查找轮图的直径、周期和边缘的程序

轮子图：轮子图是通过将单个通用顶点连接到循环的所有顶点而形成的图。特性：-

轮图是平面图。
Wheel 图中总是存在一个哈密顿循环。
如果 n 分别为奇数和偶数，则色数为 3 和 4。

问题陈述：给定轮图中的顶点数。任务是找到：

车轮图中的循环数。
车轮图中的边数。

轮图的直径。

例子：

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

示例 #1：对于 vertices = 4 Wheel Graph，总周期为 7 ：

示例 #2：对于顶点 = 5 和 7 轮图边数分别 = 8 和 12 ：

示例#3：对于顶点 = 4，直径为 1 ，因为我们可以通过仅覆盖 1 条边从任何顶点到任何顶点。

计算周期、边缘和直径的公式：-

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

以下是所需的实现：

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 Code
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

//Java Program to find the diameter, 
// cycles and edges of a Wheel Graph
import 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 Code 
if __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 Graph
using 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 Code
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));
}
}
  
// This code is contributed by inder_verma

PHP

输出：

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