Program to find the diameter, cycles and edges of a Wheel Graph
Last Updated :
26 Dec, 2022
Wheel Graph: A Wheel graph is a graph formed by connecting a single universal vertex to all vertices of a cycle.
Properties:
- Wheel graphs are Planar graphs.
- There is always a Hamiltonian cycle in the Wheel graph.
- Chromatic Number is 3 and 4, if n is odd and even respectively.
Problem Statement:
Given the Number of Vertices in a Wheel Graph. The task is to find:
- The Number of Cycles in the Wheel Graph.
- A number of edges in Wheel Graph.
- The diameter of a Wheel Graph.
Examples:
Input: vertices = 4
Output: Number of cycle = 7
Number of edge = 6
Diameter = 1
Input: vertices = 6
Output: Number of cycle = 21
Number of edge = 10
Diameter = 2
Example #1: For vertices = 4 Wheel Graph, total cycle is 7:
Example #2: For vertices = 5 and 7 Wheel Graph Number of edges = 8 and 12 respectively:
Example #3: For vertices = 4, the Diameter is 1 as We can go from any vertices to any vertices by covering only 1 edge.
Formula to calculate the cycles, edges and diameter:
Number of Cycle = (vertices * vertices) - (3 * vertices) + 3
Number of edge = 2 * (vertices - 1)
Diameter = if vertices = 4, Diameter = 1
if vertices > 4, Diameter = 2
Below is the required implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int totalCycle( int vertices)
{
int result = 0;
result = pow (vertices, 2) - (3 * vertices) + 3;
return result;
}
int Edges( int vertices)
{
int result = 0;
result = 2 * (vertices - 1);
return result;
}
int Diameter( int vertices)
{
int result = 0;
if (vertices == 4)
result = 1;
else
result = 2;
return result;
}
int main()
{
int vertices = 4;
cout << "Number of Cycle = " << totalCycle(vertices) << endl;
cout << "Number of Edges = " << Edges(vertices) << endl;
cout << "Diameter = " << Diameter(vertices);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int totalCycle( double vertices)
{
double result = 0 ;
int result1 = 0 ;
result = Math.pow(vertices, 2 ) - ( 3 * vertices) + 3 ;
result1 = ( int )(result);
return result1;
}
static int Edges( int vertices)
{
int result = 0 ;
result = 2 * (vertices - 1 );
return result;
}
static int Diameter( int vertices)
{
int result = 0 ;
if (vertices == 4 )
result = 1 ;
else
result = 2 ;
return result;
}
public static void main(String[] args)
{
int vertices = 4 ;
System.out.println( "Number of Cycle = " + totalCycle(vertices));
System.out.println( "Number of Edges = " + Edges(vertices));
System.out.println( "Diameter = " + Diameter(vertices));
}
}
|
Python3
def totalCycle(vertices):
result = 0
result = ( pow (vertices, 2 ) -
( 3 * vertices) + 3 )
return result
def Edges(vertices):
result = 0
result = 2 * (vertices - 1 )
return result
def Diameter(vertices):
result = 0
if vertices = = 4 :
result = 1
else :
result = 2
return result
if __name__ = = "__main__" :
vertices = 4
print ( "Number of Cycle =" ,
totalCycle(vertices))
print ( "Number of Edges =" , Edges(vertices))
print ( "Diameter =" , Diameter(vertices))
|
C#
using System;
class GFG
{
static int totalCycle( double vertices)
{
double result = 0;
int result1 = 0;
result = Math.Pow(vertices, 2) -
(3 * vertices) + 3;
result1 = ( int )(result);
return result1;
}
static int Edges( int vertices)
{
int result = 0;
result = 2 * (vertices - 1);
return result;
}
static int Diameter( int vertices)
{
int result = 0;
if (vertices == 4)
result = 1;
else
result = 2;
return result;
}
public static void Main()
{
int vertices = 4;
Console.WriteLine( "Number of Cycle = " +
totalCycle(vertices));
Console.WriteLine( "Number of Edges = " +
Edges(vertices));
Console.WriteLine( "Diameter = " +
Diameter(vertices));
}
}
|
PHP
<?php
function totalCycle( $vertices )
{
$result = 0;
$result = pow( $vertices , 2) -
(3 * $vertices ) + 3;
return $result ;
}
function Edges( $vertices )
{
$result = 0;
$result = 2 * ( $vertices - 1);
return $result ;
}
function Diameter( $vertices )
{
$result = 0;
if ( $vertices == 4)
$result = 1;
else
$result = 2;
return $result ;
}
$vertices = 4;
echo "Number of Cycle = " ,
totalCycle( $vertices ), "\n" ;
echo "Number of Edges = " ,
Edges( $vertices ), "\n" ;
echo "Diameter = " , Diameter( $vertices );
?>
|
Javascript
function totalCycle(vertices)
{
var result = 0;
result = Math.pow(vertices, 2) - (3 * vertices) + 3;
return result;
}
function Edges(vertices)
{
var result = 0;
result = 2 * (vertices - 1);
return result;
}
function Diameter(vertices)
{
var result = 0;
if (vertices === 4)
result = 1;
else
result = 2;
return result;
}
var vertices = 4;
console.log( "Number of Cycle = " + totalCycle(vertices) + "\n" );
console.log( "Number of Edges = " + Edges(vertices) + "\n" );
console.log( "Diameter = " + Diameter(vertices));
|
Output
Number of Cycle = 7
Number of Edges = 6
Diameter = 1
Time complexity: O(1)
Auxiliary Space: O(1)
Share your thoughts in the comments
Please Login to comment...