Group all co-prime numbers from 1 to N

Given an integer N, the task is to group numbers such that each group is mutually co-prime together with the total grouping is minimum.
Examples:

Input: N = 8
Output:
1 2 3
4 5
6 7
8

Input: N = 5
Output:
1 2 3
4 5

Approach: The key observation in this problem is two consecutive numbers are always co-prime. That is GCD(a, a+1) = 1. Another important observation is even numbers can’t be listed in one group. Because they will lead to the greatest common divisor of 2. Therefore, every consecutive even and odd numbers can be grouped into one group and 1 can be in any group because the greatest common divisor of numbers with 1 is always 1.

Below is the implementation of the above approach :

C++

// C++ implementation to group  
// mutually coprime numbers into  
// one group with minimum group possible  
#include<bits/stdc++.h> 
using namespace std; 
  
// Function to group the mutually  
// co-prime numbers into one group  
void mutually_coprime(int n) 
{ 
    if (n <= 3) 
    { 
          
        // Loop for the numbers less  
        // than the 4  
        for(int j = 1; j <= n; j++) 
        { 
            cout << j << " "; 
        } 
        cout << "\n"; 
    } 
    else
    { 
          
        // Integers 1, 2 and 3 can be  
        // grouped into one group  
        cout << "1 2 3\n"; 
          
        for(int j = 4; j < n; j += 2) 
        { 
              
            // Consecutive even and  
            // odd numbers  
            cout << j << " " << j + 1 << "\n"; 
        } 
        if(n % 2 == 0) 
            cout << n << "\n"; 
    } 
} 
  
// Driver Code          
int main() 
{ 
    int n = 9; 
      
    // Function call  
    mutually_coprime(n); 
} 
  
// This code is contributed by yatinagg 

Java

// Java implementation to group 
// mutually coprime numbers into  
// one group with minimum group possible 
class GFG{ 
      
// Function to group the mutually 
// co-prime numbers into one group 
static void mutually_coprime(int n) 
{ 
    if (n <= 3) 
    { 
          
        // Loop for the numbers less 
        // than the 4 
        for(int j = 1; j < n + 1; j++) 
           System.out.print(j + " "); 
        System.out.println(); 
    }  
    else
    { 
          
        // Integers 1, 2 and 3 can be 
        // grouped into one group 
        System.out.println("1 2 3"); 
        for(int j = 4; j < n; j += 2) 
        { 
  
           // Consecutive even and 
           // odd numbers 
           System.out.println(j + " " + (j + 1)); 
           if (n % 2 == 0) 
           System.out.println(n); 
        } 
    } 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int n = 9; 
  
    // Function Call 
    mutually_coprime(n); 
} 
} 
  
// This code is contributed by sapnasingh4991 

Python3

# Python3 implementation to group 
# mutually coprime numbers into  
# one group with minimum group possible 
  
# Function to group the mutually 
# co-prime numbers into one group 
def mutually_coprime (n):     
    if ( n <= 3): 
        # Loop for the numbers less  
        # than the 4  
        for j in range (1, n + 1): 
            print (j, end =" ") 
        print () 
    else: 
        # Integers 1, 2 and 3 can be  
        # grouped into one group 
        print (1, 2, 3) 
        for j in range ( 4, n, 2 ): 
              
            # Consecutive even and  
            # odd numbers 
            print (j, ( j + 1 )) 
        if(n % 2 == 0):          
            print (n) 
  
# Driver Code             
if __name__ == "__main__": 
    n = 9
      
    # Function Call 
    mutually_coprime (n) 

C#

// C# implementation to group 
// mutually coprime numbers into  
// one group with minimum group possible 
using System; 
  
class GFG{ 
      
// Function to group the mutually 
// co-prime numbers into one group 
static void mutually_coprime(int n) 
{ 
    if (n <= 3) 
    { 
          
        // Loop for the numbers less 
        // than the 4 
        for(int j = 1; j < n + 1; j++) 
           Console.Write(j + " "); 
             
        Console.WriteLine(); 
    }  
    else
    { 
          
        // ints 1, 2 and 3 can be 
        // grouped into one group 
        Console.WriteLine("1 2 3"); 
        for(int j = 4; j < n; j += 2) 
        { 
           // Consecutive even and 
           // odd numbers 
           Console.WriteLine(j + " " + (j + 1)); 
              
           if (n % 2 == 0) 
               Console.WriteLine(n); 
        } 
    } 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
    int n = 9; 
  
    // Function Call 
    mutually_coprime(n); 
} 
} 
// This code is contributed by sapnasingh4991

Output:

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My Personal Notes

C++

Java

Python3

C#

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