Check if Euler Totient Function is same for a given number and twice of that number

Given an integer N, the task is to check whether the Euler’s Totient Function of N and 2 * N are the same or not. If they are found to be the same, then print “Yes”. Otherwise, print “No”.

Examples:

Input: N = 9 
Output: Yes 
Explanation:
Let phi() be the Euler Totient function 
Since 1, 2, 4, 5, 7, 8 have GCD 1 with 9. Therefore, phi(9) = 6. 
Since 1, 5, 7, 11, 13, 17 have GCD 1 with 18. Therefore, phi(18) = 6. 
Therefore, phi(9) and phi(18) are equal.

Input: N = 14 
Output: No 
Explanation:
Let phi() be the Euler Totient function, then 
Since 1, 3 have GCD 1 with 4. Therefore, phi(4) = 2. 
Since 1, 3, 5, 7 have GCD 1 with 8. Therefore, phi(8) = 4. 
Therefore, phi(4) and phi(8) are not the same.

Naive Approach: The simplest approach is to find Euler’s Totient Function of N and 2 * N. If they are the same, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the Euler's
// Totient Function
int phi(int n)
{
    // Initialize result as N
    int result = 1;

    // Consider all prime factors
    // of n and subtract their
    // multiples from result
    for (int p = 2; p < n; p++) {
        if (__gcd(p, n) == 1) {
            result++;
        }
    }

    // Return the count
    return result;
}

// Function to check if phi(n)
// is equals phi(2*n)
bool sameEulerTotient(int n)
{

    return phi(n) == phi(2 * n);
}

// Driver Code
int main()
{
    int N = 13;
    if (sameEulerTotient(N))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program of
// the above approach
import java.util.*;
class GFG{

// Function to find the Euler's
// Totient Function
static int phi(int n)
{
  // Initialize result as N
  int result = 1;

  // Consider all prime factors
  // of n and subtract their
  // multiples from result
  for (int p = 2; p < n; p++)
  {
    if (__gcd(p, n) == 1)
    {
      result++;
    }
  }

  // Return the count
  return result;
}

// Function to check if phi(n)
// is equals phi(2*n)
static boolean sameEulerTotient(int n)
{
  return phi(n) == phi(2 * n);
}

static int __gcd(int a, int b)
{
  return b == 0 ? a :__gcd(b, a % b);
}

// Driver Code
public static void main(String[] args)
{
  int N = 13;
  if (sameEulerTotient(N))
    System.out.print("Yes");
  else
    System.out.print("No");
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program of the
# above approach

# Function to find the Euler's
# Totient Function
def phi(n):

    # Initialize result as N
    result = 1

    # Consider all prime factors
    # of n and subtract their
    # multiples from result
    for p in range(2, n):
        if (__gcd(p, n) == 1):
            result += 1

    # Return the count
    return result

# Function to check if phi(n)
# is equals phi(2*n)
def sameEulerTotient(n):

    return phi(n) == phi(2 * n)

def __gcd(a, b):

    return a if b == 0 else __gcd(b, a % b)

# Driver Code
if __name__ == '__main__':

    N = 13

    if (sameEulerTotient(N)):
        print("Yes")
    else:
        print("No")

# This code is contributed by Amit Katiyar

C#

// C# program of
// the above approach
using System;
class GFG{

// Function to find the Euler's
// Totient Function
static int phi(int n)
{
  // Initialize result as N
  int result = 1;

  // Consider all prime factors
  // of n and subtract their
  // multiples from result
  for (int p = 2; p < n; p++)
  {
    if (__gcd(p, n) == 1)
    {
      result++;
    }
  }

  // Return the count
  return result;
}

// Function to check if phi(n)
// is equals phi(2*n)
static bool sameEulerTotient(int n)
{
  return phi(n) == phi(2 * n);
}

static int __gcd(int a, int b)
{
  return b == 0 ? a : __gcd(b, a % b);
}

// Driver Code
public static void Main(String[] args)
{
  int N = 13;
  if (sameEulerTotient(N))
    Console.Write("Yes");
  else
    Console.Write("No");
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// JavaScript  program for
//the above approach

// Function to find the Euler's
// Totient Function
function phi(n)
{
  // Initialize result as N
  let result = 1;

  // Consider all prime factors
  // of n and subtract their
  // multiples from result
  for (let p = 2; p < n; p++)
  {
    if (__gcd(p, n) == 1)
    {
      result++;
    }
  }

  // Return the count
  return result;
}

// Function to check if phi(n)
// is equals phi(2*n)
function sameEulerTotient(n)
{
  return phi(n) == phi(2 * n);
}

function __gcd( a, b)
{
  return b == 0 ? a :__gcd(b, a % b);
}

// Driver Code

    let N = 13;
  if (sameEulerTotient(N))
    document.write("Yes");
  else
    document.write("No");

// This code is contributed by souravghosh0416.
</script>

Output:

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the key observation is that there is no need to calculate the Euler’s Totient Function as only odd numbers follow the property phi(N) = phi(2*N), where phi() is the Euler’s Totient Function.

Therefore, the idea is to check if N is odd or not. If found to be true, print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if phi(n)
// is equals phi(2*n)
bool sameEulerTotient(int N)
{
    // Return if N is odd
    return (N & 1);
}

// Driver Code
int main()
{
    int N = 13;

    // Function Call
    if (sameEulerTotient(N))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program of the above approach
class GFG{

// Function to check if phi(n)
// is equals phi(2*n)
static int sameEulerTotient(int N)
{

    // Return if N is odd
    return (N & 1);
}

// Driver code
public static void main(String[] args)
{
    int N = 13;

    // Function call
    if (sameEulerTotient(N) == 1)
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by Dewanti

Python3

# Python3 program of the above approach

# Function to check if phi(n)
# is equals phi(2*n)
def sameEulerTotient(N):

    # Return if N is odd
    return (N & 1);

# Driver code
if __name__ == '__main__':

    N = 13;

    # Function call
    if (sameEulerTotient(N) == 1):
        print("Yes");
    else:
        print("No");

# This code is contributed by Amit Katiyar

C#

// C# program of
// the above approach
using System;
class GFG{

// Function to check if phi(n)
// is equals phi(2*n)
static int sameEulerTotient(int N)
{
  // Return if N is odd
  return (N & 1);
}

// Driver code
public static void Main(String[] args)
{
  int N = 13;

  // Function call
  if (sameEulerTotient(N) == 1)
    Console.Write("Yes");
  else
    Console.Write("No");
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>


//Javascript program of the above approach

// Function to check if phi(n)
// is equals phi(2*n)
function sameEulerTotient(N)
{
    // Return if N is odd
    return (N & 1);
}

// Driver Code
var N = 13;
// Function Call
if (sameEulerTotient(N))
    document.write( "Yes");
else
    document.write("No");


</script>

Output:

Yes

Time Complexity: O(1)
Auxiliary Space: O(1)