Given the number of vertices N of a graph. The task is to determine the Edge cover.
Edge Cover: Minimum number of edges required to cover all vertex is known as Edge Cover.
Examples:
Input : N = 5 Output : 3 Input : N = 4 Output : 2
Example 1: For N = 5 vertices,
Edge Cover is: 3 (Choosing the edges marked in Red, all of the vertices will get covered)
Example 2: For N = 8 vertices,
Edge Cover is: 4 (Choosing the edges marked in Red, all of the vertices will get covered)
Formula:
Edge Cover = ceil (no. of vertices / 2)
Implementation:
C++
// C++ program to find Edge Cover #include <bits/stdc++.h> using namespace std; // Function that calculates Edge Cover int edgeCover( int n) { float result = 0; result = ceil (n / 2.0); return result; } // Driver Code int main() { int n = 5; cout << edgeCover(n); return 0; } |
Java
// Java program to find Edge Cover import java.util.*; import java.lang.*; import java.io.*; class GFG{ // Function that calculates Edge Cover static int edgeCover( int n) { int result = 0 ; result = ( int )Math.ceil(( double )n / 2.0 ); return result; } // Driver Code public static void main(String args[]) { int n = 5 ; System.out.print(edgeCover(n)); } } |
Python3
# Python 3 implementation of the above approach. import math # Function that calculates Edge Cover def edgeCover(n): result = 0 result = math.ceil(n / 2.0 ) return result # Driver code if __name__ = = "__main__" : n = 5 print ( int (edgeCover(n))) # this code is contributed by Naman_Garg |
C#
// C# program to find Edge Cover using System; class GFG { // Function that calculates Edge Cover static int edgeCover( int n) { int result = 0; result = ( int )Math.Ceiling(( double )n / 2.0); return result; } // Driver Code static public void Main () { int n = 5; Console.Write(edgeCover(n)); } } // This code is contributed by Raj |
PHP
<?php // PHP program to find Edge Cover // Function that calculates // Edge Cover function edgeCover( $n ) { $result = 0; $result = ceil ( $n / 2.0); return $result ; } // Driver Code $n = 5; echo edgeCover( $n ); // This code is contributed by Raj ?> |
Javascript
<script> // javascript program to find Edge Cover // Function that calculates Edge Cover function edgeCover(n) { var result = 0; result = parseInt( Math.ceil(n / 2.0)); return result; } // Driver Code var n = 5; document.write(edgeCover(n)); // This code contributed by gauravrajput1 </script> |
3
Time Complexity: O(1), As we are doing constant time operations only.
Auxiliary Space: O(1)
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