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 nodesint numOfways(int n, int k){ int p = 1; if (k % 2) p = -1; return (pow(n - 1, k) + p * (n - 1)) / n;}// Driven Programint 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 nodesdef numOfways(n,k): p = 1 if (k % 2): p = -1 return (pow(n - 1, k) + p * (n - 1)) / n# Driver coden = 4k = 2print (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 nodesfunction 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
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