Count of sets possible using integers from a range [2, N] using given operations that are in Equivalence Relation

Given an integer N, repeatedly choose two distinct integers from the range 2 to N and if their GCD is found to be greater than 1, insert them into the same set, as long as possible. The sets formed in Equivalence Relation. Therefore, if integers a and b are in the same set and integers b and c are in the same set, then integers a and c are also said to be in the same group. The task is to find the total number of such sets that can be formed.

Examples:

Input: N = 3 
Output: 2
Explanation: Sets formed are: {2}, {3}. They cannot be put in the same set because there GCD is 1.

Input: N = 9
Output : 3
Sets formed are : {2, 3, 4, 6, 8, 9}, {5}, {7}
As {2, 4, 6, 8} lies in same set and {3, 6, 9} also lies in same set. Hence, all these lie in one set together, because 6 is the common element in both the sets.

Approach: The idea to solve the problem is based on the following observation that all the numbers less than or equal to N/2 belong to the same set because if 2 is multiplied in them, they will be even number and has GCD greater than 1 with 2. So the remaining sets are formed by numbers greater than N/2 and are prime because if they are not prime then there is a number less than or equal to N/2 which is the divisor of that number. The prime numbers from 2 to N can be found using the Sieve of Eratosthenes.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
bool prime[100001];

// Sieve of Eratosthenes to find
// primes less than or equal to N
void SieveOfEratosthenes(int n)
{

    memset(prime, true, sizeof(prime));

    for (int p = 2; p * p <= n; p++) {

        if (prime[p] == true) {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}

// Function to find number of Sets
void NumberofSets(int N)
{
    SieveOfEratosthenes(N);

    // Handle Base Case
    if (N == 2) {
        cout << 1 << endl;
    }
    else if (N == 3) {
        cout << 2 << endl;
    }
    else {

        // Set which contains less
        // than or equal to N/2
        int ans = 1;

        // Number greater than N/2 and
        // are prime increment it by 1
        for (int i = N / 2 + 1; i <= N; i++) {

            // If the number is prime
            // Increment answer by 1
            if (prime[i]) {
                ans += 1;
            }
        }

        cout << ans << endl;
    }
}
// Driver Code
int main()
{
    // Input
    int N = 9;

    // Function Call
    NumberofSets(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

static boolean prime[] = new boolean[100001];

// Sieve of Eratosthenes to find
// primes less than or equal to N
static void SieveOfEratosthenes(int n)
{
    Arrays.fill(prime, true);

    for(int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for(int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}

// Function to find number of Sets
static void NumberofSets(int N)
{
    SieveOfEratosthenes(N);

    // Handle Base Case
    if (N == 2)
    {
        System.out.print(1);
    }
    else if (N == 3)
    {
        System.out.print(2);
    }
    else
    {

        // Set which contains less
        // than or equal to N/2
        int ans = 1;

        // Number greater than N/2 and
        // are prime increment it by 1
        for(int i = N / 2 + 1; i <= N; i++)
        {

            // If the number is prime
            // Increment answer by 1
            if (prime[i])
            {
                ans += 1;
            }
        }
        System.out.print(ans);
    }
}

// Driver Code
public static void main(String[] args)
{

    // Input
    int N = 9;

    // Function Call
    NumberofSets(N);
}
}

// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
prime = [True] * 100001

# Sieve of Eratosthenes to find
# primes less than or equal to N
def SieveOfEratosthenes(n):

    global prime

    for p in range(2, n + 1):
        if p * p > n:
            break

        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False

# Function to find number of Sets
def NumberofSets(N):

    SieveOfEratosthenes(N)

    # Handle Base Case
    if (N == 2):
        print(1)
    elif (N == 3):
        print(2)
    else:

        # Set which contains less
        # than or equal to N/2
        ans = 1

        # Number greater than N/2 and
        # are prime increment it by 1
        for i in range(N // 2, N + 1):

            # If the number is prime
            # Increment answer by 1
            if (prime[i]):
                ans += 1

        print(ans)

# Driver Code
if __name__ == '__main__':

    # Input
    N = 9

    # Function Call
    NumberofSets(N)

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{
static bool []prime = new bool[100001];

// Sieve of Eratosthenes to find
// primes less than or equal to N
static void SieveOfEratosthenes(int n)
{

    for(int i=0;i<100001;i++)
        prime[i] = true;

    for (int p = 2; p * p <= n; p++) {

        if (prime[p] == true) {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}

// Function to find number of Sets
static void NumberofSets(int N)
{
    SieveOfEratosthenes(N);

    // Handle Base Case
    if (N == 2) {
        Console.Write(1);
    }
    else if (N == 3) {
        Console.Write(2);
    }
    else {

        // Set which contains less
        // than or equal to N/2
        int ans = 1;

        // Number greater than N/2 and
        // are prime increment it by 1
        for (int i = N / 2 + 1; i <= N; i++) {

            // If the number is prime
            // Increment answer by 1
            if (prime[i]) {
                ans += 1;
            }
        }

        Console.Write(ans);
    }
}

// Driver Code
public static void Main()
{
    // Input
    int N = 9;

    // Function Call
    NumberofSets(N);
}
}

// This code is contributed by SURENDRA_GANGWAR.

Javascript

<script>
// Javascript program for the above approach

let prime = new Array(100001);

// Sieve of Eratosthenes to find
// primes less than or equal to N
function SieveOfEratosthenes(n)
{
    for(let i=0;i<prime.length;i++)
    {
        prime[i]=true;
    }

    for(let p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for(let i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
}

// Function to find number of Sets
function NumberofSets(N)
{
    SieveOfEratosthenes(N);

    // Handle Base Case
    if (N == 2)
    {
        document.write(1);
    }
    else if (N == 3)
    {
        document.write(2);
    }
    else
    {

        // Set which contains less
        // than or equal to N/2
        let ans = 1;

        // Number greater than N/2 and
        // are prime increment it by 1
        for(let i = Math.floor(N / 2) + 1; i <= N; i++)
        {

            // If the number is prime
            // Increment answer by 1
            if (prime[i])
            {
                ans += 1;
            }
        }
        document.write(ans);
    }
}

// Driver Code
// Input
let N = 9;

// Function Call
NumberofSets(N);

// This code is contributed by unknown2108
</script>

Output:

3

Time Complexity: O(N)
Auxiliary Space: O(K)