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Number of cycles formed by joining vertices of n sided polygon at the center

  • Last Updated : 15 Apr, 2021

Given an N-sided regular polygon, we have connected all the vertices at the center of the polygon, thus dividing the polygon into N equal parts. Our task is to the Count of the total number of cycles in the polygon.
Note: A Cycle is a closed-loop starting and ending on same point.
Examples: 
 

Input: N = 3 
Output:
Explanation: 
 

When a 3 sided polygon is connected by vertices at the center then we get 7 cycles possible for it as shown in the image.
Input: N = 5 
Output: 21 
Explanation: 
 



When a 5 sided polygon is connected by vertices at the center then we get 21 cycles possible for it as shown in the image. 
 

 

Approach: To the problem mentioned above we are supposed to count the total number of closed loops possible in the given polygon after division. The approach is based upon Mathematical Pattern. There will be N cycles already created due to the division of polygon. One out of N blocks will form a cycle with rest (N – 1) blocks. The remaining (N – 1) blocks will form cycle with other (N – 2) blocks. So the total cycles we have can be found out using the formula given below:
 

Total Cycles = N + 1 * (N – 1) + (N – 1) * (N – 2) 
Total Cycles = 2 * N – 1) + (N – 1) * (N – 2) 
 

Below is the implementation of the above approach:
 

C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to calculate number of cycles
int findCycles(int N)
{
    int res = 0;
    int finalResult = 0;
    int val = 2 * N - 1;
  
    // BigInteger is used here
    // if N=10^9 then multiply
    // will result into value
    // greater than 10^18
    int s = val;
  
    // BigInteger multiply function
    res = (N - 1) * (N - 2);
    finalResult = res + s;
  
    // Return the final result
    return finalResult;
}
 
// Driver code
int main()
{
    // Given N
    int N = 5;
  
    // Function Call
    cout << findCycles(N) << endl;   
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java




// Java program for the above approach
 
import java.util.*;
import java.math.*;
 
class GFG {
 
    // Function to calculate number of cycles
    static BigInteger findCycles(int N)
    {
        BigInteger res, finalResult;
        long val = 2 * N - 1;
 
        String st = String.valueOf(val);
 
        // BigInteger is used here
        // if N=10^9 then multiply
        // will result into value
        // greater than 10^18
 
        BigInteger str = new BigInteger(st);
        String n1 = String.valueOf((N - 1));
        String n2 = String.valueOf((N - 2));
 
        BigInteger a = new BigInteger(n1);
        BigInteger b = new BigInteger(n2);
 
        // BigInteger multiply function
        res = a.multiply(b);
 
        finalResult = res.add(str);
 
        // Return the final result
        return finalResult;
    }
 
    // Driver Code
    public static void
    main(String args[]) throws Exception
    {
        // Given N
        int N = 5;
 
        // Function Call
        System.out.println(findCycles(N));
    }
}

Python3




# Python3 program for the above approach
  
# Function to calculate number of cycles
def findCycles(N):
    res = 0
    finalResult = 0
    val = 2 * N - 1;
 
    # BigInteger is used here
    # if N=10^9 then multiply
    # will result into value
    # greater than 10^18
    s = val
 
    # BigInteger multiply function
    res = (N - 1) * (N - 2)
    finalResult = res + s;
 
    # Return the final result
    return finalResult;
 
# Driver Code
if __name__=='__main__':
     
    # Given N
    N = 5;
 
    # Function Call
    print(findCycles(N));
 
    # This code is contributed by pratham76

C#




// C# program for the above approach
using System;
class GFG {
 
  // Function to calculate number of cycles
  static int findCycles(int N)
  {
    int res = 0;
    int finalResult = 0;
    int val = 2 * N - 1;
 
    // BigInteger is used here
    // if N=10^9 then multiply
    // will result into value
    // greater than 10^18
    int s = val;
 
    // BigInteger multiply function
    res = (N - 1) * (N - 2);
    finalResult = res + s;
 
    // Return the final result
    return finalResult;
  }
 
  // Driver code
  static void Main()
  {
     
    // Given N
    int N = 5;
 
    // Function Call
    Console.WriteLine(findCycles(N));
  }
}
 
// This code is contributed by divyesh072019

Javascript




<script>
 
    // Javascript program for the above approach
     
    // Function to calculate number of cycles
    function findCycles(N)
    {
      let res = 0;
      let finalResult = 0;
      let val = 2 * N - 1;
 
      // BigInteger is used here
      // if N=10^9 then multiply
      // will result into value
      // greater than 10^18
      let s = val;
 
      // BigInteger multiply function
      res = (N - 1) * (N - 2);
      finalResult = res + s;
 
      // Return the final result
      return finalResult;
    }
     
    // Given N
    let N = 5;
  
    // Function Call
    document.write(findCycles(N));
     
</script>
Output: 
21

 

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