# Euler’s Totient Function

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 :

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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.

 // A simple C++ program to calculate // Euler's Totient Function  #include  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("<

 // A simple C program to calculate Euler's Totient Function #include    // 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; }

 // 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

 # 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

 // 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

 <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>

 

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]
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++ 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 code int main() {     int n;           for(n = 1; n <= 10; n++)     {         cout << "Phi" << "("              << n << ")" << " = "              << phi(n) <

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

 // 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 *= (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.

 # 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 function for n in range(1, 11) :     print("phi(", n, ") = ", phi(n))        # This code is contributed # by Nikita Tiwari.

 // 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 *= (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 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 *= (1.0 - (1.0 / $n));  return intval($result); }   // Driver Code for ($n = 1; $n <= 10; $n++) echo "phi(" .$n. ") =" . phi($n)."\n";   // This code is contributed by Sam007 Φ>  // Javascript program to calculate  // Euler's Totient Function  // using Euler's product formula function phi(n) {  // Initialize result as n  let result = n;   // Consider all prime factors  // of n and for every prime  // factor p, multiply result   // with (1 - 1/p)  for (let p = 2; p * p <= n; ++p)   {    // Check if p is  // a prime factor.  if (n % p == 0)   {    // If yes, then update  // n and result  while (n % p == 0)  n /= p;  result *= (1.0 - (1.0 / p));  }  }  // If n has a prime factor greater   // than sqrt(n) (There can be at-most  // one such prime factor)  if (n > 1)  result *= (1.0 - (1.0 / n));  return parseInt(result); } // Driver Code for (let n = 1; n <= 10; n++)  document.write(phi(${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 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++ 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 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 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 function int main() {  int n;  for (n = 1; n <= 10; n++)  printf("phi(%d) = %d\n", n, phi(n));  return 0; }  // 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 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# 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  <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 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)}
);       // 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

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 number
We know that  where k is any random number and
Total number from 1 to p = p
Number for which  is , i.e the number p itself, so subtracting 1 from p


Examples :

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

Proof :

, where a and b are prime numbers
,
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 = 35
Multiples 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 )
, removing all number with  with 

Examples :

3) For a prime number p

Proof :

 , where p is a prime number

Total numbers from 1 to
Total multiples of
Removing these multiples as with them Example :
p = 2, k = 5,  = 32
Multiples of 2 (as with them ) = 32 / 2 = 16 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32}


Examples :

4) For two number a and b

Special Case : gcd(a, b) = 1

Examples :

Special Case : ,
Normal Case : , 

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 = 6

n = 8
factors = {1, 2, 4, 8}
n =  = 1 + 1 + 2 + 4 = 8

n = 10
factors = {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.emory.edu/site/math125/chineseRemainderTheorem/