Euler’s Totient Function

Euler’s Totient function ? (n) for an input n is the count of numbers in {1, 2, 3, …, n-1} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1.

Examples :

?(1) = 1  
gcd(1, 1) is 1

?(2) = 1
gcd(1, 2) is 1, but gcd(2, 2) is 2.

?(3) = 2
gcd(1, 3) is 1 and gcd(2, 3) is 1

?(4) = 2
gcd(1, 4) is 1 and gcd(3, 4) is 1

?(5) = 4
gcd(1, 5) is 1, gcd(2, 5) is 1, 
gcd(3, 5) is 1 and gcd(4, 5) is 1

?(6) = 2
gcd(1, 6) is 1 and gcd(5, 6) is 1, 

How to compute ?(n) for an input n?
A simple solution is to iterate through all numbers from 1 to n-1 and count numbers with gcd with n as 1. Below is the implementation of the simple method to compute Euler’s Totient function for an input integer n. 

C++

// A simple C++ program to calculate
// Euler's Totient Function
#include <iostream>
using namespace std;

// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// A simple method to evaluate Euler Totient Function
int phi(unsigned int n)
{
    unsigned int result = 1;
    for (int i = 2; i < n; i++)
        if (gcd(i, n) == 1)
            result++;
    return result;
}

// Driver program to test above function
int main()
{
    int n;
    for (n = 1; n <= 10; n++)
        cout << "phi("<<n<<") = " << phi(n) << endl;
    return 0;
}

// This code is contributed by SHUBHAMSINGH10

C

// A simple C program to calculate Euler's Totient Function
#include <stdio.h>

// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// A simple method to evaluate Euler Totient Function
int phi(unsigned int n)
{
    unsigned int result = 1;
    for (int i = 2; i < n; i++)
        if (gcd(i, n) == 1)
            result++;
    return result;
}

// Driver program to test above function
int main()
{
    int n;
    for (n = 1; n <= 10; n++)
        printf("phi(%d) = %d\n", n, phi(n));
    return 0;
}

Java

// A simple java program to calculate
// Euler's Totient Function
import java.io.*;

class GFG {

    // Function to return GCD of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    // A simple method to evaluate
    // Euler Totient Function
    static int phi(int n)
    {
        int result = 1;
        for (int i = 2; i < n; i++)
            if (gcd(i, n) == 1)
                result++;
        return result;
    }

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

        for (n = 1; n <= 10; n++)
            System.out.println("phi(" + n + ") = " + phi(n));
    }
}

// This code is contributed by sunnusingh

Python3

# A simple Python3 program
# to calculate Euler's
# Totient Function

# Function to return
# gcd of a and b
def gcd(a, b):

    if (a == 0):
        return b
    return gcd(b % a, a)

# A simple method to evaluate
# Euler Totient Function
def phi(n):

    result = 1
    for i in range(2, n):
        if (gcd(i, n) == 1):
            result+=1
    return result

# Driver Code
for n in range(1, 11):
    print("phi(",n,") = ",
           phi(n), sep = "")

# This code is contributed
# by Smitha

C#

// A simple C# program to calculate
// Euler's Totient Function
using System;

class GFG {

    // Function to return GCD of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    // A simple method to evaluate
    // Euler Totient Function
    static int phi(int n)
    {
        int result = 1;
        for (int i = 2; i < n; i++)
            if (gcd(i, n) == 1)
                result++;
        return result;
    }

    // Driver code
    public static void Main()
    {
        for (int n = 1; n <= 10; n++)
        Console.WriteLine("phi(" + n + ") = " + phi(n));
    }
}

// This code is contributed by nitin mittal

PHP

<Φphp
// PHP program to calculate
// Euler's Totient Function

// Function to return
// gcd of a and b
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}

// A simple method to evaluate
// Euler Totient Function
function phi($n)
{
    $result = 1;
    for ($i = 2; $i < $n; $i++)
        if (gcd($i, $n) == 1)
            $result++;
    return $result;
}

// Driver Code
for ($n = 1; $n <= 10; $n++)
    echo "phi(" .$n. ") =" . phi($n)."\n";

// This code is contributed by Sam007
Φ>

Javascript

<script>
// Javascript program to calculate
// Euler's Totient Function

// Function to return
// gcd of a and b
function gcd(a, b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// A simple method to evaluate
// Euler Totient Function
function phi(n)
{
    let result = 1;
    for (let i = 2; i < n; i++)
        if (gcd(i, n) == 1)
            result++;
    return result;
}

// Driver Code
for (let n = 1; n <= 10; n++)
    document.write(`phi(${n}) = ${phi(n)} <br>`);

// This code is contributed by _saurabh_jaiswal

</script>

Output

phi(1) = 1
phi(2) = 1
phi(3) = 2
phi(4) = 2
phi(5) = 4
phi(6) = 2
phi(7) = 6
phi(8) = 4
phi(9) = 6
phi(10) = 4

The above code calls gcd function O(n) times. The time complexity of the gcd function is O(h) where “h” is the number of digits in a smaller number of given two numbers. Therefore, an upper bound on the time complexity of the above solution is O(N^2 log N) [How? there can be at most Log₁₀n digits in all numbers from 1 to n]

Auxiliary Space: O(log N)

Below is a Better Solution. The idea is based on Euler’s product formula which states that the value of totient functions is below the product overall prime factors p of n. 

The formula basically says that the value of ?(n) is equal to n multiplied by-product of (1 – 1/p) for all prime factors p of n. For example value of ?(6) = 6 * (1-1/2) * (1 – 1/3) = 2.
We can find all prime factors using the idea used in this post. 

1) Initialize : result = n
2) Run a loop from 'p' = 2 to sqrt(n), do following for every 'p'.
     a) If p divides n, then 
           Set: result = result  * (1.0 - (1.0 / (float) p));
           Divide all occurrences of p in n.
3) Return result  

Below is the implementation of Euler’s product formula.  

C++

// C++ program to calculate Euler's
// Totient Function using Euler's
// product formula
#include <bits/stdc++.h>
using namespace std;

int phi(int n)
{

    // Initialize result as n
    float result = n;

    // Consider all prime factors of n
    // and for every prime factor p,
    // multiply result with (1 - 1/p)
    for(int p = 2; p * p <= n; ++p)
    {

        // Check if p is a prime factor.
        if (n % p == 0)
        {

            // If yes, then update n and result
            while (n % p == 0)
                n /= p;

            result *= (1.0 - (1.0 / (float)p));
        }
    }

    // If n has a prime factor greater than sqrt(n)
    // (There can be at-most one such prime factor)
    if (n > 1)
        result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number

    return (int)result;
}

// Driver code
int main()
{
    int n;

    for(n = 1; n <= 10; n++)
    {
        cout << "Phi" << "("
             << n << ")" << " = "
             << phi(n) <<endl;
    }
    return 0;
}

// This code is contributed by koulick_sadhu

C

// C program to calculate Euler's Totient Function
// using Euler's product formula
#include <stdio.h>

int phi(int n)
{
    float result = n; // Initialize result as n

    // Consider all prime factors of n and for every prime
    // factor p, multiply result with (1 - 1/p)
    for (int p = 2; p * p <= n; ++p) {

        // Check if p is a prime factor.
        if (n % p == 0) {

            // If yes, then update n and result
            while (n % p == 0)
                n /= p;
            result *= (1.0 - (1.0 / (float)p));
        }
    }

    // If n has a prime factor greater than sqrt(n)
    // (There can be at-most one such prime factor)
    if (n > 1)
        result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number

    return (int)result;
}

// Driver program to test above function
int main()
{
    int n;
    for (n = 1; n <= 10; n++)
        printf("phi(%d) = %d\n", n, phi(n));
    return 0;
}

Java

// Java program to calculate Euler's Totient
// Function using Euler's product formula
import java.io.*;

class GFG {
    static int phi(int n)
    {
        // Initialize result as n
        float result = n;

        // Consider all prime factors of n and for
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 2; p * p <= n; ++p) {
            // Check if p is a prime factor.
            if (n % p == 0) {
                // If yes, then update n and result
                while (n % p == 0)
                    n /= p;
                result *= (1.0 - (1.0 / (float)p));
            }
        }

        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (n > 1)
            result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number

        return (int)result;
    }

    // Driver program to test above function
    public static void main(String args[])
    {
        int n;
        for (n = 1; n <= 10; n++)
            System.out.println("phi(" + n + ") = " + phi(n));
    }
}

// This code is contributed by Nikita Tiwari.

Python3

# Python 3 program to calculate
# Euler's Totient Function
# using Euler's product formula

def phi(n) :

    result = n   # Initialize result as n

    # Consider all prime factors
    # of n and for every prime
    # factor p, multiply result with (1 - 1 / p)
    p = 2
    while p * p<= n :

        # Check if p is a prime factor.
        if n % p == 0 :

            # If yes, then update n and result
            while n % p == 0 :
                n = n // p
            result = result * (1.0 - (1.0 / float(p)))
        p = p + 1


    # If n has a prime factor
    # greater than sqrt(n)
    # (There can be at-most one
    # such prime factor)
    if n > 1 :
        result -= result // n
  #Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  #if n is a prime number

    return int(result)


# Driver program to test above function
for n in range(1, 11) :
    print("phi(", n, ") = ", phi(n))


# This code is contributed
# by Nikita Tiwari.

C#

// C# program to calculate Euler's Totient
// Function using Euler's product formula
using System;

class GFG {

    static int phi(int n)
    {

        // Initialize result as n
        float result = n;

        // Consider all prime factors
        // of n and for every prime
        // factor p, multiply result
        // with (1 - 1 / p)
        for (int p = 2; p * p <= n; ++p)
        {

            // Check if p is a prime factor.
            if (n % p == 0)
            {

                // If yes, then update
                // n and result
                while (n % p == 0)
                    n /= p;
                result *= (float)(1.0 - (1.0 / (float)p));
            }
        }

        // If n has a prime factor
        // greater than sqrt(n)
        // (There can be at-most
        // one such prime factor)
        if (n > 1)
            result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number

        return (int)result;
    }

    // Driver Code
    public static void Main()
    {
        int n;
        for (n = 1; n <= 10; n++)
            Console.WriteLine("phi(" + n + ") = " + phi(n));
    }
}

// This code is contributed by nitin mittal.

PHP

<Φphp
// PHP program to calculate
// Euler's Totient Function
// using Euler's product formula
function phi($n)
{
    // Initialize result as n
    $result = $n;

    // Consider all prime factors
    // of n and for every prime
    // factor p, multiply result
    // with (1 - 1/p)
    for ($p = 2; $p * $p <= $n; ++$p)
    {

        // Check if p is
        // a prime factor.
        if ($n % $p == 0)
        {

            // If yes, then update
            // n and result
            while ($n % $p == 0)
                $n /= $p;
            $result *= (1.0 - (1.0 / $p));
        }
    }

    // If n has a prime factor greater
    // than sqrt(n) (There can be at-most
    // one such prime factor)
    if ($n > 1)
        $result -= $result / $n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number

    return intval($result);
}

// Driver Code
for ($n = 1; $n <= 10; $n++)
echo "phi(" .$n. ") =" . phi($n)."\n";

// This code is contributed by Sam007
Φ>

Javascript

// Javascript program to calculate
// Euler's Totient Function
// using Euler's product formula
function phi(n)
{
    // Initialize result as n
    let result = n;

    // Consider all prime factors
    // of n and for every prime
    // factor p, multiply result
    // with (1 - 1/p)
    for (let p = 2; p * p <= n; ++p)
    {

        // Check if p is
        // a prime factor.
        if (n % p == 0)
        {

            // If yes, then update
            // n and result
            while (n % p == 0)
                n /= p;
            result *= (1.0 - (1.0 / p));
        }
    }

    // If n has a prime factor greater
    // than sqrt(n) (There can be at-most
    // one such prime factor)
    if (n > 1)
        result -= result / n;
  //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
  //if n is a prime number

    return parseInt(result);
}

// Driver Code
for (let n = 1; n <= 10; n++)
 document.write(`phi(${n}) = ${phi(n)} <br>`);

// This code is contributed by _saurabh_jaiswal

Output

Phi(1) = 1
Phi(2) = 1
Phi(3) = 2
Phi(4) = 2
Phi(5) = 4
Phi(6) = 2
Phi(7) = 6
Phi(8) = 4
Phi(9) = 6
Phi(10) = 4

Time Complexity: O(?n log n)
Auxiliary Space: O(1)

We can avoid floating-point calculations in the above method. The idea is to count all prime factors and their multiples and subtract this count from n to get the totient function value (Prime factors and multiples of prime factors won’t have gcd as 1) 

1) Initialize result as n
2) Consider every number 'p' (where 'p' varies from 2 to ?n). 
   If p divides n, then do following
   a) Subtract all multiples of p from 1 to n [all multiples of p
      will have gcd more than 1 (at least p) with n]
   b) Update n by repeatedly dividing it by p.
3) If the reduced n is more than 1, then remove all multiples
   of n from result.

Below is the implementation of the above algorithm. 

C++

// C++ program to calculate Euler's
// Totient Function
#include <bits/stdc++.h>
using namespace std;

int phi(int n)
{
    // Initialize result as n
    int result = n;

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

        // Check if p is a prime factor.
        if (n % p == 0)
        {

            // If yes, then update n and result
            while (n % p == 0)
                n /= p;

            result -= result / p;
        }
    }

    // If n has a prime factor greater than sqrt(n)
    // (There can be at-most one such prime factor)
    if (n > 1)
        result -= result / n;

    return result;
}

// Driver code
int main()
{
    int n;
    for(n = 1; n <= 10; n++)
    {
        cout << "Phi" << "("
             << n << ")" << " = "
             << phi(n) << endl;
    }
    return 0;
}

// This code is contributed by koulick_sadhu

C

// C program to calculate Euler's Totient Function
#include <stdio.h>

int phi(int n)
{
    int result = n; // Initialize result as n

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

        // Check if p is a prime factor.
        if (n % p == 0) {

            // If yes, then update n and result
            while (n % p == 0)
                n /= p;
            result -= result / p;
        }
    }

    // If n has a prime factor greater than sqrt(n)
    // (There can be at-most one such prime factor)
    if (n > 1)
        result -= result / n;
    return result;
}

// Driver program to test above function
int main()
{
    int n;
    for (n = 1; n <= 10; n++)
        printf("phi(%d) = %d\n", n, phi(n));
    return 0;
}

Java

// Java program to calculate
// Euler's Totient Function
import java.io.*;

class GFG
{
static int phi(int n)
{
    // Initialize result as n
    int result = n;

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

        // Check if p is
        // a prime factor.
        if (n % p == 0)
        {

            // If yes, then update
            // n and result
            while (n % p == 0)
                n /= p;
            result -= result / p;
        }
    }

    // If n has a prime factor
    // greater than sqrt(n)
    // (There can be at-most
    // one such prime factor)
    if (n > 1)
        result -= result / n;
    return result;
}

// Driver Code
public static void main (String[] args)
{
    int n;
    for (n = 1; n <= 10; n++)
        System.out.println("phi(" + n +
                           ") = " + phi(n));
}
}

// This code is contributed by ajit

Python3

# Python3 program to calculate
# Euler's Totient Function
def phi(n):

    # Initialize result as n
    result = n;

    # Consider all prime factors
    # of n and subtract their
    # multiples from result
    p = 2;
    while(p * p <= n):

        # Check if p is a
        # prime factor.
        if (n % p == 0):

            # If yes, then
            # update n and result
            while (n % p == 0):
                n = int(n / p);
            result -= int(result / p);
        p += 1;

    # If n has a prime factor
    # greater than sqrt(n)
    # (There can be at-most
    # one such prime factor)
    if (n > 1):
        result -= int(result / n);
    return result;

# Driver Code
for n in range(1, 11):
    print("phi(",n,") =", phi(n));

# This code is contributed
# by mits

C#

// C# program to calculate
// Euler's Totient Function
using System;

class GFG
{

static int phi(int n)
{
// Initialize result as n
int result = n;

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

    // Check if p is
    // a prime factor.
    if (n % p == 0)
    {

        // If yes, then update
        // n and result
        while (n % p == 0)
            n /= p;
        result -= result / p;
    }
}

// If n has a prime factor
// greater than sqrt(n)
// (There can be at-most
// one such prime factor)
if (n > 1)
    result -= result / n;
return result;
}

// Driver Code
static public void Main ()
{
    int n;
    for (n = 1; n <= 10; n++)
        Console.WriteLine("phi(" + n +
                              ") = " +
                              phi(n));
}
}

// This code is contributed
// by akt_mit

PHP

<Φphp
// PHP program to calculate
// Euler's Totient Function

function phi($n)
{
    // Initialize
    // result as n
    $result = $n;

    // Consider all prime
    // factors of n and subtract
    // their multiples from result
    for ($p = 2;
         $p * $p <= $n; ++$p)
    {

        // Check if p is
        // a prime factor.
        if ($n % $p == 0)
        {

            // If yes, then
            // update n and result
            while ($n % $p == 0)
                $n = (int)$n / $p;
            $result -= (int)$result / $p;
        }
    }

    // If n has a prime factor
    // greater than sqrt(n)
    // (There can be at-most
    // one such prime factor)
    if ($n > 1)
        $result -= (int)$result / $n;
    return $result;
}

// Driver Code
for ($n = 1; $n <= 10; $n++)
    echo "phi(", $n,") =",
          phi($n), "\n";

// This code is contributed
// by ajit
Φ>

Javascript

// Javascript program to calculate
// Euler's Totient Function

function phi(n)
{
    // Initialize
    // result as n
    let result = n;

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

        // Check if p is
        // a prime factor.
        if (n % p == 0)
        {

            // If yes, then
            // update n and result
            while (n % p == 0)
                n = parseInt(n / p);
            result -= parseInt(result / p);
        }
    }

    // If n has a prime factor
    // greater than sqrt(n)
    // (There can be at-most
    // one such prime factor)
    if (n > 1)
        result -= parseInt(result / n);
    return result;
}

// Driver Code
for (let n = 1; n <= 10; n++)
    document.write(`phi(${n}) = ${phi(n)} <br>`);

// This code is contributed
// by _saurabh_jaiswal

Output

Phi(1) = 1
Phi(2) = 1
Phi(3) = 2
Phi(4) = 2
Phi(5) = 4
Phi(6) = 2
Phi(7) = 6
Phi(8) = 4
Phi(9) = 6
Phi(10) = 4

Time Complexity: O(?n log n)
Auxiliary Space: O(1)

Let us take an example to understand the above algorithm. 

n = 10. 
Initialize: result = 10

2 is a prime factor, so n = n/i = 5, result = 5
3 is not a prime factor.

The for loop stops after 3 as 4*4 is not less than or equal
to 10.

After for loop, result = 5, n = 5
Since n > 1, result = result - result/n = 4

Some Interesting Properties of Euler’s Totient Function

1) For a prime number p, 

Proof :

 , where p is any prime numberWe know that  where k is any random number and [Tex]\\[/Tex]Total number from 1 to p = p Number for which  is , i.e the number p itself, so subtracting 1 from p 

Examples : 

[Tex]\\[/Tex][Tex]\\[/Tex]

2) For two prime numbers a and b, used in RSA Algorithm

Proof :

, where a and b are prime numbers , [Tex]\\[/Tex]Total number from 1 to ab = ab Total multiples of a from 1 to ab =  = Total multiples of b from 1 to ab =  = Example:a = 5, b = 7, ab = 35Multiples of a =  = 7 {5, 10, 15, 20, 25, 30, 35}Multiples of b =  = 5 {7, 14, 21, 28, 35}Can there be any double counting ?(watch above example carefully, try with other prime numbers also for more grasp)Ofcourse, we have counted  twice in multiples of a and multiples of b so, Total multiples =  a + b - 1 (with which  with )[Tex]\phi(ab) = ab - (a + b - 1)[/Tex] , removing all number with  with  [Tex]\phi(ab) = (a - 1) \cdot (b - 1)[/Tex]

Examples :

[Tex]\\[/Tex][Tex]\\[/Tex]

3) For a prime number p, 

Proof : 

 , where p is a prime numberTotal numbers from 1 to  Total multiples of Removing these multiples as with them [Tex]\\[/Tex]Example : p = 2, k = 5,  = 32Multiples of 2 (as with them ) = 32 / 2 = 16 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32}[Tex]\phi(p ^ k) = p ^ k - p ^ {k - 1}[/Tex]

Examples : 

[Tex]\\[/Tex][Tex]\\[/Tex]

4) For two number a and b

Special Case : gcd(a, b) = 1

Examples :

Special Case : ,  [Tex]\\[/Tex][Tex]\\[/Tex] [Tex]\\[/Tex]Normal Case : , [Tex]\\[/Tex]  [Tex]\\[/Tex][Tex]\\[/Tex]

5) Sum of values of totient functions of all divisors of n is equal to n. 

Examples :

n = 6 
factors = {1, 2, 3, 6} 
n =  = 1 + 1 + 2 + 2 = 6n = 8factors = {1, 2, 4, 8}n =  = 1 + 1 + 2 + 4 = 8n = 10factors = {1, 2, 5, 10}n =  = 1 + 1 + 4 + 4 = 10

6) The most famous and important feature is expressed in Euler’s theorem : 

The theorem states that if n and a are coprime
(or relatively prime) positive integers, then

a?(n) ? 1 (mod n) 

The RSA cryptosystem is based on this theorem:
In the particular case when m is prime say p, Euler’s theorem turns into the so-called Fermat’s little theorem : 

ap-1 ? 1 (mod p) 

7) Number of generators of a finite cyclic group under modulo n addition is ?(n).

References:
http://e-maxx.ru/algo/euler_function
http://en.wikipedia.org/wiki/Euler%27s_totient_function

https://cp-algorithms.com/algebra/phi-function.html

http://mathcenter.oxford.memory.edu/site/math125/chineseRemainderTheorem/
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