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> |
Output
3
Time Complexity: O(1), As we are doing constant time operations only.
Auxiliary Space: O(1)