# Program to find the diameter, cycles and edges of a Wheel Graph

• Last Updated : 05 Sep, 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:

1. The Number of Cycles in the Wheel Graph.
2. A number of edges in Wheel Graph.
3. 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // C++ Program to find the diameter,// cycles and edges of a Wheel Graph#include using namespace std; // Function that calculates the// Number of Cycle in Wheel Graph.int totalCycle(int vertices){    int result = 0;     // calculates no. of Cycle.    result = pow(vertices, 2) - (3 * vertices) + 3;     return result;} // Function that calculates the// Number of Edges in Wheel graph.int Edges(int vertices){    int result = 0;     result = 2 * (vertices - 1);     return result;} // Function that calculates the// Diameter in Wheel Graph.int Diameter(int vertices){    int result = 0;     // calculates Diameter.    if (vertices == 4)        result = 1;    else        result = 2;     return result;} // Driver Codeint 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

 //Java Program to find the diameter,// cycles and edges of a Wheel Graphimport java.io.*; class GFG{    // Function that calculates the    // Number of Cycle in Wheel Graph.    static int totalCycle(double vertices)    {        double result = 0;        int result1 = 0;               // calculates no. of Cycle.        result = Math.pow(vertices, 2) - (3 * vertices) + 3;               result1 = (int)(result);        return result1;    }           // Function that calculates the    // Number of Edges in Wheel graph.    static int Edges(int vertices)    {        int result = 0;               result = 2 * (vertices - 1);               return result;    }           // Function that calculates the    // Diameter in Wheel Graph.    static int Diameter(int vertices)    {        int result = 0;               // calculates Diameter.        if (vertices == 4)            result = 1;        else            result = 2;               return result;    }         //Driver Code    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

 # Python3 Program to find the diameter,# cycles and edges of a Wheel Graph # Function that calculates the# Number of Cycle in Wheel Graph.def totalCycle(vertices):     result = 0     # calculates no. of Cycle.    result = (pow(vertices, 2) -             (3 * vertices) + 3)    return result # Function that calculates the# Number of Edges in Wheel graph.def Edges(vertices):     result = 0    result = 2 * (vertices - 1)    return result # Function that calculates the# Diameter in Wheel Graph.def Diameter(vertices):     result = 0     # calculates Diameter.    if vertices == 4:        result = 1    else:        result = 2     return result # Driver Codeif __name__ == "__main__":     vertices = 4     print("Number of Cycle =",           totalCycle(vertices))    print("Number of Edges =", Edges(vertices))    print("Diameter =", Diameter(vertices)) # This code is contributed by Rituraj Jain

## C#

 // C# Program to find the diameter,// cycles and edges of a Wheel Graphusing System;class GFG{// Function that calculates the// Number of Cycle in Wheel Graph.static int totalCycle(double vertices){    double result = 0;    int result1 = 0;     // calculates no. of Cycle.    result = Math.Pow(vertices, 2) -                     (3 * vertices) + 3;     result1 = (int)(result);    return result1;} // Function that calculates the// Number of Edges in Wheel graph.static int Edges(int vertices){    int result = 0;     result = 2 * (vertices - 1);     return result;} // Function that calculates the// Diameter in Wheel Graph.static int Diameter(int vertices){    int result = 0;     // calculates Diameter.    if (vertices == 4)        result = 1;    else        result = 2;     return result;} // Driver Codepublic 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));}} // This code is contributed by inder_verma

## PHP



Output

Number of Cycle = 7
Number of Edges = 6
Diameter = 1

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