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

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 <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.
?>

Javascript

<script>

// JavaScript 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)
    {
        let p = 1;

        if (k % 2 != 0)
            p = -1;

        return (Math.pow(n - 1, k)
                    + p * (n - 1)) / n;
    }

// Driver code
         let n = 4, k = 2;
        document.write(numOfways(n, k));

    // This code is contributed by code_hunt.
</script>

Output

3