# Euler’s Totient Function

• Difficulty Level : Medium
• Last Updated : 23 Jun, 2022

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

Examples :

Φ(1) = 1
gcd(1, 1) is 1

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

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

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

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

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

How to compute Φ(n) for an input nΦ
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 using namespace std; // Function to return gcd of a and bint gcd(int a, int b){    if (a == 0)        return b;    return gcd(b % a, a);} // A simple method to evaluate Euler Totient Functionint 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 functionint main(){    int n;    for (n = 1; n <= 10; n++)        cout << "phi("<

## C

 // A simple C program to calculate Euler's Totient Function#include  // Function to return gcd of a and bint gcd(int a, int b){    if (a == 0)        return b;    return gcd(b % a, a);} // A simple method to evaluate Euler Totient Functionint 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 functionint 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 Functionimport 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 bdef gcd(a, b):     if (a == 0):        return b    return gcd(b % a, a) # A simple method to evaluate# Euler Totient Functiondef phi(n):     result = 1    for i in range(2, n):        if (gcd(i, n) == 1):            result+=1    return result # Driver Codefor 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 Functionusing 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 bfunction gcd($a, $b){    if ($a == 0) return $b;    return gcd($b % $a, $a);} // A simple method to evaluate// Euler Totient Functionfunction phi($n){    $result = 1; for ($i = 2; $i < $n; $i++) if (gcd($i, $n) == 1) $result++;    return $result;} // Driver Codefor ($n = 1; $n <= 10; $n++)    echo "phi(" .$n. ") =" . phi($n)."\n"; // This code is contributed by Sam007Φ>

## Javascript

 

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 log N) [HowΦ 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 Φ(n) is equal to n multiplied by-product of (1 – 1/p) for all prime factors p of n. For example value of Φ(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 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 *= (1.0 - (1.0 / (float)n));      return (int)result;}  // Driver codeint main(){    int n;         for(n = 1; n <= 10; n++)    {        cout << "Phi" << "("             << n << ")" << " = "             << phi(n) <

## C

 // C program to calculate Euler's Totient Function// using Euler's product formula#include  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 *= (1.0 - (1.0 / (float)n));     return (int)result;} // Driver program to test above functionint 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 formulaimport 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 *= (1.0 - (1.0 / (float)n));         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 * (1.0 - (1.0 / float(n)))      return int(result)          # Driver program to test above functionfor 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 formulausing 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 *= (float)(1.0 - (1.0 / (float)n));         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.

 <Φphp// PHP program to calculate// Euler's Totient Function// using Euler's product formulafunction 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 *= (1.0 - (1.0 / $n));  return intval($result);} // Driver Codefor ($n = 1; $n <= 10; $n++)echo "phi(" .$n. ") =" . phi($n)."\n";  // This code is contributed by Sam007Φ> ## Javascript  // Javascript program to calculate// Euler's Totient Function// using Euler's product formulafunction phi(n){ // Initialize result as n let result = n;  // Consider all prime factors // of n and for every prime // factor p, multiply result // with (1 - 1/p) for (let p = 2; p * p <= n; ++p) {   // Check if p is // a prime factor. if (n % p == 0) {   // If yes, then update // n and result while (n % p == 0) n /= p; result *= (1.0 - (1.0 / p)); } }  // If n has a prime factor greater // than sqrt(n) (There can be at-most // one such prime factor) if (n > 1) result *= (1.0 - (1.0 / n));  return parseInt(result);} // Driver Codefor (let n = 1; n <= 10; n++) document.write(phi(${n}) = ${phi(n)} );  // 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 Φn). 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 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 codeint 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  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 functionint 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 Functionimport 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 Codepublic 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 Functiondef 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 Codefor n in range(1, 11): print("phi(",n,") =", phi(n));  # This code is contributed# by mits ## C#  // C# program to calculate// Euler's Totient Functionusing System; class GFG{  static int phi(int n){// Initialize result as nint result = n; // Consider all prime // factors of n and// subtract their// multiples from resultfor (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 Codestatic 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 $result;} // Driver Codefor ($n = 1; $n <= 10; $n++) echo "phi(", $n,") =",          phi($n), "\n";  // This code is contributed// by ajitΦ> ## 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 Codefor (let n = 1; n <= 10; n++) document.write(phi(${n}) = \${phi(n)}
);     // 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

aΦ(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)

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

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