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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
cycle

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



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

 f(n, k) =  frac{(n-1)^{k} + (-1)^{k}(n-1)}{n}

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 <bits/stdc++.h>
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

<?php
// PHP 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
function numOfways( $n, $k)
{
$p = 1;
  
if ($k % 2)
    $p = -1;
  
return (pow($n - 1, $k) + 
        $p * ($n - 1)) / $n;
}
  
    // Driver Code
    $n = 4;
    $k = 2;
    echo numOfways($n, $k);
      
// This code is contributed by vt_m. 
?>

Output:

3


This article is attributed to GeeksforGeeks.org

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