k):             p = 1        if (k % 2):          p = -1        return (pow(n - 1, k) +                    p * (n - 1)) / n     # Driver code  n = 4 k = 2 print (numOfways(n, k))     # This code is contributed by Sam007.

## C#

 // C# Program to find number of cycles  // of length k in a graph with n nodes.  using System;     class GFG {             // Return the Number of ways from      // a node to make a loop of size      // K in undirected complete       // connected graph of N nodes      static int numOfways(int n, int k)      {          int p = 1;                 if (k % 2 != 0)              p = -1;                 return (int)(Math.Pow(n - 1, k)                       + p * (n - 1)) / n;      }             // Driver code      static void Main()      {          int n = 4, k = 2;                     Console.Write( numOfways(n, k) );      }  }     // This code is contributed by Sam007.

## PHP

 

Output:

3


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