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Program to Calculate the Edge Cover of a Graph

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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)



Last Updated : 23 Dec, 2022
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